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Dr. Garg Learning

Chemistry

Gen Chem 1 and Gen Chem 2 · all 21 chapters · by Dr. Akash Garg
Part one · General Chemistry 1
Chemistry · Chapter 1

Essential Ideas

Chemistry is the study of matter and the changes that matter goes through. This chapter hands you the tools that the rest of the book will run on. You will learn what matter is and how its particles are packed together. You will sort samples into solids, liquids and gases, and you will tell a physical change from a chemical one. Then you will start to measure. Every number in chemistry carries a unit, and that unit carries real meaning. You will meet the SI units, learn to report what your instruments actually tell you, and practice turning one unit into another. By the end of the chapter, your numbers will be worth trusting.

By the end of this chapter you can

  • Define matter and describe how particles are arranged in the three common states.
  • Sort any sample as an element, a compound or a mixture, and explain your choice.
  • Separate physical properties and changes from chemical properties and changes.
  • Name the seven SI base units and attach the common prefixes to any of them.
  • Count significant figures and round a calculated answer the correct way.
  • Convert between units by chaining conversion factors until the units cancel.
  • Calculate a density and convert among Fahrenheit, Celsius and kelvin.

Section 1Matter and its states

What counts as matter

Matter is anything that has mass and takes up space, which covers nearly everything around you. Your chair is matter, and so is the air in the room and the water in your glass. Light and sound are not matter, because they carry energy without holding any mass or volume.

All matter is built from atoms, and an atom is the smallest piece of an element that still behaves like that element. Atoms bond together into molecules, which are small groups of joined atoms. A single water molecule is two hydrogen atoms bonded to one oxygen atom, which chemists write as H2O.

Solid, liquid and gas

The three states you meet most often are solid, liquid and gas, and what separates them is how the particles are packed and how freely they travel.

In a solid, the particles sit close together and stay in fixed positions. They still vibrate in place but cannot trade spots, so a solid holds both its own shape and its own volume.

In a liquid, the particles still touch each other, but they slide past their neighbors freely. A liquid keeps its volume while taking the shape of its container, so pouring 50 mL of water into a wide dish still leaves you with 50 mL.

In a gas, the particles fly apart in every direction and spread out to fill whatever holds them. Most of a gas sample is actually empty space, which is why you can squeeze a gas into a smaller can while a solid barely budges.

Matter heated to extreme temperatures can reach a fourth state called plasma, in which electrons are torn away from their atoms. Stars shine with plasma, and so does the gas inside a glowing neon sign.

Particle packing in the three common states of matter Solid Fixed spots, tight pattern holds shape and volume Liquid Touching but sliding keeps volume, not shape Gas Far apart, moving fast fills any container Add heat and the particles move faster, so the packing loosens from left to right.
Figure 1.1 The same particles in three states. In the solid they hold a tight pattern. In the liquid they touch but slide. In the gas they are far apart and fast, so the sample can be squeezed.

Heating or cooling moves a sample between these states. Ice melts into water, water boils into steam, and steam cools and condenses back into water again. Dry ice skips the liquid stage and turns straight into a gas, a change known as sublimation. Through every one of these changes the substance itself stays the same, and only the packing of the particles changes.

Pure substances and mixtures

A pure substance has exactly the same makeup all the way through the sample. An element is a pure substance built from only one kind of atom, such as copper or oxygen. A compound is a pure substance built from two or more elements bonded in a fixed ratio. Water always carries two hydrogen atoms for each oxygen atom, in every sample on Earth.

A mixture holds two or more substances that are not bonded together, so each one keeps its own identity. A homogeneous mixture looks completely uniform throughout, and salt water and clean air are both good examples. We also call a homogeneous mixture a solution. A heterogeneous mixture has visible parts you could pick out, like sand sitting in water or a bowl of trail mix.

Because nothing in a mixture is bonded, a mixture can always be separated by physical means. Boil a pot of salt water and the water escapes as steam, while all of the salt stays behind in the pot.

Worked example · a mixture by mass

A cook stirs 25.0 g of salt into 475 g of water. What percent of the finished solution is salt by mass?

Find the total mass

25.0 g + 475 g = 500. g of solution. Addition keeps the fewest decimal places, so this answer stops at the ones place.

Divide the part by the whole

(25.0 g ÷ 500. g) × 100 = 5.00 percent salt by mass.

Check it

Five percent of 500 g is 25 g, which matches the salt that was added, and the water makes up the other 95.0 percent.

Key points
  • Matter has both mass and volume, while energy has neither.
  • Solids keep their shape and volume, liquids keep only volume, and gases keep neither.
  • Elements and compounds are pure substances, while mixtures can be pulled apart physically.
  • A change of state rearranges the packing and never changes the substance itself.

Section 2Physical and chemical properties

Two kinds of property

A physical property is a trait you can measure without turning the sample into something new. Color, smell, hardness, melting point, boiling point and density are all physical properties. You can melt an ice cube to find its melting point and still have ordinary water at the end.

A chemical property describes the way a sample turns into a new substance, so you have to let the change happen before you can observe one. Iron rusting in damp air is a chemical property of iron, and so is the way paper burns or the way an antacid tablet fizzes in acid.

Does the amount matter?

Some properties depend on how much of the sample you happen to have. Mass, volume and length are extensive properties, so two liters of milk carry twice the mass of one liter.

Other properties stay exactly the same at any amount, and density, color, melting point and temperature are these intensive properties. Cut a copper bar in half and each piece still has a density of 8.96 g/cm3. Intensive properties help you identify an unknown sample, because they do not care how big the sample is.

PropertyKindWhy
Mass of a sugar cubeExtensiveTwo cubes weigh twice as much
Volume of a gas sampleExtensiveMore gas fills more space
Density of goldIntensive19.3 g/cm3 for a ring or a brick
Boiling point of ethanolIntensive78 °C for a drop or a drum
FlammableChemicalYou learn it only by burning the sample

Physical change and chemical change

A physical change rearranges a sample without changing what the sample is made of. Melting, boiling, dissolving, cutting and bending are all physical changes, and the molecules are identical before and after.

A chemical change produces new substances, and we also call it a chemical reaction. Bonds break, atoms trade partners, and fresh bonds form in their place. Burning methane turns CH4 and O2 into CO2 and H2O, and those products carry their own new properties.

Physical change keeps the molecules, chemical change rebuilds them Physical change · ice melts 8 water molecules, locked heat the same 8 molecules, sliding Same substance. Only the packing has changed. Chemical change · methane burns CH₄ 2 O₂ burn CO₂ 2 H₂O New substances. Same atoms, new partners, same mass. hydrogen oxygen carbon
Figure 1.2 Melting ice is a physical change, since the same H2O molecules come out the other side. Burning methane is a chemical change, since the atoms are rebuilt into different molecules.

A handful of clues can hint that a chemical change is under way. You might notice bubbles of a brand new gas, a shift in color, a solid forming inside a clear liquid, or heat and light pouring out. Treat all of these as hints rather than proof, because boiling water bubbles hard while remaining a purely physical change. The real test is whether a new substance is present at the end.

Matter is conserved

The law of conservation of matter says that matter is never created or destroyed during a chemical change. The atoms are only rearranged, and every single atom is still present at the end. That means the total mass of the products equals the total mass of the reactants, and this one rule lets you solve for a mass you never measured.

Worked example · a missing mass

Inside a sealed burner, 4.00 g of methane reacts with 16.0 g of oxygen, and the only products are carbon dioxide and water. If 11.0 g of carbon dioxide forms, what mass of water forms with it?

Add up what went in

4.00 g + 16.0 g = 20.0 g of reactants in total.

Set the products equal to that

Mass of water = 20.0 g − 11.0 g = 9.0 g of water.

Check it

Adding the products gives 11.0 g + 9.0 g = 20.0 g, so nothing has gone missing. Notice that the carbon dioxide outweighs the methane it came from, which makes sense because oxygen atoms joined the carbon.

Key points
  • Physical properties can be measured while the sample stays intact.
  • Chemical properties appear only when the sample becomes a new substance.
  • Extensive properties follow the amount, while intensive properties ignore it.
  • Bubbles, color shifts and released heat are clues to a reaction rather than proof.
  • The total mass stays the same before and after any chemical change.

Section 3Measurement and SI units

A number alone says nothing

Every measurement has two parts, a number and a unit, and neither part works alone. Asking a lab partner for five means nothing at all, because five grams, five liters and five kelvin are wildly different things. Write the unit down every single time.

Scientists in every country share one system of units, the SI units, and SI is short for the International System of Units. Seven base units hold the entire system up, and every other unit is built out of those seven.

QuantityBase unitSymbol
Lengthmeterm
Masskilogramkg
Timeseconds
TemperaturekelvinK
Amount of substancemolemol
Electric currentampereA
Luminous intensitycandelacd

Mass is the odd one out in that table, because the kilogram already carries a prefix inside its name. Grams are still what you will weigh samples in on most days in the lab.

Units built from other units

A derived unit is a combination of base units. Area is a length times a length, so area arrives in m2, and volume is a length cubed, so its SI unit is m3. A cubic meter is enormous, roughly the size of a washing machine, so laboratories use the liter instead. One liter is exactly one cubic decimeter, and one milliliter is exactly one cubic centimeter.

Speed is measured in meters per second, written m/s, and density is mass over volume, often reported in g/cm3. Watch the pattern in those names, because the word "per" always signals division.

Mass and weight are two different quantities. Mass measures how much matter a sample contains, and it does not change when you travel. Weight is the pull of gravity on that mass, so a 1.00 kg block keeps its mass on the Moon while weighing about one sixth as much.

Prefixes do the heavy lifting

Real samples range from enormous to microscopic, so SI uses prefixes instead of long strings of zeros. Each prefix stands for a power of ten. Kilo means one thousand, so 1 km is 1000 m, and milli means one thousandth, so 1 mg is 0.001 g. Centi is one hundredth, micro is one millionth, and nano is one billionth.

The SI prefix ladder, each rung a power of ten One prefix, one power of ten Climbing up multiplies the base unit. Stepping down divides it. G giga 109 1 Gm = 1,000,000,000 m M mega 106 1 MJ = 1,000,000 J k kilo 103 1 km = 1000 m — (none) 100 1 m = 1 m c centi 10-2 1 cm = 0.01 m m milli 10-3 1 mg = 0.001 g µ micro 10-6 1 µm = 0.000001 m n nano 10-9 1 nm = 0.000000001 m bigger unit smaller unit
Figure 1.3 The prefix ladder. Each rung is a power of ten away from the plain base unit in the middle. Moving up makes the unit bigger, so the number in front gets smaller.

A prefix can be glued onto whatever unit you like, so a nanosecond is a billionth of a second and a megajoule is a million joules. The rule behind the prefix never changes.

Worked example · prefixes in action

Part A. A flask holds 2.50 L of solution. How many cubic centimeters is that?

Use the two exact links

One liter is exactly 1000 mL, and one milliliter is exactly 1 cm3.

Scale up

2.50 L × 1000 = 2500 cm3, which we write as 2.50 × 103 cm3 to hold on to three significant figures.

Part B. A single bacterium measures 1.5 µm in length. Report that length in meters.

Swap the prefix for its power of ten

Micro stands for 10−6, so 1.5 µm = 1.5 × 10−6 m, which is 0.0000015 m.

Check it

A meter is vastly bigger than a bacterium, so the number of meters should come out tiny, and it does.

Key points
  • A measurement is a number plus a unit, so never leave the unit off.
  • Seven SI base units cover length, mass, time, temperature, amount, current and light.
  • Derived units are built by multiplying and dividing the base units together.
  • 1 L = 1 dm3 and 1 mL = 1 cm3, and both of those are exact.
  • A prefix is simply a power of ten attached to any unit you choose.

Section 4Uncertainty and significant figures

Every measurement carries doubt

No instrument is perfect, so a measurement is really a best value plus some uncertainty. The habit to build is straightforward. Read every digit you are certain of, then estimate one more digit past the finest mark on the scale. That final digit is the uncertain one, and you still write it down.

Picture a ruler with marks printed every centimeter. The edge of your sample sits past the 7 cm mark, about a third of the way toward 8, so you report 7.3 cm. The 7 is certain while the 3 is your estimate. A ruler marked every millimeter would let you report 7.34 cm instead.

A digital screen follows the same logic, so you trust the digits it displays while remembering that the last one still wobbles.

Accuracy is not precision

Accuracy describes how close a measurement lands to the true value, while precision describes how close repeated measurements land to each other. The two can easily come apart. A balance that always reads 2.00 g high is wonderfully precise and badly wrong. Readings scattered all around the true value can average out well while being anything but precise.

Accuracy and precision shown on four targets Accurate and precise Tight group, on the center Precise, not accurate Tight group, off the center Accurate, not precise Wide spread, centered Neither Wide spread, off the center The gold dot is the true value Precision is how close the shots are to one another. Accuracy is how close they are to the gold dot. A tool with a steady offset gives the second picture every time.
Figure 1.4 Four sets of shots at the same target. Tight groups are precise. Groups centered on the bullseye are accurate. You want both.

Counting significant figures

Significant figures are the digits in a measurement that carry real information about its size. Four rules cover nearly every case you will ever meet.

Counted objects and defined values are exact numbers, so they hold unlimited significant figures and never limit your answer. There are exactly 12 eggs in a dozen, and exactly 2.54 cm in an inch, because that is how the inch is defined.

Scientific notation ends the trailing zero problem for good. In 1.2 × 103, only the digits in front of the power of ten count. Write 1.20 × 103 for three figures and 1.200 × 103 for four. No guessing.

Rounding a calculation

Your answer can never be more certain than the data behind it, and two rules keep you honest. For multiplication and division, keep as many significant figures as the measurement that carries the fewest. For addition and subtraction, keep as many decimal places as the measurement that carries the fewest.

Save the rounding for the very end, because rounding in the middle of a long chain lets small errors pile up.

Worked example · rounding both ways

Part A. A strip of foil measures 12.11 cm long and 0.0450 cm wide. Find its area.

Multiply, then count

12.11 cm × 0.0450 cm = 0.544950 cm2. The length carries four significant figures, while the width carries three, since leading zeros never count.

Keep the smaller count

Rounding to three significant figures gives an area of 0.545 cm2.

Part B. Add three measured masses together: 13.64 g, 0.7 g and 1.456 g.

Add, then find the weakest decimal place

13.64 + 0.7 + 1.456 = 15.796 g, and the 0.7 g is known only to the tenths place.

Round to the tenths place

The reported sum is 15.8 g. Notice that addition cares about decimal places, while multiplication cares about the count of digits.

Key points
  • Report every certain digit plus exactly one estimated digit.
  • Accuracy means close to the truth, while precision means close to each other.
  • Leading zeros are never significant, and captive zeros always are.
  • Trailing zeros count only when a decimal point is present in the number.
  • Multiply and divide by the digit count, then add and subtract by decimal place.

Section 5Dimensional analysis

Multiply by one on purpose

Dimensional analysis is the habit of carrying units through a calculation and letting them steer the setup. Its main tool is a conversion factor, which is a fraction built out of an equality. Because 1 m = 100 cm, both of these fractions are equal to exactly one.

100 cm ÷ 1 m   and   1 m ÷ 100 cm

Multiplying by one never changes how much you have, and it only changes the label on the amount. So pick whichever version puts the unit you want on top and the unit you want gone on the bottom, and that unwanted unit then cancels.

The units are also your error check. If the surviving unit is wrong, then your setup was wrong, and no amount of careful arithmetic will rescue it. Flip the factor over and try the chain again.

Chain as many steps as you need

You often will not know the single factor that takes you straight to the target, and that is perfectly fine. String several factors together instead and let each unwanted unit cancel in turn. Write the whole chain first, confirm that only the target unit survives, and save the arithmetic for last.

Units cancel down a chain of conversion factors 5.00 km written in inches Each unit on top cancels the same unit on the bottom of the next fraction. 5.00 km × 1000 m 1 km × 100 cm 1 m × 1 in 2.54 cm = 1.97 × 105 in Kilometers, meters and centimeters all cancel in pairs. Inches are the only unit left, so the setup is right. Arithmetic: 5.00 × 1000 × 100 ÷ 2.54 = 196,850, rounded to three significant figures.
Figure 1.5 A chain of three conversion factors. Each unit that appears on top of one fraction and on the bottom of the next one cancels. Only inches are left standing.

Worked example · kilometers to inches

A hiking trail runs 5.00 km from end to end. How long is that trail in inches?

Plan the path

Travel from km to m, then from m to cm, then from cm to inches, using the exact link 1 in = 2.54 cm.

Write the chain

5.00 km × (1000 m per 1 km) × (100 cm per 1 m) × (1 in per 2.54 cm)

Cancel and compute

Kilometers, meters and centimeters all cancel away, leaving inches behind. The arithmetic runs 5.00 × 1000 × 100 ÷ 2.54 = 196,850 in.

Round and check

Only the 5.00 was measured, so the answer keeps three significant figures as 1.97 × 105 in. An inch is a small unit, so a large count of them makes sense.

Rates are conversion factors too

Any ratio that ties two quantities together can serve as a conversion factor. A dose of 0.750 mg per kilogram of body mass is a factor, and so is a syrup labeled 25.0 mg per milliliter. Those two factors let you walk straight from a body mass to a volume you can pour.

Worked example · a two step dose

A drug is prescribed at 0.750 mg per kilogram of body mass. The patient has a mass of 68.0 kg, and the syrup on the shelf holds 25.0 mg per milliliter. How many milliliters should be poured?

Mass to dose

68.0 kg × (0.750 mg per 1 kg) = 51.0 mg of drug needed.

Dose to volume

51.0 mg × (1 mL per 25.0 mg) = 2.04 mL of syrup.

Check it

Run the chain backward, since 2.04 mL × 25.0 mg per mL = 51.0 mg, which is the dose we wanted. Every unit except milliliters cancelled somewhere along the way.

A squared unit needs its factor twice. To change cm2 into m2, multiply by (1 m per 100 cm) two times, which divides by 10,000 rather than by 100. The same trick handles cubic units three times over.
Key points
  • A conversion factor is a fraction whose value is exactly one.
  • Arrange it so the unwanted unit cancels, then confirm the unit that survives.
  • Chain factors together freely and leave the arithmetic until the end.
  • Rates such as milligrams per kilogram serve as conversion factors as well.
  • Exact factors never limit the significant figures in your final answer.

Section 6Density and temperature

Density packs mass into space

Density tells you how much mass is crammed into a given volume, and it is simply mass divided by volume.

d = m ÷ V

Solids and liquids are usually reported in g/cm3, which is identical to g/mL, while gases are so light that we report them in g/L. Liquid water is very close to 1.00 g/cm3. It peaks near 4 °C at 1.000 g/cm3 and drops to 0.997 g/cm3 by 25 °C.

Density is an intensive property, so a gold ring and a gold brick share the same density of 19.3 g/cm3. That steadiness makes density a strong clue to the identity of an unknown sample.

Whether something floats depends on its density rather than its mass, so a sample floats in a liquid whenever its density is lower than the liquid's. Ice has a density of 0.917 g/cm3, so ice cubes bob on water, while a steel bolt sinks because steel sits near 7.8 g/cm3. An enormous ship still floats because its shape traps air, which lowers the average density of the whole hull.

A density column, with each object resting on the first liquid denser than itself Denser liquids sink to the bottom Each object stops at the first layer that is denser than the object itself. Cooking oil 0.92 g/cm³ Water 1.00 g/cm³ Glycerin 1.26 g/cm³ Honey 1.42 g/cm³ Cork 0.24 g/cm³ Wax 0.95 g/cm³ Grape 1.10 g/cm³ Steel nut 7.80 g/cm³ Wax at 0.95 g/cm³ sinks through oil but floats on water. The steel nut outweighs every layer, so it reaches the floor.
Figure 1.6 A density column. Liquids stack with the densest at the bottom. Each dropped object settles at the level where the liquid below it is denser than the object.

Worked example · naming a metal by density

A cube of metal measures 2.00 cm along every edge and has a mass of 71.7 g. Find its density and identify the metal.

Find the volume

V = 2.00 cm × 2.00 cm × 2.00 cm = 8.00 cm3.

Divide mass by volume

d = 71.7 g ÷ 8.00 cm3 = 8.9625, which rounds to 8.96 g/cm3 at three significant figures.

Match and check

Copper has a density of 8.96 g/cm3, so this cube is very likely copper. Multiplying back gives 8.96 × 8.00 = 71.7 g, which matches the measured mass.

Three temperature scales

Temperature tells you which direction heat will flow, because heat always moves from the hotter sample toward the cooler one on its own.

The Celsius scale sets 0 °C at the freezing point of water and 100 °C at its boiling point under 1 atm of pressure. That span is divided into 100 steps called degrees.

The Fahrenheit scale places those same two points at 32 °F and 212 °F. That gives 180 degrees across the identical span, so a Fahrenheit degree is smaller, and nine Fahrenheit degrees cover the same gap as five Celsius degrees.

The kelvin is the SI unit of temperature. Its steps are exactly the size of Celsius degrees, but the scale begins at absolute zero, the coldest temperature that can exist. Absolute zero is 0 K, which is −273.15 °C. Nothing can be colder, so kelvin never goes negative and it takes no degree sign at all. You say 298 kelvin rather than 298 degrees kelvin.

K = °C + 273.15     °F = (9/5 × °C) + 32     °C = 5/9 × (°F − 32)
Kelvin matters more than it first appears. Gas laws and energy equations only behave with kelvin, because a scale with negative numbers would let you double a temperature and get nonsense. Switch to kelvin early and you will dodge a long list of errors.

Worked example · across all three scales

Part A. A room sits at 25.00 °C. Report that temperature in kelvin and in degrees Fahrenheit.

Add for kelvin

25.00 + 273.15 = 298.15 K. Both values are known to the hundredths place, so the answer keeps that place.

Scale and shift for Fahrenheit

(9/5 × 25.00) + 32 = 45.00 + 32 = 77.00 °F. The 9/5 and the 32 are both exact by definition.

Part B. A thermometer reads a fever of 102.0 °F. Report that temperature in degrees Celsius.

Subtract first, then scale

102.0 − 32 = 70.0, and 5/9 × 70.0 = 38.888, which rounds to 38.9 °C.

Check it

Normal body temperature is 37.0 °C, so a fever sitting a little above that figure is exactly what we expect.

Key points
  • Density is mass over volume, and it never depends on the amount of sample.
  • A sample floats whenever it is less dense than the liquid holding it.
  • Water sits near 1.00 g/cm3, while ice is less dense at 0.917 g/cm3.
  • Kelvin steps match Celsius steps exactly, shifted along by 273.15.
  • Zero kelvin is absolute zero, which equals −273.15 °C.

The chapter in one breath

  • Matter has both mass and volume, and it reaches you as a solid, a liquid or a gas.
  • Pure substances are either elements or compounds, and everything else is a mixture.
  • Physical changes keep the substance intact, while chemical changes build brand new substances.
  • Matter is conserved, so the products of a reaction weigh exactly what the reactants weighed.
  • Seven SI base units plus a handful of prefixes can describe any measurement you will meet.
  • Significant figures record how certain a measurement is, and the rounding rules protect that record.
  • Dimensional analysis converts any unit into any other by multiplying through by one.
  • Density is mass divided by volume, and comparing densities tells you what will float.
  • Kelvin shares the step size of Celsius but begins at absolute zero, which is −273.15 °C.

ReferenceWords worth knowing

Here are the terms this chapter leans on. Each one shows up again later in the book.

Absolute zero
The coldest temperature possible, 0 K or −273.15 °C.
Accuracy
How close a measurement is to the true value.
Chemical change
A change that breaks and forms bonds, producing new substances. Also called a reaction.
Chemical property
A trait you can only observe by letting the sample turn into something new.
Compound
A pure substance built from two or more elements bonded in a fixed ratio.
Conversion factor
A fraction made from an equality, so its value is exactly one.
Density
Mass divided by volume, usually in g/cm3 for solids and liquids.
Derived unit
A unit made by multiplying or dividing base units, such as m/s.
Dimensional analysis
Carrying units through a calculation and using them to check the setup.
Element
A pure substance made of only one kind of atom.
Extensive property
A property that changes with the amount of sample, such as mass.
Heterogeneous mixture
A mixture with parts you can see and separate, like sand in water.
Homogeneous mixture
A mixture that looks uniform throughout. Also called a solution.
Intensive property
A property that stays the same at any amount, such as density.
Law of conservation of matter
The rule that matter is neither created nor destroyed in a chemical change.
Physical change
A change in form or state that leaves the substance itself alone.
Physical property
A trait you can measure while the sample stays the same substance.
Precision
How close repeated measurements are to each other.
Significant figures
The digits of a measurement that carry real information about its size.
Uncertainty
The amount of doubt that rides along with every measurement.

Check yourself

1. Is clean dry air an element, a compound or a mixture?
A homogeneous mixture. Air is mostly nitrogen and oxygen molecules that are not bonded to each other. The ratio can shift from place to place, so it is not a compound.
2. Which state keeps its own volume but takes the shape of its container?
A liquid. Its particles still touch, so the volume holds, but they slide freely, so the shape gives way. A solid keeps both and a gas keeps neither.
3. Chopping wood and burning wood. Which one is a chemical change, and how can you tell?
Burning. It makes new substances, including carbon dioxide gas, water vapor and ash. Chopping only makes smaller pieces of the very same wood.
4. How many significant figures are in 0.03080?
Four. The leading zeros only place the decimal point. The 3, the captive 0, the 8 and the trailing 0 all count, since a decimal point is shown.
5. Multiply 6.02 by 2.1 and report the answer properly.
13. The raw product is 6.02 × 2.1 = 12.642. The weaker value, 2.1, has two significant figures, so round to two: 13.
6. Add 24.681 g, 0.35 g and 1.4 g, then report the sum properly.
26.4 g. The raw sum is 24.681 + 0.35 + 1.4 = 26.431 g. The 1.4 g is known only to the tenths place, so the sum stops there.
7. Convert 3.50 kg into milligrams.
3.50 × 106 mg. Chain two factors: 3.50 kg × (1000 g per kg) × (1000 mg per g) = 3,500,000 mg. Both factors are exact, so keep the three figures from 3.50.
8. A 250.0 mL sample of liquid has a mass of 227.5 g. What is its density?
0.9100 g/mL. Divide mass by volume: 227.5 g ÷ 250.0 mL = 0.91 g/mL exactly. Both values hold four significant figures, so write it as 0.9100 g/mL. Since this is below 1.00 g/mL, the liquid would float on water.
9. A gas cylinder reads 310. K. What is that in degrees Celsius?
37 °C. Subtract: 310. − 273.15 = 36.85. The 310. value is known only to the ones place, so round there and report 37 °C. That is near body temperature.
10. Aluminum has a density of 2.70 g/cm3. Find the mass of a cube 3.00 cm on each edge.
72.9 g. The volume is 3.00 × 3.00 × 3.00 = 27.0 cm3. Then mass = density × volume = 2.70 g/cm3 × 27.0 cm3 = 72.9 g.
11. Four shots land in a tight clump near the edge of the target. Are they accurate, precise, both or neither?
Precise but not accurate. The shots agree closely with each other, which is precision. They miss the true center, which means poor accuracy. A tool with a steady offset behaves this way.
12. In a sealed flask, 10.0 g of iron reacts fully with 4.30 g of oxygen to form rust. What mass of rust forms?
14.3 g. Matter is conserved, so the product mass equals the total reactant mass: 10.0 g + 4.30 g = 14.3 g. The flask is sealed, so no oxygen can escape and none can sneak in.
Chemistry · Chapter 2

Atoms, Molecules and Ions

Everything around you is built from a short list of parts. Protons, neutrons and electrons pack into atoms. Atoms join into molecules, or trade charge and become ions. From there they stack into every material you have ever touched. This chapter hands you the parts and the rules. You will count protons to name an element. You will mix isotopes to get an atomic mass. You will read the periodic table like a map. You will balance charge to write a formula. Then you will pick up the naming system chemists share. By the end, a label like Fe(NO3)3 will read like a sentence.

By the end of this chapter you can

  • Name the three subatomic particles and give the charge and mass of each.
  • Use atomic number and mass number to identify any atom or ion.
  • Work out an average atomic mass from isotope masses and abundances.
  • Find groups, periods, metals, nonmetals and metalloids on the table.
  • Predict ion charges and write neutral ionic formulas.
  • Tell empirical, molecular and structural formulas apart.
  • Name ionic, molecular and acid compounds from their formulas.

Section 1Atomic theory and the subatomic particles

A very old idea, finally tested

People argued about matter for a long time. Could you cut a lump of gold in half forever? Or do you reach a smallest piece that will not split? For centuries nobody could test the question. Careful weighing finally settled it.

In 1808 John Dalton pulled the evidence into one theory. An atom is the smallest piece of an element that still behaves like that element. All atoms of one element act the same way in a reaction. Atoms of different elements differ. Atoms combine in fixed, whole-number ratios to make compounds. And a chemical reaction only rearranges atoms. It never creates them and never destroys them.

Two mass laws backed him up. The law of definite proportions says one compound always holds the same mass ratio of elements. Pure water is 11.19% hydrogen by mass, whether it came from a river or a flask. The law of multiple proportions compares two compounds made from the same pair of elements. In carbon monoxide, 1.00 g of carbon is joined to 1.33 g of oxygen. In carbon dioxide, 1.00 g of carbon is joined to 2.66 g of oxygen. The ratio of 2.66 to 1.33 is a clean 2 to 1. Tidy whole numbers like that point to whole particles.

Three particles, and where they sit

Dalton thought the atom had no parts. He was wrong, and the next hundred years showed why. J. J. Thomson ran a current through a thin gas and watched a beam bend toward the positive plate. He had found the electron, a very light particle with a negative charge. Ernest Rutherford then fired tiny positive bullets at thin gold foil. Most flew straight through. A few bounced almost straight back. That only happens if the positive charge is crammed into a tiny core. He named the core the nucleus. The positive particle inside it is the proton. In 1932 James Chadwick found the neutron, a nuclear particle with no charge at all.

Masses here are quoted in the atomic mass unit (u). One unit is exactly 1/12 the mass of a carbon-12 atom, or 1.6605 × 10-24 g. A proton weighs 1.0073 u. A neutron weighs 1.0087 u. An electron weighs only 0.00055 u, which is about 1/1836 of a proton. So the nucleus carries almost all of the mass.

Size runs the other way. A whole atom is roughly 10-10 m across. Its nucleus is roughly 10-15 m across. The atom is about 100,000 times wider than its own core. Blow the nucleus up to the size of a marble and the atom would stretch across a sports field. Almost all of an atom is empty space with a few electrons moving through it.

Inside an atom, and how its symbol is written A LITHIUM-7 ATOM HOW THE SYMBOL IS WRITTEN nucleus protons + neutrons electron shells 2 inside, 1 outside 7 3 Li mass number A = 7 protons plus neutrons atomic number Z = 3 protons, which name it element symbol Neutral, so it also holds 3 electrons. proton: 1+, 1.0073 u neutron: 0, 1.0087 u electron: 1−, 0.00055 u
Inside an atom. The nucleus holds the protons and neutrons and nearly all of the mass. Electrons are drawn here in simple shells. The symbol carries the mass number on top and the atomic number below.

Counting protons names the element

The atomic number (Z) is the number of protons in the nucleus. It is the atom's identity badge. Six protons means carbon, every single time. The mass number (A) is the protons plus the neutrons added together. Chemists write both on the symbol, with A above and Z below. A neutral atom holds as many electrons as protons. Change the electron count and the atom carries a charge instead.

ParticleWhere it sitsChargeMass (u)Mass vs a proton
ProtonIn the nucleus1+1.00731
NeutronIn the nucleus01.0087about 1
ElectronAround the nucleus1−0.00055about 1/1836

Worked example · Reading an atom from its particle counts

A particle holds 26 protons, 30 neutrons and 24 electrons. What is it?

Step 1 · Use the protons to name itZ = 26, so this is iron, symbol Fe. The protons alone decide this.

Step 2 · Add for the mass numberA = 26 protons + 30 neutrons = 56.

Step 3 · Compare protons with electronscharge = 26 − 24 = 2+. Two electrons are missing, so two positive charges are left over.

Step 4 · Write it56Fe2+, read as iron-56 with a 2+ charge. It is an ion, not a neutral atom.

CheckProtons still equal 26, so it is still iron. Only the charge and the neutron count changed.

Key points
  • Atoms combine in whole-number ratios, which is what the two mass laws showed.
  • Protons are 1+, neutrons are neutral, and both sit in the nucleus. Electrons are 1− and sit outside.
  • The nucleus holds nearly all the mass in nearly none of the volume.
  • Z counts protons and names the element. A counts protons plus neutrons.
  • Protons minus electrons gives the charge.

Section 2Isotopes and atomic mass

Same element, different weight

Dalton said every atom of an element has the same mass. That part needed a repair. Atoms of one element can hold different numbers of neutrons. Those versions are isotopes. They share an atomic number and differ in mass number. Hydrogen shows it most clearly. Almost every hydrogen atom is one proton with no neutron at all. Roughly one in 8,700 carries a neutron, and that form is called deuterium. A third form, tritium, carries two neutrons and breaks down over time.

Isotopes of one element react the same way. Chemistry is run by electrons, and isotopes have matching electron counts. Mass does shift a few physical traits. Heavy water, D2O, freezes at 3.8 °C instead of 0 °C. It is also about 11% denser than ordinary water.

A nuclide is one specific build of a nucleus, such as carbon-13 or uranium-235. Naming a nuclide means fixing both numbers, not just the element. Some nuclides are stable and last forever. Others come apart and release radiation.

Isotopes of hydrogen, and the weighted average mass of chlorine THREE ISOTOPES OF HYDROGEN A = 1A = 2A = 3 hydrogen-1 1 proton, 0 neutrons 99.99% of all hydrogen hydrogen-2 (deuterium) 1 proton, 1 neutron 0.01% of all hydrogen hydrogen-3 (tritium) 1 proton, 2 neutrons traces only, and unstable WHY CHLORINE AVERAGES 35.45 u 75.76% 24.24% Cl-35 34.969 u Cl-37 36.966 u average = 35.45 u atomic mass (u), lighter on the left
Top: three isotopes of hydrogen, same one proton, different neutron counts. Bottom: bar height is how common each chlorine isotope is. The average sits where the two bars would balance, so it lands near the common lighter one.

Weighing a mixture

A mass spectrometer sorts atoms by mass. It knocks an electron off each atom, speeds the ions up, then bends their path with a magnet. Heavy ions bend less than light ones. A detector counts how many land at each mass. Out comes the mass of every isotope and its percent abundance, the share of atoms that are that isotope.

The number printed under each symbol on the periodic table is the average atomic mass. It is a weighted average, not a plain one. Multiply each isotope mass by its share, then add the results:

average atomic mass = Σ (fraction × isotope mass)

Use the fraction, so divide each percent by 100 first. The answer always leans toward the common isotope. Chlorine averages 35.45 u, far nearer 35 than 37, because the lighter form is about three times as common.

Worked example · The average atomic mass of chlorine

Chlorine is 75.76% chlorine-35 at 34.96885 u and 24.24% chlorine-37 at 36.96590 u. Find its average atomic mass.

Step 1 · Turn percents into fractions75.76% → 0.7576 and 24.24% → 0.2424. They add to 1.0000, so nothing is missing.

Step 2 · Weight each mass0.7576 × 34.96885 u = 26.4924 u.
0.2424 × 36.96590 u = 8.9605 u.

Step 3 · Add the pieces26.4924 u + 8.9605 u = 35.4529 u, which rounds to 35.45 u.

CheckThe answer must fall between 34.97 and 36.97, and it does. It sits close to the lighter isotope, which is the common one. That matches the table value of 35.45 u.

An atom weighs less than its parts. Add up the pieces of a carbon-12 atom and you get 6(1.0073) + 6(1.0087) + 6(0.00055) = 12.0993 u. Yet a carbon-12 atom is exactly 12 u, because the unit was built from it. The missing 0.0993 u was released as energy when the nucleus formed. Chemists call the gap the mass defect.

Mass number is a count, atomic mass is a mass

Keep these two apart. A mass number is a whole number and nothing else. It counts protons plus neutrons. An atomic mass is a measured mass in u, and it usually arrives with decimals. Carbon-13 has mass number 13 and an atomic mass of 13.00335 u. Only carbon-12 has both numbers equal, and that is on purpose.

Key points
  • Isotopes of one element have the same Z and different A, so they differ only in neutrons.
  • Isotopes act alike chemically, because they hold the same electrons.
  • Percent abundance is the share of atoms of that isotope in a natural sample.
  • Average atomic mass = sum of (fraction × isotope mass), so it always leans toward the common isotope.
  • Mass number counts particles. Atomic mass measures mass in u.

Section 3The periodic table

A map built out of behavior

By the 1860s chemists knew dozens of elements and their masses. Dmitri Mendeleev lined them up in order of mass and spotted a repeat. Properties kept coming back around. So he broke the line into rows and stacked the look-alikes into columns. He even left holes for elements nobody had found, and guessed what they would be like. Gallium turned up in 1875 and matched his guess closely. That is when the table stopped looking like a filing trick and started looking like a law.

Today the elements are ordered by atomic number instead of mass. That fixed a few seats. Tellurium weighs more than iodine, yet tellurium comes first, because it has fewer protons. Order by protons and every look-alike lands in the right column.

Rows, columns and the long drawers

A row is a period, numbered 1 through 7. A column is a group, numbered 1 through 18. Elements in one group behave alike because they hold the same number of outer electrons. Four groups have names worth knowing. Group 1 is the alkali metals, soft metals that react hard with water. Group 2 is the alkaline earth metals. Group 17 is the halogens, which grab electrons eagerly. Group 18 is the noble gases, which mostly refuse to react with anything.

Groups 1, 2 and 13 through 18 are the main group elements. Groups 3 through 12 hold the transition metals, the familiar working metals like iron, copper and nickel. Two long rows are pulled out and printed underneath, which keeps the page from stretching too wide. They are the lanthanides, elements 57 through 71, and the actinides, elements 89 through 103. Slide them back into period 6 and period 7 and the table becomes one sheet 32 columns across.

Metals, nonmetals and the stair step

A jagged step runs down the right side of the table. Metals sit to the left of it. They shine, they bend rather than shatter, and they carry heat and current well. Nonmetals sit to the right. They are dull, brittle when solid, and poor conductors. Sitting right on the step are the metalloids, which act partly like each side. Boron, silicon, germanium, arsenic, antimony and tellurium are the usual six. Silicon's halfway conducting is exactly what computer chips need. Books draw the step in slightly different places, so polonium and astatine get sorted either way.

Hydrogen is printed above group 1, but it is a nonmetal and does its own thing. Of the 118 known elements, the first 94 turn up in nature, a few only in traces. The rest are built one atom at a time in accelerators and vanish in moments. At room temperature most elements are solid. Eleven are gases. Only two are liquid, mercury and bromine.

All 118 elements, coloured by metal, metalloid and nonmetal 1234 5678 9101112 13141516 1718 GROUP 1234 567 PERIOD 57-7189-103 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og lanthanidesactinides metals metalloids nonmetals 118 known elements, and the first 94 occur in nature
The whole table. Metals fill most of it. Nonmetals crowd the top right. The six metalloids trace the step between them. The two dashed cells mark where the lanthanide and actinide rows really belong.

Worked example · Placing an element from its atomic number

Where does selenium, Z = 34, sit? Is it a metal?

Step 1 · Count off the periodsPeriod 1 holds 2 elements, so it ends at 2. Periods 2 and 3 hold 8 each, ending at 10 and 18. Period 4 holds 18, ending at 36. Since 18 < 34 ≤ 36, selenium is in period 4.

Step 2 · Count across that periodPeriod 4 starts at Z = 19 in group 1 and fills all 18 groups in order. So the group is 34 − 18 = group 16.

Step 3 · Read off the classGroup 16, period 4 sits to the right of the step, so selenium is a nonmetal.

CheckGroup 16 also holds oxygen and sulfur, both nonmetals. Selenium belongs with them, and it does act like a heavier sulfur.

Key points
  • Periods are the 7 rows. Groups are the 18 columns.
  • One group shares outer electron count, which is why it shares behavior.
  • Group 1 alkali metals, group 2 alkaline earth metals, group 17 halogens, group 18 noble gases.
  • Groups 3 to 12 are transition metals. The lanthanides and actinides are the two rows printed below.
  • Metals sit left of the step, nonmetals right of it, and six metalloids sit on it.

Section 4Ions and ionic compounds

Move an electron and you make a charge

An atom with equal protons and electrons has no net charge. Pull an electron away and the protons now outvote the electrons, so the particle is positive. Push an extra electron on and the particle is negative. A charged atom, or a charged group of atoms, is an ion. A positive ion is a cation. A negative ion is an anion. Losing one electron gives 1+. Losing two gives 2+. Gaining one gives 1−. Note that protons never move in ordinary chemistry. Only electrons do.

The group number predicts the charge for most main group elements. Group 1 forms 1+ ions. Group 2 forms 2+. Aluminum, in group 13, forms 3+. Group 15 often takes on three electrons for 3−. Group 16 takes two for 2−. Group 17 takes one for 1−. Noble gases usually form no ion at all. Every one of those ions ends up with the same outer electron count as a noble gas, which is the pattern behind the rule.

Transition metals are less tidy. Many of them form more than one ion. Iron forms Fe2+ and Fe3+. Copper forms Cu+ and Cu2+. A handful are fixed and easy. Zinc is always Zn2+ and silver is always Ag+.

Ions built from several atoms

A polyatomic ion is a bonded group of atoms carrying one charge for the whole group. Sulfate is SO42-. Its sulfur and four oxygens travel together as a single unit, and the 2− belongs to the unit, not to any one atom. Ammonium, NH4+, is the common positive one. Nitrate is NO3- and nitrite is NO2-. In pairs like that, the -ate form always holds one more oxygen than the -ite form.

IonFormulaIonFormula
ammoniumNH4+hydroxideOH-
nitrateNO3-nitriteNO2-
sulfateSO42-sulfiteSO32-
carbonateCO32-hydrogen carbonateHCO3-
phosphatePO43-acetateC2H3O2-
chlorateClO3-perchlorateClO4-
cyanideCN-permanganateMnO4-
chromateCrO42-dichromateCr2O72-

Balancing the charge writes the formula

An ionic compound pairs cations with anions, and it is usually a metal with a nonmetal. There is no single molecule to point at. Instead the ions stack into a repeating three-dimensional grid called a lattice, which runs on for as long as the crystal does. So the formula gives the smallest whole-number ratio that cancels the charge. That ratio is called a formula unit.

The rule is one line. Total positive charge plus total negative charge must come to zero. Find the smallest counts that pull that off. One useful trick is the lowest common multiple of the two charge sizes.

Balancing charge to get a formula unit, and the lattice it builds BALANCE THE CHARGE AN IONIC LATTICE (NaCl) Al3+ Al3+ O2− O2− O2− + 2 × (3+)= 6+ 3 × (2−)= 6− 6+ and 6− cancel exactly formula unit: Al2O3 Na+ Cl− The pattern repeats, so a 1 to 1 ratio gives NaCl.
Left: two 3+ ions and three 2− ions both reach 6, so the charges cancel and the formula unit is Al2O3. Right: in a lattice every ion is ringed by ions of the opposite charge, and the grid repeats in all directions.

Worked example · Two ionic formulas from charge alone

Write the formula for aluminum oxide, and for calcium phosphate.

Step 1 · Get the two chargesAluminum is group 13, so Al3+. Oxygen is group 16, so O2-.

Step 2 · Find the lowest common multipleThe charge sizes are 3 and 2, and their lowest common multiple is 6. So you need 6+ of positive and 6− of negative.

Step 3 · Divide to get the counts6 ÷ 3 = 2 aluminum ions, and 6 ÷ 2 = 3 oxide ions. The formula is Al2O3.

Step 4 · Repeat with a polyatomic ionCalcium is Ca2+ and phosphate is PO43-. The charge sizes 2 and 3 give 6 again. That needs 3 calcium ions and 2 phosphate ions, so the formula is Ca3(PO4)2. The parentheses keep the 2 from touching the oxygen count.

Check3(2+) = 6+ and 2(3−) = 6−, which sum to zero. Also check that the counts share no common factor, because the formula must be the smallest ratio.

Key points
  • Only electrons move. Losing them makes a cation, gaining them makes an anion.
  • Main group charges follow the group: 1+, 2+, 3+, then 3−, 2−, 1−.
  • Many transition metals form more than one charge, so the charge has to be stated.
  • A polyatomic ion carries one charge for the whole bonded group.
  • An ionic formula is the smallest whole-number ratio that makes the total charge zero.

Section 5Molecular compounds and formulas

Sharing instead of handing over

Two nonmetals do not pass electrons back and forth. Neither one gives up easily, so they share instead. A shared pair of electrons is a covalent bond. A neutral group of atoms held together by covalent bonds is a molecule. A molecular compound is built out of molecules, so it comes in countable units. One water molecule is two hydrogens and one oxygen, and the next one is the same.

That is a real difference from an ionic compound. Table salt has no smallest salt particle, only a ratio that repeats through the crystal. Sugar does have one. You can point at a single sucrose molecule and say that is the whole thing. Molecular compounds also tend to melt at lower temperatures and dissolve without splitting into ions.

Some elements travel as molecules of themselves. Seven are diatomic under normal conditions: H2, N2, O2, F2, Cl2, Br2 and I2. Phosphorus goes around as P4 and sulfur as S8. A lone oxygen atom is rare and very reactive, which is why the air holds O2 and not O.

Three ways to write the same compound

A molecular formula counts every atom in one molecule. Glucose is C6H12O6, so six carbons, twelve hydrogens and six oxygens. An empirical formula gives the smallest whole-number ratio of those atoms. Divide every subscript in glucose by 6 and you get CH2O. Something is lost on the way. Formaldehyde really is CH2O, and acetic acid is C2H4O2. All three share one ratio, and all three are completely different substances.

A structural formula shows which atom is bonded to which. Lines stand in for bonds. That is often the detail that decides behavior. Ethanol and dimethyl ether are both C2H6O. Ethanol is the alcohol in drinks. Dimethyl ether is a gas used as a propellant. Compounds that share a molecular formula and differ in structure are isomers.

Models push further still. A ball-and-stick model keeps the bonds visible so you can see the angles between them. A space-filling model drops the sticks and swells the balls, so you see the room the electrons actually occupy. Each view hides something, which is why chemists keep all of them.

One compound written three ways MOLECULAR FORMULAEMPIRICAL FORMULASTRUCTURAL FORMULA C2H4O2 every atom in one molecule 2 carbon, 4 hydrogen, 2 oxygen acetic acid, in vinegar the smallest whole-number ratio each subscript divided by 2 1 carbon : 2 hydrogen : 1 oxygen CH2O CC OO HHHH condensed form: CH3COOH Glucose and formaldehyde share the empirical formula CH2O, and they are nothing alike.
One compound, three views. The molecular formula counts atoms. The empirical formula gives the ratio and loses the count. The structural formula is the only one that tells you the double bond sits on the second carbon.

Worked example · Molecular formula to empirical formula

Give the empirical formula for octane, C8H18, and for N2O4.

Step 1 · List the subscriptsOctane gives 8 and 18.

Step 2 · Find the largest number that divides bothFactors of 8 are 1, 2, 4 and 8. Factors of 18 are 1, 2, 3, 6, 9 and 18. The largest shared factor is 2.

Step 3 · Divide and write it8 ÷ 2 = 4 and 18 ÷ 2 = 9, so the empirical formula is C4H9.

Step 4 · Do the second oneN2O4 gives 2 and 4, which share a factor of 2. Dividing gives NO2.

CheckIn each answer the subscripts share no factor above 1, so each is truly the smallest ratio. Note that NO2 is also a real compound in its own right, which is exactly why an empirical formula alone cannot name a substance.

Key points
  • Nonmetals share electrons in covalent bonds and build molecules.
  • A molecular compound has a smallest unit. An ionic compound has only a repeating ratio.
  • Seven elements are diatomic, plus P4 and S8.
  • Molecular formula counts atoms. Empirical formula gives the smallest ratio. Structural formula shows the bonds.
  • Isomers share a molecular formula and differ in structure, so they differ in behavior.

Section 6Naming compounds

Two questions sort every formula

Chemists needed one naming system so that a formula reads the same in any country. The system splits by compound type, so start by deciding which type you have. Ask two questions in order. Does the formula start with hydrogen and act as an acid in water? Then name it as an acid. If not, does it start with a metal or with ammonium? If yes it is ionic, and if no it is molecular. Two questions, three roads.

How to choose a naming rule A formula Starts with H, acting as an acid? A metal or ammonium in front? YESNONOYES ACIDIONICMOLECULAR hydro- + root + -ic acid an -ate ion gives -ic acid an -ite ion gives -ous acid HCl → hydrochloric acid HNO3 → nitric acid cation name, then anion name single-atom anion takes -ide variable metal gets a numeral NaCl → sodium chloride FeCl3 → iron(III) chloride count atoms with prefixes mono, di, tri, tetra, penta second element takes -ide CO → carbon monoxide N2O4 → dinitrogen tetroxide
Pick the road first. The same two questions sort any formula into one of three naming rules. Getting the type right is most of the work, because each road has its own small set of rules.

Ionic names

Name the cation, then the anion, and never use prefixes. The metal keeps its ordinary name. A single-atom anion drops its ending and picks up -ide. So NaCl is sodium chloride and MgO is magnesium oxide. A polyatomic ion keeps whatever name it already has, so Na2SO4 is sodium sulfate.

If the metal can form more than one charge, you have to say which one. Put the charge in Roman numerals in parentheses right after the metal name. FeCl2 is iron(II) chloride and FeCl3 is iron(III) chloride. You read that charge off the formula by balancing, never by guessing. Metals with only one common charge, such as sodium or zinc, get no numeral at all. One more case: a hydrate traps water inside its crystal, and a prefix counts it. CuSO4·5H2O is copper(II) sulfate pentahydrate.

Molecular names

Here you name the subscripts out loud with prefixes. Mono is 1, di is 2, tri is 3, tetra is 4, penta is 5, hexa is 6, hepta is 7, octa is 8, nona is 9 and deca is 10. The first element keeps its name and skips mono. The second element takes the -ide ending and always gets its prefix. CO is carbon monoxide and CO2 is carbon dioxide. SF6 is sulfur hexafluoride. Drop the final vowel of a prefix before oxide, which is why you say monoxide and pentoxide.

Acid names

Acids come in two families. A binary acid is hydrogen plus one other element. Use hydro-, then the root of that element, then -ic acid. HCl in water is hydrochloric acid. An oxyacid is hydrogen plus a polyatomic ion that holds oxygen. Here the ending of the anion steers the name. An -ate anion gives -ic acid. An -ite anion gives -ous acid. So HNO3 is nitric acid and HNO2 is nitrous acid. H2SO4 is sulfuric acid and H2SO3 is sulfurous acid.

AnionAnion formulaAcidAcid name
chlorideCl-HClhydrochloric acid
sulfideS2-H2Shydrosulfuric acid
nitrateNO3-HNO3nitric acid
nitriteNO2-HNO2nitrous acid
sulfateSO42-H2SO4sulfuric acid
sulfiteSO32-H2SO3sulfurous acid
phosphatePO43-H3PO4phosphoric acid
perchlorateClO4-HClO4perchloric acid
Why sulfuric and not sulfic? A few roots keep an extra syllable so the name stays easy to say. Sulfate gives sulfuric, and phosphate gives phosphoric. The rule still holds underneath. An -ate ion gives -ic acid, and an -ite ion gives -ous acid.

Worked example · Naming three compounds

Name Fe2O3, N2O5 and H2SO3.

Step 1 · Sort each oneIron is a metal, so the first is ionic. Nitrogen and oxygen are both nonmetals, so the second is molecular. The third starts with hydrogen and is an acid.

Step 2 · Work out the iron chargeOxide is always O2-, and three of them give 3 × (2−) = 6−. The two iron ions must supply 6+, so each one is 6 ÷ 2 = 3+. That makes it iron(III) oxide.

Step 3 · Count the molecular oneTwo nitrogens give di, and five oxygens give penta. Drop the final a before oxide. That makes it dinitrogen pentoxide.

Step 4 · Follow the acid endingThe anion here is sulfite, SO32-. An -ite ion gives an -ous acid, so this is sulfurous acid.

CheckRun each name backwards. Iron(III) oxide pairs Fe3+ with O2-, which balances at Fe2O3. Dinitrogen pentoxide rebuilds N2O5. Sulfurous acid needs two hydrogens for a 2− ion, giving H2SO3.

Key points
  • Sort the formula first: acid, ionic or molecular. Each type has its own rules.
  • Ionic names use no prefixes. The anion ends in -ide unless it is polyatomic.
  • A metal with more than one charge needs a Roman numeral, read off by balancing.
  • Molecular names use prefixes on both elements, except mono on the first.
  • Binary acids are hydro...-ic. For oxyacids, -ate gives -ic acid and -ite gives -ous acid.

The chapter in one breath

  • Three particles build everything. Protons and neutrons pack the nucleus, and electrons move outside it.
  • Protons are the name tag. The atomic number Z counts them, and the mass number A adds the neutrons on.
  • Isotopes differ only in neutrons. They act the same in reactions, because they hold the same electrons.
  • The table mass is a weighted average. Multiply each isotope mass by its share, then add.
  • The table is a behavior map. Columns are groups and share behavior. Rows are periods.
  • Ions are atoms that gained or lost electrons. Group number predicts the charge for most main group elements.
  • An ionic formula is a charge balance. Find the smallest counts that sum to zero charge.
  • Molecules share electrons instead. Molecular, empirical and structural formulas each tell you a different thing.
  • Naming has three roads. Decide acid, ionic or molecular first, then apply that road's rules.

ReferenceWords worth knowing

Every term below showed up in bold somewhere above. Come back here whenever a word stops making sense.

alkali metals
Group 1 metals. Soft, very reactive, and they form 1+ ions.
anion
An ion with a negative charge, made by gaining one or more electrons.
atomic mass unit (u)
A mass unit equal to 1/12 the mass of a carbon-12 atom, or 1.6605 × 10-24 g.
atomic number (Z)
The number of protons in a nucleus. It decides which element the atom is.
average atomic mass
The isotope masses of an element weighted by their abundances and added. This is the value printed on the table.
cation
An ion with a positive charge, made by losing one or more electrons.
covalent bond
A pair of electrons shared between two atoms.
empirical formula
The smallest whole-number ratio of atoms in a compound, such as CH2O.
formula unit
The smallest whole-number ratio of ions that makes an ionic compound neutral.
group
A column of the periodic table, numbered 1 to 18. Members behave alike.
ion
An atom or bonded group of atoms that carries a net electric charge.
ionic compound
A compound of cations and anions held in a lattice, usually a metal with a nonmetal.
isotopes
Atoms of one element with different neutron counts, so the same Z and different A.
mass number (A)
Protons plus neutrons in one nucleus. Always a whole number.
metalloid
An element on the step between metals and nonmetals that behaves partly like each.
molecular formula
The actual count of each kind of atom in one molecule, such as C6H12O6.
oxyacid
An acid of hydrogen plus a polyatomic ion containing oxygen, such as HNO3.
percent abundance
The share of atoms in a natural sample that are one particular isotope.
period
A row of the periodic table, numbered 1 to 7.
polyatomic ion
A bonded group of atoms carrying one charge for the whole group, such as SO42-.

Check yourself

1. An atom holds 15 protons and 16 neutrons. Name it and give its mass number.
Phosphorus-31. Z = 15 names it phosphorus. A = 15 + 16 = 31. The neutron count never changes the name.
2. A particle holds 20 protons, 20 neutrons and 18 electrons. Write its symbol with charge.
40Ca2+ Z = 20 is calcium. A = 20 + 20 = 40. Charge = 20 − 18 = 2+, so two electrons are missing.
3. Copper is 69.15% copper-63 at 62.9296 u and 30.85% copper-65 at 64.9278 u. Find its average atomic mass.
63.55 u. 0.6915 × 62.9296 u = 43.5158 u. 0.3085 × 64.9278 u = 20.0302 u. Adding gives 63.5460 u, so 63.55 u. It leans toward copper-63, which is the common one.
4. Boron is 19.9% boron-10 at 10.0129 u and 80.1% boron-11 at 11.0093 u. Find the average atomic mass.
10.81 u. 0.199 × 10.0129 u = 1.9926 u. 0.801 × 11.0093 u = 8.8184 u. The sum is 10.8110 u, which rounds to 10.81 u. Here the heavier isotope is the common one, so the answer sits near 11.
5. Which is larger for carbon-13, the mass number or the atomic mass? Why are they not the same thing?
The atomic mass, 13.00335 u, is a little larger than the mass number of 13. A mass number is a count of particles, so it is always a whole number. An atomic mass is a measured mass in u. Only carbon-12 has the two numbers equal, because the unit was defined from it.
6. Give the group and period of selenium, Z = 34, and say whether it is a metal.
Group 16, period 4, and it is a nonmetal. Period 4 runs from Z = 19 to Z = 36, so 34 lands there. Counting across gives 34 − 18 = group 16, the same group as oxygen and sulfur.
7. Sort these as metal, nonmetal or metalloid: Si, Ca, Ne, Sb.
Si metalloid, Ca metal, Ne nonmetal, Sb metalloid. Silicon and antimony both sit on the step. Calcium is a group 2 metal. Neon is a noble gas, far to the right of the step.
8. Write the formula for the compound of Mg2+ and N3-.
Mg3N2 The charge sizes are 2 and 3, and their lowest common multiple is 6. You need 6 ÷ 2 = 3 magnesium ions and 6 ÷ 3 = 2 nitride ions. Check: 3(2+) = 6+ and 2(3−) = 6−, which cancel.
9. Write the formula for iron(III) sulfate.
Fe2(SO4)3 The numeral says Fe3+, and sulfate is SO42-. Sizes 3 and 2 give 6, so you need 2 iron ions and 3 sulfate ions. Check: 2(3+) = 6+ and 3(2−) = 6−. The parentheses keep the 3 from changing the oxygen count.
10. Give the empirical formula of C6H6 and of C4H10.
CH and C2H5. For C6H6 the largest shared factor of 6 and 6 is 6, and 6 ÷ 6 = 1 for each, giving CH. For C4H10 the largest shared factor of 4 and 10 is 2, and 4 ÷ 2 = 2 with 10 ÷ 2 = 5, giving C2H5.
11. Name these: (a) K2O, (b) Cu(NO3)2, (c) SF6, (d) HBr in water.
(a) potassium oxide, (b) copper(II) nitrate, (c) sulfur hexafluoride, (d) hydrobromic acid. In (b) the two nitrate ions carry 2− in total, so copper must be 2+. In (c) both parts are nonmetals, so prefixes apply. In (d) a binary acid takes hydro... plus -ic acid.
12. Write formulas for: (a) ammonium carbonate, (b) dinitrogen pentoxide, (c) phosphoric acid.
(a) (NH4)2CO3, (b) N2O5, (c) H3PO4. In (a) two 1+ ammonium ions balance one 2− carbonate. In (b) the prefixes give the subscripts directly. In (c) phosphoric comes from phosphate, PO43-, so three hydrogens are needed.
Chemistry · Chapter 3

Composition of Substances and Solutions

Atoms are far too small to count one at a time. A single grain of salt holds more of them than you could tally in a lifetime. So chemists count by weighing instead. This chapter hands you the tool that makes that trade work. It is called the mole, and it ties the mass on a balance to the number of particles you cannot see. From there you will find the mass of any formula, split a compound into the share of mass each element owns, and work backward from lab data to a formula. At the end you will mix and water down solutions to an exact strength.

By the end of this chapter you can

  • Use Avogadro's number to move between moles and particle counts.
  • Find the molar mass of any formula, parentheses and hydrates included.
  • Convert grams to moles and moles to grams in either direction.
  • Work out the percent by mass of every element in a compound.
  • Turn mass or percent data into an empirical formula.
  • Scale an empirical formula up to a molecular formula.
  • Mix a solution to a target molarity and dilute a stock solution to order.

Section 1The mole and Avogadro's number

Chemistry needs a counting word

You already use counting words every day. A dozen means 12. A ream of paper means 500. Each word stands for a fixed count, whatever you point it at. A dozen eggs and a dozen bricks hold the same number of things. Their masses are nothing alike.

Chemistry needs a counting word too, but a huge one. A teaspoon of water holds roughly 1.7 × 1023 molecules. Counting them one per second would take longer than the age of the universe. The mole is the counting word that makes such numbers workable. One mole is 6.022 × 1023 of whatever you are counting. The unit is written mol.

That count has a name of its own. Avogadro's number, written NA, is 6.022 × 1023 particles per mole. The exact value is now fixed by definition. One mole is 6.02214076 × 1023 particles, no more and no less. Four digits is plenty for the work in this book.

Why such a strange count

The number looks random. It was chosen with great care. It was set so that one mole of carbon-12 atoms weighs almost exactly 12 grams. That link is the whole point of the mole. It means the mass of one mole in grams matches the mass of one particle in atomic mass units. Carbon has an atomic mass of 12.011 amu, so one mole of carbon atoms weighs 12.011 g. You weigh grams on a balance. You get a count for free.

The word particle is doing real work in that definition. A mole counts whatever you aim it at, so you have to say what. A mole of neon is 6.022 × 1023 atoms. A mole of water is that many molecules. An ionic solid like NaCl has no molecules at all, so we count formula units instead. One formula unit of NaCl is one Na+ paired with one Cl-.

Crossing the bridge

Moles sit in the middle of every problem in this chapter. Mass sits on one side. Particle count sits on the other. To go from moles to particles, multiply by 6.022 × 1023 per mole. To come back, divide by it. Write the unit on every number and let the units cancel. If what is left over is not the unit you wanted, your setup is wrong, not your calculator.

The mole bridge: mass, moles and particle count Mass grams on a balance Moles the middle of every problem Particles atoms, molecules, units divide by molar mass g × (1 mol / g) multiply by molar mass mol × (g / 1 mol) multiply by 6.022 × 10²³ mol × (particles / 1 mol) divide by 6.022 × 10²³ particles × (1 mol / particles) Every trip between the two ends passes through moles.
Figure 3.1 The mole bridge. Molar mass connects grams to moles. Avogadro's number connects moles to a count. There is no shortcut from grams straight to particles.

Worked example · moles to particles and back

You have 2.50 mol of water. How many molecules is that? How many hydrogen atoms?

Step 1 · read the amountYou hold 2.50 mol of H2O. Three significant figures.

Step 2 · pick the factorOne mole holds 6.022 × 1023 molecules, so the factor is (6.022 × 1023 molecules / 1 mol).

Step 3 · multiply2.50 mol × 6.022 × 1023 molecules/mol = 1.5055 × 1024, which rounds to 1.51 × 1024 molecules.

Step 4 · check the unitsMoles cancel and molecules remain. Three figures went in, so three come out.

Step 5 · go one step furtherEach H2O carries 2 H atoms. So 2.50 mol H2O holds 2 × 2.50 = 5.00 mol H. That is 5.00 mol × 6.022 × 1023/mol = 3.01 × 1024 H atoms.

Step 6 · run it backwardA different sample holds 1.20 × 1024 iron atoms. Divide by Avogadro's number. (1.20 × 1024) / (6.022 × 1023/mol) = 1.99 mol Fe.

Key points
  • One mole is 6.022 × 1023 particles. The unit of NA is per mole.
  • A mole counts atoms, molecules or formula units. Always name which one.
  • Moles to particles, multiply by NA. Particles to moles, divide by NA.
  • Equal moles of two substances hold equal counts. They do not weigh the same.
  • The mole was built so that one mole of carbon-12 weighs 12 g.

Section 2Molar mass

The mass of one mole

Molar mass is the mass of one mole of a substance. Its unit is grams per mole, written g/mol. For a single element you can read it straight off the periodic table. Take the atomic mass and attach g/mol to it. Carbon is 12.011 g/mol. Oxygen is 15.999 g/mol. Iron is 55.845 g/mol.

For a compound you add up the atoms in the formula. Each subscript tells you how many of that atom to count. Multiply each atomic mass by its subscript, then add. The total is the formula mass in amu. It is also the molar mass in g/mol. The number is the same and only the unit changes.

Water is H2O. Two hydrogens at 1.008 plus one oxygen at 15.999 gives 18.015. So water is 18.02 g/mol. Through this book we round molar masses to two decimal places. That is already finer than any balance you will use.

Parentheses and raised dots

Some formulas carry a group inside parentheses. The subscript outside multiplies everything within the group. Ca(NO3)2 holds one Ca, two N and six O. Miss that and your answer lands far from the truth. The same rule holds for a long formula like Al2(SO4)3, which holds 2 Al, 3 S and 12 O for a total of 342.13 g/mol.

Other formulas carry a raised dot, as in CuSO4·5H2O. A solid like this is a hydrate. Water molecules sit inside the crystal in a fixed ratio. The dot does not mean multiply. It means that much water is locked in. Add the water in as five whole units of H2O. The pentahydrate comes out at 249.68 g/mol. Drive the water off with heat and the dry salt is only 159.61 g/mol. Forgetting the water throws a weighing off by more than a third.

One factor, two faces

Molar mass is a conversion factor you can flip. Written as g/mol it turns moles into grams. Flipped to mol/g it turns grams into moles. Choose the face that cancels the unit you are holding. That choice is the only decision in most of these problems.

Where the 164.09 grams in one mole of calcium nitrate sit One mole of Ca(NO₃)₂ weighs 164.09 g Bar width shows each element's share of that mass. Ca 24.4% N 17.1% O 58.5% 1 × 40.08 = 40.08 g 2 × 14.007 = 28.01 g 6 × 15.999 = 95.99 g total 164.09 g/mol Six oxygens hide inside the parentheses. They own most of the mass.
Figure 3.2 Molar mass is a sum, not a single lookup. In calcium nitrate the two nitrogens and six oxygens together outweigh the calcium by three to one.

Worked example · molar mass of calcium nitrate

Step 1 · count the atomsCa(NO3)2 gives 1 Ca, 2 N and 6 O. The 2 outside multiplies both the N and the three O inside.

Step 2 · weigh each elementCa: 1 × 40.08 = 40.08 g/mol. N: 2 × 14.007 = 28.014 g/mol. O: 6 × 15.999 = 95.994 g/mol.

Step 3 · add them40.08 + 28.014 + 95.994 = 164.088 g/mol.

Step 4 · roundThe molar mass is 164.09 g/mol.

Worked example · grams and moles, both directions

First, grams to moles. You weigh out 25.0 g of glucose, C6H12O6.

Step 1 · build the molar massC: 6 × 12.011 = 72.066. H: 12 × 1.008 = 12.096. O: 6 × 15.999 = 95.994. Sum = 180.156, so 180.16 g/mol.

Step 2 · flip the factorYou hold grams and want moles, so use (1 mol / 180.16 g).

Step 3 · divide25.0 g × (1 mol / 180.16 g) = 0.1388 mol, which rounds to 0.139 mol.

Step 4 · now go the other wayHow much does 0.350 mol of NaCl weigh? NaCl is 22.990 + 35.45 = 58.44 g/mol.

Step 5 · multiply0.350 mol × 58.44 g/mol = 20.454 g, which rounds to 20.5 g. Moles cancel and grams are left.

Key points
  • Molar mass is grams per mole. It matches the formula mass in amu.
  • Multiply each atomic mass by its subscript, then add every term.
  • A subscript outside parentheses multiplies the whole group inside.
  • A raised dot means water locked in the crystal. Count it in.
  • g/mol turns moles into grams. mol/g turns grams into moles.

Section 3Percent composition

Who owns the mass

A pure compound has a fixed recipe. Every sample of pure glucose carries the same share of carbon by mass, whether it came from a grape or a beaker. Percent composition reports that share for each element in the formula. It answers a very practical question. In 100 g of this substance, how many grams are carbon?

The method is one line long. Take exactly one mole of the compound. For each element, divide the mass that element contributes by the molar mass of the whole thing. Multiply by 100 to turn the fraction into a percent. The result for one element is its mass percent.

mass percent of X = (mass of X in one mole ÷ molar mass) × 100%

Notice that the compound's identity is doing all the work. You never need to know how big your sample is. A grain and a truckload give the same percents. That is why percent composition is printed on labels and quoted in catalogs.

A number you can reuse

Percent composition is more than a tidy fact. Once you have it, you can jump straight from the mass of a sample to the mass of one element inside it. Turn the percent into a decimal and multiply. A fertilizer bag marked 35.00% nitrogen says just that. A 50.0 kg bag holds 0.3500 × 50.0 kg = 17.5 kg of nitrogen. No moles needed.

The shares always add to 100%, give or take a little rounding. That hands you a free check. If your percents sum to 94% or 108%, something is wrong. Go back and recount the atoms in your molar mass. A missed oxygen or a dropped parenthesis is the usual culprit.

Percent composition of glucose by mass 180.16 g/mol Glucose, C₆H₁₂O₆ Oxygen 53.28% 95.994 g of the 180.156 g in one mole Carbon 40.00% 72.066 g of the 180.156 g in one mole Hydrogen 6.71% 12.096 g of the 180.156 g in one mole Twelve hydrogens, yet the lightest share. Mass is not a head count.
Figure 3.3 Glucose has twice as many hydrogen atoms as carbon atoms, but hydrogen owns under 7% of the mass. Percent composition tracks mass, never number.

Worked example · percent composition of glucose

Find the mass percent of each element in C6H12O6, then use it on a 5.00 g sample.

Step 1 · molar mass72.066 + 12.096 + 95.994 = 180.156 g/mol, or 180.16 g/mol.

Step 2 · mass from each elementCarbon 72.066 g, hydrogen 12.096 g, oxygen 95.994 g, all per mole.

Step 3 · divide and scaleC: (72.066 / 180.156) × 100 = 40.00%. H: (12.096 / 180.156) × 100 = 6.71%. O: (95.994 / 180.156) × 100 = 53.28%.

Step 4 · check the sum40.00 + 6.71 + 53.28 = 99.99%. That is 100% apart from rounding, so the work holds.

Step 5 · use itA 5.00 g sample holds 0.4000 × 5.00 g = 2.00 g of carbon.

Worked example · nitrogen in a fertilizer salt

Ammonium nitrate, NH4NO3, is sold for its nitrogen. What percent of its mass is nitrogen?

Step 1 · count carefullyThere are two N, four H and three O. Both nitrogens count, one in the ammonium and one in the nitrate.

Step 2 · molar massN: 2 × 14.007 = 28.014. H: 4 × 1.008 = 4.032. O: 3 × 15.999 = 47.997. Sum = 80.043, so 80.04 g/mol.

Step 3 · divide and scale(28.014 / 80.043) × 100 = 35.00% N.

Step 4 · checkH works out to 5.04% and O to 59.96%. The three add to 100.00%.

Key points
  • Mass percent is the element's mass in one mole divided by the molar mass, times 100.
  • Percent composition does not depend on sample size. It belongs to the compound.
  • Count every atom of an element, even when it appears in two places in the formula.
  • The percents should sum to 100%. Use that as a built-in check.
  • Percent as a decimal times sample mass gives the mass of that element.

Section 4Empirical and molecular formulas

Two kinds of formula

A molecular formula tells you exactly how many atoms sit in one molecule. An empirical formula gives the smallest whole-number ratio of those same atoms. Glucose is C6H12O6. Divide all three subscripts by 6 and you get CH2O. That is the empirical formula of glucose.

Very different compounds can share one empirical formula. Formaldehyde really is CH2O. Acetic acid is C2H4O2, which also reduces to CH2O. All three have the very same percent composition. So percent data by itself can never hand you a molecular formula. You need one more measurement, and that measurement is the molar mass.

Ionic compounds are a special case. A crystal has no molecules, so its formula is already the simplest ratio. NaCl and Fe2O3 are empirical formulas.

From lab data to a ratio

Analysis gives you masses or percents. It never hands you subscripts. The fix is to turn mass into moles, because subscripts are mole ratios. Four steps do it every time.

  1. Start with grams. If you were given percents, pretend you hold exactly 100.00 g of the compound. Then each percent is already a mass in grams.
  2. Divide each mass by that element's molar mass. Now every number is in moles.
  3. Divide every mole value by the smallest of them. The smallest becomes 1.000 and the rest fall close to whole numbers.
  4. If a value sits near .50, .33 or .25, multiply all of them by 2, 3 or 4 to clear it.

Judgment lives in step 4. A value of 1.99 is 2, since lab data is never perfect. A value of 1.50 is not 2. Rounding it would change the compound. Multiply instead.

Scaling up to the molecule

The empirical formula has a mass of its own, called the empirical formula mass. Divide the real molar mass by it. The answer is a whole number n, the number of empirical units per molecule. Multiply every subscript by n and you have the molecular formula.

n = molar mass ÷ empirical formula mass

If n comes out as 1, the two formulas are the same. Water and ammonia are both like that. If n lands on 2.98 or 3.03, treat it as 3 and move on.

From percent data to an empirical formula to a molecular formula Percents 40.00% C 6.71% H, 53.28% O Grams assume 100.00 g 40.00 g, 6.71 g, 53.28 g Moles 3.330 C, 6.657 H 3.330 O Ratio 1.000 : 1.999 : 1.000 CH₂O Molecule n = 180.16 / 30.03 = 6 C₆H₁₂O₆ percent → grams divide by atomic mass divide by the smallest multiply by n Stuck on a ratio like 1 : 1.50 or 1 : 1.33? Multiply every value by 2 or by 3. Subscripts must be whole numbers, so 1 : 1.50 becomes 2 : 3.
Figure 3.4 The route from analysis to a formula. Percent data stops at the empirical formula. Only a measured molar mass gets you the last box.

Worked example · percents to a molecular formula

A compound is 40.00% C, 6.71% H and 53.28% O by mass. A separate measurement puts its molar mass at 180.16 g/mol. Find both formulas.

Step 1 · assume 100.00 gThat gives 40.00 g C, 6.71 g H and 53.28 g O.

Step 2 · grams to molesC: 40.00 / 12.011 = 3.330 mol. H: 6.71 / 1.008 = 6.657 mol. O: 53.28 / 15.999 = 3.330 mol.

Step 3 · divide by the smallestThe smallest is 3.330. C: 3.330 / 3.330 = 1.000. H: 6.657 / 3.330 = 1.999. O: 3.330 / 3.330 = 1.000.

Step 4 · read the empirical formulaThe ratio is 1 : 2 : 1, so the empirical formula is CH2O.

Step 5 · empirical formula mass12.011 + 2(1.008) + 15.999 = 30.026, or 30.03 g/mol.

Step 6 · find nn = 180.16 / 30.03 = 5.999, which is 6.

Step 7 · scale upMultiply each subscript by 6. The molecular formula is C6H12O6.

Worked example · masses to a formula

A sample of an iron oxide is broken down and found to hold 11.17 g of Fe and 4.80 g of O. What is its formula?

Step 1 · grams to molesFe: 11.17 / 55.845 = 0.2000 mol. O: 4.80 / 15.999 = 0.3000 mol.

Step 2 · divide by the smallestFe: 0.2000 / 0.2000 = 1.000. O: 0.3000 / 0.2000 = 1.500.

Step 3 · clear the half1.500 is not close to a whole number. Multiply both by 2. Fe becomes 2 and O becomes 3.

Step 4 · write itThe formula is Fe2O3, iron(III) oxide, at 159.69 g/mol. It is ionic, so this is both its empirical and its only formula.

Key points
  • An empirical formula is the smallest whole-number atom ratio. A molecular formula is the real count per molecule.
  • Subscripts are mole ratios, so always convert mass to moles first.
  • Percents alone give the empirical formula. Molar mass is needed for the molecular one.
  • Divide by the smallest mole value, then multiply out any stubborn halves or thirds.
  • n = molar mass divided by empirical formula mass, and n is always a whole number.

Section 5Molarity

Strength is not amount

A solution is a mixture so evenly blended that it acts like one substance. The part that dissolves is the solute. The part it dissolves into is the solvent. When the solvent is water we call it an aqueous solution and tag the formula with (aq).

Picture two glasses of the same salt water. One holds a sip and one holds a liter. The amount of salt in them is very different. The strength is identical. Concentration is the word for that strength. It is the amount of solute per amount of solution, and it does not care how much you poured.

Molarity is the concentration unit chemistry leans on hardest. It is moles of solute per liter of solution. The symbol is a capital M. A bottle marked 2.0 M is read out loud as two molar.

molarity (M) = moles of solute ÷ liters of solution

Of solution, not of solvent

Read the bottom of that fraction again. It says liters of solution. It does not say liters of solvent. The difference is real work in the lab. You do not pour 1.00 L of water onto your solid. You drop the solid into a volumetric flask, swirl it into part of the water, then top up to the etched mark on the neck. The solute takes up room of its own. Filling to the mark is the only way to know your final volume.

Molarity has one soft spot. Liquids swell when warmed, so a volume shifts a little with temperature. A 0.100 M solution on a hot bench is a hair weaker than the same flask in a cold room. For ordinary lab work the drift is too small to matter.

Three ways the equation gets used

Molarity, moles and volume form a small triangle. Know any two and you can get the third. Moles = M × V. Volume = moles ÷ M. Convert the volume to liters before anything else. 250.0 mL is 0.2500 L, and 25.0 mL is 0.0250 L.

Making a solution to a known molarity in a volumetric flask etched mark 250.0 mL of solution Solid goes in first, then water up to the line. 11.7 g of NaCl, topped up to 250.0 mL 1. grams to moles 11.7 g ÷ 58.44 g/mol = 0.2002 mol 2. milliliters to liters 250.0 mL = 0.2500 L of solution 3. divide 0.2002 mol ÷ 0.2500 L = 0.8008 mol/L 0.801 M NaCl three figures, set by 11.7 g
Figure 3.5 Molarity counts liters of finished solution, not liters of water. The flask is filled to the etched mark after the solid has dissolved.

Worked example · molarity from a weighed mass

You dissolve 11.7 g of NaCl and top the flask up to 250.0 mL. What is the molarity?

Step 1 · grams to molesNaCl is 58.44 g/mol. 11.7 g × (1 mol / 58.44 g) = 0.2002 mol.

Step 2 · volume to liters250.0 mL × (1 L / 1000 mL) = 0.2500 L.

Step 3 · divide0.2002 mol ÷ 0.2500 L = 0.8008 mol/L.

Step 4 · roundThe mass had three figures, so the answer is 0.801 M.

Worked example · working back to a mass

You need 500.0 mL of 0.150 M glucose. How much glucose do you weigh out?

Step 1 · find the molesmoles = M × V = 0.150 mol/L × 0.5000 L = 0.0750 mol.

Step 2 · moles to grams0.0750 mol × 180.16 g/mol = 13.512 g, so 13.5 g.

Step 3 · make it upWeigh 13.5 g into a 500.0 mL volumetric flask. Swirl it into some water. Top up to the mark and invert to mix.

Step 4 · one more useHow many moles sit in a 25.0 mL pipette of 0.100 M HCl? 0.0250 L × 0.100 mol/L = 2.50 × 10-3 mol.

Key points
  • Molarity is moles of solute per liter of solution, written M.
  • The volume is the finished solution, measured to the mark, not the water you started with.
  • moles = M × V, and V = moles ÷ M. Put volume in liters first.
  • To build a solution from a solid, find the moles you need, then convert to grams.
  • Equal volumes of the same molarity hold equal moles of solute.

Section 6Dilution

Add water, keep the solute

Labs rarely weigh out a solid for every weak solution they need. They mix a strong one and water it down. The strong one is the stock solution. Watering it down is dilution.

One idea makes every dilution problem easy. Adding water adds no solute and removes none. The moles before equal the moles after. Only the volume grew, so the concentration fell.

Moles are M × V, so that sentence becomes an equation. Subscript 1 is before and 2 is after.

M1V1 = M2V2

Both sides of that line are moles of solute. The same volume unit appears on each side, so you may leave both volumes in milliliters. The only rule is to use one unit throughout.

Working out the pour

Usually you know the stock concentration, the one you want and the final volume. The unknown is how much stock to measure out. Rearrange for V1.

V1 = M2V2 ÷ M1

Measure out that much stock, then add solvent up to V2. With a strong acid the order matters. Acid goes into water, never water into acid, because mixing releases a lot of heat.

The dilution factor

The dilution factor is V2 ÷ V1. It also equals M1 ÷ M2, a handy second check. Take 10.0 mL up to 100.0 mL and the factor is 10. The concentration drops to one tenth. That is a 1:10 dilution.

Factors multiply when you chain steps. Two 1:10 steps in a row give a factor of 100. This is called a serial dilution. It is how a lab reaches very low concentrations without weighing out a speck of solid.

Dilution keeps the solute and grows the volume Before: the stock V₁ = 25.0 mL M₁ = 0.800 M add water no solute in, no solute out After: the dilution V₂ = 200.0 mL M₂ = 0.100 M M₁V₁ = M₂V₂ 0.800 × 25.0 = 20.0 mmol 0.100 × 200.0 = 20.0 mmol Eight dots in each beaker. Volume grew 8 times over, so molarity fell 8 times over. Dilution factor = 8.00
Figure 3.6 Both beakers hold the same eight particles of solute. Only the volume changed, so the drop in molarity matches the growth in volume exactly.

Worked example · how much stock to pour

You need 500.0 mL of 0.500 M HCl. The shelf holds 12.0 M HCl. How much stock do you measure?

Step 1 · list what you knowM1 = 12.0 M. M2 = 0.500 M. V2 = 500.0 mL.

Step 2 · rearrangeV1 = M2V2 ÷ M1.

Step 3 · put the numbers inV1 = (0.500 M × 500.0 mL) ÷ 12.0 M = 250 ÷ 12.0 = 20.83 mL, so 20.8 mL.

Step 4 · check both sidesStock side: 12.0 mol/L × 0.0208 L = 0.250 mol. Final side: 0.500 mol/L × 0.5000 L = 0.250 mol. The two agree.

Step 5 · do it safelyAdd the acid to water already in the flask, then top up to 500.0 mL.

Worked example · the new concentration

You pipette 25.0 mL of 0.800 M KCl into a flask and top up to 200.0 mL. Find the new molarity.

Step 1 · rearrange for M2M2 = M1V1 ÷ V2.

Step 2 · substituteM2 = (0.800 M × 25.0 mL) ÷ 200.0 mL = 20.0 ÷ 200.0 = 0.100 M.

Step 3 · check with the factorThe volume grew 200.0 / 25.0 = 8.00 times, so the molarity should fall 8.00 times. 0.800 ÷ 8.00 = 0.100 M. It agrees.

Key points
  • Dilution adds solvent only. The moles of solute never change.
  • M1V1 = M2V2, with before on the left and after on the right.
  • Both volumes may stay in milliliters, if you use one unit on both sides.
  • V1 = M2V2 ÷ M1 tells you how much stock to measure.
  • The dilution factor is V2/V1 and also M1/M2. Chained steps multiply.
  • Pour strong acid into water, not water into acid.

The chapter in one breath

  • A mole is 6.022 × 1023 particles. It is a counting word, like a dozen, sized for atoms.
  • Molar mass in g/mol is the sum of the atomic masses in a formula. It links grams to moles.
  • Avogadro's number links moles to a count. Every problem runs through moles in the middle.
  • Percent composition is each element's mass in one mole divided by the molar mass, times 100.
  • Percent composition belongs to the compound, so sample size never enters the math.
  • An empirical formula is the smallest whole-number atom ratio. Convert mass to moles, then divide by the smallest.
  • Molar mass divided by empirical formula mass gives n, and n scales the ratio up to the molecular formula.
  • Molarity is moles of solute per liter of solution, measured to the mark in a volumetric flask.
  • Dilution keeps the solute and grows the volume, so M1V1 = M2V2.

ReferenceWords worth knowing

These are the terms this chapter leans on. Each one shows up in the worked examples above, so read them with those numbers in mind.

Aqueous solution
A solution whose solvent is water. Written with the tag (aq) after a formula.
Avogadro's number
The number of particles in one mole, NA = 6.022 × 1023 per mole. Fixed by definition at 6.02214076 × 1023.
Concentration
The amount of solute per amount of solution. A measure of strength, not of quantity.
Dilution
Adding solvent to a solution to lower its concentration. The moles of solute stay the same.
Dilution factor
V2/V1, equal to M1/M2. A factor of 10 means a 1:10 dilution.
Empirical formula
The smallest whole-number ratio of atoms in a compound. CH2O is the empirical formula of glucose.
Formula mass
The sum of the atomic masses in a formula, in amu. Numerically equal to the molar mass in g/mol.
Formula unit
The smallest neutral group an ionic compound's formula describes, such as one Na+ with one Cl-.
Hydrate
A solid that holds water inside its crystal in a fixed ratio, shown with a raised dot, as in CuSO4·5H2O.
Mass percent
The share of a compound's mass owned by one element, as a percent. Also called percent by mass.
Molar mass
The mass of one mole of a substance, in grams per mole. The bridge between grams and moles.
Molarity
Moles of solute per liter of solution, symbol M. The most used concentration unit in this book.
Mole
The counting unit for particles. One mole is 6.022 × 1023 atoms, molecules or formula units.
Molecular formula
The actual number of atoms of each element in one molecule, such as C6H12O6.
Percent composition
The full set of mass percents for a compound. The shares add to 100%.
Solute
The substance that dissolves in a solution. It is usually the smaller part by mass.
Solution
A mixture blended so evenly that it behaves as a single substance.
Solvent
The substance that does the dissolving. Water is the most common one.
Stock solution
A concentrated solution kept on hand and diluted when a weaker one is needed.
Volumetric flask
A flask with one etched line on its neck, used to make a solution up to an exact volume.

Check yourself

1. How many moles are in 54.0 g of water?
3.00 mol Water is 2(1.008) + 15.999 = 18.015, so 18.02 g/mol. Then 54.0 g × (1 mol / 18.02 g) = 2.997 mol, which rounds to 3.00 mol. That is also 3.00 × 6.022 × 1023 = 1.81 × 1024 molecules.
2. A friend says a mole of anything weighs the same. What is wrong with that?
A mole is a count, not a mass. One mole always holds 6.022 × 1023 particles, but those particles have different masses. A mole of water weighs 18.02 g while a mole of glucose weighs 180.16 g.
3. Find the molar mass of Al2(SO4)3.
342.13 g/mol The 3 outside multiplies the whole sulfate group, so there are 2 Al, 3 S and 12 O. Al: 2 × 26.982 = 53.964. S: 3 × 32.06 = 96.18. O: 12 × 15.999 = 191.988. Sum = 342.132 g/mol.
4. What percent of the mass of CO2 is oxygen?
72.71% CO2 is 12.011 + 2(15.999) = 44.009, so 44.01 g/mol. Oxygen contributes 31.998 g per mole. (31.998 / 44.009) × 100 = 72.71%. Carbon takes the other 27.29%.
5. A manganese oxide is 69.6% Mn and 30.4% O. What is its empirical formula?
Mn2O3 Assume 100.00 g. Mn: 69.6 / 54.938 = 1.267 mol. O: 30.4 / 15.999 = 1.900 mol. Divide both by 1.267. Mn = 1.000 and O = 1.500. Multiply by 2 to clear the half, giving Mn2O3.
6. How many grams of KMnO4 make 100.0 mL of 0.0200 M solution?
0.316 g moles = 0.0200 mol/L × 0.1000 L = 2.00 × 10-3 mol. KMnO4 is 39.098 + 54.938 + 4(15.999) = 158.032, so 158.03 g/mol. Mass = 2.00 × 10-3 mol × 158.03 g/mol = 0.316 g.
7. What volume of 6.00 M NaOH gives 250.0 mL of 0.150 M NaOH?
6.25 mL Use V1 = M2V2 ÷ M1 = (0.150 × 250.0) ÷ 6.00 = 37.5 ÷ 6.00 = 6.25 mL. Measure that out, then add water up to 250.0 mL.
8. Why does molarity use liters of solution rather than liters of solvent?
Because the solute takes up room too. If you added one liter of water to a solid, the finished volume would be more than one liter. Topping up to a flask's mark is the only way to fix the volume you divide by.
9. Benzene is C6H6. What is its empirical formula, and why is that not enough to identify it?
CH, and because many compounds share it. Dividing both subscripts by 6 gives CH. Acetylene, C2H2, reduces to CH as well. Only the molar mass tells the two apart, since n = 78.11 / 13.02 = 6 for benzene and 2 for acetylene.
10. How many oxygen atoms are in 1.50 mol of Ca(NO3)2?
5.42 × 1024 atoms Each formula unit holds 6 O, so 1.50 mol × 6 = 9.00 mol O. Then 9.00 mol × 6.022 × 1023/mol = 5.42 × 1024 oxygen atoms.
11. You dilute 50.0 mL of 0.200 M KCl to 500.0 mL. How many moles of KCl are in the flask afterward?
0.0100 mol, the same as before. Dilution adds no solute. Before: 0.0500 L × 0.200 mol/L = 0.0100 mol. After: M2 = (0.200 × 50.0) / 500.0 = 0.0200 M, and 0.5000 L × 0.0200 mol/L = 0.0100 mol.
12. Formaldehyde, acetic acid and glucose all give 40.00% C, 6.71% H and 53.28% O. What extra fact separates them?
The molar mass. All three share the empirical formula CH2O, whose mass is 30.03 g/mol. Formaldehyde is 30.03 g/mol so n = 1. Acetic acid is 60.05 g/mol so n = 2. Glucose is 180.16 g/mol so n = 6.
Chemistry · Chapter 4

Stoichiometry of Chemical Reactions

A chemical equation is a recipe written in formulas. It names what goes in, what comes out, and in what ratio. This chapter turns that recipe into numbers you can trust. First you will balance an equation so that no atom appears or vanishes. Then you will sort reactions into a few families, so a new one looks familiar. You will track oxidation numbers to see which atoms gave electrons away. Last you will use the balanced equation as a bridge. Grams become moles, moles of one substance become moles of another, and moles become grams again. By the end you can predict how much product a real run should give.

By the end of this chapter you can

  • Balance a chemical equation by counting atoms on both sides.
  • Write the molecular, complete ionic and net ionic forms of a reaction in water.
  • Use solubility rules to predict whether a solid falls out of a mixture.
  • Spot the acid and base family and the redox family from an equation.
  • Assign oxidation numbers and name what was oxidized and what was reduced.
  • Convert grams to moles to moles to grams with a mole ratio.
  • Find the reactant that runs out first, then report theoretical and percent yield.

Section 1Writing and balancing equations

What a chemical equation says

A chemical equation is a short written record of a chemical change. The starting substances, called the reactants, sit on the left side. The substances that form, called the products, sit on the right side. An arrow joins the two sides, and that arrow means turns into. Here is methane burning in air.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

The letters in parentheses record the physical state of each substance. Use (s) for a solid, (l) for a liquid and (g) for a gas. Use (aq) whenever the substance is dissolved in water. That last label is short for aqueous, which simply means dissolved in water.

Two kinds of number show up in an equation, and they do completely different jobs. The large number in front of a formula is a coefficient, and it counts whole units of that substance. The small lowered number inside a formula is a subscript, and it counts atoms inside a single unit. You may change a coefficient whenever balancing requires it. You may never change a subscript, because a different subscript names a different substance. Carbon dioxide, CO2, behaves nothing like carbon monoxide, CO.

Why the two sides must match

Atoms are never created or destroyed during a chemical change, and that statement is the law of conservation of matter. Bonds break and new bonds form, but the very same atoms turn up again in the products. So every element must appear in equal numbers on both sides of the arrow. An equation that passes that test is balanced.

The reliable way to get there is an atom inventory. List every element down the page, count it on the left, then count it on the right. Change one coefficient and count again. Two habits make the process a great deal faster. Begin with an element that appears in only one formula on each side. Save the element that appears in the most formulas for last, which is usually oxygen. Finish by reducing everything to the smallest whole numbers that still work.

Worked example · balance the burning of ethane

Ethane, C2H6, burns in oxygen to give carbon dioxide and water. The skeleton is C2H6 + O2 → CO2 + H2O.

Count what one ethane unit holds

One C2H6 unit holds 2 C and 6 H. Set those two elements first.

Match carbon, then hydrogen

Two carbon atoms need 2 CO2. Six hydrogen atoms need 3 H2O.

Total the oxygen on the right

2 CO2 holds 4 O. 3 H2O holds 3 more. That is 7 O atoms, which takes 7/2 units of O2.

Clear the fraction

Half units are not allowed, so double every coefficient.

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) Check all three elements

Carbon: 4 on the left and 4 on the right. Hydrogen: 12 and 12. Oxygen: 14 on the left, and 8 + 6 = 14 on the right. It balances.

Atom inventory for methane burning in oxygen, with equal counts on both sides Same atoms, rearranged CH₄ + 2 O₂ → CO₂ + 2 H₂O REACTANTS C H H H H 1 CH₄ + O O O O 2 O₂ PRODUCTS O C O 1 CO₂ + O O 2 H₂O small open circles are H teal circles are O ATOM INVENTORY ELEMENT LEFT SIDE RIGHT SIDE MATCH C 1 in CH₄ 1 in CO₂ 1 = 1 yes H 4 in CH₄ 2 × 2 in H₂O 4 = 4 yes O 2 × 2 in O₂ 2 in CO₂ + 2 in H₂O 4 = 4 yes
Figure 4.1 Methane combustion as an atom count. The coefficients 1, 2, 1 and 2 are the only set of small whole numbers that makes every row of the inventory match.
A note on fractions. A coefficient of 7/2 is not wrong in thermochemistry, where chemists often balance to exactly one mole of fuel. For ordinary stoichiometry, clear the fraction and keep whole numbers.
Key points
  • Coefficients are yours to change. Subscripts are not.
  • Balanced means every element has the same count on both sides.
  • The labels (s), (l), (g) and (aq) record the state of each substance.
  • Start with an element in only one formula per side. Leave oxygen for last.
  • Report the smallest whole numbers that work.

Section 2Classifying reactions

One reaction, three equations

A huge share of chemistry happens in water. A reaction in water can be written three ways, and each way shows something the others hide. The molecular equation shows every compound as a whole formula. The complete ionic equation splits each dissolved salt, strong acid and strong base into its free ions. The net ionic equation keeps only the parts that change. Ions that start out free and end up free took no part at all, and those are the spectator ions you cross off.

Precipitation: a solid falls out

Pour two clear solutions together and a cloud of solid can suddenly appear. That solid is a precipitate, and the change is a precipitation reaction. To predict one in advance you need the solubility rules, which are short statements about which ionic solids dissolve well in water and which hardly dissolve at all.

Ion in the compoundUsually dissolvesWatch out for
NO3-, CH3COO-Yes, every oneNo common exceptions
Group 1 metals, NH4+Yes, every oneNo common exceptions
Cl-, Br-, I-Yes, most of themAg+, Pb2+ and Hg22+ salts stay solid
SO42-Yes, most of themBa2+, Sr2+ and Pb2+ stay solid; Ca2+ barely dissolves
OH-Only group 1 and Ba2+Ca2+ and Sr2+ barely dissolve; the rest are solid
CO32-, PO43-, S2-Only group 1 and NH4+Nearly all others stay solid

Using the table takes two simple moves. Swap the partners in the two compounds to see the possible new pairings, then test each new pair against the rules. Any pair the table calls insoluble is your precipitate.

Worked example · lead nitrate meets potassium iodide

Swap the partners

Pb2+ can pair with I-, and K+ can pair with NO3-. Balance the charges to get PbI2 and KNO3.

Test both pairs

Every nitrate dissolves, so KNO3 stays in solution. Iodides dissolve except with lead, so PbI2 is the solid.

Write the molecular equation and balance it Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) Split everything dissolved into ions Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq) → PbI2(s) + 2 K+(aq) + 2 NO3-(aq) Cross off the spectators

K+ and NO3- appear unchanged on both sides, so they drop out.

Pb2+(aq) + 2 I-(aq) → PbI2(s)

The net ionic equation is the whole story in six symbols. Charge balances too, since 2+ and two 1- ions add to zero.

Acids and bases trade a proton

An acid gives away a hydrogen ion, H+. A base accepts one. When an acid meets a base in water the two cancel out, and chemists call that an acid-base reaction, or neutralization. The products are water and a dissolved salt. A strong acid comes apart fully in water. The common ones are HCl, HBr, HI, HNO3, H2SO4 and HClO4. A weak acid only comes apart part way, like the acetic acid in vinegar.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Strip out the Na+ and Cl- spectators and a single short line remains. Every strong acid and strong base pairing boils down to exactly the same change.

H+(aq) + OH-(aq) → H2O(l)

Redox moves electrons

In the third family, electrons actually move from one atom to another. Chemists call that a redox reaction, which is short for oxidation and reduction. Burning fuel is redox, and so is iron rusting or a battery running your phone. Combustion is the rapid reaction of a fuel with oxygen. In a single displacement reaction, one element shoves another element out of its compound. Zinc metal does exactly that to the hydrogen in hydrochloric acid.

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Some reactions sit in two families at once, and the next section gives you a firm test for the redox family.

A three way test that sorts a reaction into precipitation, acid and base, or redox Three questions sort most reactions Look at the equation Does a solid appear? A product carries (s) and both starters were (aq) PRECIPITATION Pb²⁺ + 2 I⁻ → PbI₂(s) Does H⁺ move to a base? An acid and a base go in, water and a salt come out ACID AND BASE H⁺ + OH⁻ → H₂O(l) Do oxidation numbers shift? One atom climbs, another one drops REDOX Zn + 2 HCl → ZnCl₂ + H₂
Figure 4.2 Run a new equation past all three questions. A reaction may answer yes to more than one, and burning a fuel in air is both a combustion and a redox change.
Key points
  • A net ionic equation drops every spectator ion and shows only the change.
  • Swap the partners, then check the solubility rules to find a precipitate.
  • Strong acid plus strong base always nets out to H+ + OH- → H2O.
  • Redox means electrons changed hands between atoms.
  • These families overlap, so one reaction can carry two labels.

Section 3Oxidation numbers and redox

A bookkeeping charge

An oxidation number is the charge an atom would carry if every bond around it were completely ionic. You pretend each shared pair belongs entirely to the greedier atom, and then you count. The result is usually not the real charge on that atom at all. It is still worth the trouble, because any change in the number proves that electrons moved.

The rules, in order

Work down this list and stop as soon as a rule settles an atom, because the higher rules always win.

  1. An atom in a free element is 0. In O2, in Na metal and in S8, every atom is 0.
  2. A single atom ion takes its own charge. Na+ is +1 and S2- is -2.
  3. Fluorine in a compound is always -1.
  4. Group 1 metals are +1. Group 2 metals are +2.
  5. Hydrogen is +1 with a nonmetal and -1 with a metal, as in NaH.
  6. Oxygen is -2, except in a peroxide like H2O2, where it is -1.
  7. Other halogens are -1 unless bonded to oxygen or to fluorine.
  8. The numbers sum to 0 in a neutral compound. In an ion they sum to the charge.

Rule 8 is the one that does the real work, because it lets you solve for the single atom you do not know yet.

Worked example · chromium in the dichromate ion

Find the oxidation number of chromium in Cr2O72-.

Pin down the atoms you can

Oxygen is -2 here, and there are seven of them. That is 7 × (-2) = -14.

Write the sum

Let x be the number for one chromium atom. Two of them plus the oxygen must equal the ion charge, so 2x + (-14) = -2.

Solve

2x = -2 + 14 = 12, so x = +6. Each chromium atom is +6.

Check on a second ion

In MnO4-, four oxygen atoms give -8. Then x - 8 = -1, so manganese is +7. In KMnO4 the potassium is +1, so x + 1 - 8 = 0 and manganese is +7 again.

Which atom was oxidized?

Oxidation is a rise in oxidation number, which means the atom gave electrons away. Reduction is a drop in oxidation number, which means the atom took electrons in. Neither one can happen alone, since the electrons always have to go somewhere. The substance that accepts the electrons is the oxidizing agent, and the substance that hands them over is the reducing agent. Notice the twist in those names, because the oxidizing agent is itself the species that gets reduced.

You can split the whole change into two half-reactions, one for the loss and one for the gain. Zinc metal dropped into copper sulfate solution shows both halves clearly.

Zn(s) → Zn2+(aq) + 2 e-    and    Cu2+(aq) + 2 e- → Cu(s)

Worked example · smelting iron ore

Iron oxide reacts with carbon monoxide in a blast furnace.

Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) Number every atom

In Fe2O3, oxygen is -2, so 2x - 6 = 0 and iron is +3. In CO, oxygen is -2, so carbon is +2. Free iron metal is 0. In CO2, carbon is +4.

Follow each element across the arrow

Iron goes from +3 to 0, a drop of 3. Iron was reduced. Carbon goes from +2 to +4, a rise of 2. Carbon was oxidized.

Check that the electrons balance

Two iron atoms gain 3 electrons each, which is 6 electrons. Three carbon atoms lose 2 electrons each, which is also 6. The books close.

Name the agents

Fe2O3 took the electrons, so it is the oxidizing agent. CO gave them up, so it is the reducing agent.

Zinc metal transfers two electrons to a copper ion, raising zinc and lowering copper on an oxidation number scale Two electrons change hands Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) Zn strip Cu²⁺ solution 2 e⁻ Zinc atoms leave the strip as Zn²⁺. Copper ions pick up the electrons and plate out as copper metal on the surface. OXIDATION NUMBER SCALE +2 +1 0 Zn climbs 0 → +2 lost 2 electrons, so it was oxidized Zn is the reducing agent Cu falls +2 → 0 gained 2 electrons, so it was reduced Cu²⁺ is the oxidizing agent
Figure 4.3 One rise always pairs with one fall. The two electrons zinc releases are exactly the two a copper ion needs, so the total charge never changes.
Key points
  • Oxidation numbers are a count, not a measured charge.
  • Free elements are 0, and the numbers sum to the overall charge.
  • A rise is oxidation. A drop is reduction. They always come in pairs.
  • The oxidizing agent is the substance that gets reduced.
  • Electrons lost must equal electrons gained.

Section 4Mole to mole stoichiometry

The coefficients are a ratio

A balanced equation is far more than a tidy record of a change. Its coefficients give you a ratio you can actually count with, so read them as moles. In the reaction that makes ammonia, one mole of N2 reacts with three moles of H2 and yields two moles of NH3.

N2(g) + 3 H2(g) → 2 NH3(g)

Any pair of those numbers is a mole ratio. From this single equation you can pull 1 mol N2 per 3 mol H2, or 2 mol NH3 per 1 mol N2, and several others. A mole ratio is simply a conversion factor, so write it as a fraction and let the units cancel. The study of these reacting amounts is called stoichiometry.

The road map

Here is the snag. A laboratory balance weighs in grams, but the equation speaks only in moles. So almost every amount problem walks the same four stop road. Start with grams of the substance you know, then divide by its molar mass to reach moles. Multiply by the mole ratio to cross over into moles of the substance you want. Finally multiply by that second molar mass to reach grams of your answer.

Notice that the mole ratio is the only step that ever crosses the arrow, since the two outer steps never leave their own substance. If a problem hands you moles instead of grams, you simply skip a step and begin further along the road.

The four stop road map from grams of a known substance to grams of a wanted substance Grams to moles to moles to grams grams of A what the balance reads moles of A counting units now moles of B across the arrow grams of B your answer divide by molar mass of A multiply by mole ratio B over A multiply by molar mass of B only this middle step uses the equation
Figure 4.4 The road map for any amount problem. Only the middle arrow uses the balanced equation, and the two end arrows are pure molar mass work.

Worked example · mass of ammonia from 25.0 g of nitrogen

Nitrogen reacts with plenty of hydrogen. What mass of NH3 can form?

Start from the balanced equation

N2 + 3 H2 → 2 NH3. The ratio you need is 2 mol NH3 per 1 mol N2.

Get the molar masses

N2 is 2 × 14.007 = 28.014 g/mol. NH3 is 14.007 + 3 × 1.008 = 17.031 g/mol.

Grams of A to moles of A

25.0 g ÷ 28.014 g/mol = 0.8924 mol N2. The grams cancel and moles are left.

Moles of A to moles of B

0.8924 mol N2 × (2 mol NH3 / 1 mol N2) = 1.785 mol NH3.

Moles of B to grams of B

1.785 mol × 17.031 g/mol = 30.4 g NH3.

Check the significant figures

The given mass 25.0 g carries three significant figures, so the answer keeps three. The result is 30.4 g of NH3.

Worked example · a pure mole to mole step

Propane burns by C3H8 + 5 O2 → 3 CO2 + 4 H2O. You have 1.50 mol of propane.

Oxygen needed

1.50 mol C3H8 × (5 mol O2 / 1 mol C3H8) = 7.50 mol O2.

Carbon dioxide formed

1.50 mol × (3 mol CO2 / 1 mol C3H8) = 4.50 mol CO2.

Water formed

1.50 mol × (4 mol H2O / 1 mol C3H8) = 6.00 mol H2O. No molar mass was needed at all.

Key points
  • Coefficients give mole ratios, not mass ratios.
  • Write every ratio as a fraction and cancel the units.
  • Grams to moles divides by molar mass. Moles to grams multiplies by it.
  • Only the middle step of the road map crosses the arrow.
  • The given value with the fewest significant figures sets the answer.

Section 5Limiting reactant

When one ingredient runs out

Think about building sandwiches, where each one takes two slices of bread and one slice of cheese. You have 10 slices of bread and only 3 slices of cheese. The bread could cover five sandwiches, but the cheese stops you at three. Cheese is the ingredient that sets the limit, and four slices of bread are left on the counter.

Reactions behave in exactly the same way. The reactant that runs out first is the limiting reactant, and it alone decides how much product can form. Any reactant still sitting there afterward is present in excess, and that leftover material is the excess reactant.

How to find the limiter

Do not compare masses. Grams are a trap here, because a mole of one substance can weigh fifteen times as much as a mole of another. Compare what each reactant could make instead, by converting each starting amount into moles of the same product. The smallest answer is the real one, and the reactant that produced it is limiting.

Once you know the limiter, the rest is simply the road map from the previous section, using only the limiting amount. To find what remains, work out how much of the excess reactant the reaction actually consumed, then subtract that from what you started with.

Worked example · 10.0 g of hydrogen and 64.0 g of oxygen

The reaction is 2 H2(g) + O2(g) → 2 H2O(g).

Molar masses

H2 is 2.016 g/mol. O2 is 32.00 g/mol. H2O is 18.02 g/mol.

Moles of each reactant

10.0 g ÷ 2.016 g/mol = 4.96 mol H2. 64.0 g ÷ 32.00 g/mol = 2.000 mol O2.

Water each one could make

From hydrogen: 4.96 mol × (2 mol H2O / 2 mol H2) = 4.96 mol H2O. From oxygen: 2.000 mol × (2 mol H2O / 1 mol O2) = 4.000 mol H2O.

Pick the smaller one

Oxygen gives the smaller number, so oxygen is limiting. Only 4.000 mol of water can form, even though hydrogen could have supported more.

Mass of water

4.000 mol × 18.02 g/mol = 72.1 g H2O.

Leftover hydrogen

Oxygen used 2.000 mol × (2 mol H2 / 1 mol O2) = 4.000 mol H2. You began with 4.96 mol, so 4.96 - 4.00 = 0.96 mol is left. That is 0.96 mol × 2.016 g/mol = 1.9 g of H2.

Check with conservation of mass

You started with 10.0 + 64.0 = 74.0 g. You end with 72.1 g of water plus 1.9 g of hydrogen, which is 74.0 g. Nothing went missing.

Look again at the masses in that example, because there was far more oxygen on the balance, yet oxygen still ran out first. The ratio inside the equation, and not the weight on the balance, is what really matters.

Bar chart comparing the water each reactant could make, with oxygen giving the smaller amount Compare the product, not the mass 2 H₂ + O₂ → 2 H₂O, starting from 10.0 g H₂ and 64.0 g O₂ 5 mol 2.5 0 4.96 mol H₂ could make 4.96 mol water 4.96 mol H₂O 2.000 mol O₂ could make 4.000 mol water 4.000 mol H₂O the ceiling O₂ is limiting 4.000 mol × 18.02 = 72.1 g water 0.96 mol H₂ is left over unreachable
Figure 4.5 Each bar is the water that one reactant could make on its own. The lower bar sets a hard ceiling, so the shaded strip on the hydrogen bar can never be reached.
Why run an excess on purpose? In real labs the cheap reactant is often loaded in excess. Extra oxygen in a furnace pushes the fuel to burn completely, and the leftover gas simply leaves with the exhaust.
Key points
  • The limiting reactant runs out first and caps the product.
  • Never compare masses. Compare the product each reactant could make.
  • Convert both starting amounts to moles of the same product and take the smaller.
  • Leftover equals what you started with minus what the reaction used.
  • Total mass in must equal total mass out, leftovers included.

Section 6Theoretical, actual and percent yield

Three different yields

Stoichiometry tells you the most product you could possibly get, but real flasks rarely deliver it. Chemists keep three separate numbers apart so they can report how a run actually went.

Theoretical yield is the amount the limiting reactant allows, calculated from the road map, so it is a prediction. Actual yield is what you really collect, dry and weigh, so it is a measurement. Actual yield is almost always the smaller of the two. Some product clings to the glassware. Some reactant never finds a partner. A side reaction can steal material, or the reaction may reach equilibrium before it finishes. Percent yield puts the two numbers side by side.

percent yield = (actual yield ÷ theoretical yield) × 100%

Both yields must be expressed in the same unit before you divide. Grams over grams works, and moles over moles works, but you cannot mix the two.

Worked example · how the ammonia run really went

Section 4 predicted 30.4 g of NH3 from 25.0 g of N2. The flask gave 24.8 g.

Name each number

Theoretical yield is 30.4 g. Actual yield is 24.8 g.

Divide and scale

24.8 g ÷ 30.4 g = 0.8158. Multiply by 100% to get 81.6%.

Say what it means

You captured about 82 parts in every 100 that the equation promised. The other 18 parts stayed behind or never reacted.

Working the formula backward

The same relationship answers two other useful questions. If you know the usual percent yield, you can predict the real haul before you even begin. If you need a set amount of product, you can work out how much starting material to weigh.

Worked example · planning around a known yield

A synthesis step reliably runs at 72.0%. Stoichiometry says the batch should give 45.0 g.

Predict the actual yield

45.0 g × 0.720 = 32.4 g of product is what you should expect.

Turn the question around

Suppose you need 32.4 g in hand. Then the theoretical yield has to be 32.4 g ÷ 0.720 = 45.0 g, which tells you how much reactant to weigh out.

Chain two steps

A second step at 80.0% follows the first. The overall yield is 0.720 × 0.800 = 0.576, or 57.6%. Losses multiply, so long routes bleed material fast.

One warning before you go. A percent yield above 100% is not a triumph, because it means the measurement itself is wrong. The product is probably still damp with solvent, or some unreacted solid came along with it. Dry the sample thoroughly and then weigh it again.

Two bars comparing a theoretical yield of 30.4 grams with an actual yield of 24.8 grams, giving 81.6 percent What the equation promised, and what the flask gave theoretical yield 30.4 g NH₃ 100% actual yield 24.8 g NH₃ 81.6% 5.6 g lost bar length is proportional to mass 24.8 ÷ 30.4 = 0.8158, so the percent yield is 81.6% WHERE IT GOES left on the glass never reacted lost to a side reaction stopped at equilibrium
Figure 4.6 Percent yield is one bar divided by the other. The dashed gap is product that the equation allowed but the run never delivered.
Atom economy. Percent yield grades the run. Atom economy grades the route. It asks what fraction of the mass of all the reactants ends up in the product you want. A route with high atom economy wastes less, even before you step into the lab.
Key points
  • Theoretical yield is predicted. Actual yield is weighed.
  • Percent yield is actual divided by theoretical, times 100%.
  • Both yields must share the same unit before you divide.
  • Multiply the yields of chained steps to get the overall yield.
  • Over 100% means a measurement problem, such as a damp product.

The chapter in one breath

  • A balanced equation keeps every element equal on both sides, because atoms are never lost.
  • Change coefficients to balance. Never touch a subscript.
  • A net ionic equation drops the spectator ions and shows only the real change.
  • Solubility rules predict a precipitate. Acid plus base nets out to water. Redox shifts oxidation numbers.
  • An oxidation number rise is oxidation. A drop is reduction. The two always pair up.
  • Coefficients are mole ratios, so they convert moles of one substance into moles of another.
  • Every mass problem walks the same road: grams, moles, moles, grams.
  • The limiting reactant is the one that makes the least product, and it sets the theoretical yield.
  • Percent yield is actual divided by theoretical, times 100%, and it is almost never 100.

ReferenceWords worth knowing

These are the terms this chapter leans on. Come back to the list whenever an equation stops making sense.

acid-base reaction
A reaction in which an acid transfers H+ to a base, usually giving water and a dissolved salt. Also called neutralization.
actual yield
The mass or moles of product you really collect and weigh at the end of a run.
balanced equation
A chemical equation whose coefficients give every element the same atom count on both sides of the arrow.
coefficient
The whole number written in front of a formula. It counts units of that substance and may be changed to balance an equation.
combustion
The rapid reaction of a fuel with oxygen, releasing heat and light. Hydrocarbon combustion gives CO2 and H2O.
excess reactant
A reactant still present after the reaction stops, because the limiting reactant ran out first.
limiting reactant
The reactant that is used up first. It sets the maximum amount of product that can form.
mole ratio
A conversion factor built from two coefficients of a balanced equation, such as 2 mol NH3 per 1 mol N2.
net ionic equation
An equation in which spectator ions are removed, leaving only the species that change.
oxidation
An increase in oxidation number, which means the atom lost electrons.
oxidation number
The charge an atom would have if every bond in its compound were fully ionic. A bookkeeping device, not a measured charge.
oxidizing agent
The substance that accepts electrons in a redox reaction. It is the species that gets reduced.
percent yield
Actual yield divided by theoretical yield, multiplied by 100%. A measure of how well a run performed.
precipitate
An insoluble solid that forms and separates when two solutions are mixed.
reducing agent
The substance that donates electrons in a redox reaction. It is the species that gets oxidized.
reduction
A decrease in oxidation number, which means the atom gained electrons.
solubility rules
A short set of statements that predict whether a given ionic compound dissolves well in water.
spectator ion
An ion that is present in solution before and after the reaction and takes no part in the change.
stoichiometry
The use of a balanced equation to relate amounts of reactants and products, in moles, mass or volume.
theoretical yield
The maximum product the limiting reactant allows, calculated from the balanced equation.

Check yourself

1. Why may you change a coefficient while balancing, but never a subscript?
A coefficient counts units; a subscript defines the substance. Changing 2 CO2 to 3 CO2 just takes more of the same gas. Changing CO2 to CO3 invents a different substance that was never in the reaction.
2. Balance this equation: Fe + O2 → Fe2O3.
4 Fe + 3 O2 → 2 Fe2O3 One Fe2O3 needs 3 O, so you need 3/2 O2. Double everything to clear the fraction: Fe is 4 = 2 × 2, and O is 3 × 2 = 6 = 2 × 3.
3. Write the net ionic equation for AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq), and name the spectators.
Ag+(aq) + Cl-(aq) → AgCl(s); Na+ and NO3- are spectators. NaNO3 stays fully dissolved, so its two ions appear unchanged on both sides and cancel out.
4. Mix Ba(NO3)2(aq) with Na2SO4(aq). Does a solid form?
Yes. BaSO4 precipitates. Most sulfates dissolve, but barium sulfate is one of the exceptions. The other new pair, NaNO3, is soluble, so the net ionic equation is Ba2+ + SO42- → BaSO4(s).
5. What is the oxidation number of sulfur in SO42-?
+6 Each of the four oxygen atoms is -2, giving -8. The sum must equal the ion charge, so x + (-8) = -2. Then x = -2 + 8 = +6.
6. What is the oxidation number of nitrogen in NH4+?
-3 Hydrogen bonded to a nonmetal is +1, and four of them give +4. The sum must equal +1, so x + 4 = +1 and x = -3.
7. In Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g), what is oxidized and what is the oxidizing agent?
Zinc is oxidized; HCl is the oxidizing agent. Zn goes from 0 to +2, a rise of 2, so it lost electrons. Hydrogen drops from +1 to 0, so it was reduced, which makes HCl the oxidizing agent. Chlorine stays at -1 throughout.
8. How many moles of O2 are needed to burn 2.40 mol of CH4?
4.80 mol O2 The balanced equation is CH4 + 2 O2 → CO2 + 2 H2O. So 2.40 mol CH4 × (2 mol O2 / 1 mol CH4) = 4.80 mol O2.
9. What mass of CO2 forms when 15.0 g of propane, C3H8, burns completely?
44.9 g CO2 C3H8 + 5 O2 → 3 CO2 + 4 H2O. Molar mass of C3H8 is 44.10 g/mol, so 15.0 ÷ 44.10 = 0.3401 mol. Then 0.3401 × 3 = 1.020 mol CO2, and 1.020 × 44.01 g/mol = 44.9 g.
10. You mix 3.00 mol Al with 4.00 mol Cl2 by 2 Al + 3 Cl2 → 2 AlCl3. Which reactant limits, and how much AlCl3 forms?
Cl2 limits; 2.67 mol AlCl3 forms. Aluminum could make 3.00 × (2/2) = 3.00 mol AlCl3. Chlorine could make 4.00 × (2/3) = 2.67 mol. The smaller value wins. Aluminum used is 4.00 × (2/3) = 2.67 mol, so 0.33 mol Al is left over.
11. A run has a theoretical yield of 8.50 g and you collect 7.23 g. What is the percent yield?
85.1% Divide actual by theoretical: 7.23 g ÷ 8.50 g = 0.8506. Multiply by 100% to get 85.1%, keeping three significant figures.
12. Your calculation gives a percent yield of 104%. What should you suspect?
The product is impure or still wet, so the mass is too high. Nothing can beat the amount the limiting reactant allows. Solvent left in the solid, or unreacted starting material mixed in, both add mass. Dry the sample and weigh it again.
Chemistry · Chapter 5

Thermochemistry

Every chemical change moves energy. Burn methane and the room gets warm. Open a cold pack and your hand gets cold. Thermochemistry is the part of chemistry that tracks that energy. In this chapter you learn to say how much heat a reaction gives up or takes in. You learn to measure it with a cup of water and a thermometer. Then you learn three ways to predict it on paper. One adds up known steps. One looks up each compound in a table. One counts the bonds that break and form. By the end, a number like −890.5 kJ will tell you a whole story.

By the end of this chapter you can

  • Tell heat apart from work, and give each one the right sign.
  • Use the first law, ΔE = q + w, on a system you choose yourself.
  • Find heat with q = mcΔT, and read the result from a calorimeter.
  • Write a thermochemical equation, then scale its ΔH to any mass.
  • Flip, scale and add known steps with Hess's law to reach a target.
  • Apply products minus reactants to standard formation enthalpies.
  • Estimate a reaction enthalpy from average bond energies, and say why it is only an estimate.

Section 1Energy, heat and work

Two kinds of energy

Energy is the ability to do work or to warm something up. It shows up in two broad kinds. Kinetic energy is the energy of motion. A rolling ball has it. So does a single molecule darting across a flask. You find it with KE = ½mv2, so speed counts twice over. Potential energy is stored energy. It depends on where a thing sits, or on what it is made of. A book on a high shelf holds it. So do the bonds in a tank of gasoline.

Heat is not a third kind of energy that a thing owns. Thermal energy is the energy of random particle motion inside a sample. Temperature reports how fast those particles move, on average. Heat is thermal energy on the move, from a warm place to a cooler one. A hot pan does not hold heat. It holds thermal energy, and it gives off heat.

Draw a box around the problem

Before you can count energy, you must say what you are counting. The system is the part you care about. The surroundings are everything else in the universe. For a reaction in a beaker, the chemicals are the system. The glass, the air and your hand are the surroundings. A closed system can trade energy but not matter. An isolated system trades neither one.

Energy is never destroyed. The law of conservation of energy says it only moves or changes form. Whatever the system gains, the surroundings gave up. Written for a chemical system, that idea is the first law of thermodynamics.

ΔE = q + w

Here ΔE is the change in the system's internal energy, the total of all its kinetic and potential energy. The letter q stands for heat. The letter w stands for work, which in chemistry is usually a gas pushing on its container or getting squeezed.

Getting the signs right

Both q and w carry a sign, and the sign always describes the system. Think of a bank account. Money coming in is positive. Money going out is negative.

The SI unit of energy is the joule, J. One joule is one kg·m2/s2, which is tiny. A calorie is larger, and it equals 4.184 J by definition. The Calorie on a food label is really a kilocalorie, so one of them is 4184 J.

Worked example · signs and the first law

A gas takes in 55 J of heat. It expands and does 25 J of work on the air outside. What is ΔE for the gas?

Step 1 · Name the system.The gas is the system. The air outside is the surroundings.

Step 2 · Set each sign.Heat flows in, so q = +55 J. The gas pushes outward, so work leaves it and w = −25 J.

Step 3 · Add them.ΔE = (+55 J) + (−25 J) = +30 J.

Step 4 · Check it.The gas took in more than it spent. Its internal energy should rise, and +30 J says it rose by 30 J.

System and surroundings with signed heat and work arrows SURROUNDINGS SYSTEM ΔE = q + w what the system keeps q > 0  heat in endothermic q < 0  heat out exothermic w > 0  work on it the gas gets squeezed w < 0  work by it the gas expands Arrows into the box are positive. Arrows out of the box are negative.
Figure 5.1 Heat and work cross the boundary in both directions. Arrows pointing into the system are positive. Arrows pointing out are negative. Add the two signed amounts and you have ΔE.

Worked example · food Calories to joules

A snack bar lists 240 Calories. How many joules is that?

Step 1 · Fix the unit.A food Calorie is a kilocalorie. So the bar holds 240 kcal.

Step 2 · Convert.240 kcal × 4184 J/kcal = 1,004,160 J.

Step 3 · Round.Three significant figures gives 1.00 × 106 J, or 1.00 × 103 kJ.

Key points
  • Heat is energy in transit. No sample stores heat, it stores thermal energy.
  • Choose the system first. Every sign in the problem hangs on that choice.
  • ΔE = q + w. Energy in is positive, energy out is negative.
  • 1 cal = 4.184 J exactly, and one food Calorie equals 1 kcal, or 4184 J.

Section 2Heat capacity and calorimetry

Why water fights back

Put the same amount of heat into a steel pan and into a pot of water. The pan gets hot fast. The water barely budges. That difference has a name. Heat capacity, C, is the heat needed to raise a whole object by one degree. Its unit is J/°C. It depends on how much stuff you have, so a big pot has a larger value than a small one.

Specific heat capacity, c, is the heat needed to raise one gram by one degree. Its unit is J/g·°C. This one is a property of the substance itself, so you can look it up. Molar heat capacity is the same idea per mole instead of per gram.

Water has a specific heat of 4.184 J/g·°C. That is high for a liquid. Hydrogen bonds soak up energy before the molecules speed up much. The result shapes daily life. Lakes warm and cool slowly. Your blood holds your body near 37 °C. Car engines use water to carry heat away.

Substancec (J/g·°C)What that means
Water (liquid)4.184Hard to heat, hard to cool
Ethanol2.44Warms about twice as fast as water
Aluminum0.897Light metal, moderate for a metal
Iron0.449A pan heats quickly
Copper0.385Good at moving heat along
Silver0.233Heats fast
Lead0.130Very small, so it heats very fast
Gold0.129The smallest on this list

The equation that does the work

Heat, mass, specific heat and temperature change all tie together in one line.

q = mcΔT, where ΔT = Tfinal − Tinitial

Write ΔT as final minus initial every time. Then the sign takes care of itself. A sample that warms up has a positive ΔT and a positive q, so it gained heat. A sample that cools has a negative ΔT and a negative q. One more bonus: a change of one degree Celsius is the same size as a change of one kelvin. So you may use either unit for ΔT.

Measuring heat in a cup

Calorimetry is the practice of measuring heat by watching a temperature change. The device is a calorimeter. The simplest one is a foam cup of water with a lid and a thermometer. Foam is a poor conductor, so very little heat escapes. The cup sits open to the room, so pressure stays constant.

A bomb calorimeter is the heavy-duty version. A sealed steel vessel holds the sample and pure oxygen. The whole vessel sits in a water bath. Because the volume cannot change, it measures heat at constant volume instead.

Either way the logic is the same. If no heat leaks out, the heat one part loses is the heat the other part gains.

qhot = −qcold

Worked example · heating water

How much heat warms 125 g of water from 22.0 °C to 95.0 °C?

Step 1 · Find ΔT.ΔT = 95.0 °C − 22.0 °C = 73.0 °C.

Step 2 · Build the product.mc = (125 g)(4.184 J/g·°C) = 523.0 J/°C.

Step 3 · Multiply.q = (523.0 J/°C)(73.0 °C) = 38,179 J.

Step 4 · Round and report.Three significant figures gives q = 3.82 × 104 J, or 38.2 kJ. The grams and degrees cancel, so joules are left. The value is positive, which fits a sample that warmed up.

A coffee cup calorimeter holding water, a hot metal block and a thermometer thermometer metal foam cup, poor conductor heat flows metal → water What you put in metal: 50.0 g at 98.0 °C water: 100.0 g at 21.0 °C final reading: 24.4 °C What you work out q = mcΔT q(water) = (100.0)(4.184)(3.4) = +1423 J q(metal) = −1423 J c(metal) = 0.39 J/g·°C
Figure 5.2 A coffee cup calorimeter. The hot metal cools, the water warms, and the thermometer settles at one shared temperature. Heat lost by the metal equals heat gained by the water, with the sign flipped.

Worked example · finding a specific heat

A 50.0 g block of metal is heated to 98.0 °C. It is dropped into 100.0 g of water at 21.0 °C. The water settles at 24.4 °C. What is the metal's specific heat?

Step 1 · Heat gained by the water.ΔT = 24.4 − 21.0 = 3.4 °C. q = (100.0 g)(4.184 J/g·°C)(3.4 °C) = 1423 J.

Step 2 · Heat lost by the metal.qmetal = −1423 J. Its ΔT = 24.4 − 98.0 = −73.6 °C.

Step 3 · Solve q = mcΔT for c.c = q ÷ (m ΔT) = (−1423 J) ÷ [(50.0 g)(−73.6 °C)] = 1423 ÷ 3680 = 0.387 J/g·°C.

Step 4 · Round honestly.The water's ΔT of 3.4 °C has only two significant figures, so the answer does too. c = 0.39 J/g·°C. Copper is 0.385 J/g·°C, so the block is very likely copper.

Worth knowing. Small temperature changes are the weak link in calorimetry. A 3.4 °C rise read on a cheap thermometer limits you to two significant figures, no matter how carefully you weighed the sample.
Key points
  • Heat capacity belongs to an object. Specific heat belongs to a substance.
  • q = mcΔT, and ΔT is always final minus initial.
  • Water's 4.184 J/g·°C is high, which is why water steadies temperatures.
  • In a calorimeter, heat lost by one part equals heat gained by the other, with opposite signs.

Section 3Enthalpy and thermochemical equations

A quantity built for the lab bench

Most reactions you run sit in an open container. The pressure stays near 1 atm the whole time. If a gas forms, it has to push the air back, and that costs energy as work. So ΔE is awkward to measure in an open cup, because part of the energy slipped away as work you never saw. Chemists get around this with a quantity called enthalpy, H.

Enthalpy is defined as H = E + PV. You will never need to find H itself, and that is fine. What matters is the change. At constant pressure, the enthalpy change is exactly the heat you measure.

ΔH = Hproducts − Hreactants = qp

That is a gift. The thermometer reading from an open cup gives you ΔH with no extra work term to chase. The sign rules match the ones you already know. A negative ΔH means heat left the system, so the reaction is exothermic. A positive ΔH means heat went in, so the reaction is endothermic. Enthalpy is an extensive property, which means the amount of substance changes the size of the answer.

Thermochemical equations

A thermochemical equation is a balanced equation with two extra pieces of care. Every substance carries its physical state. And a value of ΔH is written beside it for the equation exactly as balanced.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)    ΔH = −890.5 kJ

Read that as a package deal. Burn one mole of methane with two moles of oxygen, and make liquid water, and 890.5 kJ of heat leaves. States are not decoration. If the water comes off as vapor instead, the same reaction gives only 802.5 kJ. The missing 88 kJ stayed in the vapor, because turning 2 mol of liquid water into vapor costs about 44.0 kJ per mole.

A value marked ΔH° is a standard enthalpy change. The little circle means every substance was in its standard state: a pressure of 1 bar for gases, a concentration of 1 M for solutions, and a stated temperature that is almost always 25 °C.

Scaling and reversing

Two rules let one equation answer many questions. Multiply the whole equation by a number, and ΔH gets multiplied by the same number. Reverse the equation, and ΔH keeps its size but flips its sign. Watch all three forms below say the same thing.

2H2(g) + O2(g) → 2H2O(l)    ΔH = −571.6 kJ
H2(g) + ½O2(g) → H2O(l)    ΔH = −285.8 kJ
2H2O(l) → 2H2(g) + O2(g)    ΔH = +571.6 kJ

The last line is worth a pause. Splitting water back into its elements takes in exactly the heat that forming it gave off. Nature keeps a tidy ledger.

An exothermic enthalpy diagram beside an endothermic one EXOTHERMIC enthalpy CH₄(g) + 2O₂(g) CO₂(g) + 2H₂O(l) ΔH = −890.5 kJ heat leaves the system ENDOTHERMIC CaCO₃(s) CaO(s) + CO₂(g) ΔH = +179.2 kJ heat goes in Products low means heat out. Products high means heat in.
Figure 5.3 Two enthalpy diagrams. On the left the products sit lower, so heat leaves and ΔH is negative. On the right the products sit higher, so heat must go in and ΔH is positive. The height of the arrow is the size of ΔH.

Worked example · heat from a given mass

How much heat is released when 25.0 g of methane burns to carbon dioxide and liquid water?

Step 1 · Find the molar mass.CH4 is 12.011 + 4(1.008) = 16.043, so 16.04 g/mol.

Step 2 · Convert grams to moles.n = 25.0 g ÷ 16.04 g/mol = 1.559 mol.

Step 3 · Use ΔH as a conversion factor.The equation gives −890.5 kJ per mole of CH4. So q = (1.559 mol)(−890.5 kJ/mol) = −1388 kJ.

Step 4 · Round and read the sign.q = −1.39 × 103 kJ. The minus sign says the heat left the system, so 1.39 × 103 kJ warmed the room.

Key points
  • At constant pressure, ΔH equals the heat you measure, so ΔH = qp.
  • Negative ΔH is exothermic. Positive ΔH is endothermic.
  • A thermochemical equation needs states and a ΔH for the equation as written.
  • Scale the equation and ΔH scales with it. Reverse the equation and ΔH flips sign.

Section 4Hess's law

Enthalpy does not care how you got there

Imagine hiking to a lookout 300 m above the car park. You could go straight up the steep face. You could take a long switchback trail instead. Either way you climbed 300 m. Altitude depends only on where you start and where you stop. Enthalpy behaves the same way, which makes it a state function. Its change depends on the starting state and the final state, never on the route between them.

That single fact is Hess's law. If a reaction can be written as the sum of two or more steps, then the reaction's ΔH is the sum of the steps' ΔH values. The steps do not even have to be real. They only have to add up on paper.

Hess's law is how chemists get numbers they cannot measure. Take the reaction of carbon with oxygen to make carbon monoxide. You cannot measure it cleanly, because some of the CO always burns on to CO2. So you measure two reactions you can control, then combine them.

Three moves and nothing else

Every Hess's law problem uses the same three moves on the steps you were given.

Then stack the adjusted steps and add. Anything that appears on the left in one step and on the right in another will cancel. What is left should be the target equation. If it is not, go back and adjust a factor.

A good habit is to work from the target, not from the steps. Find a substance that appears in the target. Hunt for it in the given steps. Ask two questions. Is it on the correct side? Is the coefficient correct? Flip or scale until both answers are yes, then move to the next substance.

Worked example · building a reaction from two steps

Target: C(s) + ½O2(g) → CO(g). Find ΔH from these two measured reactions.

(1) C(s) + O2(g) → CO2(g)   ΔH1 = −393.5 kJ
(2) 2CO(g) + O2(g) → 2CO2(g)   ΔH2 = −566.0 kJ

Step 1 · Place the carbon.The target has one C(s) on the left. Step (1) already has it there, with a coefficient of 1. Keep step (1) as written.

Step 2 · Place the CO.The target has CO on the right. Step (2) has CO on the left, and two of them. Reverse step (2) and halve it.

Step 3 · Adjust the enthalpy with it.Reversing flips the sign to +566.0 kJ. Halving gives +283.0 kJ. The adjusted step is CO2(g) → CO(g) + ½O2(g).

Step 4 · Stack and cancel.C(s) + O2(g) → CO2(g) sits above CO2(g) → CO(g) + ½O2(g). One CO2 cancels. One O2 on the left meets ½O2 on the right, leaving ½O2 on the left. The sum is C(s) + ½O2(g) → CO(g).

Step 5 · Add the enthalpies.ΔH = (−393.5 kJ) + (+283.0 kJ) = −110.5 kJ.

Step 6 · Check it.Tables list the formation enthalpy of CO(g) as −110.5 kJ/mol. The two agree exactly.

Enthalpy shelves showing the direct route to carbon dioxide and the two step route through carbon monoxide enthalpy C(s) + O₂(g) 0 kJ CO(g) + ½O₂(g) −110.5 kJ CO₂(g) −393.5 kJ step 1 −110.5 kJ step 2 −283.0 kJ direct  −393.5 kJ −110.5 kJ plus −283.0 kJ equals −393.5 kJ, so both routes agree.
Figure 5.4 Two routes from carbon and oxygen down to carbon dioxide. The direct drop is −393.5 kJ. The two step route stops at carbon monoxide, falling −110.5 kJ and then −283.0 kJ. Both routes total the same, so the first leg must be −110.5 kJ.
Worth knowing. Fractional coefficients look odd in a balanced equation, but Hess's law allows them. A half mole of O2 is a perfectly real amount, and it keeps the target equation from being doubled for no reason.
Key points
  • Enthalpy is a state function, so ΔH ignores the path taken.
  • Hess's law: if steps add up to the target, their ΔH values add up too.
  • Reversing a step flips the sign. Scaling a step scales the value.
  • Work from the target substance by substance, checking side and coefficient.

Section 5Standard enthalpies of formation

One table, thousands of reactions

Hess's law is powerful, but hunting for the right steps gets old. There is a shortcut that works for almost any reaction. Pick one special step and tabulate it for every compound. That step is formation from the elements.

The standard enthalpy of formation, written ΔHf°, is the enthalpy change when one mole of a substance forms from its elements, with everything in its standard state. The standard state of an element is its most stable form at 1 bar and the stated temperature, usually 25 °C. For carbon that is graphite, not diamond. For oxygen it is O2(g), not ozone. For bromine it is a liquid.

Two details follow right away. First, the equation must make exactly one mole of the product, so fractional coefficients are normal. Second, an element already in its standard state has nothing to form, so its ΔHf° is zero.

H2(g) + ½O2(g) → H2O(l)    ΔHf° = −285.8 kJ/mol

Products minus reactants

Once every compound has a number, any reaction enthalpy is one subtraction.

ΔH°rxn = Σ nΔHf°(products) − Σ nΔHf°(reactants)

Hess's law is why this works. Think of a two leg route. First take the reactants apart into their elements, which is the reverse of forming them, so you subtract their values. Then build the products from those elements, which adds their values. The elements act like sea level, a shared zero that both sides are measured from.

Three cautions save most of the lost points. Multiply each value by the coefficient from the balanced equation. Match the state in the table to the state in your equation, since H2O(l) and H2O(g) differ by 44.0 kJ/mol. And subtract the whole reactant sum, not just the first term.

SubstanceΔHf° (kJ/mol)SubstanceΔHf° (kJ/mol)
H2O(l)−285.8NH3(g)−45.9
H2O(g)−241.8NO(g)+90.25
CO2(g)−393.5NO2(g)+33.2
CO(g)−110.5SO2(g)−296.8
CH4(g)−74.6SO3(g)−395.7
C2H2(g)+227.4NaCl(s)−411.2
C2H5OH(l)−277.6CaCO3(s)−1207.6
C6H12O6(s)−1273.3CaO(s)−634.9
H2O2(l)−187.8Fe2O3(s)−824.2
Any element in its standard state0Al2O3(s)−1675.7
Formation enthalpies measured from a shared zero set by the elements Elements in standard states: Ca(s) + C(graphite) + 3/2 O₂(g) shared zero  ·  ΔH°f = 0 CaCO₃(s) −1207.6 kJ CaO(s) + CO₂(g) −1028.4 kJ take apart +1207.6 kJ build −1028.4 kJ ΔH° = +179.2 kJ Go up to the elements, then down to the products. Vertical spacing is not to scale.
Figure 5.5 The elements set a shared zero. Limestone sits 1207.6 kJ below it. Lime plus carbon dioxide sit 1028.4 kJ below it. The gap between the two shelves is the reaction enthalpy, +179.2 kJ.

Worked example · breaking down limestone

A lime kiln runs CaCO3(s) → CaO(s) + CO2(g). Find ΔH°.

Step 1 · Sum the products.Σ products = (1)(−634.9) + (1)(−393.5) = −1028.4 kJ.

Step 2 · Sum the reactants.Σ reactants = (1)(−1207.6) = −1207.6 kJ.

Step 3 · Subtract.ΔH° = (−1028.4 kJ) − (−1207.6 kJ) = +179.2 kJ.

Step 4 · Read the sign.Positive, so the reaction is endothermic. That is why a kiln has to burn fuel all day to keep making lime.

Worked example · burning ethanol

Find ΔH° for C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l).

Step 1 · Sum the products with coefficients.2(−393.5) + 3(−285.8) = −787.0 + (−857.4) = −1644.4 kJ.

Step 2 · Sum the reactants.1(−277.6) + 3(0) = −277.6 kJ. Oxygen gas is an element in its standard state, so it contributes nothing.

Step 3 · Subtract.ΔH° = (−1644.4 kJ) − (−277.6 kJ) = −1366.8 kJ.

Step 4 · Sanity check.Burning a fuel should be exothermic, and the answer is strongly negative. One mole of ethanol weighs 46.07 g, so that is about 29.7 kJ per gram.

Key points
  • ΔHf° is for one mole of a substance made from its elements in standard states.
  • An element in its own standard state has ΔHf° = 0.
  • ΔH°rxn = sum of products minus sum of reactants, each value times its coefficient.
  • States matter. Liquid and gaseous water sit 44.0 kJ/mol apart.

Section 6Bond energies

Breaking costs, making pays

Formation enthalpies tell you about whole compounds. Bond energies let you look inside them. A bond dissociation energy, D, is the energy needed to break one mole of a given bond in the gas phase, pulling the pieces apart into free atoms. Breaking a bond always takes energy in, so D is always positive. Forming that same bond gives back the same amount, so it always counts as negative.

There is a wrinkle. The strength of a bond shifts a little depending on what sits nearby. The first C–H bond in methane takes about 439 kJ/mol to break. The four of them average out to about 415 kJ/mol. Tables list that kind of average bond energy, pooled across many molecules. So answers from bond energies are good estimates, not exact values. Expect to land within a few percent.

Bond order matters a lot. More shared pairs pull the atoms closer and hold them tighter. A C–C single bond is 347 kJ/mol. The double bond is 614, and the triple is 839. Notice the double bond is not twice the single, and the triple is not three times it. Each extra pair adds less than the one before.

BondD (kJ/mol)BondD (kJ/mol)
H–H436C–C347
C–H415C=C614
N–H390C≡C839
O–H464C–O350
H–F569C=O (in CO2)799
H–Cl431C–Cl339
F–F159N–N160
Cl–Cl242N≡N945
O–O140O=O498

Counting a whole reaction

Picture a reaction taking the long way around. First tear every reactant apart into loose atoms. That climb costs the sum of all the bonds you broke. Then let those atoms snap together into the products. That drop pays back the sum of all the bonds you made. The difference between the climb and the drop is the reaction enthalpy.

ΔH ≈ Σ D(bonds broken) − Σ D(bonds formed)

This is Hess's law again, with free atoms as the stopping point. If the new bonds are stronger than the old ones, the drop beats the climb and ΔH comes out negative. That is the real reason fuels release heat. Carbon dioxide and water hold unusually strong bonds, stronger than the ones in the fuel and in O2.

One shortcut helps on paper. Bonds that appear unchanged on both sides cancel out. Still, counting every bond is safer while you are learning, because a missed bond is easy to spot in a full list.

Energy climbs while bonds break and falls further while new bonds form 2 H + 2 Cl  free atoms +678 kJ H₂ + Cl₂ 0 kJ 2 HCl −184 kJ break H–H, Cl–Cl +678 kJ in form 2 H–Cl −862 kJ out net ΔH ≈ −184 kJ 678 − 862 = −184
Figure 5.6 The long way around. Breaking H–H and Cl–Cl lifts the system 678 kJ to free atoms. Forming two H–Cl bonds drops it 862 kJ. The net change is −184 kJ, so the reaction gives off heat.

Worked example · hydrogen plus chlorine

Estimate ΔH for H2(g) + Cl2(g) → 2HCl(g).

Step 1 · List the bonds broken.One H–H at 436 kJ/mol and one Cl–Cl at 242 kJ/mol. Sum = 436 + 242 = 678 kJ.

Step 2 · List the bonds formed.Each HCl has one H–Cl bond, and there are two molecules. Sum = 2(431) = 862 kJ.

Step 3 · Subtract.ΔH ≈ 678 kJ − 862 kJ = −184 kJ.

Step 4 · Compare.Formation data give 2(−92.3) = −184.6 kJ. The estimate is off by less than 1 kJ, which is better than usual.

Worked example · burning methane the long way

Estimate ΔH for CH4(g) + 2O2(g) → CO2(g) + 2H2O(g).

Step 1 · Bonds broken.Four C–H bonds at 415 and two O=O bonds at 498. Sum = 4(415) + 2(498) = 1660 + 996 = 2656 kJ.

Step 2 · Bonds formed.CO2 has two C=O bonds at 799. Each water has two O–H bonds, so four at 464. Sum = 2(799) + 4(464) = 1598 + 1856 = 3454 kJ.

Step 3 · Subtract.ΔH ≈ 2656 kJ − 3454 kJ = −798 kJ.

Step 4 · Compare.Formation data give −802.5 kJ for this version with water vapor. The estimate is within 5 kJ, or about half a percent.

Key points
  • Breaking bonds takes energy in. Forming bonds gives energy out.
  • ΔH ≈ sum of bonds broken minus sum of bonds formed.
  • Stronger bonds in the products mean a negative ΔH, which is why fuels burn hot.
  • Tabulated values are averages, so bond energy answers are estimates.

The chapter in one breath

  • Pick the system first. Every sign in a thermochemistry problem is written from the system's point of view.
  • ΔE = q + w. Energy entering the system is positive. Energy leaving it is negative.
  • q = mcΔT turns a thermometer reading into joules, and ΔT is always final minus initial.
  • Water's specific heat of 4.184 J/g·°C is high, which is why water makes such a steady heat sink.
  • At constant pressure, ΔH is the heat. Negative means exothermic, positive means endothermic.
  • Thermochemical equations scale. Double the equation and double ΔH. Reverse it and flip the sign.
  • Hess's law adds steps. Flip, scale and stack known reactions until they equal the target.
  • Formation enthalpies give a shortcut: products minus reactants, each value times its coefficient.
  • Bond energies explain the heat. Break bonds, form stronger ones, and the difference comes out as heat.

ReferenceWords worth knowing

These are the words this chapter leans on. Come back to the list whenever a problem uses one you are not sure of.

Bomb calorimeter
A sealed steel vessel used to burn a sample in oxygen at constant volume, sitting in a measured water bath.
Bond dissociation energy (D)
The energy needed to break one mole of a given bond in the gas phase. Always a positive number.
Calorimetry
Measuring heat by tracking the temperature change of a known mass.
Endothermic
A change that takes heat in from the surroundings, so q and ΔH are positive.
Enthalpy (H)
A thermodynamic quantity defined as H = E + PV. Its change is what chemists measure at constant pressure.
Enthalpy change (ΔH)
The enthalpy of the products minus the enthalpy of the reactants. At constant pressure it equals the heat exchanged.
Exothermic
A change that gives heat out to the surroundings, so q and ΔH are negative.
First law of thermodynamics
Energy is conserved. For a system, ΔE = q + w.
Heat (q)
Thermal energy moving because of a temperature difference. It flows from warmer to cooler.
Heat capacity (C)
The heat needed to raise a whole object by one degree. Measured in J/°C, and it depends on size.
Hess's law
If a reaction is the sum of several steps, its ΔH is the sum of those steps' ΔH values.
Internal energy (E)
The total kinetic and potential energy stored inside a system.
Joule (J)
The SI unit of energy, equal to 1 kg·m2/s2. One calorie is 4.184 J.
Specific heat capacity (c)
The heat needed to raise one gram of a substance by one degree, in J/g·°C. A property of the substance.
Standard enthalpy of formation (ΔHf°)
The enthalpy change for making one mole of a substance from its elements, all in standard states.
Standard state
The reference condition for tabulated values: 1 bar for gases, 1 M for solutions, and the element's most stable form, at a stated temperature.
State function
A property whose change depends only on the starting and final states, never on the path taken.
Surroundings
Everything outside the system that can trade energy or matter with it.
System
The part of the universe you choose to study, such as the contents of a beaker.
Work (w)
Energy transferred by a force acting through a distance. In chemistry it is usually a gas expanding or being squeezed.

Check yourself

1. A system gives off 45 J of heat. At the same time the surroundings do 12 J of work on it. What is ΔE?
ΔE = −33 J. Heat out makes q = −45 J. Work done on the system makes w = +12 J. So ΔE = (−45 J) + (+12 J) = −33 J. The system ends up with 33 J less than it started with.
2. How much heat is needed to warm 250.0 g of water from 20.0 °C to 80.0 °C?
62.8 kJ. ΔT = 80.0 − 20.0 = 60.0 °C. Then mc = (250.0 g)(4.184 J/g·°C) = 1046.0 J/°C, and q = (1046.0)(60.0) = 62,760 J. Rounded to three significant figures that is 6.28 × 104 J, or 62.8 kJ.
3. You add 500. J of heat to 100.0 g of copper, and the same 500. J to 100.0 g of water. How much does each one warm up?
Copper rises 13.0 °C. Water rises only 1.20 °C. Rearrange to ΔT = q ÷ mc. For copper, ΔT = 500. ÷ [(100.0)(0.385)] = 500. ÷ 38.5 = 13.0 °C. For water, ΔT = 500. ÷ [(100.0)(4.184)] = 500. ÷ 418.4 = 1.20 °C. Water's larger specific heat means a smaller temperature change for the same heat.
4. A 15.0 g piece of aluminum at 85.0 °C is dropped into 75.0 g of water at 22.0 °C. What final temperature do they reach? Use c = 0.897 J/g·°C for aluminum.
About 24.6 °C. Set heat lost equal to heat gained. For aluminum, mc = (15.0)(0.897) = 13.455 J/°C. For water, mc = (75.0)(4.184) = 313.80 J/°C. Solving 13.455(T − 85.0) + 313.80(T − 22.0) = 0 gives T = (1143.675 + 6903.6) ÷ 327.255 = 24.6 °C. The water barely moves because it has the far bigger mc.
5. Given 2H2(g) + O2(g) → 2H2O(l) with ΔH = −571.6 kJ, what is ΔH for H2O(l) → H2(g) + ½O2(g)?
ΔH = +285.8 kJ. Two changes apply. Halving the equation halves the value to −285.8 kJ. Reversing it flips the sign to +285.8 kJ. Splitting water takes in exactly the heat that forming it gave off.
6. Methane burns by CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔH = −890.5 kJ. How much heat comes from burning 5.00 g of methane?
278 kJ is released. The molar mass of CH4 is 16.04 g/mol, so n = 5.00 ÷ 16.04 = 0.3117 mol. Then q = (0.3117 mol)(−890.5 kJ/mol) = −277.6 kJ. To three significant figures, 278 kJ leaves the system.
7. Use these two steps to find ΔH for N2(g) + 2O2(g) → 2NO2(g). Step A: N2(g) + O2(g) → 2NO(g), ΔH = +180.5 kJ. Step B: 2NO(g) + O2(g) → 2NO2(g), ΔH = −114.1 kJ.
ΔH = +66.4 kJ. Both steps already point the right way with the right coefficients, so just add them. The 2NO cancels, one O2 from each step gives 2O2 on the left, and ΔH = (+180.5) + (−114.1) = +66.4 kJ.
8. Why is the standard enthalpy of formation of O2(g) equal to zero, while the value for O3(g) is not?
O2(g) is oxygen's standard state, so there is nothing to form. Formation enthalpy is measured from the elements in their most stable forms. At 1 bar and 25 °C that form is O2(g), so making O2 from O2 involves no change. Ozone is a different, less stable form, so building it from O2 does cost energy.
9. Find ΔH° for 2SO2(g) + O2(g) → 2SO3(g). Use ΔHf° = −296.8 kJ/mol for SO2(g) and −395.7 kJ/mol for SO3(g).
ΔH° = −197.8 kJ. Products: 2(−395.7) = −791.4 kJ. Reactants: 2(−296.8) + 1(0) = −593.6 kJ, since O2 is an element in its standard state. Subtracting gives (−791.4) − (−593.6) = −197.8 kJ.
10. Estimate ΔH for N2(g) + 3H2(g) → 2NH3(g) from bond energies. Use N≡N = 945, H–H = 436 and N–H = 390 kJ/mol.
About −87 kJ. Bonds broken: one N≡N plus three H–H, so 945 + 3(436) = 945 + 1308 = 2253 kJ. Bonds formed: each NH3 has three N–H bonds, so six in all, 6(390) = 2340 kJ. Then ΔH ≈ 2253 − 2340 = −87 kJ. Tables give −91.8 kJ, so the estimate is close.
11. A cold pack gets cold when ammonium nitrate dissolves in water. Which way does heat flow, and what is the sign of ΔH?
Heat flows from your hand into the pack, and ΔH is positive. The dissolving is the system. It takes heat in, which makes it endothermic with a positive ΔH of about +25.7 kJ/mol. Your hand is part of the surroundings, and it loses that heat, which is why it feels cold.
12. Why does a bond energy calculation usually miss the tabulated ΔH by a few kilojoules?
Because tabulated bond energies are averages. A bond's real strength shifts with the atoms around it. The C–H bond in methane is about 439 kJ/mol, but the table lists 415 kJ/mol as an average over many compounds. Bond energies also assume gas phase species, so any liquid or solid in the equation adds more error.
Chemistry · Chapter 6

Electronic Structure and Periodic Properties

Light carries energy in small packets. An atom can take in or give off those packets only in fixed amounts. That one fact explains the red glow of a neon sign and the orange of a sodium streetlight. It also explains the shape of the periodic table. In this chapter you follow the electron. You will see where it can sit, how much energy each seat holds, and what happens when it moves. Then you will use that map to write the electron layout of any element. By the end the table will look less like a chart and more like a record of how electrons stack up.

By the end of this chapter you can

  • Link wavelength, frequency and photon energy using c = λν and E = hν.
  • Use the Bohr energy levels to predict the four visible lines of hydrogen.
  • Name the four quantum numbers and say what each one fixes.
  • Sketch an s, p, d or f orbital and count the nodes it carries.
  • Write the electron configuration of any element, including the common exceptions.
  • Draw an orbital diagram that obeys Pauli and Hund, then count unpaired electrons.
  • Predict radius, ionization energy, electron affinity and electronegativity across the table.

Section 1Light and quantized energy

Light is a wave you can measure

Light is a moving pair of electric and magnetic fields. In empty space it always travels at one speed. That speed is c = 2.998 × 108 m/s. Every wave has a wavelength, written λ. Wavelength is the distance from one crest to the next crest. Every wave also has a frequency, written ν. Frequency counts how many crests pass a fixed point each second. Its unit is the hertz, which is just s-1.

The speed of light ties those two numbers together.

c = λν

Because c never changes, λ and ν must trade off. A short wave has to come fast. A long wave has to come slowly. Blue light has a shorter wavelength than red light. So blue light must have the higher frequency. The amplitude is the height of the wave. Amplitude sets how bright the light looks. It does not set the color.

Visible light is a thin slice of the whole electromagnetic spectrum. Red light sits near 700 nm. Violet light sits near 400 nm. Radio waves can be meters long. X-rays are smaller than an atom. All of them are the same kind of wave at different sizes.

The parts of a light wave and the electromagnetic spectrum λ = one wavelength amplitude crest trough Frequency ν crests passing one point each second, in s -1 c = λν = 2.998 × 10 8 m/s radio microwave infrared ultraviolet X-ray gamma longer λ, lower energy → ← shorter λ, higher energy 400 nm 550 nm 700 nm the visible slice, opened up
One wave, two measurements. Wavelength is a distance and frequency is a count per second. Their product is always the speed of light. The gold band on the spectrum bar is the narrow range your eye can detect.

Energy comes in packets

For a long time light looked purely like a wave. Then two experiments broke that picture. Hot objects do not glow the way wave theory predicts. Light shining on a metal knocks electrons loose only above a threshold color. Max Planck and Albert Einstein fixed both problems with one idea. Light energy comes in countable pieces.

One piece of light energy is a photon. The energy of a photon depends only on frequency.

E = hν = hc / λ

Here h is Planck's constant, 6.626 × 10-34 J·s. The number is tiny, so one photon carries very little energy. Still, the rule is strict. A photon of 400 nm light carries almost twice the energy of a 700 nm photon.

Energy that comes in fixed pieces is quantized. Think about coins. You can pay 25 cents or 50 cents. You cannot pay 25.3 cents, because no coin that size exists. Light works the same way. Brighter light means more photons, not bigger ones.

Worked example · the energy in green laser light

A green laser pointer puts out light at 532 nm. Find its frequency, the energy of one photon, and the energy of a mole of those photons.

Step 1 · put the wavelength in meters

λ = 532 nm = 532 × 10-9 m = 5.32 × 10-7 m

Step 2 · use c = λν to get frequency

ν = c / λ = (2.998 × 108 m/s) ÷ (5.32 × 10-7 m) = 5.635 × 1014 s-1

Step 3 · use E = hν for one photon

E = (6.626 × 10-34 J·s)(5.635 × 1014 s-1) = 3.734 × 10-19 J

Step 4 · scale up to one mole

(3.734 × 10-19 J)(6.022 × 1023 /mol) = 2.249 × 105 J/mol = 225 kJ/mol

Check the units

J·s times s-1 leaves J, so step 3 is set up right. And 225 kJ/mol is the right size. It sits close to the strength of a chemical bond, which is why light can drive reactions.

Key points
  • Wavelength and frequency are locked together by c = λν.
  • Short wavelength means high frequency and high photon energy.
  • E = hν with h = 6.626 × 10-34 J·s gives the energy of one photon.
  • Amplitude controls brightness. Frequency controls color and energy.
  • Energy in packets is quantized. There is no half photon.

Section 2The Bohr model and line spectra

A glowing gas gives only certain colors

Put hydrogen gas in a glass tube and run current through it. The tube glows pink. Send that glow through a prism and the pink splits apart. You do not get a smooth rainbow. You get four sharp colored lines on a dark field. That pattern is a line spectrum.

Hydrogen's four visible lines sit at 656 nm, 486 nm, 434 nm and 410 nm. Every hydrogen sample in the universe gives the same four. Sodium gives a different set. Helium gives another. A line spectrum is a fingerprint for an element. That is how we know what stars are made of.

A smooth rainbow would mean an electron can hold any energy at all. Sharp lines mean something else. The electron can only hold certain fixed energies. When it drops from a higher one to a lower one, the leftover energy leaves as a single photon. One jump gives one color.

Bohr's ladder of allowed energies

In 1913 Niels Bohr built a model that got hydrogen right. He let the electron circle the nucleus only in certain orbits. Each orbit carries a fixed energy set by a whole number n, called the principal quantum number.

En = −(2.179 × 10-18 J) / n2

The energies are negative because the electron is trapped. Zero energy means a free electron, far from the nucleus. The lowest rung, n = 1, is the ground state. Any higher rung is an excited state. Notice how fast the rungs crowd together. The gap from n = 1 to n = 2 is huge. The gap from n = 5 to n = 6 is small.

To find the photon, subtract the two energies.

ΔE = Efinal − Einitial

A drop gives a negative ΔE. The atom lost that energy, so a photon of that size left. A climb gives a positive ΔE, which means the atom took a photon in. The four visible lines all end on n = 2. Jumps that end on n = 1 release far more energy, so they land in the ultraviolet.

Hydrogen energy levels and the four visible emission lines Allowed energies of the hydrogen electron n = ∞ 0 J (free) n = 6, 5 n = 4 n = 3 n = 2 n = 1 −0.242 −0.545 −2.179 units of 10 -18 J 122 nm ultraviolet, 2 → 1 656 nm (3 → 2) 486 nm (4 → 2) Rung spacing follows 1/n 2 , so high rungs crowd. What the prism shows 410 434 486 656 wavelength, nm 6 → 2 gives 410 nm. 5 → 2 gives 434 nm. 4 → 2 gives 486 nm. 3 → 2 gives 656 nm. Four jumps, four lines, no colors between.
One jump, one line. Every allowed energy is a rung on a ladder, and only the gaps between rungs can show up as light. The four visible hydrogen lines all come from jumps that land on n = 2.

Worked example · where the red line comes from

An excited hydrogen electron drops from n = 3 to n = 2. Find the photon's energy and wavelength.

Step 1 · find each energy

E3 = −(2.179 × 10-18 J)/9 = −2.421 × 10-19 J

E2 = −(2.179 × 10-18 J)/4 = −5.448 × 10-19 J

Step 2 · subtract, final minus initial

ΔE = (−5.448 × 10-19) − (−2.421 × 10-19) = −3.026 × 10-19 J

The sign is negative, so the atom gave off a photon of 3.026 × 10-19 J.

Step 3 · turn energy into wavelength

hc = (6.626 × 10-34 J·s)(2.998 × 108 m/s) = 1.986 × 10-25 J·m

λ = hc/E = (1.986 × 10-25 J·m) ÷ (3.026 × 10-19 J) = 6.56 × 10-7 m = 656 nm

Check it

656 nm is deep red, and red is the brightest hydrogen line by eye. The same steps with n = 4, 5 and 6 give 486, 434 and 410 nm.

Push an electron all the way off a ground-state hydrogen atom and you must supply the full 2.179 × 10-18 J. Multiply by Avogadro's number and that is 1312 kJ/mol. That number is the measured ionization energy of hydrogen. A model built to explain colored lines also predicts how tightly the atom holds its electron.
Key points
  • A line spectrum shows that electron energies are quantized.
  • En = −(2.179 × 10-18 J)/n2 for hydrogen, and the sign means bound.
  • ΔE = Efinal − Einitial. Negative means a photon came out.
  • The visible lines at 656, 486, 434 and 410 nm all end on n = 2.
  • Bohr's model nails hydrogen but fails for atoms with more electrons.

Section 3Quantum numbers and orbital shapes

From orbits to clouds

Bohr's neat circles cannot be right. An electron is small and fast, and the Heisenberg uncertainty principle says we cannot pin down both its place and its motion. So we stop asking where the electron is. We ask where it is likely to be. Erwin Schrödinger wrote an equation whose answers give exactly that. Each answer is an orbital.

An orbital is a region around the nucleus where one electron is likely to be found. Picture a swarm of gnats around a porch light. No single gnat has a path you can trace. Still, the swarm has a clear shape and a clear center. The darker part of the cloud means higher probability density.

Four numbers name every orbital

Each orbital is labeled by three quantum numbers. A fourth one describes the electron sitting in it.

So s subshells hold 1 orbital, p hold 3, d hold 5 and f hold 7. Each orbital takes at most 2 electrons. A full shell therefore holds 2n2 electrons.

What the shapes look like

An s orbital is a ball centered on the nucleus. A p orbital is two lobes on opposite sides, like a long balloon pinched in the middle. The three p orbitals point along x, y and z. A d orbital usually has four lobes in a clover pattern, and the fifth d shape has two lobes with a ring around the middle. The seven f shapes are more involved still.

Most orbitals have places where the electron is never found. Those places are nodes. The count is fixed. Angular nodes equal l, so a p orbital has one flat nodal plane through the nucleus. Radial nodes equal n − l − 1, and those are thin shells of empty space. Add them and any orbital has n − 1 nodes in all. More nodes go with more energy.

Shapes and nodes of 1s, 2s, 2p and 3d orbitals 1s one ball, no nodes l = 0, one orbital radial node 2s bigger ball with a hollow shell inside nodal plane 2p two lobes, one flat node l = 1, three orbitals 3d four lobes, two flat nodes l = 2, five orbitals Shape comes from l. Node count comes from n and l. Shading shows where the electron is most likely to be. Gold dashes mark nodes, where it is never found.
Four orbital shapes side by side. The s cloud is a ball, the p cloud is two lobes split by a flat node, and this d cloud is a four-leaf pattern split by two nodes. Angular nodes equal l and radial nodes equal n − l − 1.

Worked example · counting the fourth shell

How many subshells, orbitals and electrons fit in the shell with n = 4?

Step 1 · list the allowed l values

l goes from 0 to n − 1, so l = 0, 1, 2, 3. That is four subshells: 4s, 4p, 4d and 4f.

Step 2 · count orbitals with 2l + 1

4s gives 1, 4p gives 3, 4d gives 5 and 4f gives 7. Total = 1 + 3 + 5 + 7 = 16 orbitals.

Step 3 · two electrons per orbital

2 × 16 = 32 electrons.

Check with the shortcut

2n2 = 2(4)2 = 2(16) = 32. The two routes agree.

Key points
  • An orbital is a probability cloud, not a path.
  • n sets size, l sets shape, ml sets direction, ms sets spin.
  • l must be less than n, and ml must lie between −l and +l.
  • s, p, d and f subshells hold 1, 3, 5 and 7 orbitals.
  • A shell holds 2n2 electrons, and every orbital carries n − 1 nodes.

Section 4Electron configurations

Fill the cheapest seats first

An atom at rest holds its electrons in the lowest energy arrangement it can. The Aufbau principle says to add electrons one at a time, always to the lowest open subshell. The written result is the electron configuration. Oxygen has 8 electrons, so oxygen is 1s2 2s2 2p4. The superscripts must add up to the atomic number.

The filling order is not simply 1s, 2s, 3s. Once several electrons crowd an atom, each one feels the pull of the nucleus plus the push of the others. That reshuffles the subshell energies. The useful order is this.

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

Look at 4s and 3d. The 4s subshell fills first, even though its n is larger. The diagonal chart below builds that order for you. Follow each gold arrow from its tail to its head, then start the next one.

The diagonal chart that gives the orbital filling order Read along each gold arrow, then drop to the next arrow 1s 2s2p 3s3p3d 4s4p4d4f 5s5p5d5f 6s6p6d 7s7p Order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p 4s comes before 3d, and 6s comes before 4f. The arrows put them in that order for you. Each row is one value of n. Each column is one shape: s, then p, then d, then f.
The diagonal chart. Writing the subshells in a grid and sweeping down the diagonals reproduces the measured filling order, including the swap that puts 4s ahead of 3d.

Shorthand, core and valence

Long configurations get clumsy, so we use the last noble gas as a stand-in. Calcium is 1s2 2s2 2p6 3s2 3p6 4s2, which we shorten to [Ar] 4s2. The electrons hidden in the bracket are the core electrons. The ones written out are the valence electrons, and those are the ones that do chemistry. Elements in the same group share a valence pattern, which is exactly why they behave alike.

The exceptions worth knowing

In the first row of transition metals, 4s and 3d sit very close in energy. Two elements break the simple pattern because of that.

ElementPattern predictsMeasuredWhy
Cr (24)[Ar] 4s2 3d4[Ar] 4s1 3d5Half-filled d set, six unpaired spins
Cu (29)[Ar] 4s2 3d9[Ar] 4s1 3d10Full d set of ten electrons
Mo (42)[Kr] 5s2 4d4[Kr] 5s1 4d5Same half-filled gain, one row down
Ag (47)[Kr] 5s2 4d9[Kr] 5s1 4d10Same full-set gain as copper

Moving one s electron into the d set costs almost nothing here. Spreading the electrons out lowers repulsion, so the swapped version wins. Chromium ends with six unpaired electrons, which is more than any other element in its row.

Ions follow one more rule. When an atom loses electrons, the highest n goes first. Iron is [Ar] 4s2 3d6. Fe2+ is [Ar] 3d6, not [Ar] 4s2 3d4. Fe3+ is [Ar] 3d5. Anions just add electrons to the next open spot, so Cl- is [Ne] 3s2 3p6, the same as argon.

Worked example · arsenic, element 33

Write the full and short configurations for arsenic.

Step 1 · fill in order until you reach 33

1s2 (2) → 2s2 (4) → 2p6 (10) → 3s2 (12) → 3p6 (18) → 4s2 (20) → 3d10 (30) → 4p3 (33)

Step 2 · write it out

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3

Step 3 · use the noble gas shorthand

Argon is element 18, so arsenic is [Ar] 4s2 3d10 4p3.

Check the sum

2 + 2 + 6 + 2 + 6 + 2 + 10 + 3 = 33. Arsenic has five valence electrons, which matches group 15.

Key points
  • Add electrons to the lowest open subshell, and the superscripts must total the atomic number.
  • The order 1s 2s 2p 3s 3p 4s 3d 4p 5s comes from the diagonal chart.
  • [Noble gas] shorthand hides the core and shows the valence electrons.
  • Cr is [Ar] 4s1 3d5 and Cu is [Ar] 4s1 3d10. Mo and Ag match them.
  • Cations lose the highest n electrons first, so Fe3+ is [Ar] 3d5.

Section 5Orbital diagrams and Hund's rule

Boxes and arrows

A configuration like 2p4 tells you the count. It does not tell you how those four electrons are spread over the three 2p orbitals. An orbital diagram shows that. Draw one box for each orbital and one arrow for each electron. An up arrow is spin +½ and a down arrow is spin −½. Two rules decide where the arrows go.

Pauli caps every box at two

The Pauli exclusion principle says no two electrons in one atom can share all four quantum numbers. An orbital already fixes n, l and ml. Only spin is left, and spin has two settings. So a box holds at most two electrons, and they must point opposite ways. Those two are a paired set. This single rule is why s holds 2, p holds 6, d holds 10 and f holds 14.

Hund spreads them out first

Orbitals in the same subshell have equal energy. We call them degenerate. Hund's rule says to put one electron in each degenerate box, all spinning the same way, before doubling up anywhere. Electrons repel each other, so two in one box cost extra energy. Keeping them in separate boxes with parallel spins keeps them farther apart and lowers the total energy.

Nitrogen shows the rule plainly. Its 2p3 electrons go one per box, all three arrows up. Oxygen has one more. That fourth electron must pair somewhere, so oxygen ends with two unpaired electrons.

Unpaired electrons are measurable. An atom or ion with unpaired electrons is paramagnetic, and a magnet pulls on it. With every electron paired it is diamagnetic, and a magnet pushes it away very weakly. Zinc, [Ar] 4s2 3d10, is diamagnetic. Iron, with four unpaired 3d electrons, is strongly drawn to a magnet.

Orbital diagrams for nitrogen and oxygen showing Pauli and Hund One box per orbital, one arrow per electron 1s2s2p N O 1s 2 2s 2 2p 3 3 unpaired, paramagnetic Hund: spread out, spins parallel 1s 2 2s 2 2p 4 2 unpaired, paramagnetic Pauli: the pair must be opposite Up arrow is spin +1/2. Down arrow is spin −1/2. A box never holds two arrows pointing the same way.
Nitrogen and oxygen, box by box. Nitrogen keeps its three 2p electrons apart with parallel spins, as Hund's rule requires. Oxygen's fourth 2p electron has to pair up, and Pauli forces it to take the opposite spin.

Worked example · unpaired electrons in sulfur and iron

How many unpaired electrons does a sulfur atom have? What about an iron atom?

Step 1 · write sulfur's configuration

Sulfur is element 16, so it is [Ne] 3s2 3p4.

Step 2 · place the 3p electrons

Three boxes, four electrons. One per box uses three. The fourth pairs with one of them.

Boxes read paired, single, single. That gives 2 unpaired electrons, so sulfur is paramagnetic.

Step 3 · do the same for iron

Iron is element 26, so it is [Ar] 4s2 3d6. The 4s pair is paired.

Five d boxes take one each, which uses five. The sixth pairs up.

Boxes read paired, single, single, single, single. That gives 4 unpaired electrons.

Why it matters

Those four unpaired spins are why iron sticks to a magnet. Zinc, with 3d10, has none and does not.

Key points
  • An orbital diagram shows one box per orbital and one arrow per electron.
  • Pauli: no two electrons share all four quantum numbers, so a box holds two at most.
  • Paired electrons in one box must have opposite spins.
  • Hund: fill degenerate boxes singly with parallel spins before pairing any.
  • Unpaired electrons mean paramagnetic. All paired means diamagnetic.

One idea runs all four trends

Every trend in this section comes from a tug of war. The nucleus pulls the outer electrons in. The inner electrons get in the way and push back. The net pull that an outer electron actually feels is the effective nuclear charge, written Zeff. The blocking by inner electrons is shielding.

Move right across a period and protons pile up fast. Shielding barely changes, because the new electrons go into the same shell. So Zeff climbs and the outer electrons get held tighter. Move down a group and the valence electrons start a brand new shell. They sit farther out, with a thick layer of core electrons in between. The grip loosens.

Atomic radius

The atomic radius shrinks left to right and grows top to bottom. Across period 2 it falls from 167 pm at lithium to 42 pm at fluorine. Down group 1 it rises from 167 pm at lithium to 298 pm at cesium. Adding a shell always wins over adding protons.

Ions break the pattern in a useful way. A cation lost electrons, so the remaining ones feel a stronger pull per electron. A sodium atom is near 190 pm, while Na+ is about 102 pm. An anion gained electrons, which adds repulsion. Cl- is about 181 pm, much wider than the chlorine atom it came from. In an isoelectronic set, every ion has the same electron count. Then the one with the most protons is smallest, so the order runs Na+ < Ne < F-.

Atomic radius across period 2 and down group 1, with trend directions Atomic radius in picometers Across period 2, left to right LiBeBC NOFNe 1671128767 56484238 Down group 1, top to bottom LiNaKRbCs 167190243265298 up and to the right Z eff , ionization energy, electron affinity and electronegativity all rise down and to the left Atomic radius rises Radii come from one calculated set, so they can be compared with each other.
Two slices of the table. Going right across period 2, radius drops by a factor of four. Going down group 1, each new shell pushes the radius out again. The box at the bottom gives the direction of all four trends.

Ionization energy

The ionization energy is the energy needed to pull the loosest electron off a gaseous atom. It rises left to right and falls top to bottom. Sodium needs 496 kJ/mol, while argon in the same period needs 1521 kJ/mol. Cesium, far down group 1, needs only 376 kJ/mol.

Two small dips break the rise, and both are worth knowing. Boron (801 kJ/mol) is lower than beryllium (899 kJ/mol), because boron's electron leaves a higher-energy 2p orbital. Oxygen (1314 kJ/mol) is lower than nitrogen (1402 kJ/mol), because oxygen already has a pair crowding one 2p orbital. Those dips are the orbital diagrams showing up in measured data.

Removing a second electron always costs more. Sodium's first ionization energy is 496 kJ/mol. Its second is 4562 kJ/mol, roughly nine times as much. The second electron has to come out of the neon-like core. That huge jump is why sodium forms Na+ and never Na2+.

Worked example · from kJ per mole to joules per atom

Hydrogen's ionization energy is 1312 kJ/mol. How much energy is that for a single atom?

Step 1 · convert kilojoules to joules

1312 kJ/mol = 1.312 × 106 J/mol

Step 2 · divide by Avogadro's number

(1.312 × 106 J/mol) ÷ (6.022 × 1023 /mol) = 2.179 × 10-18 J per atom

Step 3 · compare with section 2

That is exactly the Bohr ground-state energy, 2.179 × 10-18 J. Lifting the electron from n = 1 to n = ∞ costs the same energy as ionizing the atom, because those are the same event.

Electron affinity and electronegativity

The electron affinity is the energy change when a gaseous atom accepts an electron. A negative value means energy is released, so the atom wants the electron. Chlorine is the champion at −349 kJ/mol, with fluorine close behind at −328 kJ/mol. Beryllium and the noble gases have positive values, because their subshells are already full and a new electron has nowhere good to sit.

Finally, electronegativity measures how hard an atom pulls on electrons it is sharing inside a bond. The Pauling scale runs from 0.79 at cesium to 3.98 at fluorine. It rises toward the upper right, just like ionization energy. Oxygen is 3.44, nitrogen is 3.04, carbon is 2.55 and hydrogen is 2.20. You will use those four numbers constantly in the next chapter to decide whether a bond is polar.

All four trends point the same way for a reason. Each one asks a version of the question “how tightly does this atom hold electrons?” So if you remember only one thing, remember that Zeff rises toward fluorine and falls toward cesium. Then you can rebuild every trend on the spot.
Key points
  • Zeff rises across a period and barely changes down a group.
  • Radius falls across a period and rises down a group. Cations shrink, anions swell.
  • Ionization energy rises across and falls down, with dips at group 13 and group 16.
  • Electron affinity is most negative near chlorine and fluorine.
  • Electronegativity runs from 0.79 at cesium to 3.98 at fluorine.

The chapter in one breath

  • Light is a wave with c = λν, and it delivers energy in photons of size E = hν.
  • Line spectra prove that an atom's electron energies are quantized, not continuous.
  • Bohr's levels, En = −(2.179 × 10-18 J)/n2, give hydrogen's lines at 656, 486, 434 and 410 nm.
  • Four quantum numbers name an orbital and the electron in it: n, l, ml and ms.
  • Shapes follow l: s is a ball, p is two lobes, d is mostly four lobes, f is more complex.
  • Configurations follow the diagonal order, with Cr and Cu taking 4s1 to gain a half or full d set.
  • Pauli caps an orbital at two opposite spins, and Hund spreads electrons out before pairing them.
  • Unpaired electrons make an atom paramagnetic, which is a property you can measure with a magnet.
  • Zeff drives every periodic trend, so radius, ionization energy and electronegativity all follow one map.

ReferenceWords worth knowing

These are the terms this chapter leans on most. Come back to the list whenever a problem uses a word you want to pin down.

amplitude
The height of a wave. It controls brightness, not color or photon energy.
Aufbau principle
The rule that electrons fill the lowest open subshell first.
core electrons
The inner electrons hidden inside the noble gas bracket. They rarely take part in bonding.
degenerate orbitals
Orbitals in the same subshell, which all have equal energy.
effective nuclear charge
Written Zeff. The net pull an outer electron feels after inner electrons shield it.
electron affinity
The energy change when a gaseous atom gains one electron. Negative means energy is released.
electron configuration
The list of subshells with electron counts, such as 1s2 2s2 2p4.
electronegativity
How strongly an atom pulls on electrons it shares in a bond. Fluorine leads at 3.98.
frequency
Written ν. The number of wave crests passing a point each second, in s-1.
ground state
The lowest energy arrangement of an atom's electrons.
Hund's rule
Fill degenerate orbitals singly with parallel spins before any orbital gets a pair.
ionization energy
The energy needed to remove the loosest electron from a gaseous atom.
isoelectronic
Having the same number of electrons. Na+, Ne and F- are an isoelectronic set.
node
A place inside an orbital where the electron is never found.
orbital
A region around a nucleus where one electron is likely to be. It holds two electrons at most.
paramagnetic
Having one or more unpaired electrons, so a magnet attracts the sample.
Pauli exclusion principle
No two electrons in an atom may share all four quantum numbers.
photon
One packet of light energy, with E = hν.
quantized
Allowed only in fixed amounts, like coins rather than a smooth stream.
wavelength
Written λ. The distance from one wave crest to the next.

Check yourself

1. A blue photon has a wavelength of 450 nm. What is its frequency?
6.66 × 1014 s-1. First convert: 450 nm = 4.50 × 10-7 m. Then ν = c/λ = (2.998 × 108 m/s) ÷ (4.50 × 10-7 m) = 6.66 × 1014 s-1.
2. What is the energy of one of those 450 nm photons?
4.41 × 10-19 J. E = hν = (6.626 × 10-34 J·s)(6.66 × 1014 s-1) = 4.41 × 10-19 J. Scaled to a mole that is 266 kJ/mol.
3. The 434 nm hydrogen line comes from a drop that ends on n = 2. Where did it start?
At n = 5. The photon energy is hc/λ = (1.986 × 10-25 J·m) ÷ (4.34 × 10-7 m) = 4.577 × 10-19 J. Divide by 2.179 × 10-18 J to get 0.210, and 0.210 = 1/4 − 1/25. So the start was n = 5.
4. How many orbitals and how many electrons fit in the n = 3 shell?
9 orbitals and 18 electrons. l = 0, 1, 2 gives 3s, 3p and 3d. Orbitals: 1 + 3 + 5 = 9. Electrons: 2 × 9 = 18, which matches 2n2 = 2(9) = 18.
5. Is the set n = 3, l = 3, ml = 0 allowed?
No. The rule is l ≤ n − 1, so with n = 3 the only choices are l = 0, 1 or 2. There is no 3f subshell.
6. Write chromium's electron configuration and count its unpaired electrons.
[Ar] 4s1 3d5, with 6 unpaired electrons. Check the count: 18 + 1 + 5 = 24, and chromium is element 24. One electron sits alone in 4s and one sits in each of the five 3d orbitals, so all six spins are parallel.
7. Copper is element 29. Why is it [Ar] 4s1 3d10 instead of [Ar] 4s2 3d9?
A full 3d set is worth the move. The 4s and 3d levels are nearly equal in energy. Shifting one 4s electron into 3d completes the ten-electron d set and lowers repulsion, so the swapped version is lower in energy.
8. What is the electron configuration of Fe3+?
[Ar] 3d5. Iron is [Ar] 4s2 3d6. Electrons leave the highest n first, so both 4s electrons go, then one 3d electron. That leaves five 3d electrons, all unpaired.
9. How many unpaired electrons does a phosphorus atom have?
Three. Phosphorus is [Ne] 3s2 3p3. Hund's rule puts one electron in each of the three 3p orbitals with parallel spins, so phosphorus is paramagnetic.
10. Rank Mg, Si and Ca from largest to smallest atomic radius.
Ca > Mg > Si. Calcium is one period below magnesium, so it has an extra shell. Silicon sits to the right of magnesium in the same period, where higher Zeff pulls the cloud in tighter.
11. Oxygen's first ionization energy is 1314 kJ/mol, below nitrogen's 1402 kJ/mol. Why?
Oxygen already has a crowded 2p orbital. Nitrogen's 2p3 has one electron in each orbital. Oxygen's 2p4 forces one pair to share a box, and that repulsion raises the electron's energy. The crowded electron is easier to remove.
12. Hydrogen's ionization energy is 1312 kJ/mol. How many joules is that per atom?
2.179 × 10-18 J per atom. 1312 kJ/mol = 1.312 × 106 J/mol. Divide by Avogadro's number: (1.312 × 106) ÷ (6.022 × 1023) = 2.179 × 10-18 J. That is the Bohr ground-state energy, which is the point.
Chemistry · Chapter 7

Chemical Bonding and Molecular Geometry

Atoms almost never travel alone. The way they join sets the rules for everything that happens next. This chapter follows two ways of joining. In the first, one atom hands an electron over for good. In the second, two atoms share. Then we get to shape, because a bonded group of atoms is not a shapeless blob. It has angles you can predict, and you predict them by counting. By the end you can take a formula like SF4, draw it, name its shape, and say whether its pull is lopsided or even.

By the end of this chapter you can

  • Show how an ionic bond forms, and say what makes lattice energy large or small
  • Use Pauling electronegativity to sort a bond as nonpolar, polar, or ionic
  • Draw a Lewis structure from a formula, and name the three ways the octet rule breaks
  • Write resonance forms, and use formal charge to pick the best one
  • Count electron domains, then name both the domain shape and the molecular shape
  • Give the bond angle for every common shape, including the squeezed cases
  • Add bond dipoles as arrows to tell a polar molecule from a nonpolar one

Section 1Ionic bonding and lattice energy

One atom gives, the other takes

Picture sodium next to chlorine. Sodium has one electron in its outer shell, while chlorine is one electron short of a full shell. So sodium hands its electron over. Sodium is left with 11 protons and 10 electrons, so it becomes Na+. A positive ion like this is a cation. Chlorine gains the electron and becomes Cl−. A negative ion is an anion. Opposite charges pull on each other, and that pull is an ionic bond.

There are no tidy pairs here. Each Na+ tugs on every Cl− nearby, and each Cl− tugs back on every Na+. The ions stack into a repeating solid called a lattice. In table salt, every sodium ion has six chloride ions around it, and every chloride ion has six sodium ions around it. So NaCl is a ratio, not a tiny two-atom unit you could pull out with tweezers.

The sodium chloride lattice in cross-section Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ Na⁺ Cl⁻ r₀ = 283 pm Charges alternate in every row and every column. Six neighbors each Every ion is held by the six ions of opposite charge touching it. r₀ sets the pull r₀ is the gap between two neighboring ion centers.
Figure 7.1 Salt is a repeating grid, not a molecule. The ratio 1:1 in the formula NaCl comes from the pattern, and r0 is the center-to-center gap between neighbors.

Lattice energy measures the glue

Lattice energy (ΔHlattice) is the energy it takes to pull one mole of the solid apart into free gas-phase ions. A large value means a tough solid. For NaCl it is about 769 kJ/mol, while for MgO it is about 3795 kJ/mol. So MgO takes nearly five times as much energy to break up. That shows up in the melting points. Salt melts at 801 °C, while MgO holds out past 2800 °C.

Two things set the size, and both come from Coulomb's law. Charge is the first, and the pull grows with the two charges multiplied together. Distance is the second, and the pull fades as the ion centers move apart.

ΔHlattice  ∝  Q1Q2 / r0

Here Q1 and Q2 are the ion charges, and r0 is the gap between the centers. Double both charges and the energy roughly quadruples, while swapping in a fatter ion makes the energy drop. MgO wins on both counts. Mg2+ and O2− each carry twice the charge of the salt ions, and both are smaller, so r0 is shorter.

Worked example · predict the lattice energy of NaF

Problem. NaCl has a lattice energy near 769 kJ/mol. Use ionic radii to estimate the value for NaF.

Step 1 · get r₀ for each solid Ionic radii: Na+ 102 pm, Cl− 181 pm, F− 133 pm. Add the two radii to get the center-to-center gap.
NaCl: r0 = 102 pm + 181 pm = 283 pm.
NaF: r0 = 102 pm + 133 pm = 235 pm.

Step 2 · check the charges Both solids pair a 1+ ion with a 1− ion. So Q1Q2 is the same for both, and only the distance changes.

Step 3 · scale by the distance Energy goes as 1/r0, so the smaller gap gives the bigger number.
769 kJ/mol × (283 pm / 235 pm) = 769 × 1.204 = 926 kJ/mol.

Step 4 · check it against the real value The measured lattice energy of NaF is about 910 kJ/mol. The estimate is high by 16 kJ/mol, or about 2 percent. Good enough to rank solids, and a reminder that the model is only the leading term.

Key points
  • Metals lose electrons to form cations. Nonmetals gain them to form anions.
  • An ionic solid is a lattice of ions, so its formula gives a ratio.
  • Lattice energy is the energy to split one mole of solid into gas-phase ions.
  • It rises with bigger charges and falls with a bigger r0.
  • MgO beats NaCl on both counts, so its lattice energy is about five times larger.
Worth knowing. Lattice energy is also why salt dissolves in water but not in oil. Water molecules can surround each ion and pay back most of that 769 kJ/mol. Oil cannot, so the lattice stays put.

Section 2Covalent bonds, electronegativity and polarity

Sharing instead of handing over

Two chlorine atoms cannot both win. Each one needs a single electron to fill its shell, and neither will give one up. So they share a pair instead. A shared pair of electrons holding two atoms together is a covalent bond. The pair sits in the space between the two nuclei, and both nuclei pull on it. That shared pull is the bond.

Sharing is rarely fair, because some atoms tug harder on a shared pair than others. Electronegativity is the number that scores that tug. Linus Pauling built the scale we still use, and he put fluorine at the top at 3.98, which makes it the greediest atom there is. Cesium sits near the bottom at 0.79. Electronegativity rises as you go right across a row and falls as you go down a column. Those two trends track the same cause, because nuclear charge grows to the right and the outer shell drifts farther from the nucleus as you go down.

ElementPauling valueElementPauling value
F3.98Si1.90
O3.44Al1.61
Cl3.16Be1.57
N3.04Mg1.31
Br2.96Li0.98
I2.66Na0.93
S2.58Ca1.00
C2.55K0.82
H2.20Cs0.79
P2.19B2.04

The gap in the numbers tells you the bond type

Subtract the smaller value from the larger one. That difference, written ΔEN, sorts bonds along a smooth range. A tiny gap means the pair sits in the middle, which is a nonpolar covalent bond. A medium gap drags the pair toward the greedier atom, which is a polar covalent bond, and one end picks up a partial negative charge written δ−. The other end goes δ+. A huge gap means the electron is taken, not shared, and the bond counts as ionic.

ΔENBond typeExampleΔEN for it
below 0.4Nonpolar covalentC−H in CH42.55 − 2.20 = 0.35
0.4 to 1.8Polar covalentO−H in H2O3.44 − 2.20 = 1.24
above 1.8IonicNa−Cl in salt3.16 − 0.93 = 2.23

Treat those cutoffs as road signs, not walls. H−F has ΔEN = 1.78 and still behaves as a very polar covalent bond. NaI has ΔEN = 1.73 yet is a true ionic solid. The better habit is to read ΔEN as a dial. Larger gap, more lopsided bond.

Nonpolar, polar and ionic bonding compared Nonpolar covalent Polar covalent Ionic Cl–Cl  ΔEN = 0 H–Cl  ΔEN = 0.96 Na–Cl  ΔEN = 2.23 ClCl HCl δ+δ− dipole points to Cl NaCl 1+1− The cloud sits deadcenter. No dipole. The cloud leans towardthe greedier atom. The electron has movedacross. Full charges.
Figure 7.2 One smooth range, not three separate worlds. As ΔEN grows, the shared cloud slides from the middle toward one atom, and at the far end it stops being shared at all.

Putting a number on the lopsidedness

A polar bond has a dipole moment, written μ. It equals the size of the partial charge times the distance between the charge centers. The unit is the debye (D), where 1 D = 3.336 × 10−30 C·m. Chemists often report the same idea as percent ionic character. To get it you compare the real dipole moment with the one you would see if a whole electron had moved across.

Worked example · how ionic is the H−Cl bond?

Problem. HCl has a measured dipole moment of 1.08 D and a bond length of 127 pm. What fraction of a full charge has shifted?

Step 1 · find μ for a full transfer Use μ = Qr with one whole electron charge moved across the bond.
μ = (1.602 × 10−19 C)(127 × 10−12 m) = 2.03 × 10−29 C·m.

Step 2 · change the unit to debye Divide by the size of one debye.
(2.03 × 10−29 C·m) / (3.336 × 10−30 C·m per D) = 6.10 D.

Step 3 · compare the real value to that limit 1.08 D / 6.10 D = 0.177.

Step 4 · read the answer The bond is about 18 percent ionic. Only about one fifth of an electron charge has shifted toward chlorine. The ΔEN of 0.96 told us to expect a clearly polar bond, and this number says how polar.

Key points
  • A covalent bond is a shared pair of electrons held by two nuclei.
  • Electronegativity scores how hard an atom tugs on a shared pair.
  • ΔEN below 0.4 is nonpolar, 0.4 to 1.8 is polar, above 1.8 is ionic.
  • A polar bond carries δ+ on one atom and δ− on the other.
  • The dipole moment μ puts a measured number on that split.
Worth knowing. No bond is perfectly ionic. Even in NaCl, a little electron density stays shared. That leftover sharing is why the simple charge-and-distance model misses the measured lattice energy by a few percent.

Section 3Lewis structures

Dots for electrons, lines for bonds

A Lewis structure is a map of where the outer electrons sit. Only valence electrons appear, because only they do the bonding. A shared pair is drawn as a line and called a bonding pair. A pair that belongs to one atom alone is drawn as two dots and called a lone pair. Two lines between the same two atoms make a double bond, and three lines make a triple bond. The number of shared pairs between two atoms is the bond order.

Most atoms stop when their valence shell holds eight electrons. That habit is the octet rule, and it comes straight from the s and p orbitals of the outer shell. Hydrogen is the standing exception, since its shell is full at two.

Five steps that always work

Build every structure the same way. First, add up the valence electrons from all the atoms. Add one more for each unit of negative charge, and take one away for each unit of positive charge. Second, sketch a skeleton with one line per bond. The atom with the lowest electronegativity usually goes in the middle, and hydrogen is never in the middle. Third, give every outer atom a full octet using lone pairs. Fourth, park any electrons that are left on the central atom. Fifth, if the central atom is still short of eight, pull a lone pair in from a neighbor to make a double or triple bond.

Keep a running count the whole time, because every line you draw uses two electrons and every pair of dots uses two. When your total matches the number from step one, the structure is complete. That count is the one thing that catches most mistakes.

Worked example · draw methanal, CH2O

Problem. Give a complete Lewis structure for CH2O.

Step 1 · count valence electrons C has 4, each H has 1, O has 6.
4 + 2(1) + 6 = 12 valence electrons, which is 6 pairs.

Step 2 · draw the skeleton Hydrogen is never in the middle, and carbon is less electronegative than oxygen. So carbon goes in the middle. Both H atoms and the O attach to it. Three single bonds use 3 × 2 = 6 electrons. Six are left.

Step 3 · fill the outer atoms Each H is already full with its two shared electrons. Oxygen needs 6 more, so it gets 3 lone pairs. That uses the last 6 electrons. Running total: 6 + 6 = 12. All 12 are placed.

Step 4 · check the central atom Carbon has only 3 bonds, so it holds 6 electrons. It is 2 short of an octet.

Step 5 · make a multiple bond Shift one lone pair from oxygen into the C−O bond. Now that bond is a double bond. Carbon has 4 bonds, so 8 electrons. Oxygen has 2 bonds plus 2 lone pairs, which is 4 + 4 = 8 electrons. The total is still 12, and both atoms have an octet.

Answer. H2C=O, with two lone pairs on oxygen and a C=O bond order of 2.

Three ways the octet rule breaks

The rule is a strong habit, not a law. It fails in three tidy families.

Some molecules have an odd electron count, so no structure can pair everything up. Nitrogen monoxide, NO, has 5 + 6 = 11 valence electrons. One electron has to stay alone, and the nitrogen ends up with seven. A molecule with an unpaired electron is a free radical, and radicals like this tend to be reactive.

Some central atoms settle for less than eight, and beryllium and boron do this often. BF3 has 3 + 3(7) = 24 valence electrons. Three B−F bonds use 6, and 18 fill the lone pairs on fluorine. Boron is left with only 6 electrons, and the molecule is stable anyway. That gap is why BF3 grabs a lone pair from almost anything.

Atoms in period 3 and below can hold more than eight. This is an expanded valence shell, and it needs a central atom big enough to fit the extra neighbors. PCl5 has 5 + 5(7) = 40 valence electrons, and phosphorus carries 10. SF6 has 6 + 6(7) = 48, and sulfur carries 12. XeF4 has 8 + 4(7) = 36. Four bonds use 8, the fluorines take 24 in lone pairs, and the last 4 sit on xenon as two lone pairs. Xenon ends up with 12 electrons. Second-row atoms such as C, N, O and F never do this, because they have no room for it.

Three families that break the octet rule Too few Expanded to 10 Expanded to 12 B F F F P Cl Cl Cl Cl Cl S F F F F F F BF₃, 24 electrons total. Boron keeps just 6. Still a stable molecule. PCl₅, 40 electrons total. Phosphorus holds 10. Five bonds, no lone pairs. SF₆, 48 electrons total. Sulfur holds 12. Six bonds, no lone pairs.
Figure 7.3 Lone pairs on the outer atoms are left off so the central counts stand out. Boron settles for six electrons. Phosphorus and sulfur take on more than eight, which only period 3 and heavier atoms can do.
Key points
  • Count valence electrons first, adjusting for the charge on an ion.
  • Lines are shared pairs, dots are lone pairs, and both use two electrons.
  • Fill the outer atoms, then the central atom, then add multiple bonds if needed.
  • Odd electron counts, such as NO with 11, cannot pair up at all.
  • B and Be often stop short of eight. Period 3 and below can go past eight.

Section 4Resonance and formal charge

When one drawing is not enough

Ozone, O3, has 18 valence electrons. Follow the five steps and you get a bent molecule with one double bond and one single bond. Now ask which side gets the double bond. Nothing in the rules picks a side, and you can draw it either way. Measurement settles the argument, because both bonds in ozone are 127.8 pm long. A true O=O double bond is about 121 pm and a true O−O single bond is about 148 pm. The real bonds are identical and sit between the two.

So neither drawing is the molecule. The two drawings are called resonance forms, and the real molecule is a single blend of them called the resonance hybrid. The double-headed arrow between forms is important. It does not mean the molecule flips back and forth, but rather that one structure is not enough to hold the truth, so you average the set. In ozone each bond is one and a half bonds, which fits the measured length exactly.

Average the shared pairs to get the real bond order, so ozone, with 3 bonding pairs spread over 2 bond positions, comes out at 3/2 = 1.5. Carbonate, CO32−, has three resonance forms and 4 bonding pairs over 3 positions, so its bond order is 4/3 = 1.33. All three carbonate bonds measure about 129 pm, which again sits between a single bond near 143 pm and a double bond near 123 pm. Spreading electrons out this way also lowers the energy, which is why these ions are so stable.

The two resonance forms of ozone and the blend they average to Resonance form AResonance form B The real molecule O O O +1 0 −1 O O O +1 −1 0 O O O 1.5 1.5 One double bond on theleft, one single on the right. The same drawing with thedouble bond swapped over. Two matching bonds, each127.8 pm, bond order 1.5. Lone pairs are left off to keep the charges readable. The central O carries +1 in both forms.
Figure 7.4 The arrow between forms A and B is not a flip. The real ozone molecule is the average on the right, with two identical bonds of order 1.5.

Formal charge keeps the bookkeeping honest

Formal charge is the charge an atom would carry if every bonding pair were split exactly down the middle. It is a bookkeeping tool rather than a real measured charge, and it is still the fastest way to rank competing structures.

formal charge = valence electrons − lone pair electrons − ½(bonding electrons)

Three habits make the tool reliable. The formal charges must add up to the overall charge on the species. The best structure keeps those charges as close to zero as it can. When a negative charge has to land somewhere, it belongs on the more electronegative atom.

Worked example · which CO2 drawing is better?

Problem. Both O=C=O and O≡C−O use all 16 valence electrons and give every atom an octet. Use formal charge to choose.

Step 1 · score the symmetric form, O=C=O Carbon: 4 − 0 − ½(8) = 4 − 0 − 4 = 0.
Each oxygen: 6 − 4 − ½(4) = 6 − 4 − 2 = 0.
Sum: 0 + 0 + 0 = 0, which matches a neutral molecule.

Step 2 · score the lopsided form, O≡C−O Carbon: 4 − 0 − ½(8) = 0.
Triple-bonded oxygen: 6 − 2 − ½(6) = 6 − 2 − 3 = +1.
Single-bonded oxygen: 6 − 6 − ½(2) = 6 − 6 − 1 = −1.
Sum: 0 + 1 − 1 = 0, so this one is legal too.

Step 3 · compare the scores Add the sizes of the charges. The symmetric form totals 0. The lopsided form totals 2.

Step 4 · pick the winner O=C=O wins. Its charges are all zero, and the loser puts a positive charge on oxygen, the second greediest atom on the scale. Measured CO2 has two equal bonds of 116 pm, which agrees.

Key points
  • Resonance forms differ only in where electrons sit, never in where atoms sit.
  • The real molecule is one blend of the forms, not a switch between them.
  • Bond order = bonding pairs divided by bond positions, so ozone is 1.5.
  • Formal charge = valence − lone pair electrons − half the bonding electrons.
  • Best structure: charges near zero, and any negative charge on the greedier atom.
Worth knowing. Equal resonance forms share the load equally, as in ozone and carbonate. When the forms are not equal, the better one counts for more. The blend leans toward whichever form scores best on formal charge.

Section 5VSEPR and molecular shapes

Electron groups push each other apart

A Lewis structure is flat on the page. Real molecules are not. To get the shape, use VSEPR, which stands for valence shell electron pair repulsion. The whole idea fits in one line: electron groups around a central atom are all negative, so they spread out as far from each other as they can.

Count groups, not atoms. Each group is an electron domain, and one single bond counts as one domain. A double or triple bond is still one domain, because those pairs sit in the same region. A lone pair is one domain too. Add them up, and the number of domains fixes the electron-domain geometry for you.

Two domains spread to 180°, which is linear. Three settle at 120° in a flat triangle, which is trigonal planar. Four go to 109.5° and point at the corners of a tetrahedron. Five give a trigonal bipyramid, with three equatorial positions 120° apart and two axial positions at 90° to that belt. Six give an octahedron, with every neighbor at 90°.

The five electron-domain geometries and their angles 2 domains3 domains4 domains 5 domains6 domains LinearTrigonal planarTetrahedral TrigonalbipyramidalOctahedral 180°120°109.5° 90° and 120°90° Gold is the central atom. Teal marks each electron domain.
Figure 7.5 Five domain counts, five patterns. These angles hold when every domain is a bond. Lone pairs bend them, as the text below explains.

Two shapes, one molecule

Here is the step that is easiest to skip. The domain geometry counts every domain, including lone pairs. The molecular geometry counts only the atoms, because lone pairs are invisible to a camera. So one molecule gets two names.

Water makes the point. Oxygen has 2 bonds and 2 lone pairs, so 4 domains, and the domain geometry is tetrahedral. The molecular geometry is bent, because you only see the two hydrogen atoms. Ammonia has 3 bonds and 1 lone pair, so also 4 domains. Its domain geometry is tetrahedral and its molecular shape is trigonal pyramidal.

Lone pairs also squeeze the angles. A lone pair is held by one nucleus instead of two, so it spreads wider and pushes harder on its neighbors. Methane has 4 bonds and no lone pairs, and its angle is the ideal 109.5°. Ammonia loses a little, down to 107°, and water loses more, down to 104.5°. Each lone pair shaves off about 2 to 3 degrees. For five domains, lone pairs always take equatorial spots, where there is more elbow room.

Bonds + lone pairsDomain geometryMolecular geometryAnglesExample
2 + 0LinearLinear180°CO2
3 + 0Trigonal planarTrigonal planar120°BF3
2 + 1Trigonal planarBent119°SO2
4 + 0TetrahedralTetrahedral109.5°CH4
3 + 1TetrahedralTrigonal pyramidal107°NH3
2 + 2TetrahedralBent104.5°H2O
5 + 0Trigonal bipyramidalTrigonal bipyramidal90°, 120°PCl5
4 + 1Trigonal bipyramidalSeesaw102°, 173°SF4
3 + 2Trigonal bipyramidalT-shaped87.5°ClF3
2 + 3Trigonal bipyramidalLinear180°XeF2
6 + 0OctahedralOctahedral90°SF6
5 + 1OctahedralSquare pyramidal84.8°BrF5
4 + 2OctahedralSquare planar90°XeF4

Worked example · name both shapes of SF4

Problem. Give the domain geometry, the molecular geometry and the angles for SF4.

Step 1 · count valence electrons S has 6 and each F has 7.
6 + 4(7) = 34 valence electrons, which is 17 pairs.

Step 2 · place them Four S−F bonds use 8 electrons. Giving each fluorine 3 lone pairs uses 24 more. Running total: 8 + 24 = 32. That leaves 34 − 32 = 2 electrons, so sulfur gets 1 lone pair.

Step 3 · count domains Sulfur has 4 bonds plus 1 lone pair, so 5 domains. Five domains means a trigonal bipyramidal domain geometry. Sulfur is in period 3, so holding 10 electrons is allowed.

Step 4 · drop the lone pair and name the shape The lone pair takes an equatorial spot. That leaves 2 axial and 2 equatorial fluorines. The molecular geometry is a seesaw.

Step 5 · adjust the angles Start from the ideal 120° and 90°, then let the lone pair push. The measured equatorial angle is about 102° and the axial pair closes from 180° to about 173°.

Key points
  • Count domains. A multiple bond is one domain, and a lone pair is one domain.
  • 2, 3, 4, 5 and 6 domains give linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.
  • The ideal angles are 180°, 120°, 109.5°, then 90° with 120°, then 90°.
  • Domain geometry counts lone pairs. Molecular geometry counts only atoms.
  • Lone pairs push harder, so each one trims the angle by a few degrees.
  • With five domains, lone pairs always sit in the equatorial belt.
Worth knowing. Shape decides function. Water is bent, so it clings to its neighbors and stays liquid at room temperature. If it were linear, the pulls would cancel, and water would boil far below 0 °C.

Section 6Molecular polarity

Arrows that can cancel

Every polar bond has a little arrow on it. The arrow points toward the greedier atom, and its length tracks ΔEN. That arrow is the bond dipole, and a whole molecule is polar only if those arrows fail to cancel. Arrows are vectors, so you add them head to tail, the same way you add forces. The leftover arrow is the molecular dipole moment.

Carbon dioxide shows the catch. Both C=O bonds are strongly polar, since ΔEN = 3.44 − 2.55 = 0.89. But CO2 is linear, so the two arrows point in exactly opposite directions, which means they cancel and the measured dipole moment is 0 D. A molecule can be built entirely from polar bonds and still be nonpolar overall.

Water has the same two polar bonds and a bigger ΔEN of 1.24. It is bent at 104.5°, so the two arrows do not oppose each other, and they add to a strong net arrow pointing from the hydrogen side toward the oxygen. Water's dipole moment is 1.85 D, and that one number explains much of its behavior.

Bond dipoles cancelling in carbon dioxide and adding in water CO₂ · linear H₂O · bent C O O ΔEN = 0.89 on each bond Equal arrows, opposite ways. They add to nothing. μ = 0 D · nonpolar O H H 104.5° net Bent, so the arrows add. μ = 1.85 D · polar
Figure 7.6 Same kind of polar bond, opposite outcome. Shape decides whether the bond dipoles cancel. The gold arrows are bond dipoles and the teal arrow is the sum.

A two-question test

You can usually judge polarity without any math. Ask two questions. Are all the outer atoms the same? Are they placed symmetrically around the center, with no lone pairs on the central atom? If both answers are yes, the arrows cancel and the molecule is nonpolar. Any no, and it is polar.

That test sorts the common cases fast. CH4, CCl4, BF3, CO2, PCl5 and SF6 all pass, so all are nonpolar. Swap one atom and symmetry dies. CHCl3 has a dipole moment of 1.04 D because one corner is now hydrogen, and a lone pair usually breaks symmetry just as well. SO2 is bent, and its dipole moment is 1.63 D, while linear CO2 sits at zero. XeF2 and XeF4 are the exception. Their lone pairs sit exactly opposite each other, so the bond arrows still cancel.

Worked example · how big is one O−H bond dipole?

Problem. Water has μ = 1.85 D and a bond angle of 104.5°. Find the dipole of a single O−H bond.

Step 1 · set up the vector sum The two bond arrows are equal in size. Each leans half the bond angle away from the net arrow. Only the part along the net direction survives, so the sum is
μnet = 2 μbond cos(θ/2).

Step 2 · put in the angle θ/2 = 104.5° / 2 = 52.25°, and cos 52.25° = 0.6124.

Step 3 · solve for the bond dipole μbond = μnet / (2 cos 52.25°)
μbond = 1.85 D / (2 × 0.6124) = 1.85 / 1.2248 = 1.51 D.

Step 4 · sanity check A single O−H bond is more polar than the whole water molecule, which sounds odd until you look at the arrows. The two bonds point partly against each other, so some of each one is lost. Compare 1.51 D with 1.08 D for H−Cl. The O−H bond has the larger ΔEN, 1.24 against 0.96, so the larger dipole fits.

Key points
  • A bond dipole is an arrow pointing at the greedier atom, sized by ΔEN.
  • Add the arrows as vectors. The leftover is the molecular dipole moment.
  • Polar bonds in a symmetric shape cancel, as in CO2 and CCl4.
  • Matching outer atoms and no lone pairs on the center means nonpolar.
  • For a bent AX2 molecule, μnet = 2μbond cos(θ/2).
Worth knowing. NH3 and NF3 have the same shape, yet their dipole moments are 1.47 D and 0.23 D. In ammonia the bond arrows point at nitrogen, the same way the lone pair leans. In NF3 the arrows point out at the fluorines, against the lone pair, so most of the pull cancels.

The chapter in one breath

  • Ionic bonds move an electron across. The solid is a lattice, so the formula is a ratio.
  • Lattice energy climbs with bigger ion charges and shrinks as r0 grows.
  • Covalent bonds share a pair, and ΔEN scores how fairly: under 0.4 nonpolar, 0.4 to 1.8 polar, over 1.8 ionic.
  • Lewis structures start with a valence electron count and end with every octet checked.
  • The octet rule bends three ways: odd counts, too few electrons, and expanded shells.
  • Resonance blends two or more drawings, which is why ozone has two 1.5 bonds.
  • Formal charge ranks competing structures, and charges near zero win.
  • VSEPR turns a domain count into two shape names and a bond angle.
  • A molecule is polar only when its bond dipoles fail to cancel.

ReferenceWords worth knowing

Twenty terms carry this chapter. Read the list once now, then come back to it when you are drawing structures and naming shapes.

anion
An ion with a negative charge, made when an atom gains one or more electrons.
bond dipole
The arrow on one bond, pointing to the more electronegative atom. Its length tracks ΔEN.
bond order
The number of shared pairs between two atoms. Resonance can make it a fraction, such as 1.5 in ozone.
cation
An ion with a positive charge, made when an atom loses one or more electrons.
covalent bond
A bond made of one or more electron pairs shared between two atoms.
dipole moment (μ)
The measured size of a charge separation, in debye. 1 D = 3.336 × 10−30 C·m.
electron domain
One group of valence electrons around a central atom. A single, double or triple bond counts as one, and so does a lone pair.
electronegativity
A number for how strongly an atom pulls on a shared pair. On the Pauling scale F is 3.98 and Cs is 0.79.
expanded valence shell
A central atom holding more than eight valence electrons, as phosphorus does in PCl5. Only period 3 and heavier atoms can.
formal charge
Valence electrons minus lone pair electrons minus half the bonding electrons. A bookkeeping score used to rank structures.
ionic bond
The pull between oppositely charged ions after an electron has moved from one atom to another.
lattice energy (ΔHlattice)
The energy needed to pull one mole of an ionic solid apart into gas-phase ions. About 769 kJ/mol for NaCl.
Lewis structure
A drawing of the valence electrons, with lines for shared pairs and dots for lone pairs.
lone pair
A valence pair that belongs to one atom and is not shared.
molecular geometry
The shape traced by the atoms alone, with lone pairs left out. Water is bent.
nonpolar covalent bond
A bond with ΔEN under about 0.4, where the shared pair sits near the middle.
octet rule
The habit of ending up with eight valence electrons. Hydrogen stops at two.
polar covalent bond
A bond with ΔEN of about 0.4 to 1.8, giving δ+ on one atom and δ− on the other.
resonance
Two or more valid Lewis structures for one species. The real molecule is their blend, not a switch between them.
VSEPR
Valence shell electron pair repulsion. The model that predicts shape by spreading electron domains as far apart as possible.

Check yourself

1. How many valence electrons go into the Lewis structure of the chlorate ion, ClO3−?
26 electrons. Cl brings 7 and each O brings 6, so 7 + 3(6) = 25. The 1− charge adds one more electron: 25 + 1 = 26, which is 13 pairs.
2. NaCl has a lattice energy near 769 kJ/mol. Estimate the value for NaBr, given radii Na+ 102 pm, Cl− 181 pm and Br− 196 pm.
About 730 kJ/mol. r0(NaCl) = 102 + 181 = 283 pm and r0(NaBr) = 102 + 196 = 298 pm. The charges match, so scale by distance: 769 × (283/298) = 769 × 0.9497 = 730 kJ/mol. The measured value is about 732 kJ/mol.
3. Find ΔEN for the Si−O bond and classify it. Use Si 1.90 and O 3.44.
ΔEN = 1.54, a polar covalent bond. 3.44 − 1.90 = 1.54, which lands between 0.4 and 1.8. Oxygen takes the δ− end. This is the bond that builds sand and glass.
4. Work out the formal charge on the central oxygen in ozone. It has one double bond, one single bond and one lone pair.
+1. Oxygen has 6 valence electrons. The lone pair is 2 nonbonding electrons. Three bonds mean 6 bonding electrons, and half of that is 3. So 6 − 2 − 3 = +1. The single-bonded outer oxygen comes out at 6 − 6 − 1 = −1, so the molecule is neutral overall.
5. What is the bond order in the nitrate ion, NO3−?
1.33. Nitrate has 5 + 3(6) + 1 = 24 valence electrons. Each resonance form has one N=O and two N−O bonds, so 4 bonding pairs are spread over 3 bond positions: 4/3 = 1.33. All three bonds are the same length.
6. SO2 has μ = 1.63 D and a bond angle of 119°. How big is one S−O bond dipole?
About 1.61 D. Use μnet = 2μbond cos(θ/2). Here θ/2 = 59.5° and cos 59.5° = 0.5075. So μbond = 1.63 / (2 × 0.5075) = 1.63 / 1.015 = 1.61 D.
7. Give both geometry names and the approximate bond angle for ClF3.
Trigonal bipyramidal domains, T-shaped molecule, about 87.5°. ClF3 has 7 + 3(7) = 28 valence electrons. Three bonds use 6 and the fluorine lone pairs use 18, leaving 4 for two lone pairs on chlorine. So 3 bonds + 2 lone pairs = 5 domains, and both lone pairs take equatorial spots.
8. Both CO2 and H2O are built from polar bonds. Why is only one of them polar?
Shape decides it. CO2 is linear, so its two bond dipoles point exactly opposite and cancel to μ = 0 D. Water is bent at 104.5°, so its two arrows add to 1.85 D pointing toward oxygen.
9. Which of NO, CO2 and CH4 cannot possibly follow the octet rule, and why?
NO. It has 5 + 6 = 11 valence electrons, an odd number, so one electron cannot be paired. Nitrogen ends up with seven, which makes NO a reactive free radical.
10. Name the electron-domain geometry and the molecular geometry of water, and say why they differ.
Tetrahedral domains, bent molecule. Oxygen has 2 bonds and 2 lone pairs, so 4 domains spread tetrahedrally. The domain name counts lone pairs and the molecular name does not, so you see only the bent H−O−H frame at 104.5°.
11. MgO has a lattice energy near 3795 kJ/mol against 769 kJ/mol for NaCl. Account for the difference using charges and radii. Use Mg2+ 72 pm, O2− 140 pm.
Charge does most of it, and the shorter distance adds the rest. The charge product goes from (1)(1) to (2)(2), a factor of 4. r0 falls from 283 pm to 72 + 140 = 212 pm, a factor of 283/212 = 1.33. Together the model predicts 4 × 1.33 = 5.3 times, and the real ratio is 3795/769 = 4.9 times.
12. A classmate says ozone spends half its time as form A and half as form B. What is wrong with that?
There is no switching. Ozone is one molecule with one structure, the blend of both forms. Both bonds are 127.8 pm at all times, with a bond order of 1.5. If it really switched, you would catch one short bond and one long bond.
Chemistry · Chapter 8

Advanced Theories of Covalent Bonding

Lewis dots tell you which atoms share pairs. They do not tell you why a shared pair holds, why a double bond refuses to twist, or why liquid oxygen sticks to a magnet. For that you need orbitals. In this chapter you will build two pictures of the same bond. The first keeps each atom's own orbitals and lets them overlap. The second mixes orbitals across the whole molecule. Each one pays off somewhere the other falls short. Together they explain shape, stiffness, bond strength and magnetism with the same small set of rules.

By the end of this chapter you can

  • Explain a covalent bond as two orbitals that overlap and share a spin‑paired pair of electrons.
  • Read an energy curve to find a bond length and a bond energy.
  • Count regions of electron density and name the hybrid set: sp, sp2, sp3, sp3d or sp3d2.
  • Tell a sigma bond from a pi bond, and count both in a molecule.
  • Say why a double bond locks rotation while a single bond spins.
  • Fill a molecular orbital diagram for any period 1 or period 2 diatomic.
  • Work out bond order, then predict bond length and paramagnetic or diamagnetic behaviour.

Section 1Valence bond theory and orbital overlap

A bond is a shared pair in a shared space

Valence bond theory keeps the atoms you already know. Each atom brings its own orbitals to the meeting. A bond forms when an orbital on one atom overlaps an orbital on the other. Two electrons then move into that shared space. Both nuclei pull on that pair at once. The double pull is the glue.

The key idea is orbital overlap. Overlap means two orbitals share some of the same space. More overlap piles more electron density between the nuclei. More density there makes a stronger bond. Overlap also has a direction, because orbitals point. A p orbital points along one axis. The bond it builds points along that axis too. That is why this theory predicts shapes so well.

One rule limits every bond to a pair. The two shared electrons must have opposite spins. Spin‑paired electrons in overlapping orbitals lower the energy. The same two electrons with matching spins raise it instead. So a full orbital has nothing to offer a new partner. That is why two He atoms meet and bounce apart.

How the energy of two hydrogen atoms changes with the distance between them 0 74 pm −436 Nuclei repel. Energy climbs fast. Far apart. No overlap, no bond. good overlap no overlap bond length = 74 pm the distance at the bottom of the dip bond energy = 436 kJ/mol Distance between the two nuclei (pm) Energy (kJ/mol)
The dip is the bond. Two H atoms fall to lower energy as their 1s orbitals overlap. Past the dip the two nuclei shove each other away. The bottom of the dip gives both the bond length and the bond energy.

Why atoms stop at one best distance

Slide two H atoms together and track the energy. Far apart they barely notice each other, so we call that energy zero. As they close in, each nucleus starts to pull on the other atom's electron. The energy falls. Lower energy means a steadier pair.

Keep pushing and the fall stops. Both nuclei carry positive charge, so they begin to shove. The energy climbs fast. Between the two trends sits a dip. The bottom of the dip is the bond length, the distance the atoms really keep. The depth of the dip is the bond energy, the price of pulling them apart again.

For H2 that dip sits at 74 pm and runs 436 kJ/mol deep. A picometre is 10−12 m, so 74 pm is a very short reach. Note that 436 kJ/mol is a mole of bonds, not one bond. The next example turns it into the cost of a single bond.

Worked example · the energy of one bond

The H–H bond energy is 436 kJ/mol. How much energy does it take to break just one H–H bond?

Step 1 · put the energy in joules436 kJ/mol × 1000 J/kJ = 436,000 J/mol.

Step 2 · divide by the number of bonds in a moleOne mole holds NA = 6.022 × 1023 bonds.

436,000 J/mol ÷ (6.022 × 1023 /mol) = 7.24 × 10−19 J

Step 3 · check it backwards7.24 × 10−19 J × 6.022 × 1023 = 4.36 × 105 J, which is 436 kJ. The units cancel to joules per bond, so the answer is the energy of one bond.

That is a tiny number, and it should be. One bond is one pair of atoms. The big number only shows up because a mole is such a huge count.

Key points
  • A covalent bond is two orbitals overlapping with a spin‑paired pair of electrons inside.
  • More overlap means more density between the nuclei, so a stronger bond.
  • Orbitals point, so bonds point. Overlap sets the shape of the molecule.
  • The dip in an energy curve gives the bond length at its bottom and the bond energy as its depth.
  • For H2 those values are 74 pm and 436 kJ/mol.
Worth knowing: the overlap of two 1s orbitals at 74 pm is close to 0.75 on a scale where 1 means the two orbitals sit right on top of each other. The interactive below reports that number as you slide the atoms.

Section 2Hybrid orbitals

The shape problem that forces mixing

Carbon has one 2s orbital and three 2p orbitals in its valence shell. The 2s is round. The three 2p orbitals point along x, y and z, at 90° to each other. Yet CH4 has four identical bonds at 109.5°. Plain s and p orbitals cannot give that. Something has to change.

The fix is hybridization. Orbitals on the same atom are waves, and waves on the same atom can add together. Mix one s with three p orbitals and you get four new hybrid orbitals. Each one has the same shape, the same energy, and one fat lobe pointing out. Four fat lobes spread as far apart as they can. That lands them at 109.5°, which is exactly what CH4 shows.

Mixing never creates or destroys orbitals. Put four orbitals in and you take four out. The names just record the recipe. An sp3 set comes from one s plus three p. An sp2 set comes from one s plus two p, leaving one p orbital untouched. An sp set comes from one s plus one p, leaving two p orbitals untouched. Those leftover p orbitals matter a lot in section 3.

The sp, sp2 and sp3 hybrid sets and the angles between their lobes 180° sp 1 s + 1 p → 2 hybrids linear · 50% s · 2 p left over 120° sp2 1 s + 2 p → 3 hybrids trigonal planar · 33% s · 1 p left over 109.5° sp3 1 s + 3 p → 4 hybrids tetrahedral · 25% s · 0 p left over Every hybrid has one big lobe (teal) and one small back lobe (gold) pale lobe points back behind the page
Same atom, new orbitals. Each hybrid set spreads its lobes as far apart as the count allows. Two lobes go to 180°, three go to 120° and four go to 109.5°. The small gold back lobe is why a hybrid points so much better than a plain p orbital.

Counting your way to the answer

You never have to guess a hybrid set. Count regions of electron density on the central atom. A region is one sigma bond or one lone pair. A double or triple bond still counts as one region, because the whole group points one way. The number of regions equals the number of hybrid orbitals you need.

Two regions give sp. Three give sp2. Four give sp3. Five and six need more room, so the bookkeeping adds d orbitals: sp3d for five and sp3d2 for six. The s character of a hybrid is the share of s in the mix. For sp it is 1 of 2, or 50%. For sp2 it is 1 of 3, near 33%. For sp3 it is 1 of 4, or 25%. More s character pulls the pair in closer, so those bonds run shorter and stronger.

Worked example · naming two hybrid sets

Find the hybrid set on the oxygen in H2O and on the sulfur in SF4.

Step 1 · count regions on OTwo O–H sigma bonds plus two lone pairs gives 2 + 2 = 4 regions.

Step 2 · match the countFour regions need four hybrids, so oxygen is sp3. Its s character is 1/4 = 25%.

Step 3 · count regions on SFour S–F sigma bonds plus one lone pair gives 4 + 1 = 5 regions.

Step 4 · match againFive regions need five hybrids, so sulfur is sp3d. Its s character is 1/5 = 20%.

Water bends to 104.5° rather than 109.5°. A lone pair spreads wider than a bonding pair, so it squeezes the two O–H bonds together. The hybrid set still tells you where to start.

Key points
  • Hybrid orbitals are mixes of s and p orbitals on one atom.
  • Orbitals in equals orbitals out: 1 s + 3 p gives 4 sp3 hybrids.
  • Count regions of electron density, then read off the set: 2 → sp, 3 → sp2, 4 → sp3, 5 → sp3d, 6 → sp3d2.
  • A multiple bond counts as one region, not two or three.
  • s character runs 50%, 33% and 25% for sp, sp2 and sp3. More s means shorter bonds.
Worth knowing: careful calculations show that d orbitals take very little real part in SF4 or SF6. Labels like sp3d2 stay useful because they count regions and give the right geometry. Treat them as good bookkeeping rather than a photograph.

Section 3Sigma and pi bonds

Two ways for orbitals to meet

Overlap can happen in two shapes. Orbitals can meet end to end, along the line joining the nuclei. That line is the internuclear axis. End‑to‑end overlap builds a sigma bond, written σ. The electron density piles up right on the axis, between the two nuclei. Look down the axis and the bond looks like a plain round tube. Nothing changes as you spin it.

Orbitals can also meet side by side. Two p orbitals that stay parallel overlap above the axis and below it at the same time. That is a pi bond, written π. A pi bond puts no density on the axis at all. The axis sits inside a nodal plane, a flat sheet where the chance of finding the electron is zero. So a pi bond is really one bond in two slabs, one over and one under.

Sigma overlap is the better deal. Pointing two lobes straight at each other gets them as close as they can get. Side‑on lobes have to lean past one another, so they share less space. Less overlap means a weaker bond. Every single bond you have ever drawn is a sigma bond. Pi bonds only show up as extras, on top of a sigma bond that is already there.

Head-on overlap gives a sigma bond and side-on overlap gives a pi bond overlap sits on the axis internuclear axis σ bond · head‑on round about the axis · spins freely bigger overlap · stronger nodal plane one slab above one slab below π bond · side‑on zero density on the axis · cannot twist
Shape of the meeting. Head‑on lobes share one region centred on the axis, so the sigma bond is round about that axis. Parallel lobes share two regions, one above and one below, with a nodal plane through both nuclei.
FeatureSigma (σ) bondPi (π) bond
Overlapend to end, along the axisside by side, parallel lobes
Density on the axishighest therezero, a nodal plane runs through it
Built froms, p or hybrid orbitalsunhybridized p orbitals
Relative strengthstronger, more overlapweaker, less overlap
Rotation about the axisfreeblocked
Where it appearsin every covalent bondonly on top of a σ bond

Counting sigma and pi bonds

The counting rule is short. A single bond is 1 σ. A double bond is 1 σ + 1 π. A triple bond is 1 σ + 2 π. The second pi bond in a triple bond uses the other pair of p orbitals, at right angles to the first. The two pi bonds wrap the sigma bond from four sides.

Worked example · sigma and pi in acetic acid

Acetic acid gives vinegar its bite. Its formula is CH3COOH. Count every σ and π bond.

Step 1 · list the bondsThree C–H bonds, one C–C bond, one C=O bond, one C–O bond and one O–H bond.

Step 2 · count sigmaEvery bond on that list holds one sigma bond: 3 + 1 + 1 + 1 + 1 = 7 σ.

Step 3 · count piOnly the C=O bond is a double bond, so it adds 1 π. Nothing else does.

Step 4 · check the total7 σ + 1 π = 8 bonding pairs, which is 16 bonding electrons. A Lewis structure of CH3COOH shows 8 lines, and the count agrees.

Key points
  • Sigma bonds come from head‑on overlap and hold their density on the internuclear axis.
  • Pi bonds come from side‑on overlap of parallel p orbitals, above and below a nodal plane.
  • Sigma overlap is larger, so a sigma bond is stronger than a pi bond.
  • Single = 1 σ. Double = 1 σ + 1 π. Triple = 1 σ + 2 π.
  • A pi bond never stands alone. It always rides on a sigma bond.

Section 4Multiple bonds and rotation

Why a double bond will not twist

Take ethene, C2H4. Each carbon is sp2, so each one keeps one unhybridized p orbital. Those two p orbitals stand up at right angles to the flat part of the molecule. When they line up parallel, they overlap side by side and make the pi bond. The whole molecule goes flat and stays flat.

Now try to twist one CH2 group. The sigma bond does not care, because it is round about the axis. The pi bond cares a great deal. Twisting tips one p orbital out of line with the other. Their overlap shrinks. It falls roughly as the cosine of the twist angle. At 90° the two p orbitals sit at right angles and share nothing. The pi bond is simply gone.

That is restricted rotation. To spin a double bond you must first pay for breaking a pi bond. For ethene that bill is near 270 kJ/mol. Room temperature cannot cover it, so the molecule stays locked. A single bond, as in ethane, only has to slide past a small bump of about 12 kJ/mol. Ethane spins millions of times a second.

Twisting a double bond destroys the side-on overlap of the two p orbitals twist 0° π overlap 100% cost 0 kJ/mol flat and locked twist 45° π overlap 71% cost 78 kJ/mol straining hard twist 90° π overlap 0% cost 267 kJ/mol π bond broken
The lock. Teal and gold mark the two halves of each p orbital. At 0° the lobes match and the pi bond is whole. By 90° the orbitals cross at right angles, the shaded overlap is gone, and the pi bond has been paid off.

What the lock buys you

Restricted rotation is not a nuisance. It is the reason shape can carry information. Two compounds can share a formula and still differ, because the groups on a double bond are pinned in place. Cis means two chosen groups sit on the same side. Trans means they sit on opposite sides. Cis‑2‑butene and trans‑2‑butene are different substances, with different melting points and different boiling points.

Your eyes run on this. A molecule called retinal sits in the back of your eye with one cis double bond. Light hits it and that bond flips to trans. The shape change starts the nerve signal. Without restricted rotation the molecule would wobble all day and you would see nothing.

Bond data backs the picture. A C–C single bond is 154 pm long. A C=C double bond is 134 pm. A C≡C triple bond is 120 pm. More shared pairs pull the nuclei closer. The bonds get stronger too, but not in equal steps, as the next example shows.

Worked example · how strong is a pi bond?

A C–C bond takes 347 kJ/mol to break. A C=C bond takes 614 kJ/mol. A C≡C bond takes 839 kJ/mol. Find the strength of each pi bond.

Step 1 · first pi bondA double bond is one sigma plus one pi. So the first pi bond is the extra:

614 kJ/mol − 347 kJ/mol = 267 kJ/mol

Step 2 · second pi bondA triple bond adds one more pi on top of the double bond:

839 kJ/mol − 614 kJ/mol = 225 kJ/mol

Step 3 · read the resultBoth pi bonds come out weaker than the 347 kJ/mol sigma bond. Side‑on overlap really is the poorer deal. Double is not twice single: 2 × 347 = 694 kJ/mol, well above the measured 614 kJ/mol.

Step 4 · compare with heatAt 298 K the thermal energy RT is (8.314 J/mol·K)(298 K) = 2477 J/mol, or 2.48 kJ/mol. The pi bond costs 267 ÷ 2.48 ≈ 108 times that. Room temperature has no chance.

Key points
  • A pi bond needs its two p orbitals parallel, so twisting destroys it.
  • Pi overlap falls off with the cosine of the twist angle and hits zero at 90°.
  • Rotating a C=C bond costs near 270 kJ/mol. Rotating a C–C bond costs about 12 kJ/mol.
  • The lock makes cis and trans isomers real, separate substances.
  • More shared pairs means shorter bonds: 154 pm, 134 pm and 120 pm for single, double and triple carbon bonds.

Section 5Molecular orbital theory

Spread the electrons over the whole molecule

Valence bond theory has one blind spot. It says O2 has a tidy double bond with every electron paired. Hang liquid oxygen between the poles of a magnet and it clings there. Only unpaired electrons do that. So O2 must have some, and the shared‑pair picture misses them.

Molecular orbital theory fixes this by letting go of the idea that an electron belongs to one bond. Atomic orbitals are waves, so add them across the whole molecule. Adding two atomic orbitals in step builds up density between the nuclei. That gives a bonding molecular orbital, lower in energy than either atomic orbital. Adding them out of step cancels density between the nuclei. That gives an antibonding molecular orbital, higher in energy, with a node right between the atoms. The method is called the linear combination of atomic orbitals.

Two atomic orbitals always give two molecular orbitals, one down and one up. Electrons then fill those molecular orbitals by the same rules you already use. Lowest energy first. Two electrons per orbital, with opposite spins. Equal‑energy orbitals each take one electron before any of them takes a second. Orbitals at the same energy are called degenerate.

Period 1 and period 2 diatomics

For H2 and He2 only 1s orbitals matter. They give σ1s and σ*1s. The star marks antibonding. H2 puts both electrons in σ1s and holds together. He2 has to fill the star as well, which undoes the gain, so He2 does not exist.

Period 2 brings the 2s and 2p orbitals. The 2s pair gives σ2s and σ*2s. The six 2p orbitals give six molecular orbitals: σ2p and σ*2p from the pair pointing along the axis, plus two degenerate π2p and two degenerate π*2p from the two side‑on pairs.

Here is the twist that trips people up. The order of those levels is not the same across the row. For B2, C2 and N2 the π2p pair sits below σ2p. For O2, F2 and Ne2 the σ2p sits below the π2p pair. The reason is mixing. The 2s and 2p orbitals lie close in energy on the left of the row, so the sigma orbitals from 2s and 2p push on each other and σ2p gets shoved up. Moving right, the nuclear charge grows and the 2s drops far below the 2p. The mixing fades and the normal order returns.

Order of levels, low to highHolds forWhy
σ2s, σ*2s, π2p, σ2p, π*2p, σ*2pLi2, Be2, B2, C2, N2strong 2s–2p mixing lifts σ2p above π2p
σ2s, σ*2s, σ2p, π2p, π*2p, σ*2pO2, F2, Ne22s sits far below 2p, so mixing is small
Molecular orbital diagram for O2, filled with twelve valence electrons O2 · 12 valence electrons · O/F ordering atom O molecular orbitals atom O 2s 2p 2s 2p σ2s σ*2s σ2p π2p (two) π*2p (two) σ*2p two unpaired, same spin Bonding electrons = 2 + 2 + 4 = 8   Antibonding electrons = 2 + 2 = 4 Bond order = (8 − 4) / 2 = 2   Unpaired electrons = 2   So O2 is paramagnetic
Where the magnet comes from. The last two electrons have to go into the two degenerate π*2p orbitals. Each takes one, with matching spins, so O2 carries two unpaired electrons. The bond order still comes out 2, which matches the double bond you draw.

Worked example · filling O2

Fill the diagram for O2 and state the magnetism.

Step 1 · count valence electronsOxygen is in group 16, so each atom brings 6. That makes 6 + 6 = 12 electrons to place.

Step 2 · pick the orderOxygen uses the O/F order: σ2s, σ*2s, σ2p, π2p, π*2p, σ*2p.

Step 3 · fill from the bottomσ2s takes 2. σ*2s takes 2. σ2p takes 2. The two π2p orbitals take 4. That is 10 placed, with 2 left.

Step 4 · obey the half‑filling ruleThe last 2 go into the two degenerate π*2p orbitals, one each, with the same spin. Two unpaired electrons, so O2 is paramagnetic.

Step 5 · bond orderBonding = 8, antibonding = 4, so the bond order is (8 − 4)/2 = 2. The magnet and the double bond both come out of one diagram.

Key points
  • Molecular orbitals spread over the whole molecule, not over one bond.
  • Two atomic orbitals give one bonding orbital (lower) and one antibonding orbital (higher, marked with a star).
  • Fill lowest first, two per orbital with opposite spins, and spread singly across degenerate orbitals first.
  • B2, C2 and N2 put π2p below σ2p. O2 and F2 do the opposite.
  • That ordering is why O2 ends up with two unpaired electrons.

Section 6Bond order and magnetism

One number that ties it together

Bond order is the net number of bonding pairs holding two atoms together. Count the electrons in bonding molecular orbitals. Count the electrons in antibonding ones. Subtract, then halve.

bond order = (bonding electrons − antibonding electrons) ÷ 2

Antibonding electrons cancel bonding ones, so they carry a minus sign. A bond order of 0 means no bond at all. That is the whole story of He2, Be2 and Ne2. They never form, because their stars fill as fast as their bonding orbitals do. Bond order can also land on a half, such as 2.5. Nothing is wrong. It just means one lone electron is pulling its weight in a bonding or antibonding orbital.

Bond order sets length and strength

Bond order is not just bookkeeping. It predicts two things you can measure. Raise the bond order and the two nuclei pull closer, so the bond length shrinks. Raise the bond order and the bond also gets harder to break, so the bond energy climbs. The row from N2 to F2 shows it clearly. N2 has bond order 3, a length of 110 pm, and needs 945 kJ/mol. O2 has bond order 2, a length of 121 pm, and needs 498 kJ/mol. F2 has bond order 1, a length of 142 pm, and needs only 159 kJ/mol.

Bond length of the four oxygen species against their bond order Higher bond order, shorter bond O2+ O2 O2− O22− BO 2.5 BO 2 BO 1.5 BO 1 112 pm121 pm 133 pm149 pm 1 unpaired · paramagnetic 2 unpaired · paramagnetic 1 unpaired · paramagnetic 0 unpaired · diamagnetic 70100 130160 190 Bond length (pm)
Four relatives, one trend. Pulling an electron out of O2 empties part of an antibonding orbital, so the bond order rises and the bond shortens. Adding electrons fills more of the star, so the bond order falls and the bond stretches.

Reading the magnet

Magnetism is the other prize. A substance with one or more unpaired electrons is paramagnetic. Each unpaired electron acts like a tiny magnet, so the sample is drawn into a magnetic field. A substance with every electron paired is diamagnetic. Paired spins cancel, and the sample is pushed out of the field, very weakly. So the test is simple. Finish the diagram, then look for a lone arrow. One or more lone arrows means paramagnetic.

This is where molecular orbital theory earns its keep. Pour liquid oxygen between the poles of a strong magnet and it hangs there in a bridge. Liquid nitrogen in the same gap pours straight through. The diagrams say why. O2 keeps two lone arrows in its π* level. N2 has none.

Worked example · the whole O2 family

Find the bond order and magnetism of O2+, O2, O2− and O22−, then rank them by bond length.

Step 1 · count electronsO2 has 12 valence electrons. Taking one away gives 11. Adding one gives 13. Adding two gives 14.

Step 2 · place themEvery one of these fills σ2s2 σ*2s2 σ2p2 π2p4 first. That is 10 electrons and 8 of them are bonding. Whatever is left goes into the two π*2p orbitals.

Step 3 · do the four sumsBonding stays at 8 every time. Antibonding is 2 from σ*2s plus whatever lands in π*.

O2+: π* holds 1, so antibonding = 3. (8 − 3)/2 = 2.5, with 1 unpaired.

O2: π* holds 2, so antibonding = 4. (8 − 4)/2 = 2, with 2 unpaired.

O2−: π* holds 3, so antibonding = 5. (8 − 5)/2 = 1.5, with 1 unpaired.

O22−: π* holds 4, so antibonding = 6. (8 − 6)/2 = 1, with 0 unpaired.

Step 4 · rank the lengthsBond order falls in the order O2+ > O2 > O2− > O22−, so bond length must rise in that same order. Measurement agrees: 112 pm, 121 pm, 133 pm and 149 pm.

Step 5 · magnetismThe first three carry at least one lone arrow, so all three are paramagnetic. Only O22−, the peroxide ion, is diamagnetic.

Key points
  • Bond order = (bonding electrons − antibonding electrons) ÷ 2.
  • Bond order 0 means the molecule does not form. He2, Be2 and Ne2 all land there.
  • Half‑integer bond orders are fine and show up in ions such as O2+.
  • Higher bond order gives a shorter and stronger bond.
  • Any unpaired electron makes a substance paramagnetic. All pairs means diamagnetic.
  • N2, O2 and F2 run 110, 121 and 142 pm, and 945, 498 and 159 kJ/mol.
Worth knowing: bond order compares bonds between the same kinds of atoms best. A bond order of 1 in Li2 is worth only about 110 kJ/mol, while a bond order of 1 in F2 is 159 kJ/mol and in H2 is 436 kJ/mol. The size of the orbitals matters too.

The chapter in one breath

  • Valence bond theory builds a bond from two overlapping orbitals holding one spin‑paired pair.
  • The dip in an energy curve gives the bond length at its lowest point and the bond energy as its depth.
  • Hybrid orbitals mix s and p on one atom so the lobes point where the molecule needs them.
  • Count regions of electron density: 2 → sp, 3 → sp2, 4 → sp3, 5 → sp3d, 6 → sp3d2.
  • Sigma bonds come from head‑on overlap, pi bonds from side‑on overlap of leftover p orbitals.
  • A pi bond needs parallel p orbitals, so a double bond cannot twist. That is what makes cis and trans real.
  • Molecular orbital theory spreads electrons over the molecule in bonding and antibonding levels.
  • B2, C2 and N2 keep π2p below σ2p. O2, F2 and Ne2 swap them.
  • Bond order = (bonding − antibonding)/2, and it predicts length, strength and magnetism in one go.

ReferenceWords worth knowing

These are the words this chapter leans on. Each one is defined the way it is used here.

antibonding molecular orbital
A molecular orbital that is higher in energy than the atomic orbitals that built it. It has a node between the nuclei and weakens the bond. Its symbol carries a star, as in σ*2p.
bond energy
The energy needed to break one mole of a given bond. It is the depth of the dip in the energy curve. For H2 it is 436 kJ/mol.
bond length
The average distance between two bonded nuclei. It sits at the bottom of the energy dip. For H2 it is 74 pm.
bond order
Half the difference between bonding and antibonding electrons. It gives the net number of bonds and can be a half number.
bonding molecular orbital
A molecular orbital lower in energy than the atomic orbitals that built it. It piles electron density between the nuclei and holds the atoms together.
degenerate orbitals
Two or more orbitals with exactly the same energy, such as the pair of π2p orbitals. Electrons spread out singly across them before pairing up.
diamagnetic
Having every electron paired. A diamagnetic sample is pushed out of a magnetic field, and only very weakly.
hybrid orbital
A new orbital on one atom built by mixing that atom's own s and p orbitals. It has one large lobe and one small back lobe.
hybridization
The mixing of atomic orbitals on the same atom to make an equal number of hybrid orbitals that point where the bonds go.
linear combination of atomic orbitals
The method of adding and subtracting atomic orbital waves to build molecular orbitals. Adding in step gives a bonding orbital, out of step gives an antibonding one.
molecular orbital
An orbital that spreads over a whole molecule rather than one atom. Electrons in it belong to the molecule.
molecular orbital theory
The model that builds molecular orbitals from atomic orbitals and then fills them with all the electrons of the molecule.
node
A place where the chance of finding the electron is zero. A pi bond has a nodal plane through both nuclei.
orbital overlap
Two orbitals sharing some of the same space. More overlap puts more density between the nuclei and makes a stronger bond.
paramagnetic
Having one or more unpaired electrons. A paramagnetic sample is pulled into a magnetic field. Liquid O2 does this.
pi bond
A bond from side‑on overlap of two parallel p orbitals. Its density sits above and below the internuclear axis, never on it.
region of electron density
One sigma bond or one lone pair on an atom. A double or triple bond still counts as a single region.
restricted rotation
The block on twisting about a double or triple bond. Twisting would break a pi bond, which costs far too much energy.
sigma bond
A bond from head‑on overlap along the internuclear axis, the straight line joining the two nuclei. Its density is highest on that axis and it spins freely.
valence bond theory
The model in which a bond is two overlapping atomic or hybrid orbitals sharing one pair of electrons with opposite spins.

Check yourself

1. Work out the bond order of N2 from its molecular orbital diagram.
Bond order 3. Nitrogen is group 15, so 5 + 5 = 10 valence electrons. N2 uses the B/C/N order, so fill σ2s2 σ*2s2 π2p4 σ2p2. Bonding = 2 + 4 + 2 = 8. Antibonding = 2. (8 − 2)/2 = 3. All electrons are paired, so N2 is diamagnetic.
2. Find the bond order of B2 and say whether it is paramagnetic.
Bond order 1, and paramagnetic. Boron brings 3 valence electrons each, so 3 + 3 = 6. The B/C/N order puts π2p next: σ2s2 σ*2s2 π2p2. Those last two electrons go one into each degenerate π orbital. Bonding = 2 + 2 = 4, antibonding = 2, so (4 − 2)/2 = 1. Two unpaired electrons make it paramagnetic. The other level order would have paired them in σ2p and predicted no magnetism, which is wrong.
3. Calculate the bond order of O2− and compare its bond length with O2.
Bond order 1.5, and the bond is longer than in O2. O2 has 12 valence electrons, so O2− has 13. Bonding stays at 8. Antibonding is 2 in σ*2s plus 3 in π*2p, which is 5. (8 − 5)/2 = 1.5. Lower bond order means a weaker pull, so the bond stretches from 121 pm to 133 pm.
4. What is the s character of an sp2 hybrid orbital, and how does it compare with sp3?
About 33% s, more than the 25% in sp3. An sp2 set mixes 1 s with 2 p, giving 3 hybrids. The s is shared over 3 orbitals: 1/3 = 0.333, or 33.3%. An sp3 set shares 1 s over 4 orbitals: 1/4 = 25%. More s character holds the pair closer to the nucleus, so sp2 bonds are shorter than sp3 bonds between the same atoms.
5. A C–C bond needs 347 kJ/mol and a C=C bond needs 614 kJ/mol. How strong is the pi bond?
About 267 kJ/mol. A double bond is one sigma plus one pi, so the pi bond is the difference: 614 kJ/mol − 347 kJ/mol = 267 kJ/mol. That is less than the sigma bond, which fits the smaller side‑on overlap. Note that a double bond is not twice a single bond, since 2 × 347 = 694 kJ/mol, not 614 kJ/mol.
6. The N≡N bond energy is 945 kJ/mol. How much energy does one N2 bond take?
1.57 × 10−18 J. First convert: 945 kJ/mol × 1000 = 945,000 J/mol. Then divide by NA: 945,000 ÷ (6.022 × 1023) = 1.569 × 10−18 J. Check it back: 1.569 × 10−18 × 6.022 × 1023 = 9.45 × 105 J, which is 945 kJ.
7. Give the hybridization of the carbon in CO2, then count the sigma and pi bonds.
sp, with 2 σ and 2 π. The structure is O=C=O. Carbon has two regions of electron density, one for each double bond, so 2 regions means 2 hybrids and sp hybridization at 180°. Each double bond is 1 σ + 1 π, so two double bonds give 2 σ + 2 π. Carbon keeps two unhybridized p orbitals, one for each pi bond.
8. Why can a single bond spin freely while a double bond cannot?
A sigma bond is round about its axis, but a pi bond needs its p orbitals parallel. Spinning a sigma bond does not change the overlap at all, so the cost is tiny, near 12 kJ/mol in ethane. Spinning a double bond tips the two p orbitals out of line and destroys the pi overlap. That costs about 270 kJ/mol, so the bond stays locked at room temperature.
9. Which level order does C2 use, and is C2 paramagnetic?
The B/C/N order, and C2 is diamagnetic with bond order 2. Carbon gives 4 + 4 = 8 valence electrons. Filling σ2s2 σ*2s2 π2p4 uses all 8. Bonding = 2 + 4 = 6, antibonding = 2, so (6 − 2)/2 = 2. Both π2p orbitals are full, so nothing is unpaired. Note that the two bonds here are both pi bonds, which valence bond theory would never guess.
10. Use bond order to explain why helium gas is made of single atoms.
He2 would have bond order 0. Two He atoms bring 4 electrons. Two fill σ1s and two fill σ*1s. Bond order is (2 − 2)/2 = 0. The antibonding pair cancels the bonding pair, so there is no net bond and the atoms drift apart. He2+ is different: it has only 3 electrons, so its bond order is (2 − 1)/2 = 0.5 and it can be detected.
11. Count the sigma and pi bonds in HCN, and name the hybridization of the carbon.
2 σ, 2 π, and the carbon is sp. The structure is H–C≡N. The H–C single bond gives 1 σ. The C≡N triple bond gives 1 σ + 2 π. Totals: 1 + 1 = 2 σ and 2 π. Carbon has two regions of electron density, so it is sp hybridized and the molecule is linear.
12. A Lewis structure of O2 shows every electron paired. Why is O2 still pulled into a magnet?
Its last two electrons sit alone in two degenerate π*2p orbitals. Oxygen has 12 valence electrons. Ten of them fill σ2s, σ*2s, σ2p and both π2p orbitals. The final two go into the two equal‑energy π* orbitals, one each with matching spins. Two unpaired electrons make O2 paramagnetic, while the bond order is still (8 − 4)/2 = 2. The Lewis picture gets the bond count right and the magnetism wrong.
Chemistry · Chapter 9

Gases

Gases are the easiest state of matter to predict. A gas has no shape and almost no structure. Its particles fly far apart and rarely touch. Because of that, one simple equation describes almost every gas you will meet. In this chapter you will learn what pressure really is and how we measure it. You will meet the simple gas laws, then fold them into the ideal gas law. You will find the volume of a gas a reaction makes. You will split the pressure of a mixture among its gases. At the end you will look at the moving particles themselves and see where the simple picture breaks down.

By the end of this chapter you can

  • Define pressure and convert among atm, torr, mmHg, kPa and bar.
  • Read a closed-end or open-end manometer and report the gas pressure.
  • Apply Boyle's, Charles's, Gay-Lussac's and Avogadro's laws, and the combined gas law.
  • Solve PV = nRT for any one missing variable, with the right value of R.
  • Use the molar volume 22.4 L/mol at STP, and find gas density and molar mass.
  • Find the volume of a gas made or used by a reaction.
  • Split a total pressure into partial pressures, and correct for water vapor.
  • Link temperature to particle speed, compare effusion rates, and say when a real gas strays from ideal.

Section 1Gas pressure and how we measure it

Pressure is a push spread over an area

A gas has no shape of its own. It spreads out to fill whatever holds it. Inside the container, the particles fly in every direction. Each one strikes the wall, bounces, and flies off again. Every hit is a tiny push. Add up all those pushes over a whole wall and you get pressure. Pressure is force divided by the area the force acts on.

P = F / A

The same force feels stronger on a small area. That is why a thumbtack sinks into wood but your thumb does not. The SI unit of pressure is the pascal (Pa), which is one newton per square meter. One pascal is a very small push, so we usually work in kilopascals.

You live at the bottom of an ocean of air. That air has weight, and it presses on you from all sides. This push is atmospheric pressure. At sea level it runs near 101 kPa. On a mountain top there is less air above you, so the push is smaller.

How a barometer and a manometer measure pressure BAROMETER 760 mm at sea level vacuum air presses on the open dish CLOSED-END gas vacuum Δh is the gas pressure in torr OPEN-END gas open air mercury higher on the open side, so Pgas = Patm + Δh
Three ways mercury reports a push. The barometer balances the whole atmosphere against a column of mercury. A closed-end manometer has a vacuum above the far arm, so the height gap is the gas pressure itself. An open-end manometer weighs the gas against the outside air.

Reading a barometer and a manometer

A barometer measures atmospheric pressure. Fill a long tube with mercury and tip it upside down into a dish. Some mercury drains out, but a tall column stays up. Air pressing on the dish is what holds it there. At sea level that column stands about 760 mm tall.

That height became a unit. One torr is the push of a mercury column 1 mm high, so 760 torr means the same thing as 760 mmHg. One atmosphere (atm) is defined as exactly 760 torr. Here are the units side by side.

UnitOne atm equalsWhere you meet it
atm1Gas law problems
torr or mmHg760Manometers, blood pressure
kPa101.325The SI choice, weather maps
Pa101,325Physics equations
bar1.01325Scuba, industry
psi14.696Tires in the United States

A manometer measures the pressure of gas trapped in a flask. A U-shaped tube of mercury joins the flask to the far side. In a closed-end manometer the far arm holds a vacuum. Nothing pushes back, so the height gap between the two mercury levels is the gas pressure in torr.

In an open-end manometer the far arm is open to the room. Now the gas and the air push against each other. If the mercury stands higher on the open side, the gas is stronger. Add the gap to air pressure. If the mercury stands higher on the gas side, the gas is weaker, so subtract the gap instead.

Worked example · pressure units and a manometer

A weather station reports 745 torr. Convert it to atm and to kPa. Then a flask of gas is joined to an open-end manometer in that same room. The mercury stands 68 mm higher on the open arm. What is the gas pressure in atm?

Step 1 · torr to atm

745 torr × (1 atm / 760 torr) = 0.980 atm

Step 2 · torr to kPa

745 torr × (101.325 kPa / 760 torr) = 99.3 kPa

Step 3 · read the manometer

Mercury is higher on the open arm, so the gas wins. Add the gap.

Pgas = 745 torr + 68 torr = 813 torr

Step 4 · convert the answer

813 torr × (1 atm / 760 torr) = 1.07 atm

Check it

The gas is a little above room pressure, so a value just over 1 atm makes sense. Going back the other way, 1.07 atm × 760 torr/atm = 813 torr.

Why mercury? Mercury is about 13.6 times denser than water. A water barometer would need a tube more than 10 m tall to balance the atmosphere. Mercury does the same job in under a meter.
Key points
  • Pressure is force per unit area, and gas pressure comes from particles striking the walls.
  • 1 atm = 760 torr = 760 mmHg = 101.325 kPa = 1.01325 bar.
  • A closed-end manometer reads the gas pressure straight off the height gap.
  • An open-end manometer adds the gap when mercury is higher on the open side, and subtracts it when mercury is higher on the gas side.

Section 2The simple gas laws

Change one thing and watch another respond

A sample of gas has four numbers that matter. Pressure P, volume V, temperature T and amount in moles n. The simple gas laws each hold two of those steady and watch the other two trade. Every one of them was found by experiment before anyone knew why it worked.

First, a rule you cannot skip. Every gas law uses temperature in kelvin, never in degrees Celsius. Add 273.15 to a Celsius reading to get kelvin. The kelvin scale starts at absolute zero, the coldest point there is, so its numbers are never negative. Celsius numbers can be negative, and a negative volume is nonsense.

Boyle, Charles, Gay-Lussac and Avogadro

Boyle's law holds temperature and moles fixed. Squeeze a gas into half the space and its pressure doubles. The particles now hit each wall twice as often. Pressure and volume are inversely related, so their product stays the same.

P1V1 = P2V2

Charles's law holds pressure and moles fixed. Warm a gas and it expands. Hotter particles move faster and hit harder, so the walls move out until the push matches again. Volume divided by kelvin temperature stays the same.

V1 / T1 = V2 / T2

Gay-Lussac's law holds volume and moles fixed. Warm a gas in a sealed steel can and the pressure climbs. This is why a spray can carries a warning about heat.

P1 / T1 = P2 / T2

Avogadro's law holds pressure and temperature fixed. Pump in twice as many moles and the volume doubles. A surprising part follows. At the same pressure and temperature, equal volumes of any two gases hold equal numbers of particles. A liter of helium and a liter of carbon dioxide hold the same count, even though their masses differ.

V1 / n1 = V2 / n2
The shapes of the three simple gas laws BOYLE · P vs V at fixed T volume pressure PV is constant CHARLES · V vs T at fixed P 0 K 273 K kelvin temperature volume a straight line aimed at 0 K GAY-LUSSAC · P vs T at fixed V 0 K kelvin temperature pressure P/T is constant
Each law has its own shape. Boyle's law curves, because P and V are inverses. Charles's law and Gay-Lussac's law are straight lines that point back to 0 K. That shared aim is why gas work must use the kelvin scale.

One equation for all four

Stack Boyle, Charles and Gay-Lussac together and you get the combined gas law. Use it whenever the amount of gas does not change.

P1V1 / T1 = P2V2 / T2

Any one of the simple laws drops out of it. Hold temperature steady and the T terms cancel, leaving Boyle's law. That is worth noticing. You only have to remember one relationship.

Worked example · the combined gas law

A weather balloon holds 15.0 L of helium at 1.00 atm and 295 K. It rises until the pressure around it is 0.350 atm and the temperature is 240 K. What is its new volume?

Step 1 · solve for V2

V2 = V1 × (P1 / P2) × (T2 / T1)

Step 2 · put the numbers in

V2 = 15.0 L × (1.00 atm / 0.350 atm) × (240 K / 295 K)

Step 3 · multiply

V2 = 15.0 L × 2.857 × 0.8136 = 34.9 L

Check it

Two effects pull against each other. The lower pressure lets the balloon swell a lot. The colder air shrinks it a little. Swelling wins, so a bigger answer is right. Testing the ratio both ways: 1.00 × 15.0 / 295 = 0.0508, and 0.350 × 34.9 / 240 = 0.0509.

Key points
  • Always convert temperature to kelvin first. Add 273.15 to the Celsius value.
  • Boyle: P and V are inverse. Charles: V and T are direct. Gay-Lussac: P and T are direct.
  • Avogadro's law says equal volumes of any gases hold equal moles at the same P and T.
  • The combined gas law, P1V1/T1 = P2V2/T2, contains all three of them.

Section 3The ideal gas law

Four variables, one equation

The combined gas law needs a before and an after. The ideal gas law needs neither. It ties all four gas variables together at one moment in time.

PV = nRT

Here P is pressure, V is volume in liters, n is moles and T is kelvin temperature. The letter R is the ideal gas constant. It is the number that makes the units agree, and its value depends on the pressure unit you choose.

Value of RUnitsUse it when
0.08206L·atm/mol·KP in atm, V in liters
8.314J/mol·KYou want energy, or P in Pa and V in m3
62.36L·torr/mol·KP already in torr

Most gas problems use 0.08206 L·atm/mol·K. Pick that one and the rest of your units are decided for you. Pressure must be in atm, volume in liters, temperature in kelvin. Convert anything that does not fit before you start.

An ideal gas is a gas that obeys this equation exactly. Real gases follow it closely at ordinary room conditions. The match gets worse at very high pressure and very low temperature, which is the last section of this chapter.

Molar volume and STP

Because equal volumes hold equal moles, one mole of any gas takes up the same space under the same conditions. Chemists picked a reference point so they could compare. Standard temperature and pressure, or STP, means 0 °C (273.15 K) and 1 atm. Put one mole into the ideal gas law at STP and out comes the molar volume.

V = (1 mol)(0.08206 L·atm/mol·K)(273.15 K) / (1 atm) = 22.4 L

So 22.4 L/mol at STP is a shortcut worth memorizing. One mole of oxygen, one mole of neon and one mole of methane each fill about 22.4 L at STP. That is roughly the volume of three basketballs.

One mole of any gas fills 22.4 liters at STP SAME VOLUME · SAME COUNT · DIFFERENT MASS He 22.4 L · 1 mol 4.00 g density 0.179 g/L N2 22.4 L · 1 mol 28.02 g density 1.25 g/L CO2 22.4 L · 1 mol 44.01 g density 1.96 g/L
Equal volumes, equal counts. At STP each box holds one mole, which is 6.022 × 1023 particles, in 22.4 L. Only the mass on the balance tells the three gases apart, so density rises with molar mass.

Worked example · solving for the missing variable

A steel cylinder with a volume of 2.00 L holds 0.0820 mol of argon. The gauge reads 1.10 atm. What is the temperature inside, in kelvin and in degrees Celsius?

Step 1 · rearrange

T = PV / nR

Step 2 · check the units

Pressure is in atm and volume is in liters. So use R = 0.08206 L·atm/mol·K.

Step 3 · compute

T = (1.10 atm)(2.00 L) / [(0.0820 mol)(0.08206 L·atm/mol·K)]

T = 2.20 / 0.006729 = 327 K

Step 4 · convert to Celsius

327 K − 273.15 = 54 °C

Check it

Feed 327 K back in. P = nRT/V = (0.0820)(0.08206)(327)/2.00 = 1.10 atm. It matches.

Two standards, two numbers. Older tables set standard pressure at 1 atm, which gives 22.4 L/mol. Modern IUPAC tables set it at exactly 1 bar, which gives 22.7 L/mol. This book uses 1 atm and 22.4 L/mol, so check which standard a table assumes before you trust it.
Key points
  • PV = nRT describes one gas sample at one moment, with no before and after needed.
  • R = 0.08206 L·atm/mol·K demands atm, liters, moles and kelvin. R = 8.314 J/mol·K is for energy work.
  • STP is 0 °C and 1 atm. At STP one mole of any gas fills 22.4 L.
  • Rearrange the one equation rather than memorizing four versions of it.

Section 4Gas density, molar mass and stoichiometry

Density follows from PV = nRT

Density is mass divided by volume. For a gas, both of those change with conditions, so gas density is not a fixed property the way it is for a solid. You can still predict it. Start from the ideal gas law and swap moles for mass.

Moles equal mass divided by molar mass, so n = m/M. Put that in place of n and rearrange until mass over volume stands alone.

PV = (m/M)RT  →  d = m/V = PM / RT

Read that equation like a sentence. Heavier gas particles make a denser gas. More pressure packs them closer, so density rises. Higher temperature spreads them out, so density falls. Warm air rising above a radiator is this equation at work.

Turn it around and you have a way to weigh a molecule. Measure the density of an unknown gas at a known pressure and temperature, and its molar mass drops out.

M = dRT / P

Worked example · density, then molar mass

Part one: find the density of carbon dioxide at 1.00 atm and 25 °C. Part two: an unknown gas has a density of 1.34 g/L at 1.00 atm and 0 °C. Find its molar mass.

Step 1 · molar mass of CO2

12.01 + 2(16.00) = 44.01 g/mol

Step 2 · temperature in kelvin

25 + 273.15 = 298.15 K

Step 3 · use d = PM/RT

d = (1.00 atm)(44.01 g/mol) / [(0.08206 L·atm/mol·K)(298.15 K)]

d = 44.01 / 24.47 = 1.80 g/L

Step 4 · now the unknown, using M = dRT/P

M = (1.34 g/L)(0.08206 L·atm/mol·K)(273.15 K) / (1.00 atm) = 30.0 g/mol

Check it

The second gas is at STP, so the shortcut works: M = d × 22.4 L/mol = 1.34 × 22.4 = 30.0 g/mol. A molar mass of 30.0 g/mol fits nitrogen monoxide, NO.

Volumes as mole ratios

Avogadro's law gives gas stoichiometry a shortcut. At the same temperature and pressure, the volume ratio of two gases equals their mole ratio. Look at ammonia synthesis.

N2(g) + 3 H2(g) → 2 NH3(g)

One liter of nitrogen needs three liters of hydrogen and makes two liters of ammonia. You can read the coefficients as liters directly, with no mole math at all. This only works when every substance is a gas at the same conditions.

When a solid or liquid is involved, take the long road. Convert the known amount to moles. Cross the balanced equation with the mole ratio. Then use PV = nRT to turn moles of gas into a volume.

The road map from grams of reactant to liters of gas GRAMS → MOLES → MOLES → LITERS grams Zn what you weigh out ÷ 65.38 moles Zn use molar mass 1 : 1 moles H2 use the mole ratio PV = nRT liters H2 the answer Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) Only the middle arrow needs the balanced equation. The last arrow needs P and T. At STP you may replace the last arrow with a simple × 22.4 L/mol.
The gas stoichiometry road map. Mass and volume never connect directly. Both must pass through moles, and the mole ratio from the balanced equation is the bridge in the middle.

Worked example · volume of a gas from a reaction

Zinc metal reacts with hydrochloric acid to make hydrogen gas. You use 5.00 g of zinc and plenty of acid. The hydrogen is collected at 745 torr and 298 K. What volume do you get?

Step 1 · the balanced equation

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Step 2 · grams of zinc to moles

5.00 g ÷ 65.38 g/mol = 0.0765 mol Zn

Step 3 · mole ratio

The ratio of Zn to H2 is 1 to 1, so you make 0.0765 mol H2.

Step 4 · convert the pressure

745 torr ÷ 760 torr/atm = 0.980 atm

Step 5 · use PV = nRT

V = nRT/P = (0.0765)(0.08206)(298) / 0.980 = 1.91 L

Check it

At STP that many moles would fill 0.0765 × 22.4 = 1.71 L. The real sample is warmer and at slightly lower pressure, so a bit more than 1.71 L is just what you expect.

Key points
  • d = PM/RT, so gas density rises with molar mass and pressure and falls with temperature.
  • M = dRT/P lets a density measurement identify an unknown gas.
  • At STP, density equals molar mass divided by 22.4 L/mol.
  • For gas to gas, coefficients act as volume ratios. Otherwise go grams to moles to moles to liters.

Section 5Dalton's law of partial pressures

Each gas acts as if it were alone

Gas particles are far apart and ignore each other. So a gas in a mixture does not care what else is in the flask. It pushes on the walls exactly as hard as it would if the other gases were removed. That push is its partial pressure.

John Dalton turned this idea into a law. The law of partial pressures says the total pressure of a mixture is the sum of the partial pressures of its gases.

Ptotal = PA + PB + PC + ...

Air is the everyday example. At 1.00 atm, nitrogen supplies about 0.78 atm and oxygen about 0.21 atm. Argon, carbon dioxide and the rest make up the last sliver. Your lungs respond to the partial pressure of oxygen, not to the total.

Mole fractions split the total

There is a faster route when you know the amounts. The mole fraction of a gas, written X, is its moles divided by the total moles. It has no units and all the mole fractions in a mixture add to 1.

XA = nA / ntotal   and   PA = XA × Ptotal

So a gas that is one quarter of the particles owns one quarter of the pressure. Mass does not enter into it. Four grams of helium and four grams of oxygen are very different counts, so they claim very different shares.

Worked example · splitting a total pressure

You seal 2.00 g of helium and 8.00 g of oxygen in a 10.0 L flask at 298 K. Find each partial pressure, the total pressure and the mole fraction of helium.

Step 1 · moles of each gas

He: 2.00 g ÷ 4.003 g/mol = 0.500 mol

O2: 8.00 g ÷ 32.00 g/mol = 0.250 mol

Step 2 · each partial pressure from PV = nRT

PHe = (0.500)(0.08206)(298) / 10.0 = 1.22 atm

PO2 = (0.250)(0.08206)(298) / 10.0 = 0.611 atm

Step 3 · add them

Ptotal = 1.22 + 0.611 = 1.83 atm

Step 4 · mole fraction of helium

XHe = 0.500 / (0.500 + 0.250) = 0.667

Check it

Use the mole fraction the other way. 0.667 × 1.83 atm = 1.22 atm, which is the helium value from step 2. Notice that helium is only a fifth of the mass but two thirds of the pressure.

Collecting a gas over water and correcting for water vapor COLLECTING A GAS OVER WATER water trough H2 + H2O vapor gas from the flask bottle filled with water, then pushed out by gas SPLITTING THE TOTAL Level the water inside and out, and the total inside equals room pressure. P of the dry gas water vapor Pgas = Ptotal − Pwater At 25 °C water vapor adds 23.8 torr, about 3% of 1 atm. Warmer water adds more, so the correction grows.
Water always sends some of itself along. A gas bubbled up through water carries water vapor with it. The vapor owns part of the measured pressure, so subtract its tabulated value to get the dry gas pressure.

Gas collected over water

Many lab gases are collected by bubbling them into an upturned bottle full of water. The gas pushes the water out and takes its place. One detail is easy to miss. Liquid water keeps evaporating into the space above it, so the trapped sample is a mixture.

The water contributes its vapor pressure, a value that depends only on temperature. Tables list it, and it climbs fast as water warms. At 20 °C it is 17.5 torr. At 25 °C it is 23.8 torr. Subtract it to get the pressure of the gas you actually want.

Pgas = Ptotal − Pwater

Worked example · moles of a gas collected over water

You collect 0.250 L of hydrogen over water at 25 °C. The barometer reads 755 torr, and the water levels are even. How many moles of hydrogen do you have?

Step 1 · subtract the water vapor

PH2 = 755 torr − 23.8 torr = 731 torr

Step 2 · convert to atm

731 torr ÷ 760 torr/atm = 0.962 atm

Step 3 · solve for moles

n = PV/RT = (0.962)(0.250) / [(0.08206)(298.15)]

n = 0.2405 / 24.47 = 9.83 × 10−3 mol

Check it

Skipping the water step would have given 1.02 × 10−2 mol. That is about 3% too high, which is exactly the share water vapor was holding.

Key points
  • Each gas in a mixture pushes as if it were alone, and the pushes add up.
  • PA = XA × Ptotal, where XA is moles of A over total moles.
  • Mole fractions split pressure by particle count, never by mass.
  • A gas collected over water shares the space with water vapor, so subtract the tabulated vapor pressure.

Section 6Kinetic molecular theory, effusion and real gases

What the particles are really doing

The gas laws say what happens. Kinetic molecular theory says why. It is a short list of claims about the particles themselves.

Those five claims explain every law in this chapter. Shrink the volume and each particle reaches a wall sooner, so hits come more often and pressure rises. That is Boyle. Raise the temperature and the particles move faster, so they hit harder and more often. That is Charles and Gay-Lussac. Add more particles and you add more hits. That is Avogadro.

Speed, mass and temperature

The last claim hides something important. At a given temperature, every gas has the same average kinetic energy per mole.

KEavg = (3/2)RT

Kinetic energy is one half mass times speed squared. So if the energy is equal but the masses differ, the speeds cannot be equal. Light particles must move faster to carry the same energy. Solve for speed and you get the root-mean-square speed.

urms = √(3RT/M)

Use R = 8.314 J/mol·K here, and put molar mass in kilograms per mole. Then the answer comes out in meters per second. This is the one place in the chapter where 0.08206 is the wrong constant.

Not every particle moves at that speed. The speeds spread out into a curve called the Maxwell-Boltzmann distribution. Heat the gas and the peak slides right and flattens out, because the range of speeds widens. Swap in a heavier gas at the same temperature and the peak slides left and gets taller and narrower.

How temperature and molar mass reshape the speed distribution THE SPREAD OF MOLECULAR SPEEDS 500 1000 1500 2000 2500 3000 particle speed (m/s) share of particles N2 at 300 K · urms = 517 m/s N2 at 1000 K · urms = 943 m/s He at 300 K · urms = 1367 m/s Dotted lines mark urms for each curve. Every curve covers the same total area.
Same gas, hotter: the peak slides right and flattens. Lighter gas, same temperature: the peak slides much further right. Each curve encloses the same area, because it still describes one whole sample. The tail on the right is the small group of very fast particles.

Effusion and Graham's law

Effusion is the escape of a gas through a tiny hole into a vacuum. A faster gas finds the hole more often, so it leaks out faster. Since speed goes as one over the square root of molar mass, so does the effusion rate. That is Graham's law.

rateA / rateB = √(MB / MA)

Watch the subscripts. The lighter gas is on top of the rate ratio but on the bottom inside the square root. A helium balloon goes limp in a day for this reason, while an air-filled one lasts much longer.

Worked example · speed and effusion

Find the root-mean-square speed of nitrogen at 298 K. Then compare how fast helium and oxygen effuse.

Step 1 · molar mass in kg/mol

N2 is 28.02 g/mol, which is 0.02802 kg/mol.

Step 2 · use the 8.314 value of R

urms = √[3(8.314 J/mol·K)(298 K) / 0.02802 kg/mol]

urms = √(265,300) = 515 m/s

Step 3 · now Graham's law

rateHe / rateO2 = √(32.00 / 4.003) = √7.994 = 2.83

Check it

Helium effuses 2.83 times faster than oxygen. Oxygen is about eight times heavier, and the square root of 8 is close to 2.8. The speed answer of 515 m/s is faster than a jet, which fits how quickly a smell crosses a room.

Where real gases stray

No gas is truly ideal. Two of the five claims are not quite true. Particles do take up space, and they do attract each other a little. At ordinary conditions those errors are too small to notice. Squeeze hard or cool far down and they show up.

Chemists track the error with the compressibility factor, Z = PV/nRT. An ideal gas has Z = 1 at every condition. Attractions pull particles away from the walls, so pressure falls below ideal and Z drops under 1. At very high pressure the particles' own volume takes over, so Z climbs above 1.

The van der Waals equation patches both problems. The constant a adds back the pressure lost to attraction. The constant b removes the space the particles themselves fill.

(P + an2/V2)(V − nb) = nRT

Both constants are measured for each gas. Large, sticky molecules get large values. Carbon dioxide has a = 3.592 L2·atm/mol2 and b = 0.04267 L/mol. Helium, which is tiny and barely sticky, has a = 0.0346 and b = 0.0238.

Worked example · how far off is ideal?

Squeeze 1.00 mol of carbon dioxide into 0.500 L at 273.15 K. Compare the ideal pressure with the van der Waals pressure.

Step 1 · the ideal answer

P = nRT/V = (1.00)(0.08206)(273.15) / 0.500 = 44.8 atm

Step 2 · the van der Waals form, solved for P

P = nRT/(V − nb) − an2/V2

Step 3 · the volume term

0.500 L − (1.00)(0.04267) = 0.4573 L, so 22.41 / 0.4573 = 49.0 atm

Step 4 · the attraction term

3.592 / (0.500)2 = 3.592 / 0.250 = 14.4 atm

Step 5 · combine

P = 49.0 − 14.4 = 34.6 atm

Check it

The real pressure is about 23% below the ideal guess. Attraction wins at this squeeze, so Z = 34.6/44.8 = 0.77, well under 1.

A useful rule of thumb. Gases behave most ideally when they are hot and spread out. They behave least ideally near the temperature and pressure at which they would turn into a liquid.
Key points
  • Average kinetic energy depends only on kelvin temperature, so lighter gases move faster.
  • urms = √(3RT/M), with R = 8.314 J/mol·K and M in kg/mol.
  • Heating widens and flattens the speed curve. Extra mass narrows it and shifts it left.
  • Graham's law: rateA/rateB = √(MB/MA).
  • Real gases stray at high pressure and low temperature. The van der Waals constants a and b correct for attraction and particle volume.

The chapter in one breath

  • Pressure is particles drumming on a wall, and 1 atm = 760 torr = 101.325 kPa.
  • Manometers compare two pushes. A closed end reads the gas pressure outright, while an open end adds or subtracts the room.
  • Kelvin or nothing. Every gas law needs absolute temperature, so add 273.15 to a Celsius value first.
  • The combined gas law holds the simple ones. Boyle, Charles and Gay-Lussac all fall out of P1V1/T1 = P2V2/T2.
  • One equation does most of the work. PV = nRT with R = 0.08206 L·atm/mol·K handles any single state.
  • At STP one mole fills 22.4 L, whatever the gas, so density at STP is molar mass over 22.4.
  • Gas stoichiometry is normal stoichiometry with PV = nRT bolted on at the end.
  • Pressure splits by count. PA = XAPtotal, and a gas over water loses the water's vapor pressure.
  • Temperature sets energy, mass sets speed. Light gases move faster, effuse faster, and all gases stray from ideal when cold and squeezed.

ReferenceWords worth knowing

These are the terms this chapter leans on. Come back to them when a problem uses a word you are not sure of.

Avogadro's law
At fixed pressure and temperature, gas volume is proportional to moles. Equal volumes hold equal counts.
barometer
A tube of mercury inverted in a dish, used to measure atmospheric pressure from the height of the column.
Boyle's law
At fixed temperature and amount, pressure and volume are inversely related, so PV stays constant.
Charles's law
At fixed pressure and amount, volume is proportional to kelvin temperature.
combined gas law
The single relation P1V1/T1 = P2V2/T2, used when the amount of gas does not change.
Dalton's law of partial pressures
The total pressure of a gas mixture equals the sum of the partial pressures of its gases.
effusion
The escape of gas through a very small opening into a vacuum.
Gay-Lussac's law
At fixed volume and amount, pressure is proportional to kelvin temperature.
Graham's law
Effusion rate varies as one over the square root of molar mass, so rateA/rateB = √(MB/MA).
ideal gas constant (R)
The proportionality constant in PV = nRT. It is 0.08206 L·atm/mol·K or 8.314 J/mol·K.
kinetic molecular theory
The particle-level model of a gas: tiny, far apart, in constant random motion, with elastic collisions and energy set by temperature.
manometer
A U-shaped mercury tube that reports the pressure of a gas sample, either against a vacuum or against room air.
Maxwell-Boltzmann distribution
The curve showing how many particles in a sample hold each speed at a given temperature.
molar volume
The volume one mole of gas fills. At STP it is 22.4 L/mol.
mole fraction (X)
Moles of one gas divided by total moles in the mixture. It sets that gas's share of the pressure.
partial pressure
The pressure one gas in a mixture would exert if it alone filled the container.
root-mean-square speed
The speed matching the average kinetic energy of a sample, equal to √(3RT/M).
standard temperature and pressure (STP)
The reference state 0 °C and 1 atm, where one mole of gas fills 22.4 L.
van der Waals equation
A corrected gas equation that adds a for particle attraction and subtracts b for particle volume.
vapor pressure
The pressure of vapor above a liquid at a given temperature. Water's value must be subtracted from a gas collected over water.

Check yourself

1. A gauge reads 0.875 atm. Give that pressure in torr and in kPa.
665 torr and 88.7 kPa. Multiply by the conversion factor each time. 0.875 atm × 760 torr/atm = 665 torr. 0.875 atm × 101.325 kPa/atm = 88.7 kPa.
2. A closed-end manometer shows mercury 148 mm higher on the closed arm than on the gas arm. What is the gas pressure in torr and in atm?
148 torr, or 0.195 atm. The closed arm holds a vacuum, so nothing pushes back and the height gap is the pressure itself. Then 148 torr ÷ 760 torr/atm = 0.195 atm.
3. A syringe holds 350 mL of air at 1.00 atm. You push the plunger until the pressure reads 2.50 atm, keeping the temperature steady. What is the new volume?
140 mL. Use Boyle's law. V2 = P1V1/P2 = (1.00)(350)/2.50 = 140 mL. Pressure went up by a factor of 2.5, so volume dropped by the same factor.
4. A balloon holds 1.75 L at 295 K. You warm it to 350 K at constant pressure. What is the new volume?
2.08 L. Charles's law gives V2 = V1T2/T1 = 1.75 L × (350/295) = 1.75 × 1.186 = 2.08 L.
5. Why do gas laws fail if you put temperature in degrees Celsius?
Because the Celsius zero is not a real zero. Volume and pressure are proportional to temperature measured from absolute zero. Going from 10 °C to 20 °C does not double anything, since in kelvin that is 283 K to 293 K. Celsius values can also be negative, which would predict a negative volume.
6. What volume does 0.500 mol of neon fill at 2.00 atm and 350 K?
7.18 L. V = nRT/P = (0.500)(0.08206)(350)/2.00. The top is 14.36, and 14.36 ÷ 2.00 = 7.18 L.
7. What volume does 3.00 mol of oxygen fill at STP?
67.2 L. At STP the molar volume is 22.4 L/mol, so 3.00 mol × 22.4 L/mol = 67.2 L. The gas identity never enters the calculation.
8. Find the density of oxygen gas at STP.
1.43 g/L. One mole of O2 has a mass of 32.00 g and fills 22.4 L. So d = 32.00 g ÷ 22.4 L = 1.43 g/L. Room air is near 1.2 g/L, and oxygen is a little heavier than average air.
9. A flask holds 0.200 mol of nitrogen and 0.300 mol of oxygen. The total pressure is 2.50 atm. What is the partial pressure of nitrogen?
1.00 atm. First the mole fraction: X = 0.200 / 0.500 = 0.400. Then P = 0.400 × 2.50 atm = 1.00 atm. The oxygen holds the other 1.50 atm.
10. You collect oxygen over water at 25 °C. The total pressure is 762 torr, and the vapor pressure of water at that temperature is 23.8 torr. What is the partial pressure of the oxygen?
738 torr. Subtract the water vapor: 762 torr − 23.8 torr = 738.2 torr, which rounds to 738 torr. Skipping this step would overstate your oxygen by about 3%.
11. Methane (CH4, 16.04 g/mol) and sulfur dioxide (SO2, 64.06 g/mol) sit beside the same pinhole. How much faster does the methane effuse?
About 2.00 times faster. Graham's law gives rateCH4/rateSO2 = √(64.06/16.04) = √3.994 = 2.00. Sulfur dioxide is four times heavier, and the square root of 4 is 2.
12. Two samples of carbon dioxide sit at 1 atm and 300 K, and at 200 atm and 220 K. Which one strays further from PV = nRT, and why?
The cold, squeezed sample at 200 atm and 220 K. High pressure crowds the particles, so their own volume stops being negligible. Low temperature makes them slow, so weak attractions get a chance to pull. Both effects push Z = PV/nRT away from 1.
Chemistry · Chapter 10

Liquids and Solids

A gas fills any room you put it in. A liquid and a solid do not. Pour water into a glass and it stays in the glass. Set an ice cube on a plate and it keeps its shape. Something holds those particles close together. This chapter is about that something. You will meet the weak pulls that act between molecules. You will see how they set boiling points, thickness and crystal shape. You will walk a sample from ice to steam and count the heat each step takes. Then you will read the map of the three states and find the odd one out, which turns out to be water.

By the end of this chapter you can

  • Name the four main pulls between particles and rank their strength.
  • Predict which of two liquids boils at the higher temperature, and say why.
  • Explain thickness, surface tension and capillary rise in terms of those pulls.
  • Find the heat a phase change takes, using heat of fusion and heat of vaporization.
  • Read a heating curve and say what each flat stretch means.
  • Use vapor pressure to find a boiling point at any outside pressure.
  • Read a phase diagram, point to its triple point and critical point, and count the atoms in a unit cell.

Section 1Intermolecular forces

Inside a molecule, and between molecules

Bonds hold the atoms inside one molecule. Weaker pulls act between whole molecules. Those weaker pulls are the intermolecular forces, and they run this chapter. Break the bonds in water and you get hydrogen and oxygen. Break the forces between water molecules and you get steam. It is still water. The molecules are just spread out. That is why boiling a pot of water takes far less energy than splitting it would.

Dispersion forces

Every molecule holds electrons, and electrons move. For an instant, more of them crowd one side. That side turns slightly negative and the far side turns slightly positive. The molecule now has a brief dipole, which is a split of charge. It tugs on the electrons next door and sets up a matching dipole there. The two then attract. This is the dispersion force, and every substance has it.

A big, loose electron cloud is easy to push out of shape. We say such a molecule is more polarizable. So dispersion grows as the electron count grows. The noble gases show it cleanly. They boil at 4 K, 27 K, 87 K, 120 K and 165 K going from helium down to xenon. Shape counts too. A long thin molecule lies against its neighbor along its whole length, so it sticks better than a round one of the same mass.

Dipole-dipole forces

Some molecules are lopsided all the time. One atom pulls the shared electrons harder, so one end stays slightly negative. These molecules line up plus to minus and hold on. That pull is the dipole-dipole force. It adds to dispersion. It never replaces it.

Hydrogen bonds

One case is so strong that it gets its own name. Tie an H atom to N, O or F and the H is left very exposed. Its one electron is pulled away, and no inner shell hides it. That bare H can reach a lone pair on an N, O or F atom next door. The link is a hydrogen bond. In water it is worth about 20 kJ/mol. The O–H bond inside the molecule is about 464 kJ/mol, so a hydrogen bond is nowhere near a real bond. It is still several times stronger than a plain dipole pull of the same size.

Water shows the payoff. H2O boils at 100 °C. H2S has the same bent shape and almost twice the mass, and it boils at −60 °C. Sulfur is not electronegative enough to make hydrogen bonds. That one missing force is worth 160 °C.

Ion-dipole forces

Stir table salt into water and the ions come loose. Each Na+ gathers the negative ends of water molecules around itself. Each Cl− gathers the positive ends. That pull is the ion-dipole force, and it runs from 40 to 600 kJ/mol. It is the strongest of the four. It is also what pays the bill for taking a crystal apart.

Dispersion, dipole-dipole and hydrogen bonding compared DISPERSION every substance · 0.05 to 40 kJ/mol δ− δ+ δ− δ+ a flicker on the left forces a match on the right DIPOLE–DIPOLE polar molecules · 5 to 25 kJ/mol H Cl δ+ δ− H Cl δ+ δ− the minus end of one meets the plus end of the next HYDROGEN BOND H on N, O or F · about 20 kJ/mol in water O H H O H H O to O: about 2.8 Å a bare H reaches the lone pair on the next O
Figure 10.1 The three forces between neutral molecules, drawn to the same idea. Dispersion is a flicker that spreads. A dipole-dipole pull is permanent. A hydrogen bond is a special, stronger case of that pull, and it only happens when H sits on N, O or F.

Ranking them

Compare like with like. For molecules of about the same size, the order is clear. Hydrogen bonding beats dipole-dipole, which beats dispersion alone. But size wins when the mismatch is large. I2 has no dipole at all and boils at 184 °C. HCl is polar and boils at −85 °C. Iodine has far more electrons, so its dispersion forces swamp the dipoles in HCl.

Worked example · Rank three liquids by boiling point

Put CH4, CH3F and CH3OH in order from lowest boiling point to highest.

Step 1 · name the forcesCH4 is a symmetric nonpolar molecule, so it has dispersion only. CH3F is polar, so it has dispersion plus dipole-dipole. CH3OH has an H on an O, so it adds hydrogen bonding on top of both.

Step 2 · check the massesThe molar masses are 16.04, 34.03 and 32.04 g/mol. The last two are nearly a tie, so their dispersion forces are close. Mass will not decide between them.

Step 3 · rankMore forces and stronger forces mean a higher boiling point. So CH4 < CH3F < CH3OH.

Step 4 · check itThe measured values are −161 °C, −78 °C and 65 °C. The order holds, and the jump to the alcohol is the biggest one.

Key points
  • Intermolecular forces act between molecules. Bonds act inside them.
  • Dispersion is in every substance and grows with the electron count.
  • Dipole-dipole needs a permanent lopsided molecule.
  • A hydrogen bond needs H joined to N, O or F, and a lone pair to reach.
  • Ion-dipole is the strongest of the four and drives salts into water.
  • Rank by comparing molecules of similar size. Otherwise dispersion can win on mass alone.
A water molecule can make four hydrogen bonds. It gives two through its H atoms and takes two through the lone pairs on its O. In liquid water the average is a bit under four at any instant. That dense web is why water holds so much heat and boils so high for a molecule of mass 18.02 g/mol.

Section 2Properties of liquids

Thickness, in a word: viscosity

Viscosity measures how much a liquid resists flowing. Water pours. Honey crawls. Two things raise viscosity. Stronger forces between molecules make them harder to slide past each other. Long molecules also tangle like cooked spaghetti. Glycerol has three OH groups on a three-carbon chain, so it does both at once. At 25 °C its viscosity is about 934 mPa·s. Water at the same temperature is 0.890 mPa·s. Glycerol is more than a thousand times thicker.

Heat thins a liquid. Faster molecules spend less time in each other's grip. Water drops from 1.79 mPa·s at 0 °C to 0.890 at 25 °C, and down to 0.282 at 100 °C. That is why warm syrup pours and cold syrup will not.

Surface tension

A molecule deep in a liquid is pulled on from every side. The pulls cancel. A molecule at the top surface has no neighbors above it, so the leftover pull aims sideways and down. The surface behaves like a stretched skin. Surface tension is the energy it costs to make new surface area. We report it in mN/m, which is the same as mJ/m2.

Water comes in at 72.0 mN/m at 25 °C, which is high for a liquid. Ethanol is only 22.1 mN/m, since it makes fewer hydrogen bonds per molecule. Mercury reaches 486 mN/m, because metal atoms hold each other with a sea of shared electrons. High surface tension pulls a drop into a bead and lets a water strider stand on a pond without sinking.

Cohesion, adhesion and the meniscus

Cohesive forces pull a liquid toward itself. Adhesive forces pull it toward the walls of its container. Which one wins sets the curve at the edge, called the meniscus. Water sticks to glass better than to itself, so it climbs the wall and the curve dips in the middle. Read the bottom of that dip. Mercury does the reverse. It holds itself far better than it holds glass, so the curve humps up. Read the top.

Capillary action

In a narrow tube the climb does not stop at the rim of the meniscus. The whole column rises. That lift is capillary action. The surface pulls up on the ring where liquid meets glass, and the weight of the raised column pulls back down. The two balance at one height.

h = 2T cos θ / (r ρ g)

Here T is surface tension, θ is the contact angle, r is the tube radius, ρ is density and g is 9.81 m/s2. The radius sits on the bottom, so a thinner tube lifts the liquid higher. Paper towels and plant stems are packed with very thin channels, and that is the whole trick.

Meniscus shape and capillary rise WATER IN GLASS sticking beats self-pull read here ↓ curve dips in the middle MERCURY IN GLASS self-pull beats sticking read here ↑ curve humps in the middle CAPILLARY RISE IN WATER halve the radius and the column doubles open dish 2.9 cm r = 0.50 mm 5.9 cm r = 0.25 mm
Figure 10.2 Water wets glass, so its meniscus curves up at the wall and the column rises in a thin tube. Mercury does not wet glass, so its meniscus humps and the column is pushed down instead. The two rise heights shown are the worked example below.

Worked example · How high does water climb?

Find the rise of water at 25 °C in a clean glass tube of radius 0.25 mm.

Step 1 · collect the valuesT = 72.0 mN/m = 0.0720 N/m. Water wets clean glass, so θ is near 0 and cos θ = 1. ρ = 1.00 × 103 kg/m3. r = 0.25 mm = 2.5 × 10−4 m.

Step 2 · build the bottom rρg = (2.5 × 10−4)(1.00 × 103)(9.81) = 2.45 N/m2 per m.

Step 3 · divideh = 2(0.0720)(1) / 2.45 = 0.144 / 2.45 = 0.0587 m, which is 5.9 cm.

Step 4 · test the patternDouble the radius to 0.50 mm and the bottom doubles to 4.90. Then h = 0.144 / 4.90 = 0.0294 m, or 2.9 cm. Twice the width, half the climb.

Key points
  • Stronger intermolecular forces mean higher viscosity and higher surface tension.
  • Heating a liquid always lowers its viscosity.
  • Cohesion is the pull to itself. Adhesion is the pull to the wall.
  • Water wets glass and curves down in the middle. Mercury does not wet glass and humps up.
  • Capillary rise goes as 1/r, so thin tubes lift liquid the furthest.
Soap drops water's surface tension from about 72 mN/m to near 30 mN/m. The soap molecules crowd the surface and get in the way of the hydrogen bonds. Low surface tension lets water spread into cloth and creep into the gaps in dirt instead of sitting on top in beads.

Section 3Phase changes and heating curves

Six changes, three in each direction

Matter moves between states in six named ways. Melting, vaporizing and subliming all take heat in. Freezing, condensing and depositing all give that heat back. Sublimation is the jump from solid straight to gas, with no liquid in between. Dry ice does it in a warm room. So does old snow on a cold dry day, which shrinks without ever making a puddle.

Heat that does not raise the temperature

Here is the surprise of this section. Keep heating ice that has reached 0 °C and the thermometer stops moving. The energy is still going in. It is being spent to pull molecules out of their fixed spots, not to speed them up. The reading holds at 0 °C until the last crystal is gone.

The heat of fusion is the energy to melt a set amount at the melting point. For water it is 334 J/g, which is 6.01 kJ/mol. The heat of vaporization is the energy to boil a set amount at the boiling point. For water that is 2,260 J/g, or 40.7 kJ/mol. Freezing and condensing return the same amounts with the sign flipped.

Look at the gap. Boiling water takes almost seven times the energy of melting it. Melting only loosens the grip. The molecules stay packed and touching. Boiling has to break the hold completely and send each molecule off alone. That is the expensive step.

Reading a heating curve

Put a cold sample on a steady flame. Plot temperature up and heat added across. You get five stretches: three slopes and two flats. On a slope only the temperature changes, and the heat is q = mcΔT. On a flat only the state changes, and the heat is the mass times the heat of fusion or vaporization. The specific heats for water differ by state: ice is 2.09 J/g·°C, liquid water is 4.184, and steam is 1.84.

The flat at 0 °C is melting. The flat at 100 °C is boiling, and it is far wider on the page. Width on this plot means energy, so the widest stretch is the most expensive change. A cooling curve is the same picture run backward, with heat leaving instead of entering.

Heating curve for water, with both plateaus to scale °C 0 100 melting, 0 °C 8,350 J boiling, 100 °C · 56,500 J, the widest stretch ice warms 784 J water warms 10,460 J steam warms 690 J heat added → total 76,784 J for 25.0 g Flat stretches are drawn to scale, so you can see that boiling costs almost three quarters of the whole bill.
Figure 10.3 A heating curve for 25.0 g of water, from ice at −15 °C to steam at 115 °C. Each stretch is labeled with its own energy. The two flats are phase changes, where heat goes in and the temperature holds still.

Worked example · Ice at −15 °C to steam at 115 °C

How much heat does 25.0 g of water need for the whole trip? Use 2.09, 4.184 and 1.84 J/g·°C for ice, liquid and steam.

Step 1 · warm the ice q = (25.0 g)(2.09 J/g·°C)(15.0 °C) = 784 J.

Step 2 · melt it q = (25.0 g)(334 J/g) = 8,350 J.

Step 3 · warm the water q = (25.0 g)(4.184 J/g·°C)(100.0 °C) = 10,460 J.

Step 4 · boil it q = (25.0 g)(2,260 J/g) = 56,500 J.

Step 5 · warm the steam q = (25.0 g)(1.84 J/g·°C)(15.0 °C) = 690 J.

Step 6 · add and look 784 + 8,350 + 10,460 + 56,500 + 690 = 76,784 J, or 76.8 kJ. Step 4 alone is 73.6% of that. Phase changes, not temperature changes, own the energy budget.

Key points
  • During a phase change the temperature holds still while heat keeps flowing.
  • On a slope use q = mcΔT. On a flat use mass times the heat of fusion or vaporization.
  • Water: 334 J/g to melt and 2,260 J/g to boil. That is 6.01 and 40.7 kJ/mol.
  • Boiling costs almost seven times what melting costs for the same mass.
  • Freezing and condensing release the same heat they took in.
This is why steam burns are so much worse than hot water burns. A gram of steam at 100 °C that condenses on your skin dumps 2,260 J before its temperature has dropped at all. Add the heat of fusion to the heat of vaporization and you get the heat of sublimation, near 47 kJ/mol. The measured value for ice is about 51 kJ/mol, because vaporizing takes more energy at 0 °C than it does at 100 °C.

Section 4Vapor pressure and boiling

Where vapor pressure comes from

Molecules in a liquid do not all move at the same speed. A few at the top surface are fast enough to break free. In a sealed flask those escapees build up in the space above the liquid. Some of them hit the surface and fall back in. Escape and return speed up until they match. At that point the gas pressure above the liquid stops climbing and holds. That steady value is the vapor pressure.

Vapor pressure depends on just two things: which liquid it is, and how warm it is. Raise the temperature and more molecules clear the bar, so the pressure climbs fast. Water sits at 23.8 torr at 25 °C, 92.5 torr at 50 °C, 355 torr at 80 °C and 760 torr at 100 °C. Note the shape of that run. It nearly quadruples from 25 to 50 °C, then nearly quadruples again by 80 °C.

Weak forces mean a high vapor pressure at any given temperature. A liquid like that is called volatile, and you can smell it from across the room. Diethyl ether reaches 533 torr at 25 °C, which is more than twenty times water's value at the same temperature.

Boiling

Warming a liquid in an open beaker does not make bubbles until one condition is met. The vapor pressure must match the pressure pressing down from above. Only then can a bubble inside the liquid hold its own shape instead of being squeezed flat. So boiling is a tug of war that the vapor wins at one exact temperature.

The normal boiling point is the temperature where vapor pressure reaches 1 atm, which is 760 torr. For water that is 100 °C. Drop the outside pressure and the liquid boils sooner. Denver sits near 625 torr, and water there boils at about 95 °C. On the summit of Everest the pressure is near 253 torr and water boils at about 72 °C. Food takes longer to cook up there, because the water bathing it is cooler. A pressure cooker runs the trade the other way. At 2 atm water will not boil until 121 °C, and the hotter water cooks beans in a third of the time.

Vapor pressure against temperature for three liquids torr 760 torr = 1 atm 400 200 ether 34.6 °C ethanol 78.4 °C water 100 °C 0 25 50 75 100 temperature (°C) Where a curve meets the gold line, that liquid boils in open air at sea level.
Figure 10.4 Vapor pressure rises steeply with temperature. The three curves come from measured data for diethyl ether, ethanol and water. Each one crosses 760 torr at its own normal boiling point. Lower the gold line and every crossing slides left.

The Clausius-Clapeyron idea

Those curves look hopeless to work with, but one trick straightens them. Plot the natural log of vapor pressure against 1 over the Kelvin temperature. The points fall on a line. Its slope is −ΔHvap/R. Two points on that line give the useful form:

ln(P2/P1) = −(ΔHvap/R)(1/T2 − 1/T1)

A steep line means a large heat of vaporization, which means strong forces. So the slope of this plot is a direct read on the forces from section 1. The equation also runs backward. Give it a target pressure and it hands you the boiling point at that pressure.

Worked example · Predict a vapor pressure

Water's vapor pressure at 60 °C is 149 torr. Estimate it at 80 °C. Use ΔHvap = 40.7 kJ/mol and R = 8.314 J/mol·K.

Step 1 · go to KelvinT1 = 333.15 K and T2 = 353.15 K.

Step 2 · the bracket1/353.15 − 1/333.15 = 0.0028317 − 0.0030017 = −1.700 × 10−4 K−1.

Step 3 · the front factorΔHvap/R = 40,700 / 8.314 = 4,895 K.

Step 4 · multiplyln(P2/149) = −(4,895)(−1.700 × 10−4) = 0.832. So P2/149 = e0.832 = 2.30.

Step 5 · finishP2 = 149 × 2.30 = 343 torr.

Step 6 · judge itThe measured value is 355 torr, so the estimate is 3% low. The equation assumes the heat of vaporization never changes. Near 70 °C water's real value is about 42.4 kJ/mol, and that number gives 355 torr. Short jumps in temperature stay accurate. Long ones drift.

Key points
  • Vapor pressure is the pressure of vapor sitting over its own liquid at balance.
  • It depends only on the liquid and the temperature, not on the amount of liquid.
  • Boiling happens when vapor pressure equals the outside pressure.
  • Lower outside pressure gives a lower boiling point. Higher pressure raises it.
  • ln P against 1/T is a straight line with slope −ΔHvap/R.
Vapor pressure does not care how much liquid you have. A teaspoon of water and a full bathtub both sit at 23.8 torr at 25 °C. Surface area changes how fast balance arrives, not where it lands. Add a wider dish and you get there sooner, at the very same pressure.

Section 5Phase diagrams

A map of the three states

A phase diagram plots pressure up the side and temperature across the bottom. Every point on it is one set of conditions. The region a point lands in names the state that is stable there. Three lines divide the map. The solid-liquid line is melting and freezing. The liquid-gas line is boiling and condensing, and it is exactly the vapor pressure curve from section 4. The solid-gas line is sublimation. Crossing a line means changing state.

The triple point

The three lines meet at one point and only one. At the triple point solid, liquid and gas all sit together in balance. Ice, water and vapor coexist, and none of them is winning. For water it is 0.01 °C and 0.006 atm, which is 4.58 torr. Below that pressure liquid water cannot exist at any temperature. Warm ice at 0.001 atm and it goes straight to vapor. Freeze drying lives in that gap. Pump a sealed chamber down below 0.006 atm and the ice in frozen food leaves as vapor, with no soggy liquid stage on the way.

The critical point

Now follow the liquid-gas line up and to the right. It does not run forever. It stops dead at the critical point, which for water is 374 °C and 218 atm. Why would a line just end? Because at that temperature the hot, dense gas and the hot, thin liquid have the same density. There is nothing left to tell them apart, and no surface divides them. Past that point the substance is a supercritical fluid. It flows and spreads like a gas but dissolves things like a liquid. Supercritical CO2 is how coffee beans get decaffeinated without a drop of solvent left behind.

Water leans the wrong way

Look at the solid-liquid line on almost any other substance and it tilts to the right. Squeeze that substance and you help it freeze. Water's line tilts left instead. Press harder on ice held at 0 °C and it melts.

The reason is density. Ice is 0.917 g/cm3 and liquid water is 1.00 g/cm3, so ice floats. The hydrogen bond network in ice locks molecules into an open cage with gaps in it. Melting collapses part of that cage, and the pieces pack tighter. Pressure always favors whichever phase takes up less room, so here pressure favors the liquid. The effect is real but small. It takes roughly 135 atm to pull water's melting point down by a single degree.

Phase diagram of water, with carbon dioxide for contrast pressure (atm) 0.0001 0.01 1 100 1000 how almost every other substance leans SOLID LIQUID GAS (vapor) triple point 0.01 °C, 0.006 atm critical point 374 °C, 218 atm supercritical 0 °C 100 °C the 1 atm line, where everyday melting and boiling happen −80 0 100 200 300 400 temperature (°C) · water, pressure on a log scale CO₂ for contrast 1 atm SOL LIQ GAS triple point 5.11 atm sits above the 1 atm line, so no liquid. Sketch only.
Figure 10.5 Water's phase diagram, drawn from measured data with pressure on a log scale. The solid-liquid line really does lean left, though only slightly. The grey dashed line shows the right-leaning direction nearly every other substance takes. The small sketch at the right shows why CO2 has no liquid state at 1 atm.

CO2, the useful contrast

Carbon dioxide's triple point is −56.6 °C at 5.11 atm. One atmosphere is below that pressure, so the 1 atm line never touches the liquid region at all. Solid CO2 sublimes at −78.5 °C, which is why dry ice smokes away and leaves the floor dry. Its critical point is 31.0 °C and 72.9 atm. A CO2 cylinder in a cool room holds liquid and gas together near 57 atm. Let the room warm past 31 °C and the liquid surface vanishes into a supercritical fluid.

Worked example · Walk two paths on the map

Start with ice at −10 °C. Warm it to 50 °C at steady pressure. Do it once at 0.002 atm and once at 1 atm.

Step 1 · place the first path0.002 atm is below the triple point pressure of 0.006 atm. The path runs under the triple point for its whole length.

Step 2 · count the crossingsIt leaves the solid region and crosses the solid-gas line only. The ice sublimes. No liquid forms at any moment.

Step 3 · place the second path1 atm is above 0.006 atm, so the path runs over the triple point.

Step 4 · count againIt crosses the solid-liquid line at 0 °C, then sits in the liquid region. At 50 °C it is still liquid, since boiling waits until 100 °C. So the answer is ice to water, and the same sample took two different routes for one reason: pressure.

Key points
  • A phase diagram names the stable state for every pressure and temperature pair.
  • At the triple point all three states sit in balance at once.
  • Below the triple point pressure, a solid sublimes and the liquid never appears.
  • The liquid-gas line ends at the critical point, where liquid and gas become one fluid.
  • Water's solid-liquid line leans left because ice is less dense than water.
  • CO2 has a triple point at 5.11 atm, above 1 atm, so it has no liquid in open air.
The triple point of water is so reproducible that it was used to define the kelvin for decades. The scale was set by calling that one point exactly 273.16 K. The 2019 redefinition moved the anchor to the Boltzmann constant, but the triple point still sits at 273.16 K to within a hair.

Section 6Solids and crystal lattices

Four families of solid

Solids differ by what sits at the lattice points and what holds those pieces together. Ionic solids stack positive and negative ions. The pulls are strong in every direction, so they melt high and shatter rather than bend. NaCl melts at 801 °C. Metallic solids park metal atoms in a shared pool of loose electrons, which is why they conduct heat and current and can be hammered flat.

Covalent network solids join every atom to its neighbors with real bonds, so melting one means breaking bonds. Diamond does not melt below 3,500 °C. Molecular solids hold whole molecules in place using only the forces from section 1. They melt low. Ice goes at 0 °C and solid CO2 never melts at 1 atm at all.

One more split cuts across all four. A crystalline solid repeats one pattern over long distances and melts at a sharp temperature. An amorphous solid has order only in small patches. Glass is the classic case. It softens over a range of temperatures instead of melting at a single one.

The unit cell

The smallest repeating box in a crystal is the unit cell. Stack copies of it in all three directions and you rebuild the whole crystal. Three cubic cells cover a great many metals, so they are worth knowing cold.

Counting atoms in a cell takes one rule: share them. An atom sitting on a corner is shared by the eight cells that meet there, so this cell owns 1/8 of it. An atom on a face is shared by two cells, so this cell owns 1/2. An atom fully inside belongs to this cell alone.

Where the spheres touch sets a link between the edge length a and the radius r. In a simple cubic cell they touch along the edge, so a = 2r. In a body-centered cell they touch along the body diagonal, giving 4r = √3 a. In a face-centered cell they touch along the face diagonal, giving 4r = √2 a.

The three cubic unit cells and how their atoms are counted SIMPLE CUBIC 8 corners × 1/8 = 1 atom touches 6 · fills 52% · a = 2r example: polonium BODY-CENTERED 8 × 1/8 + 1 = 2 atoms touches 8 · fills 68% · 4r = √3 a example: iron, tungsten, sodium FACE-CENTERED 8 × 1/8 + 6 × 1/2 = 4 atoms touches 12 · fills 74% · 4r = √2 a example: copper, aluminum, nickel
Figure 10.6 The three cubic unit cells. Teal spheres sit on corners and each one is shared by eight cells. The gold sphere in the middle cell belongs to that cell alone. The gold spheres on the right sit on faces and are split between two cells each.

From a cell to a density

Here is the payoff. X-rays measure the edge length of a unit cell. From that one number you can predict the density of the bulk metal, and the match is a real test of the model. Density is mass over volume, so take one cell. Its mass is the Z atoms it owns, each weighing M/NA. Its volume is a3.

ρ = Z M / (NA a3)

Worked example · The density of copper from one length

Copper is face-centered cubic with a = 361.5 pm and M = 63.55 g/mol. Find its density and the radius of a copper atom.

Step 1 · count the atomsFace-centered means Z = 4.

Step 2 · get the volume in cm3 a = 361.5 pm = 3.615 × 10−8 cm. a3 = 4.724 × 10−23 cm3.

Step 3 · get the mass of one cell (4)(63.55 g/mol) / (6.022 × 1023 /mol) = 4.221 × 10−22 g.

Step 4 · divideρ = 4.221 × 10−22 g / 4.724 × 10−23 cm3 = 8.94 g/cm3. The handbook value is 8.96 g/cm3.

Step 5 · find the radiusSpheres touch along the face diagonal, so 4r = √2 a and r = a/2√2 = 361.5/2.828 = 127.8 pm. The listed radius of copper is 128 pm.

Key points
  • Ionic, metallic, covalent network and molecular solids differ in what holds them together.
  • Crystalline solids melt sharply. Amorphous solids soften over a range.
  • A corner atom counts as 1/8, a face atom as 1/2, a body atom as 1.
  • Simple cubic holds 1 atom, body-centered holds 2, face-centered holds 4.
  • ρ = ZM/(NAa3) turns one measured edge into a density.
Table salt is two face-centered lattices laid on top of each other, one of Na+ and one of Cl−, offset by half a cell edge. Count each one and you get 4 ions of each kind per cell. The ratio is 4 to 4, which is the 1 to 1 you already write as NaCl.

The chapter in one breath

  • Dispersion forces act in every substance and grow with the number of electrons.
  • Dipole-dipole needs a lopsided molecule, and a hydrogen bond needs H on N, O or F.
  • Ion-dipole is the strongest of the four, and it is what pulls salts into water.
  • Stronger forces mean higher boiling point, higher viscosity and higher surface tension.
  • A phase change takes heat without changing temperature: 334 J/g to melt water, 2,260 J/g to boil it.
  • Vapor pressure rises steeply with temperature, and boiling comes when it matches the outside pressure.
  • ln P against 1/T is a line whose slope gives the heat of vaporization.
  • A phase diagram holds a triple point and a critical point, and water's melting line leans left.
  • Unit cells hold 1, 2 or 4 atoms, and one edge length gives you the density.

ReferenceWords worth knowing

These are the words this chapter leans on. Each one is defined the way it is used here.

adhesive force
The attraction between a liquid and the surface it touches. It competes with cohesion to set the shape of the meniscus.
body-centered cubic
A cubic cell with atoms on the 8 corners plus 1 in the middle. It owns 2 atoms and fills 68% of space.
capillary action
The rise of a liquid in a narrow tube, driven by adhesion to the walls. The height goes as 1 over the radius.
Clausius-Clapeyron equation
The relation ln(P2/P1) = −(ΔHvap/R)(1/T2 − 1/T1), linking vapor pressure to temperature.
critical point
The end of the liquid-gas line. Above it liquid and gas have the same density and no surface separates them. Water's is 374 °C and 218 atm.
dipole-dipole attraction
The pull between permanently lopsided molecules lining up plus to minus. Usually 5 to 25 kJ/mol.
dispersion force
The pull from a brief, flickering split of charge in an electron cloud. Present in all matter and stronger for bigger clouds.
face-centered cubic
A cubic cell with atoms on the 8 corners plus 1 on each of the 6 faces. It owns 4 atoms and fills 74% of space.
heat of fusion
The energy a set amount of solid needs to melt at its melting point. Water needs 334 J/g, or 6.01 kJ/mol.
heat of vaporization
The energy a set amount of liquid needs to boil at its boiling point. Water needs 2,260 J/g, or 40.7 kJ/mol.
hydrogen bond
A strong dipole pull between an H atom bonded to N, O or F and a lone pair on a nearby N, O or F. About 20 kJ/mol in water.
intermolecular force
Any attraction acting between whole molecules rather than inside one. It sets melting point, boiling point and most liquid behavior.
ion-dipole attraction
The pull between an ion and a polar molecule. At 40 to 600 kJ/mol it is the strongest force in this chapter, and it drives dissolving.
phase diagram
A plot of pressure against temperature that names the stable state at every point, with lines marking where states change.
sublimation
The change from solid straight to gas, with no liquid stage. Dry ice does it at room temperature and 1 atm.
surface tension
The energy cost of making new surface, in mN/m. Water is 72.0 mN/m at 25 °C and mercury is 486 mN/m.
triple point
The one pressure and temperature where solid, liquid and gas all sit in balance. Water's is 0.01 °C and 0.006 atm.
unit cell
The smallest repeating box of a crystal. Stacking it in three directions rebuilds the whole solid.
vapor pressure
The pressure of vapor over its own liquid once escape and return are balanced. It depends on the liquid and the temperature only.
viscosity
A liquid's resistance to flow. It rises with stronger forces and with molecule length, and it falls as temperature rises.

Check yourself

1. Water’s heat of fusion is 6.01 kJ/mol. Show that this is 334 J/g.
334 J/g. 6.01 kJ/mol is 6,010 J/mol. Divide by the molar mass: 6,010 J/mol ÷ 18.02 g/mol = 333.5 J/g, which rounds to 334 J/g. The same trick on 40.7 kJ/mol gives 40,700 ÷ 18.02 = 2,259 J/g, the 2,260 J/g used in this chapter.
2. How much heat does 45.0 g of ice at 0 °C need to melt completely?
15.0 kJ. Melting is a flat stretch, so use mass times the heat of fusion. q = (45.0 g)(334 J/g) = 15,030 J = 15.0 kJ. The temperature never leaves 0 °C while this happens.
3. Find the heat needed to take 50.0 g of liquid water from 20.0 °C to steam at 100.0 °C.
1.30 × 105 J, or 130 kJ. Two steps. Warming: q = (50.0)(4.184)(80.0) = 16,736 J. Boiling: q = (50.0)(2,260) = 113,000 J. Total = 129,736 J = 130 kJ. Boiling is 87% of the bill even though the warming covers 80 degrees.
4. At what temperature does water boil when the outside pressure is 608 torr (0.800 atm)? Use 760 torr at 373.15 K and ΔHvap = 40.7 kJ/mol.
About 93.8 °C. ln(608/760) = ln(0.800) = −0.2231. And ΔHvap/R = 40,700/8.314 = 4,895 K. So −0.2231 = −4,895(1/T2 − 1/373.15). That gives 1/T2 − 0.0026799 = 4.558 × 10−5, so 1/T2 = 0.0027255 and T2 = 366.9 K = 93.8 °C. Steam tables give 93.9 °C, so the estimate lands right on it.
5. Iron is body-centered cubic with an edge length of 286.6 pm and a molar mass of 55.85 g/mol. Find its density.
7.88 g/cm3. Body-centered means Z = 2. a = 2.866 × 10−8 cm, so a3 = 2.354 × 10−23 cm3. Cell mass = (2)(55.85)/(6.022 × 1023) = 1.855 × 10−22 g. Divide: 1.855 × 10−22 / 2.354 × 10−23 = 7.88 g/cm3. The handbook value is 7.87 g/cm3.
6. How high does water climb in a clean glass tube of radius 0.10 mm? Use T = 0.0720 N/m, cos θ = 1 and ρ = 1.00 × 103 kg/m3.
About 15 cm. h = 2Tcosθ/(rρg). The bottom is (1.0 × 10−4)(1.00 × 103)(9.81) = 0.981. The top is 2(0.0720) = 0.144. So h = 0.144/0.981 = 0.147 m = 14.7 cm. That is 2.5 times the 5.9 cm from the worked example, because the radius is 2.5 times smaller.
7. Count the atoms in a face-centered cubic unit cell, showing the shares.
4 atoms. The 8 corner atoms each give 1/8, for 8 × 1/8 = 1. The 6 face atoms each give 1/2, for 6 × 1/2 = 3. Add them: 1 + 3 = 4 atoms per cell. Compare simple cubic at 1 and body-centered at 2.
8. Rank argon, HCl and water from lowest boiling point to highest, and name the deciding force.
Ar < HCl < H2O. Argon has dispersion only and boils at −186 °C. HCl adds dipole-dipole and boils at −85 °C. Water adds hydrogen bonding and boils at 100 °C. All three have similar electron counts, so the type of force decides, not the mass.
9. Why can you not pour liquid CO2 out of a bottle in an open room?
Its triple point pressure is above 1 atm. CO2’s triple point is −56.6 °C at 5.11 atm. At 1 atm the sample never enters the liquid region, no matter the temperature. Solid CO2 sublimes at −78.5 °C instead.
10. Why does water’s solid-liquid line lean left when nearly every other substance leans right?
Because ice is less dense than liquid water. Pressure pushes a sample toward whichever phase takes less room. Ice is 0.917 g/cm3 and liquid water is 1.00 g/cm3, so squeezing favors the liquid and the melting point drops. For most substances the solid is denser, so squeezing favors freezing.
11. Glycerol is about 1,000 times more viscous than water at 25 °C. Give two reasons, and say what warming it does.
Three OH groups and a longer chain, and warming thins it. Glycerol makes several hydrogen bonds per molecule instead of water’s few, and its longer chain tangles with its neighbors. Heating gives molecules enough energy to slip out of each grip sooner, so viscosity always falls as temperature rises.
12. Freeze drying holds frozen food below 0.006 atm. Name the phase change it uses and say why the low pressure matters.
Sublimation, below the triple point pressure. Water’s triple point is 0.01 °C and 0.006 atm. Below that pressure the liquid region is gone from the map entirely. Ice warmed there turns straight to vapor, so the food dries without ever getting wet.
Chemistry · Chapter 11

Solutions and Colloids

Stir a spoonful of salt into water and the salt seems to vanish. Nothing was destroyed. The salt is still in the glass, split into ions and wrapped in water. That kind of mixture is a solution, and solutions run most of chemistry. Blood, seawater, soda, eye drops and brass are all solutions. This chapter shows you why some things dissolve and some refuse. You will follow the energy that dissolving costs and pays back. You will measure how much solute a mixture holds, five different ways. Then you will watch a pinch of solute move the freezing point and the boiling point of water.

By the end of this chapter you can

  • Describe dissolving as a trade of old attractions for new ones, and name the forces involved.
  • Use “like dissolves like” to predict whether a solute and a solvent will mix.
  • Add the three energy steps of dissolving to get ΔHsoln, and say what its sign means.
  • Work out molarity, molality, mole fraction, percent by mass and ppm, and convert between them.
  • Read a solubility curve, and use Henry's law to find how much gas dissolves at a given pressure.
  • Find the new freezing point, boiling point and osmotic pressure of a solution, using the van 't Hoff factor.
  • Tell a colloid apart from a true solution and a suspension, and explain the Tyndall effect.

Section 1The dissolving process

What counts as a solution

A solution is a mixture that looks the same all the way through. Take a drop from the top and a drop from the bottom and you get the same thing. The part present in the largest amount is the solvent. Anything dissolved in it is a solute. When water is the solvent we call the mixture aqueous and write (aq) after the formula.

Solutions are not always liquid. Air is a gas solution of oxygen and argon in nitrogen. Brass is a solid solution of zinc in copper. Water solutions matter most in a lab, so water is where we will spend our time.

Old attractions traded for new ones

Picture a grain of salt sitting in water. Inside the grain, Na+ and Cl- ions hold each other hard. In the water, molecules hold each other through hydrogen bonds. For the salt to dissolve, both of those grips must loosen. In return, water molecules get to grab the ions.

That trade is the whole story of dissolving. It happens when the new attractions pay for the ones that broke. A water molecule has a slightly negative end at oxygen and a slightly positive end at each hydrogen. So the oxygen end swings toward Na+, and the hydrogen ends swing toward Cl-. This pull between an ion and a polar molecule is an ion-dipole force.

Water molecules lift sodium and chloride ions off a crystal face and wrap a shell around each one1 · The crystal face comes apartNa+Cl-Na+Cl-Cl-Na+Cl-Na+Na+Cl-Na+Cl-Cl-Na+Cl-Na+solid NaCl: every ion is locked inwater pullsNa+Cl-O end turns toward Na+H ends turn toward Cl-O (partly negative)H (partly positive)
Figure 11.1 Water molecules crowd the crystal face. The oxygen end turns toward Na+ and the hydrogen ends turn toward Cl-. Each freed ion leaves with a shell of water around it.

The shell of solvent that forms around each dissolved particle is called solvation. When the solvent is water, the same thing gets its own name, hydration. A hydrated Na+ ion moves through the liquid with about six water molecules facing it oxygen-first. Those water molecules are not bonded to the ion. They swap places with others many times a second.

Like dissolves like

One short rule predicts most of what you will see at the bench. Like dissolves like means that polar solvents dissolve polar and ionic solutes, and nonpolar solvents dissolve nonpolar solutes.

Water is polar, so it handles salt and sugar well. Sugar has many O-H groups, and each one can hydrogen bond to water. Oil is nonpolar and is held together only by weak dispersion forces. To make room for oil, water would have to give up strong hydrogen bonds and would get very little back. So oil and water stay in layers. Hexane is nonpolar too, and it dissolves oil with ease.

Two liquids that mix in any amount are miscible. Water and ethanol are miscible, because the O-H group on ethanol hydrogen bonds to water. Water and oil are immiscible. Keep in mind that the rule is a guide, not a law. Some polar solutes still fail in water because their own lattice is simply too strong.

Key points
  • A solution is one phase: solvent plus solute, even all the way through.
  • Dissolving trades solute-solute and solvent-solvent attractions for solute-solvent attractions.
  • Ion-dipole forces hold hydrated ions. Hydrogen bonds hold sugar in water.
  • Like dissolves like. Polar goes with polar and ionic. Nonpolar goes with nonpolar.

Worked example · counting the ions that come free

You dissolve 0.250 mol of CaCl2 in water. How many moles of ions end up in the solution?

Step 1 · write what the solid doesCaCl2(s) → Ca2+(aq) + 2 Cl-(aq)

Step 2 · read the ratioOne mole of the solid gives 1 mol of Ca2+ plus 2 mol of Cl-. That is 3 mol of ions per mole of solid.

Step 3 · scale it0.250 mol CaCl2 × (3 mol ions / 1 mol CaCl2) = 0.750 mol ions

Answer0.750 mol of ions, made of 0.250 mol Ca2+ and 0.500 mol Cl-. The 2+ charge on calcium pulls water harder than Na+ does, so it carries a tighter shell. Hold on to this count. Section 5 shows that the number of free particles, not their name, sets the freezing point.

Worth knowing. Dissolving is not melting and it is not a reaction. The ions in salt water are the same ions that sat in the crystal. Boil the water away and the crystal comes back.

Section 2Energy of solution formation

Three steps in your head, one event in the beaker

In the beaker, dissolving happens all at once. To track the energy, though, it helps to break the event into three imagined steps. The steps are a bookkeeping trick. Enthalpy does not care about the path, so any path that starts and ends in the right place gives the right total.

Step 1 pulls the solute apart. Its particles must be separated from each other, and that always costs energy. So ΔH1 is positive. For an ionic solid, step 1 is the lattice energy paid in reverse.

Step 2 pushes the solvent apart to open a space. Water molecules give up some hydrogen bonds to make the cavity. This costs energy too, so ΔH2 is positive.

Step 3 lets solute and solvent reach each other. New attractions form, so energy comes back out. ΔH3 is negative. Add all three and you have the enthalpy of solution.

ΔHsoln = ΔH1 + ΔH2 + ΔH3
Enthalpy diagram for dissolving sodium chloride, showing the two uphill steps and the one downhill step2 · Three steps add up to ΔH of solutionenthalpystartNaCl(s) + waterafter step 1free ions + waterafter step 2ions + opened waterfinishNa+(aq) + Cl-(aq)step 1 uphill+787 kJ/molstep 2 uphill+40 kJ/molstep 3 downhill-824 kJ/molnet: +787 + 40 - 824 = +3 kJ/mol
Figure 11.2 Two uphill steps and one downhill step for NaCl. Steps 2 and 3 are measured as one value, the hydration enthalpy of -784 kJ/mol. The split shown here is an estimate of how much of that is cavity work.

Why the sign is hard to guess

For a salt in water, steps 2 and 3 are measured as one number. That number is the hydration enthalpy, the heat released when free gas-phase ions are surrounded by water. So the total is a race between two giants.

Breaking one mole of the NaCl lattice into free ions costs +787 kJ/mol. Hydrating that mole of Na+ and Cl- gives back about -784 kJ/mol. What is left over is only about +3 kJ/mol. The measured value for NaCl is +3.9 kJ/mol. Both big numbers carry a few kJ of doubt, so a small leftover like this is always rough.

Tiny gaps decide the sign. NH4NO3 lands at +25.7 kJ/mol and chills the water, which is how an instant cold pack works. CaCl2 lands at -81.3 kJ/mol and warms the water, which is how a hot pack works. Both are salts in water. The sign flips on a difference of a percent or two.

A positive value does not block dissolving

NH4NO3 dissolves freely even though it soaks up heat. Energy is only half the vote. Spreading solute through solvent also raises the disorder of the mixture, a quantity called entropy. That gain can carry the process on its own. Chapter 16 puts real numbers on the balance of the two.

Key points
  • ΔHsoln = ΔH1 + ΔH2 + ΔH3. The first two are uphill, the third is downhill.
  • For salts, steps 2 and 3 together are the hydration enthalpy.
  • ΔHsoln is a small difference of two large numbers, so its sign is hard to predict.
  • Negative ΔHsoln warms the mixture. Positive ΔHsoln cools it.
  • A positive ΔHsoln can still dissolve, because entropy rises when solute spreads out.

Worked example · how cold does a cold pack get?

You stir 25.0 g of NH4NO3 into 150.0 g of water at 22.0 °C. For this salt ΔHsoln = +25.7 kJ/mol. Take the specific heat of the mixture as 4.18 J/g·°C. What is the final temperature?

Step 1 · moles of saltMolar mass of NH4NO3 = 14.01(2) + 1.008(4) + 16.00(3) = 80.05 g/mol
n = 25.0 g ÷ 80.05 g/mol = 0.3123 mol

Step 2 · heat taken inq = 0.3123 mol × 25.7 kJ/mol = 8.03 kJ = 8030 J

Step 3 · where that heat comes fromThe mixture itself gives it up. Its mass is 25.0 g + 150.0 g = 175.0 g.
ΔT = -q ÷ (m × c) = -8030 J ÷ (175.0 g × 4.18 J/g·°C)
ΔT = -8030 J ÷ 731.5 J/°C = -11.0 °C

AnswerThe mixture falls to 22.0 - 11.0 = 11.0 °C. Check the units: joules divided by joules per degree leaves degrees, which is what we want. The minus sign is right, because heat flowing into the dissolving salt has to leave the water.

Section 3Concentration units

Molarity: per liter of solution

Concentration says how crowded a solution is. The unit you pick depends on the job. Molarity (M) is moles of solute per liter of solution.

M = mol solute ÷ L solution

Molarity is the lab favorite, because you measure volume with a pipette in seconds. It has one flaw. Volume grows a little when a liquid warms, so the molarity of a bottle drifts as the room heats up.

Molality: per kilogram of solvent

Molality (m) is moles of solute per kilogram of solvent. Note the word solvent, not solution.

m = mol solute ÷ kg solvent

Mass never changes with temperature, so molality holds steady. That is why the freezing point and boiling point math in Section 5 uses molality and not molarity.

Molarity divides by the volume of the whole solution while molality divides by the mass of solvent alone3 · Same sample, two different bottoms of the fractionMolarity: mol per liter of solutionMolality: mol per kilogram of solventwhole solutionvolume is easy to pour, but it grows when warmedwater onlymass never shifts with temperatureto cross between themyou need densitymass of solutionvolume of solution= 1 mol of solute
Figure 11.3 Molarity divides by the volume of the whole solution. Molality divides by the mass of solvent only. Density is the bridge between them.

Mole fraction, percent by mass and ppm

The mole fraction of a part is its moles divided by the total moles of everything present. Mole fractions have no units, and all of them in a sample add up to 1.

XA = nA ÷ ntotal

Percent by mass is the mass of solute divided by the mass of the whole solution, times 100. For traces, scale up instead of down. Parts per million (ppm) uses a factor of 106, and parts per billion uses 109.

percent by mass = (mass solute ÷ mass solution) × 100 ppm = (mass solute ÷ mass solution) × 106

For dilute water solutions there is a handy shortcut. A liter of such a solution weighs close to 1000 g, so 1 ppm is about 1 mg per liter. Drinking water limits are usually written this way.

Crossing between units

Percent, molality and mole fraction all run on mass, so you can swap among them with molar masses alone. To reach molarity you also need density, because molarity needs a volume. The trick is to choose a sample size that makes the numbers easy. For a percent, pick 100.0 g of solution.

Key points
  • Molarity is per liter of solution. Molality is per kilogram of solvent.
  • Molality and mass percent ignore temperature. Molarity does not.
  • Mole fraction is unitless, and the fractions in a sample sum to 1.
  • In dilute water solutions, 1 ppm is roughly 1 mg per liter.
  • You cannot reach molarity from a mass-based unit without the density.

Worked example · one solution in five units

A brine is 12.0% NaCl by mass. Its density is 1.086 g/mL. Find the molarity, the molality, the mole fraction of NaCl, and the concentration in ppm.

Step 1 · pick a sampleTake 100.0 g of solution. Then it holds 12.0 g of NaCl and 100.0 - 12.0 = 88.0 g of water.

Step 2 · moles of each partn(NaCl) = 12.0 g ÷ 58.44 g/mol = 0.2053 mol
n(H2O) = 88.0 g ÷ 18.02 g/mol = 4.883 mol

Step 3 · molarity needs a volumeV = 100.0 g ÷ 1.086 g/mL = 92.08 mL = 0.09208 L
M = 0.2053 mol ÷ 0.09208 L = 2.23 M

Step 4 · molality uses the water massm = 0.2053 mol ÷ 0.0880 kg = 2.33 m

Step 5 · mole fractionX(NaCl) = 0.2053 ÷ (0.2053 + 4.883) = 0.2053 ÷ 5.088 = 0.0404

Step 6 · ppm(12.0 g ÷ 100.0 g) × 106 = 120 000 ppm

Answer2.23 M, 2.33 m, X = 0.0404, 12.0% by mass, 120 000 ppm. Molality beats molarity here, as it always does for a solution denser than water. The water mass, 0.0880 kg, is smaller than the solution volume in liters, 0.09208 L, so the same moles get divided by less.

Worth knowing. Molality is written with a lower case m and molarity with a capital M. In handwriting that is a thin line to lean on, so label your work with words as well as symbols.

Section 4Solubility, temperature and pressure

Saturated, unsaturated and supersaturated

Solubility is the most solute that will dissolve in a fixed amount of solvent at a fixed temperature. A saturated solution holds exactly that much. Extra solid sits on the bottom, and it is not idle. Particles leave the solid and rejoin it at the same rate, so the amount in solution holds still.

Below the limit the solution is unsaturated and will take more. Careful cooling can leave a solution holding more than its limit. That state is supersaturated, and it is fragile. Drop in one seed crystal and the extra crashes out in seconds. A hand warmer full of sodium acetate works on exactly this trick.

Temperature: solids go both ways

Most solids dissolve better in hot water. KNO3 climbs from 13.3 g per 100 g of water at 0 °C to 246 g at 100 °C. Some barely move. NaCl only goes from 35.7 g to 39.2 g over the same range. A few run backward. Ce2(SO4)3 falls from 21.4 g at 0 °C to under 2 g at 100 °C.

There is no shortcut rule for solids. You read a solubility curve, which is a plot of solubility against temperature built from measured points.

Solubility curves for five salts in water from zero to one hundred degrees Celsius4 · Solubility curves: every solid writes its own line04080120160200240020406080100temperature (°C)grams per 100 g waterKNO3NaNO3KClNaCl (nearly flat)Ce2(SO4)3 (falls)cool 60 °C → 20 °C110 - 31.6 = 78.4 g drops out
Figure 11.4 Measured solubility of five salts in water. Points are table values at 0, 20, 40, 60, 80 and 100 °C, joined by straight lines.

Gases are different, and they are consistent. Warm water holds less dissolved gas. Set a glass of cold tap water on the counter and bubbles form on the inside wall as it warms. The same rule has teeth outside the lab. A river warmed by a power plant carries less oxygen, and fish feel it first.

Pressure and Henry's law

Pressure hardly touches the solubility of solids and liquids. For a gas it matters a great deal. Push harder on the gas above a liquid and more of it is forced in. Henry's law puts that in one line.

C = kH P

Here C is the dissolved concentration, P is the partial pressure of that gas above the liquid, and kH is the Henry's law constant for that gas in that solvent at that temperature. The law works well for gases that stay as molecules in the liquid and do not react with it.

Key points
  • A saturated solution is at balance: solid dissolves and re-forms at equal rates.
  • A supersaturated solution holds more than its limit and dumps the excess when seeded.
  • Solids mostly dissolve better hot, but you must read the curve for each one.
  • Gases always dissolve less in warmer liquid.
  • Henry's law, C = kHP, works for gas solubility, not for solids.

Worked example · crystals from a cooling solution

You saturate 200.0 g of water with KNO3 at 60 °C, then cool it slowly to 20 °C. From the curve, solubility is 110 g per 100 g of water at 60 °C and 31.6 g per 100 g at 20 °C. How much solid appears?

Step 1 · what 200.0 g of water holds when hot110 g / 100 g water × 200.0 g water = 220 g KNO3

Step 2 · what it holds when cool31.6 g / 100 g water × 200.0 g water = 63.2 g KNO3

Step 3 · subtract220 g - 63.2 g = 157 g

AnswerAbout 157 g of KNO3 crystals. The grams of water cancel in both steps, leaving grams of salt. This is recrystallization, the standard way to clean up a solid product.

Worked example · oxygen in a lake, and gas in a soda

At 25 °C, kH for O2 in water is 1.3 × 10-3 mol/L·atm. Air is 21% O2 by volume. How much O2 dissolves from air at 1.00 atm?

Step 1 · partial pressure of oxygenP = 0.21 × 1.00 atm = 0.21 atm

Step 2 · apply Henry's lawC = (1.3 × 10-3 mol/L·atm)(0.21 atm) = 2.7 × 10-4 mol/L

Step 3 · put it in mg/L2.7 × 10-4 mol/L × 32.00 g/mol = 8.7 × 10-3 g/L = 8.7 mg/L

That is close to the 8 to 9 mg/L a meter reads in cool, air-saturated water. Now the soda. For CO2, kH = 3.4 × 10-2 mol/L·atm at 25 °C.

Step 4 · sealed can at 2.5 atm of CO2C = (3.4 × 10-2)(2.5) = 0.085 mol/L

Step 5 · opened can, where air holds only 3.9 × 10-4 atm of CO2C = (3.4 × 10-2)(3.9 × 10-4) = 1.3 × 10-5 mol/L

AnswerOpening the can cuts the gas it can hold by a factor of 0.085 ÷ 1.3 × 10-5, which is about 6500. The drink cannot lose that much in an instant, so it fizzes for a while instead.

Section 5Colligative properties

Properties that count heads, not names

A colligative property depends only on how many solute particles are in the solvent. It does not care what they are. One mole of sugar and one mole of urea shift water's freezing point by the same amount. One mole of NaCl shifts it twice as far, because it breaks into two ions.

Vapor pressure lowering

Solute particles sit at the surface and take up room that water molecules would have used to escape. Fewer escape, so the vapor pressure falls. Raoult's law states the size of the drop.

Psolution = Xsolvent × P°solvent

Here P° is the vapor pressure of the pure solvent. Pure water at 25 °C is 23.8 torr. A solution with Xwater = 0.9091 reads 0.9091 × 23.8 torr = 21.6 torr. Lower vapor pressure is the root cause of the next two effects.

Boiling point up, freezing point down

A liquid boils when its vapor pressure reaches the outside pressure. Lower the vapor pressure and you must heat the liquid further to get there. That is boiling point elevation. Solute also gets in the way of ice forming, so the liquid must be chilled further before it freezes. That is freezing point depression.

ΔTb = i Kb m     ΔTf = i Kf m

The constants belong to the solvent, not the solute. For water, Kb = 0.512 °C/m and Kf = 1.86 °C/m. Notice that Kf is more than three times larger. That is why salting a road is worth doing and salting a pot of water barely helps it boil faster.

A temperature line showing how one molal sodium chloride pushes the freezing point down and the boiling point up5 · Solute stretches the liquid range at both endspure waterliquid, 100.00 °C wide0.00100.00down 3.72up 1.021.00 m NaCli = 2liquid, 104.74 °C wide-3.72101.02-20020406080100temperature (°C)ΔTf = i Kf m = 2 × 1.86 × 1.00 = 3.72 °C    ΔTb = i Kb m = 2 × 0.512 × 1.00 = 1.02 °Cvapor pressure at 25 °C falls 23.8 → 23.0 torrosmotic pressure climbs to about 48 atm
Figure 11.5 One molal NaCl in water. The freezing point drops 3.72 °C and the boiling point climbs 1.02 °C, so the liquid range widens by 4.74 °C.

The van 't Hoff factor

The van 't Hoff factor i is the number of particles one formula unit releases. Sucrose and other nonelectrolytes give i = 1. NaCl gives 2, CaCl2 gives 3, and FeCl3 gives 4. Real measurements fall a bit short of those whole numbers, because oppositely charged ions pair up for brief moments and travel as one particle. In dilute solution, measured i for NaCl is about 1.9, and for MgSO4 it is near 1.3. The gap widens as the solution gets more crowded.

Osmotic pressure

Put a solution on one side of a membrane that passes water but not solute. Water moves toward the crowded side. That flow is osmosis. The pressure you must apply to stop it is the osmotic pressure, written Π.

Π = i M R T

Use R = 0.08206 L·atm/mol·K and T in kelvin. Osmotic pressure grows fast, which makes it the best of the four for measuring the molar mass of something rare. It also explains why a plant wilts in salty soil, and why saline for an IV drip is matched to blood at about 0.154 M NaCl.

Key points
  • Colligative properties count dissolved particles, whatever they are.
  • Raoult's law: the solvent's vapor pressure falls in step with its mole fraction.
  • ΔTb = iKbm and ΔTf = iKfm, with molality, not molarity.
  • Water: Kb = 0.512 °C/m, Kf = 1.86 °C/m.
  • i is 1 for a nonelectrolyte and equals the ions per formula unit for a salt, a little less in practice.
  • Π = iMRT, and it is the most sensitive of the four effects.

Worked example · antifreeze, then road salt

You mix 62.0 g of ethylene glycol, C2H6O2, 62.07 g/mol, into 250.0 g of water. Find the new freezing and boiling points. Glycol is a nonelectrolyte, so i = 1.

Step 1 · moles and molalityn = 62.0 g ÷ 62.07 g/mol = 0.999 mol
m = 0.999 mol ÷ 0.2500 kg = 3.996 m

Step 2 · freezing pointΔTf = (1)(1.86 °C/m)(3.996 m) = 7.43 °C
freezing point = 0.00 - 7.43 = -7.43 °C

Step 3 · boiling pointΔTb = (1)(0.512 °C/m)(3.996 m) = 2.05 °C
boiling point = 100.00 + 2.05 = 102.05 °C

Now repeat with road salt. Use 5.85 g of NaCl, 58.44 g/mol, in the same 250.0 g of water, with i = 2.

Step 4 · moles and molalityn = 5.85 g ÷ 58.44 g/mol = 0.1001 mol
m = 0.1001 mol ÷ 0.2500 kg = 0.400 m

Step 5 · both shifts at onceΔTf = (2)(1.86)(0.400) = 1.49 °C → -1.49 °C
ΔTb = (2)(0.512)(0.400) = 0.410 °C → 100.41 °C

AnswerThe glycol sample freezes at -7.43 °C and boils at 102.05 °C. The salt sample freezes at -1.49 °C and boils at 100.41 °C. Check the units: °C/m times m leaves °C. Ten times less glycol than you might guess would do nothing, and one tenth the molality of salt still earns a 2× bonus from i.

Worked example · a molar mass from a freezing point, and the pressure in blood

You dissolve 1.50 g of an unknown nonelectrolyte in 75.0 g of water. The mixture freezes at -0.620 °C. What is the molar mass?

Step 1 · molality from the shiftm = ΔTf ÷ (iKf) = 0.620 °C ÷ (1 × 1.86 °C/m) = 0.3333 m

Step 2 · moles presentn = 0.3333 mol/kg × 0.0750 kg = 0.0250 mol

Step 3 · grams per moleM = 1.50 g ÷ 0.0250 mol = 60.0 g/mol

Next, the osmotic pressure of a saline drip: 0.154 M NaCl at body temperature, 37 °C, with i = 2.

Step 4 · convert the temperatureT = 37 + 273.15 = 310.15 K

Step 5 · apply Π = iMRTΠ = (2)(0.154 mol/L)(0.08206 L·atm/mol·K)(310.15 K) = 7.8 atm

Answer60.0 g/mol, and about 7.8 atm. Urea fits that molar mass. The 7.8 atm figure sits right on the measured osmotic pressure of blood, which is why this strength of saline leaves red cells unharmed.

Section 6Colloids and suspensions

Three mixtures, sorted by particle size

Not every cloudy liquid is a solution. Sort mixtures by how big the scattered bits are. In a true solution the particles are single ions or molecules, under about 1 nm across. In a suspension such as muddy water, the bits are larger than about 1000 nm. They settle out and filter paper catches them.

Between those two sits the colloid. Its particles run from about 1 nm to 1000 nm. They are too small to settle and too large to slip through a membrane. Milk, fog, jelly, whipped cream and paint are all colloids. The scattered material is the dispersed phase and the material it floats in is the dispersion medium.

Colloids stay up for two reasons. Constant random kicks from solvent molecules, called Brownian motion, keep knocking the particles around. Many colloid particles also carry the same surface charge, so they push each other apart instead of clumping.

A light beam through three tubes: invisible in a true solution, a bright streak in a colloid, blocked and settling in a suspension6 · The Tyndall effect sorts the three mixturestrue solutionparticles under 1 nmlampbeam is invisiblenever settlespasses a membranecolloidparticles 1 to 1000 nmlampbeam shows as a streaknever settlesheld back by a membranesuspensionparticles over 1000 nmlampsettled layerbeam is blockedsettles on standingcaught by filter paper
Figure 11.6 The same lamp aimed at three mixtures. Only the middle tube, the colloid, shows the beam as a bright streak.

The Tyndall effect

Shine a lamp through a true solution and the beam stays invisible from the side. The particles are far smaller than a wavelength of light, so they barely scatter it. Shine the same lamp through a colloid and the beam shows as a bright streak. This scattering by colloid-sized particles is the Tyndall effect. Sunbeams through dusty air and headlights in fog are the same effect outdoors. It is the quickest bench test for a colloid.

Emulsions and the agents that hold them

An emulsion is liquid droplets spread through another liquid that they do not mix with. Shake oil and water and you make one for a moment. The droplets then find each other, merge and split into layers again. An emulsifying agent stops that. Its molecules have a polar head and a nonpolar tail, so they coat each droplet with the tail inward and the head facing the water.

Egg yolk does this job in mayonnaise, using a molecule called lecithin. Soap does it when it lifts grease off a pan. Milk is an oil-in-water emulsion held by proteins. Heating, freezing or adding acid can strip the coating, and then the emulsion breaks.

Colloids come in other pairs too. A liquid spread in a gas is an aerosol, as in fog. A gas in a liquid is a foam, as in whipped cream. A solid in a liquid is a sol, as in paint, and a liquid trapped in a solid network is a gel, as in jelly.

Key points
  • Solution under 1 nm, colloid 1 to 1000 nm, suspension over 1000 nm.
  • Colloids do not settle. Suspensions do.
  • Only colloids show the Tyndall effect, so a lamp tells them apart in seconds.
  • Brownian motion and like surface charges keep colloid particles spread out.
  • An emulsion needs an emulsifying agent with a polar head and a nonpolar tail.

Worked example · why colloids have so much surface

Cut a cube of solid 1.00 cm on each edge into cubes 100 nm on each edge. How much surface area do you end up with?

Step 1 · one set of units100 nm = 1.00 × 10-5 cm, and 1.00 cm = 1.00 × 107 nm.

Step 2 · count the small cubesEach edge holds 1.00 × 107 ÷ 100 = 1.00 × 105 of them.
number = (1.00 × 105)3 = 1.00 × 1015 cubes

Step 3 · area of one small cubeA = 6 × (1.00 × 10-5 cm)2 = 6.00 × 10-10 cm2

Step 4 · total area(1.00 × 1015)(6.00 × 10-10 cm2) = 6.00 × 105 cm2 = 60.0 m2

Answer60.0 m2, up from 6.00 cm2 for the whole cube. That is 100 000 times more surface, on the same mass. Surface is where a colloid does its work, which is why foams, inks and catalysts are built at this size.

Worth knowing. Add salt to a colloid and you can wreck it. The ions screen the surface charges, so the particles stop repelling and clump together. River silt drops out where fresh water meets the sea for exactly this reason, and that is how a delta is built.

The chapter in one breath

  • A solution is one even phase. Dissolving swaps old attractions for new ones, and like dissolves like.
  • Water hydrates ions with ion-dipole forces, turning its oxygen end to cations and its hydrogen ends to anions.
  • ΔHsoln is two uphill steps plus one downhill step. For salts it is a small leftover from two huge numbers.
  • A positive ΔHsoln still dissolves, because spreading out raises entropy.
  • Molarity is per liter of solution and molality is per kilogram of solvent. Density bridges them.
  • Solids follow their own solubility curve with temperature. Gases always dissolve less when warm.
  • Henry's law, C = kHP, sets how much gas dissolves at a given partial pressure.
  • Solute lowers vapor pressure, so ΔTb = iKbm and ΔTf = iKfm. Water: 0.512 and 1.86 °C/m.
  • Colloid particles span 1 to 1000 nm, never settle, and show the Tyndall effect.

ReferenceWords worth knowing

These are the words this chapter leans on. Say each one out loud with its meaning before you try the questions below.

boiling point elevation
The rise in a solvent's boiling point when solute is added. ΔTb = iKbm, and Kb for water is 0.512 °C/m.
colligative property
A property set by how many solute particles are present, not by what they are. Vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.
colloid
A mixture whose dispersed particles measure about 1 nm to 1000 nm. It does not settle and it scatters light.
emulsion
A colloid of liquid droplets in another liquid, held apart by an emulsifying agent with a polar head and a nonpolar tail.
freezing point depression
The drop in a solvent's freezing point when solute is added. ΔTf = iKfm, and Kf for water is 1.86 °C/m.
Henry's law
C = kHP. The amount of a gas dissolved in a liquid is proportional to that gas's partial pressure above the liquid.
hydration
Solvation when the solvent is water. A shell of water molecules surrounds each dissolved particle.
ion-dipole force
The attraction between an ion and the charged end of a polar molecule. It is what holds a hydrated ion together.
miscible
Able to mix in any proportion, said of two liquids. Liquids that form layers instead are immiscible.
molality
Moles of solute per kilogram of solvent, symbol m. It does not change when the sample warms.
molarity
Moles of solute per liter of solution, symbol M. Easy to measure, but it shifts slightly with temperature.
mole fraction
The moles of one part divided by the total moles present. It has no units, and all the fractions in a sample add to 1.
osmosis
The flow of solvent through a membrane that blocks solute, running toward the more crowded side.
osmotic pressure
The pressure needed to stop osmosis, Π = iMRT. The most sensitive of the colligative properties.
parts per million
Mass of solute divided by mass of solution, times 106. In dilute water solutions 1 ppm is about 1 mg per liter.
Raoult's law
Psolution = XsolventP°solvent. The solvent's vapor pressure falls in step with its mole fraction.
saturated solution
A solution holding all the solute it can at that temperature, with dissolving and re-forming in balance. Holding more than that is supersaturated and unstable.
solvation
The wrapping of solvent molecules around a dissolved solute particle.
Tyndall effect
The scattering of a light beam by colloid-sized particles, which makes the beam visible from the side.
van 't Hoff factor
Symbol i, the number of particles one formula unit of solute releases. It is 1 for a nonelectrolyte, 2 for NaCl, and a little below the whole number in practice.

Check yourself

1. Iodine, I2, barely dissolves in water but dissolves well in hexane. Explain in terms of forces.
I2 is nonpolar. Hexane is nonpolar too, so only weak dispersion forces have to be traded on both sides. Water would have to break strong hydrogen bonds and would get almost nothing back, so like dissolves like predicts the split.
2. What is the molarity of a solution made by dissolving 4.25 g of NaCl in enough water to reach 250.0 mL?
0.291 M n = 4.25 g ÷ 58.44 g/mol = 0.0727 mol. V = 250.0 mL = 0.2500 L. M = 0.0727 mol ÷ 0.2500 L = 0.291 mol/L.
3. A cold pack uses NH4NO3 with ΔHsoln = +25.7 kJ/mol. Why does the pack feel cold, and why does the salt dissolve at all?
It absorbs heat, and entropy pays the bill. A positive ΔHsoln means the dissolving salt pulls heat from the water, so the mixture cools. Spreading the ions through the water raises entropy enough to drive the change even though it costs energy.
4. You mix 90.0 g of ethylene glycol, 62.07 g/mol, into 500.0 g of water. What is the freezing point? Glycol is a nonelectrolyte.
-5.39 °C n = 90.0 g ÷ 62.07 g/mol = 1.450 mol. m = 1.450 mol ÷ 0.5000 kg = 2.900 m. ΔTf = (1)(1.86 °C/m)(2.900 m) = 5.39 °C, so the new freezing point is 0.00 - 5.39 = -5.39 °C.
5. Find the boiling point of a solution of 11.7 g of NaCl in 200.0 g of water. Use i = 2.
101.02 °C n = 11.7 g ÷ 58.44 g/mol = 0.200 mol. m = 0.200 mol ÷ 0.2000 kg = 1.00 m. ΔTb = (2)(0.512 °C/m)(1.00 m) = 1.02 °C. Boiling point = 100.00 + 1.02 = 101.02 °C.
6. At 25 °C, kH for N2 in water is 6.0 × 10-4 mol/L·atm. Air at 1.00 atm is 78% N2. How much N2 dissolves?
4.7 × 10-4 mol/L P(N2) = 0.78 × 1.00 atm = 0.78 atm. C = kHP = (6.0 × 10-4 mol/L·atm)(0.78 atm) = 4.7 × 10-4 mol/L. In mass terms that is about 13 mg/L.
7. Seawater is about 3.5% salt by mass. Write that in ppm.
35 000 ppm ppm = (mass solute ÷ mass solution) × 106. A mass percent of 3.5% is 3.5 g in 100 g, so (3.5 ÷ 100) × 106 = 3.5 × 104 = 35 000 ppm.
8. What is the osmotic pressure of a 0.0150 M glucose solution at 25 °C? Glucose is a nonelectrolyte.
0.367 atm T = 25 + 273.15 = 298.15 K. Π = iMRT = (1)(0.0150 mol/L)(0.08206 L·atm/mol·K)(298.15 K) = 0.367 atm. Units: mol/L times L·atm/mol·K times K leaves atm.
9. A 2.40 g sample of an unknown nonelectrolyte in 60.0 g of water freezes at -1.24 °C. What is its molar mass?
60.0 g/mol m = ΔTf ÷ (iKf) = 1.24 °C ÷ (1 × 1.86 °C/m) = 0.667 m. n = 0.667 mol/kg × 0.0600 kg = 0.0400 mol. M = 2.40 g ÷ 0.0400 mol = 60.0 g/mol.
10. A clear solution sits over a pile of undissolved crystals and the pile never shrinks. Is the solution saturated, unsaturated or supersaturated?
Saturated. The solid is still dissolving, but solute is rejoining the crystals at the same rate. Those equal rates are what a saturated solution means. A supersaturated solution holds extra solute and has no solid left sitting in it.
11. You aim a flashlight at two clear glasses. One beam is invisible from the side and the other shows as a bright streak. What is in each glass?
Solution, then colloid. Particles in a true solution are far smaller than a wavelength of light and scatter very little. Colloid particles, 1 nm to 1000 nm across, scatter enough to make the beam show. That is the Tyndall effect.
12. Fish struggle in a pond during a heat wave even when the water is not polluted. Explain using gas solubility.
Warm water holds less oxygen. Gas solubility always falls as temperature rises, because warmer dissolved molecules have the energy to break away and leave. Henry's law also applies, so the partial pressure of oxygen in the air above the pond sets the ceiling.
Part two · General Chemistry 2
Chemistry · Chapter 12

Kinetics

Some reactions are over before you can blink. Others creep along for years without seeming to move. Kinetics is the study of how fast a reaction runs, and why. A balanced equation tells you what you get. It says nothing about how long you wait. Here you will learn to measure a rate, write a rate law, and pull a rate constant out of real data. You will see why a small jump in temperature can double a rate. You will see how a catalyst opens a shortcut. By the end you can look at a reaction and predict what will speed it up.

By the end of this chapter you can

  • State the rate of a reaction from the change in any one species.
  • Build a rate law from initial-rate data and name the order.
  • Use the integrated rate laws to find a concentration at any time.
  • Pick the order by seeing which of three plots comes out straight.
  • Find a half-life for zero, first and second order, and say when it shifts.
  • Explain a rate with collisions, energy and the angle of the hit.
  • Get Ea from two rate constants, and judge a mechanism against the measured rate law.

Section 1Reaction rates

What a rate measures

A rate is a change divided by a time. For a car that is miles per hour. For a flask it is moles per liter each second. A reaction rate tracks how fast concentration shifts. Reactant concentration falls, and product concentration climbs at the same time. We measure concentration in molarity, so most rates here carry units of M/s.

Watch one species and you can state two things. The rate of disappearance of a reactant is how fast it gets used up. The rate of appearance of a product is how fast it builds. Both of them are quoted as positive numbers by habit. So a minus sign goes in front of the reactant change.

rate of loss of A = −Δ[A] / Δt

One rate for the whole reaction

Species in the same flask do not change at the same speed. Take 2 N2O5 → 4 NO2 + O2. Here NO2 shows up four times as fast as O2. That makes a single rate awkward to report. So we divide each change by its own coefficient. The result is one shared number, the rate of reaction.

rate = −(1/2) Δ[N2O5]/Δt = (1/4) Δ[NO2]/Δt = Δ[O2]/Δt

Now every species gives the same answer. Report that one number and nobody has to ask which species you watched.

Average, instantaneous, initial

Plot concentration against time and you get a curve, not a line. Its slope keeps changing from one moment to the next. An average rate uses two points far apart. It is the slope of the straight line joining them. An instantaneous rate is the slope of the tangent at one moment. The initial rate is the instantaneous rate right at the start.

Why does the curve bend? It bends because the reactants are running low. Fewer particles means fewer useful crashes each second. So nearly every reaction fades as it goes. The initial rate is the one we trust most. At that instant no product has built up yet.

Reading a rate off a concentration curve0.00.20.40.60.81.00246810concentration (M)time (s)tangent at t = 0initial rate = 0.250 M/sstraight line, 2 s to 8 saverage rate = 0.0785 M/sproductreactant
Reading a rate off the curve. A single reactant turns into a single product. The dashed tangent at time zero has a slope of 0.250 M/s, which is the initial rate. The dotted line joining the 2 s and 8 s points has a slope of 0.0785 M/s, the average rate over that stretch. The curve bends because the reactant is running out.

Getting the numbers in the lab

You need a signal that follows concentration. Color is a handy one for any species that soaks up visible light. A colored species soaks up light, so a spectrometer reads it right through the glass. A gas reaction can be watched by pressure instead. An acid or base reaction can be watched with a pH meter. For a slow reaction you can pull small samples, chill them to stop the change, then titrate later.

Worked example · one rate, three species

In a flask, [N2O5] falls from 0.0200 M to 0.0160 M in 100 s. The reaction is 2 N2O5 → 4 NO2 + O2. Find the rate of reaction. Then find how fast NO2 and O2 appear.

Step 1 · the change in N2O5

Δ[N2O5] = 0.0160 M − 0.0200 M = −0.0040 M

Step 2 · rate of loss

−(−0.0040 M) / 100 s = 4.0 × 10−5 M/s

Step 3 · divide by the coefficient

N2O5 has a coefficient of 2. So rate = (1/2)(4.0 × 10−5 M/s) = 2.0 × 10−5 M/s.

Step 4 · scale back up for the products

NO2 has a coefficient of 4, so it appears four times faster than the reaction rate. Δ[NO2]/Δt = 4(2.0 × 10−5) = 8.0 × 10−5 M/s. For O2 the coefficient is 1, so it appears at 2.0 × 10−5 M/s.

Check it a second way

The balanced equation makes four NO2 for every one O2. Our answers agree. 8.0 × 10−5 is exactly four times 2.0 × 10−5.

Watch the convention. A rate constant only means something once you say which rate it belongs to. The rate of loss of N2O5 and the rate of the reaction differ by the coefficient 2, so the two constants differ by 2 as well. Good data tables always say which one they mean, and so should you.
Key points
  • A rate is a concentration change divided by a time. The usual units are M/s.
  • Divide each change by its coefficient and every species gives the same rate.
  • An average rate spans an interval. An instantaneous rate is the tangent slope at one instant.
  • The initial rate is the cleanest number, because no product is in the way yet.
  • Rates nearly always drop as the reactants thin out.

Section 2Rate laws and reaction order

Writing a rate law

Concentration is the lever that drives the rate. A rate law is the equation that says exactly how. For a reaction between A and B it takes this shape.

rate = k[A]m[B]n

The letter k is the rate constant. It does not shift when you change a concentration. It does shift when you change the temperature. The exponents m and n are the reaction orders. Add them up and you get the overall order.

Here is the trap that catches almost everyone once. Those exponents are not the coefficients from the balanced equation. You cannot read them off the page. They come from experiment, every single time. Now and then they happen to match the coefficients. Often they do not match at all. An order can even be zero, a fraction, or below zero.

What each order feels like

Zero order in A means [A] has no grip on the rate at all. Double it and nothing happens. First order in A means the rate tracks [A] straight across. Double it and the rate doubles. Second order in A means the rate follows the square. Double it and the rate goes up four times. Triple it and the rate goes up nine times.

What each reaction order does to the rate13579relative rate11[A]12[A]13[A]zero order11[A]22[A]33[A]first order11[A]42[A]93[A]second orderconcentration of A, as a multiple of the first run
What the order does. Each group shows the rate when [A] is raised to one, two or three times its first value. Zero order never moves. First order follows the concentration. Second order follows its square, so three times the concentration gives nine times the rate.
A reaction order is not a count of molecules. An order of 1.5 is perfectly real, and so is an order of −1. A negative order means the species slows the reaction down, usually because it ties up a catalyst or pushes an early step backward. Fractional orders are a hint that free radicals or a surface are involved.

The units of k are a useful check. They depend on the overall order. A zero-order k comes out in M/s. A first-order k is in s−1. A second-order k is in M−1s−1. A third-order k is in M−2s−1. The pattern behind those units is simple. Each step up in order strips one M off the top.

The method of initial rates

This is the standard way to find the orders. You run the reaction several times at the same temperature. Start each run with a different mix. Measure the initial rate of each one. Then compare two runs that differ in just one reactant. Any rate change you see belongs to that reactant alone.

The math behind it is nothing more than a ratio. Say you multiply [A] by f and the rate goes up by a factor r. Then fm = r. Most of the time you can finish that in your head. If [A] doubles and the rate quadruples, m is 2.

Initial-rate runs for 2 NO + O2 → 2 NO2, all at the same temperature.
Run[NO]0 (M)[O2]0 (M)Initial rate (M/s)
10.01000.01002.00 × 10−3
20.02000.01008.00 × 10−3
30.01000.03006.00 × 10−3

Worked example · order from three runs

Use the three runs in the table. Find the rate law and the value of k.

Step 1 · find the order in NO

Runs 1 and 2 hold [O2] fixed. [NO] doubles. The rate climbs from 2.00 × 10−3 to 8.00 × 10−3 M/s. That is a factor of 4. So 2m = 4, which makes m = 2.

Step 2 · find the order in O2

Runs 1 and 3 hold [NO] fixed. [O2] triples. The rate climbs from 2.00 × 10−3 to 6.00 × 10−3 M/s. That is a factor of 3. So 3n = 3, which makes n = 1.

Step 3 · write the rate law

rate = k[NO]2[O2]. The overall order is 2 + 1 = 3.

Step 4 · solve for k with any run

Use run 1. k = rate / ([NO]2[O2]) = (2.00 × 10−3 M/s) / [(0.0100 M)2(0.0100 M)]. The bottom is 1.00 × 10−6 M3. So k = 2.00 × 103 M−2s−1.

Check it with a different run

Try run 3. k[NO]2[O2] = (2.00 × 103)(1.00 × 10−4)(0.0300) = 6.00 × 10−3 M/s. The table says 6.00 × 10−3 M/s. The units also land on M/s, so the order is right.

Key points
  • A rate law links rate to concentration through k and the orders.
  • Orders come from data, never from the balanced coefficients.
  • Change one reactant at a time and the rate change gives that order.
  • Solve fm = r for the order, then plug any run back in to get k.
  • The units of k tell you the overall order, so they make a fast check.

Section 3Integrated rate laws and half-life

From a rate to a clock

A rate law gives the speed at this instant. It does not tell you where you will be in ten minutes. For that you need the integrated rate law. Calculus turns each rate law into one equation linking concentration and time. You do not have to do the calculus. You only need the three results and the shape of each one.

The three forms you need. In each one [A]0 is the starting concentration.
OrderRate lawIntegrated formStraight plotSlopeHalf-life
Zerorate = k[A] = [A]0 − kt[A] vs t−k[A]0 / 2k
Firstrate = k[A]ln[A] = ln[A]0 − ktln[A] vs t−k0.693 / k
Secondrate = k[A]21/[A] = 1/[A]0 + kt1/[A] vs t+k1 / k[A]0

Three plots, one straight line

Look at the middle column of the table. Each form is already in the shape of a straight line. That shape is the whole trick behind finding an order. Take your data and draw three graphs. Plot [A] against time. Plot ln[A] against time. Plot 1/[A] against time. Only one of the three comes out straight. That plot names the order.

Three plots of one data set, and only one is straight[A] against tcurves downtime (s)ln[A] against tstill curvedtime (s)1/[A] against ta straight linetime (s)
One data set, three plots. These points come from a second-order run with k = 0.0500 M−1s−1 and [A]0 = 0.500 M. Only 1/[A] against time falls on a line. Its slope is k, and its intercept is 1/[A]0 = 2.00 M−1.

The data drawn above is second order. Watch what happens to each of the three plots. The plain concentration plot curves downward the whole way. The log plot still curves, just more gently. Only the plot of 1/[A] lines up against a ruler. A graph of real data will not be perfect, so you compare how straight the three lines are and pick the best.

Worked example · a first-order run

A reactant A decays by a first-order path with k = 2.50 × 10−3 s−1. You start at [A]0 = 0.800 M. What is [A] after 400 s? What is the half-life?

Step 1 · build the exponent

kt = (2.50 × 10−3 s−1)(400 s) = 1.00. The seconds cancel, so the exponent has no units. It should not have any.

Step 2 · use the first-order form

ln[A] = ln(0.800) − 1.00 = −0.223 − 1.00 = −1.223

Step 3 · undo the log

[A] = e−1.223 = 0.294 M

Step 4 · the half-life

t1/2 = 0.693 / (2.50 × 10−3 s−1) = 277 s

Check it a second way

400 s is 1.44 half-lives, since 400/277 = 1.44. Half of 0.800 is 0.400, and half again is 0.200. Our 0.294 M sits between those two, just as it should. You can also note that kt = 1 always leaves 1/e of the start, and 0.800/e = 0.294 M.

Half-life

The half-life is the time it takes for half of a reactant to vanish. We write it t1/2. First order is the special case worth memorizing. Its half-life does not care how much you start with. It is always 0.693/k. That is why radioactive decay has one fixed half-life you can quote on a chart.

Zero and second order behave differently from each other. For zero order, t1/2 = [A]0/2k. Start with more and the first half takes longer. For second order, t1/2 = 1/k[A]0. Start with more and the first half goes faster. In both cases each later half-life differs from the one before.

Worked example · a second-order half-life

NO2 breaks down by 2 NO2 → 2 NO + O2. The loss of NO2 is second order in NO2, with k = 0.540 M−1s−1 for that loss. Start at 0.0100 M. Find the first half-life. Then find the time to reach 0.00250 M.

Step 1 · the first half-life

t1/2 = 1 / k[A]0 = 1 / [(0.540 M−1s−1)(0.0100 M)] = 1 / (0.00540 s−1) = 185 s

Step 2 · set up the integrated form

1/[A] − 1/[A]0 = kt. Here 1/0.00250 = 400 M−1 and 1/0.0100 = 100 M−1.

Step 3 · solve for the time

t = (400 − 100) M−1 / (0.540 M−1s−1) = 300 / 0.540 = 556 s

Check it a second way

Walk down in halves instead. The drop to 0.00500 M takes 185 s. Now restart at 0.00500 M, so the next half-life is 1/[(0.540)(0.00500)] = 370 s. Add them: 185 + 370 = 555 s, which matches 556 s apart from rounding. Notice the second half-life is twice the first. That is the mark of second order.

Key points
  • Zero order: [A] = [A]0 − kt, and a plot of [A] against time is straight.
  • First order: ln[A] = ln[A]0 − kt, and a plot of ln[A] against time is straight.
  • Second order: 1/[A] = 1/[A]0 + kt, and a plot of 1/[A] against time is straight.
  • A first-order half-life is 0.693/k and never changes during the run.
  • Zero- and second-order half-lives depend on how much you started with.

Section 4Collision theory and activation energy

Nothing happens without contact

Collision theory starts from one simple idea about contact. Particles must meet before they can react. Molecules in a gas or a solution never stop moving. They crash into each other billions of times a second. Almost every crash does nothing at all. Only a thin slice of them make new bonds.

Three things have to line up for a crash to work.

That is why rate rises with concentration. Pack more particles into the same flask and the hits come faster. It is also why a powder burns faster than a solid block.

The barrier

The activation energy, written Ea, is the least energy a crash needs to work. Picture a hill between reactants and products. At the top sits the transition state, also called the activated complex. Up there the old bonds are half broken and the new ones are half made. It survives for about one molecular wobble.

Ea is not the same thing as ΔH. The value of ΔH is the height gap from start to finish. The value of Ea is the height of the hill you must climb on the way. An exothermic reaction can still sit behind a tall hill. Paper and air release energy when they react, yet a book on a shelf does not burst into flame. It needs a match to get over the barrier.

The reverse reaction climbs the very same hill from the other side. So Ea for the reverse equals Ea forward minus ΔH.

Activation energy, the transition state and the heat of reactionpotential energyreaction progressEaforward = 50 kJ/molEareverse = 80 kJ/molΔH= −30reactantsproductstransition state
The hill between start and finish. The forward barrier here is 50 kJ/mol, and the products land 30 kJ/mol below the reactants. So the reverse barrier is 50 − (−30) = 80 kJ/mol. The dot at the peak is the transition state, which exists for about one bond vibration.

Only a few molecules make the cut

Particles do not all carry the same energy. The spread is wide, and it shifts with temperature. At any instant a small tail of the sample has more than Ea. That tail is the share that can react. Its size is given by a short expression.

fraction over the barrier = e−Ea/RT

How many molecules clear the barrier at two temperaturesactivation energynumber of moleculeskinetic energycoolerhottershaded tails: these can reactThe axis has no scale here. The shape and the size of the tail are the point.
Why heat helps so much. Both curves hold the same number of molecules. The hotter one is flatter and shifted right. The shaded area past the dashed line is the share with enough energy, and it grows far faster than the temperature does.

Look at the shaded tails in the figure. Heat the sample and the whole curve flattens and slides right. The peak drops, but the tail past the line grows a lot. A tiny change in that tail is a big change in rate.

Angle matters too

Even a hard, fast hit can fail on geometry alone. Imagine Cl striking the molecule NOCl. Hit the Cl end and the atoms swap cleanly. Hit the O end and the pieces bounce apart. Most molecules have only a small target area that works.

Worked example · how thin is the tail?

A reaction has Ea = 50.0 kJ/mol. Find the share of collisions with enough energy at 298 K. Then do it again at 308 K. Use R = 8.314 J/mol·K.

Step 1 · match the units

Ea = 50.0 kJ/mol = 5.00 × 104 J/mol, because R is in joules.

Step 2 · the exponent at 298 K

RT = (8.314 J/mol·K)(298 K) = 2478 J/mol. So Ea/RT = 50000/2478 = 20.18.

Step 3 · the fraction at 298 K

e−20.18 = 1.72 × 10−9. That is about 2 collisions in a billion.

Step 4 · now at 308 K

RT = (8.314)(308) = 2561 J/mol, so Ea/RT = 50000/2561 = 19.52. Then e−19.52 = 3.32 × 10−9.

Step 5 · compare them

(3.32 × 10−9) / (1.72 × 10−9) = 1.9. Ten degrees nearly doubled the useful fraction.

Check it a second way

The ratio should also equal e raised to the change in the exponent. That change is 20.18 − 19.52 = 0.66, and e0.66 = 1.9. The two routes agree.

Key points
  • A crash must bring enough energy and the right angle, or nothing happens.
  • Ea is the hill between reactants and products, not the energy released.
  • Ea for the reverse step equals Ea forward minus ΔH.
  • The share of molecules past the barrier is e−Ea/RT, which is often very small.
  • Raising the temperature fattens the energy tail, and that is what speeds things up.

Section 5Temperature and the Arrhenius equation

Putting temperature into the rate constant

Temperature does not change a concentration, so it must work somewhere else. It works on the rate constant k. Svante Arrhenius found the link in 1889, and it still carries his name. The Arrhenius equation ties the rate constant to Ea and to the temperature.

k = A e−Ea/RT

The piece on the right is the fraction of crashes with enough energy. The letter A is the frequency factor. It gathers two things: how often particles meet, and how often they meet facing the right way. So A is the ceiling and the exponent is the toll gate.

Two habits of this equation are worth holding on to. A bigger Ea makes k smaller, since the exponent grows more negative. A higher T makes k larger, since the exponent moves toward zero. Temperature sits in the bottom of a fraction, inside an exponent. That is why small changes in T do so much.

The straight-line version

Take the natural log of both sides and the equation straightens out.

ln k = ln A − (Ea/R)(1/T)

This is a line if you plot ln k on the vertical axis and 1/T on the horizontal. The slope is −Ea/R. The intercept is ln A. A graph like this is called an Arrhenius plot. Multiply the slope by −R and you have Ea straight from the data.

Often you only have two measurements. Write the line twice and subtract, and the intercept drops out.

ln(k2/k1) = (Ea/R)(1/T1 − 1/T2)

Worked example · Ea from two rate constants

A first-order reaction has k = 1.00 × 10−3 s−1 at 300 K. At 320 K it has k = 8.00 × 10−3 s−1. Find Ea and then A.

Step 1 · the log of the ratio

k2/k1 = 8.00, and ln(8.00) = 2.079.

Step 2 · the temperature term

1/300 − 1/320 = 20/96000 = 2.083 × 10−4 K−1

Step 3 · solve for Ea

Ea = R × 2.079 / (2.083 × 10−4) = (8.314)(9981) = 8.30 × 104 J/mol = 83.0 kJ/mol

Step 4 · now get A

ln A = ln k + Ea/RT. At 300 K, RT = 2494 J/mol and Ea/RT = 83000/2494 = 33.28. So ln A = −6.908 + 33.28 = 26.37, and A = 2.84 × 1011 s−1.

Check it a second way

Put A and Ea back in at 320 K. There RT = 2660 J/mol and Ea/RT = 31.20. Then k = (2.84 × 1011)e−31.20 = 8.0 × 10−3 s−1. That matches the second measurement. Also note A carries the same units as k, which is right for a first-order reaction.

Why ten degrees matters so much

You may have heard that a rate doubles for every 10 °C. It is a rough guide, not a law. It holds near room temperature for an Ea of about 50 kJ/mol. With a bigger barrier the same 10 °C does more than double the rate. With a small barrier it does less. This is also why a refrigerator works. Cooling food from 25 °C to 4 °C does not stop spoilage reactions. It just slows them by a large factor.

Key points
  • k = A e−Ea/RT, so k climbs with T and falls as Ea grows.
  • A holds the collision rate and the share of hits aimed the right way.
  • A plot of ln k against 1/T is a line of slope −Ea/R.
  • Two values of k at two temperatures are enough to get Ea.
  • Keep Ea in joules per mole when R is 8.314 J/mol·K.

Section 6Mechanisms and catalysis

Reactions happen in steps

A balanced equation is a summary, not a story. Very few reactions happen in one shot. A reaction mechanism is the real list of events from start to finish. Each event in it is an elementary step, meaning one real collision or one real break.

We sort steps by how many particles take part. A unimolecular step has one particle falling apart. A bimolecular step has two particles meeting. A termolecular step needs three at once, which is so unlikely that most mechanisms avoid it.

Here is the payoff. For an elementary step, and only there, the rate law does come from the coefficients. A step A + B → C has rate = k[A][B]. A step 2 A → D has rate = k[A]2.

Species that appear in the middle and are gone by the end are reaction intermediates. They are made in one step, used in a later one, and absent from the overall equation.

The slow step sets the pace

Steps in a mechanism rarely run at the same speed. The slowest one is the rate-determining step. Think of a line of shoppers that must pass one slow register. The whole line moves at the pace of that bottleneck.

So the rate law of the slow step is usually the rate law of the reaction. That gives you a test. Derive the law the mechanism predicts and compare it with the measured one. If they disagree, the mechanism is wrong. If they agree, it is allowed but not proven.

Two steps, one slow, and the shortcut a catalyst openspotential energyreaction progresstallest hillstep 1 sets the paceintermediatereactantsproductscatalyzed path: lower hills, same products
Two steps, and a shortcut. The solid line is a mechanism with two elementary steps. The first hill is taller, so step 1 is the rate-determining step. The dip between the hills is the intermediate. The dashed line is the same reaction on a catalyst: lower hills, identical start and finish.

Worked example · test a mechanism, then add a catalyst

At moderate temperature, NO2 + CO → NO + CO2 is second order in NO2 and zero order in CO. Test this idea. Step 1, slow: 2 NO2 → NO3 + NO. Step 2, fast: NO3 + CO → NO2 + CO2.

Step 1 · do the steps add up?

Add and cancel. One NO2 and one NO3 drop out, leaving NO2 + CO → NO + CO2.

Step 2 · name the intermediate

NO3 is made in step 1 and used in step 2, so it is the intermediate.

Step 3 · derive the rate law

Step 1 sets the rate, and it is bimolecular in NO2. So rate = k1[NO2]2.

Step 4 · compare with the data

That law is second order in NO2 and has no CO in it, so it is zero order in CO. The mechanism passes.

Step 5 · what a catalyst would do

Say a catalyst drops Ea from 83.0 to 55.0 kJ/mol at 300 K. The speed-up is e raised to ΔEa/RT = 28000/2494 = 11.23. So e11.23 = 7.5 × 104, about 75,000 times faster.

Check it a second way

Near 300 K each 5.74 kJ/mol removed buys a factor of 10. We removed 28.0 kJ/mol, or 4.88 such chunks, so 104.88 = 7.6 × 104. That agrees.

Matching rate laws is evidence, not proof. Two different mechanisms can predict the same rate law, so kinetics alone can never settle the question. Chemists add other tests: trapping an intermediate, labeling an atom with a heavier isotope, or watching the reaction with a laser pulse short enough to catch the transition state.

Catalysts

A catalyst speeds a reaction up without being used up. It takes part in one step and comes back out in a later one. So it never appears in the overall equation, even though the reaction needs it.

A catalyst works by opening a path with a lower Ea. Look again at the dashed line in the figure above. Same start, same finish, shorter hills. But it leaves ΔH and the position of equilibrium exactly where they were, because it speeds both directions by the same factor.

In homogeneous catalysis the catalyst shares the phase of the reactants. Chlorine atoms wrecking ozone are a gas-phase case. First Cl + O3 → ClO + O2, then ClO + O → Cl + O2. The two add up to O3 + O → 2 O2, and the Cl atom is back to do it again.

In heterogeneous catalysis the catalyst is a different phase, usually a solid. Gases stick to its surface, react there, and leave. A car's catalytic converter is the everyday case. Platinum and rhodium turn CO and NO into CO2 and N2.

Enzymes are the catalysts your body runs on. Each one is a protein with a pocket called the active site. The reacting molecule settles in at the right angle, with the doomed bond already strained.

Key points
  • A mechanism is a list of elementary steps that add up to the overall equation.
  • For an elementary step alone, the orders match the coefficients.
  • The slowest step sets the pace, so its rate law is usually the measured one.
  • A catalyst lowers Ea, is remade, and does not shift ΔH or the equilibrium.

The chapter in one breath

  • A rate is a change over a time. Divide each species change by its coefficient and the whole reaction gets one number.
  • Rate laws come from data. The orders in rate = k[A]m[B]n are measured, not copied from the coefficients.
  • Initial rates find the orders. Change one reactant, see how the rate answers, and solve fm = r.
  • One of three plots is straight. [A] means zero order, ln[A] means first order, 1/[A] means second order.
  • Half-life depends on order. First order gives 0.693/k every time, while the other two shift with [A]0.
  • Collisions need energy and aim. Only the share e−Ea/RT of them brings enough energy to clear the hill.
  • Temperature works on k. A plot of ln k against 1/T is a line, and its slope is −Ea/R.
  • The slow step rules. The rate-determining step usually hands you the rate law, which is how a mechanism gets tested.
  • Catalysts lower the hill. They are remade each cycle and they leave ΔH and the equilibrium untouched.

ReferenceWords worth knowing

These are the words this chapter leans on. Come back to the list when a term stops feeling solid.

activated complex
The arrangement of atoms at the top of the energy hill, where old bonds are partly broken and new ones partly formed. Also called the transition state.
activation energy (Ea)
The least energy a collision must carry for the reaction to happen. It is the height of the hill, measured up from the reactants.
Arrhenius equation
k = A e−Ea/RT. It links the rate constant to the barrier and the temperature.
bimolecular step
An elementary step in which two particles collide. Its rate is first order in each of them.
catalyst
A substance that speeds a reaction by offering a lower-barrier path, and is remade by the end. Enzymes are the catalysts in living cells.
collision theory
The picture of reaction rates built on particles meeting, with enough energy and in the right orientation.
elementary step
A single molecular event in a mechanism. Its rate law can be read straight from its coefficients.
frequency factor (A)
The front term in the Arrhenius equation. It bundles how often particles collide with how often they are aimed correctly.
half-life (t1/2)
The time for half of a reactant to be used up. It is constant only for a first-order reaction.
heterogeneous catalysis
Catalysis in which the catalyst is a different phase from the reactants, usually a solid surface holding gases or liquids.
homogeneous catalysis
Catalysis in which the catalyst is mixed into the same phase as the reactants.
initial rate
The instantaneous rate at the moment the reaction starts, before any product has built up.
instantaneous rate
The rate at one moment, equal to the slope of the tangent to the concentration curve. An average rate instead uses two points spread apart.
integrated rate law
The equation that gives concentration as a function of time. Each order has its own form and its own straight-line plot.
rate constant (k)
The number in front of a rate law. It does not depend on concentration, but it does depend on temperature.
rate-determining step
The slowest elementary step in a mechanism. It sets the pace, so its rate law is usually the overall rate law.
rate law
An equation giving rate in terms of concentrations raised to measured powers.
reaction intermediate
A species made in one step and used up in a later one. It is missing from the overall balanced equation.
reaction mechanism
The ordered set of elementary steps that add up to the overall reaction.
reaction order
The power on a concentration in the rate law. Add the powers and you get the overall order.

Check yourself

1. For 2 N2O5 → 4 NO2 + O2, O2 appears at 3.0 × 10−5 M/s. How fast does N2O5 disappear, and how fast does NO2 appear?
N2O5: 6.0 × 10−5 M/s. NO2: 1.2 × 10−4 M/s. O2 has a coefficient of 1, so the rate of reaction is 3.0 × 10−5 M/s. Multiply by 2 for N2O5: 2(3.0 × 10−5) = 6.0 × 10−5 M/s. Multiply by 4 for NO2: 4(3.0 × 10−5) = 1.2 × 10−4 M/s.
2. In two runs [B] is held fixed. When [A] doubles, the rate goes up eight times. What is the order in A?
Third order in A. Set 2m = 8. Since 23 = 8, m = 3. Nothing about B can be learned from this pair, because [B] never changed.
3. A first-order reaction has a half-life of 120 s. Find k, then find the fraction left after 360 s.
k = 5.78 × 10−3 s−1, and 12.5% is left. k = 0.693/120 s = 5.78 × 10−3 s−1. Then 360 s is exactly three half-lives, since 360/120 = 3. Half three times is (1/2)3 = 1/8, which is 0.125 or 12.5%.
4. A second-order reaction has k = 0.250 M−1s−1 and starts at 0.200 M. What is [A] after 100 s?
[A] = 0.0333 M. Use 1/[A] = 1/[A]0 + kt. Here 1/0.200 = 5.00 M−1 and kt = (0.250)(100) = 25.0 M−1. Adding gives 30.0 M−1, so [A] = 1/30.0 = 0.0333 M.
5. A rate constant doubles when the temperature rises from 300 K to 310 K. Find Ea.
About 53.6 kJ/mol. ln(2) = 0.693. Also 1/300 − 1/310 = 10/93000 = 1.075 × 10−4 K−1. So Ea = (8.314)(0.693)/(1.075 × 10−4) = 5.36 × 104 J/mol. This is the barrier size behind the old rule of thumb about 10 degrees.
6. A rate constant is reported as 0.45 M−2s−1. What is the overall order?
Third order overall. The units of k are M1−orders−1. Setting 1 − order = −2 gives order = 3. A quick check: rate = k[A]3 has units of (M−2s−1)(M3) = M/s, which is right.
7. Which plot comes out straight for data that follow rate = k[A], and what does its slope give you?
ln[A] against time, with slope −k. The first-order integrated form is ln[A] = ln[A]0 − kt. That is a line in ln[A] and t. A straight plot of [A] itself would mean zero order, and a straight plot of 1/[A] would mean second order.
8. A reaction has Ea = 40.0 kJ/mol. What fraction of collisions carries enough energy at 300 K?
About 1.1 × 10−7. RT = (8.314)(300) = 2494 J/mol. Then Ea/RT = 40000/2494 = 16.04, and e−16.04 = 1.1 × 10−7. So about one collision in ten million has the energy needed.
9. Forward Ea is 75 kJ/mol and ΔH is −40 kJ/mol. What is Ea for the reverse reaction?
115 kJ/mol. The reverse climbs the same peak, but from lower down. Ea(reverse) = Ea(forward) − ΔH = 75 − (−40) = 115 kJ/mol. The reverse barrier is taller because the products sit 40 kJ/mol below the reactants.
10. A catalyst lowers Ea by 20.0 kJ/mol at 298 K. By what factor does k grow?
About 3.2 × 103. The ratio is e raised to ΔEa/RT. Here RT = (8.314)(298) = 2478 J/mol, so the exponent is 20000/2478 = 8.07. Then e8.07 = 3.2 × 103, since A is taken as unchanged.
11. A mechanism is step 1 slow, 2 A → C, then step 2 fast, C + B → D. Predict the rate law and name the intermediate.
rate = k1[A]2, and C is the intermediate. The slow step is rate determining, and it is bimolecular in A, so its rate law is k1[A]2. B is absent from the rate law, so the reaction is zero order in B. C is made in step 1 and used in step 2, so it is the intermediate.
12. A catalyst doubles the forward rate of a reaction at equilibrium. What happens to the reverse rate and to K?
The reverse rate also doubles, and K does not change. A catalyst lowers the same barrier for both directions, so both rate constants grow by the same factor. Since K is the ratio of those constants, the ratio holds steady. A catalyst gets you to equilibrium sooner, not to a different equilibrium.
Chemistry · Chapter 13

Fundamental Equilibrium Concepts

Some reactions never finish. They slow down, stop changing, and sit there holding both reactants and products at once. Nothing has stalled or died. The forward reaction and the reverse reaction are both still running, at the same speed. That balance is called equilibrium, and it rules a huge slice of chemistry. It sets how much ammonia a factory can make. It sets how much oxygen your blood can carry. In this chapter you will write the equilibrium constant, compare it with the reaction quotient, and predict how a mixture answers back when you push on it.

By the end of this chapter you can

  • Explain why an equilibrium is busy, not dead, and what has to stay constant.
  • Write Kc and Kp from a balanced equation, and leave out what does not belong.
  • Convert between Kc and Kp using Δn.
  • Find Q for any mixture and say which way the reaction must go.
  • Predict the shift when you add, remove, squeeze, heat, or add a catalyst.
  • Set up an ICE table and solve for every equilibrium amount.
  • Use the small-x shortcut, then test whether it was allowed.

Section 1Dynamic equilibrium

Two reactions in one tube

Seal a little N2O4 gas in a glass tube. The gas starts out colorless. Within seconds the tube turns brown. The brown gas is NO2. Then the color stops changing. It holds that same shade for as long as you watch.

The tube did not run out of material. Crack it open later and you still find both gases. What stopped was the net change. This is a reversible reaction, one that can run in either direction. We write it with a double arrow.

N2O4(g) ⇌ 2 NO2(g)

At the very start there is no NO2 at all. Only the forward reaction can happen, and it runs fast. As NO2 piles up, the reverse reaction switches on. Its rate climbs while the forward rate falls. At some moment the two rates match. From then on, every N2O4 that splits apart is replaced by two NO2 joining back up.

Forward and reverse rates meeting at equilibrium time → rate forward rate reverse rate the rates are now equal amounts hold steady from here on only N₂O₄ can react at first splitting apart joining back up
Equilibrium is where the two curves meet. The forward rate drops as reactant is used up. The reverse rate climbs as product builds. Once they are equal, the measured amounts stop moving.

Equal rates, not equal amounts

Dynamic equilibrium is that state. The two rates are equal, so the amounts stay put. Dynamic means moving. Molecules keep reacting both ways the whole time. Only the totals hold still. Many people picture a dead flask. Picture a busy escalator instead, with as many riders stepping on as stepping off.

Equal rates do not mean equal amounts. One side can hold far more material than the other. A rate depends on the amounts and on the rate constant. A sluggish reverse step needs a big pile of product to keep pace with a brisk forward step.

Equilibrium also needs a closed system. If the NO2 can leak away, the reverse rate never catches up, and the reaction keeps creeping forward. One more thing is worth knowing. You reach the same resting mixture from either side. Start with pure NO2 instead, hold the same temperature, and you land in the same place.

Worked example 1 · Rates that match

Hydrogen and iodine form hydrogen iodide: H2(g) + I2(g) ⇌ 2 HI(g). Treat each direction here as a single step, so rate = k × the concentrations. The forward constant is kf = 1.0 × 10-2 L/mol·s. The reverse constant is kr = 2.0 × 10-4 L/mol·s. At equilibrium [H2] = [I2] = 0.0100 M. Show that the two rates really are equal.

Step 1 · find the ratio of the constantsSetting the rates equal gives kf[H2][I2] = kr[HI]2. Rearranged, [HI]2 ÷ ([H2][I2]) = kf ÷ kr. That ratio is (1.0 × 10-2) ÷ (2.0 × 10-4) = 50. The units L/mol·s cancel, so the 50 is a bare number.

Step 2 · get the amount of product[HI]2 = 50 × (0.0100)(0.0100) = 5.0 × 10-3 M2. Take the square root: [HI] = 0.0707 M.

Step 3 · check both ratesForward rate = (1.0 × 10-2 L/mol·s)(0.0100 M)(0.0100 M) = 1.0 × 10-6 mol/L·s. Reverse rate = (2.0 × 10-4 L/mol·s)(0.0707 M)2 = (2.0 × 10-4)(5.0 × 10-3) = 1.0 × 10-6 mol/L·s.

AnswerThe rates match at 1.0 × 10-6 mol/L·s, even though HI is seven times more plentiful than either starting gas. That ratio of 50 is the equilibrium constant, and the next section gives it a name.

Key points
  • A reversible reaction runs both ways, so it settles instead of finishing.
  • At equilibrium the forward and reverse rates are equal, not zero.
  • Measured amounts stop changing, but molecules keep reacting.
  • Equal rates say nothing about equal amounts.
  • The system must be closed, and both starting points lead to the same mixture.

Section 2The equilibrium constant

Products over reactants

Equilibrium is more than a yes or no. It has a number attached. That number is the equilibrium constant, written K. For the general reaction

a A + b B ⇌ c C + d D

the constant is one fraction. Products go on top. Reactants go on the bottom. Each concentration is raised to its own coefficient from the balanced equation.

Kc = [C]c[D]d ÷ [A]a[B]b

This rule is the law of mass action. The small c in Kc means concentrations in mol/L. Square brackets always mean molarity. K values are reported with no units at all. They really come from activities, which are ratios against a standard state, so every unit cancels.

K depends on temperature and on nothing else. Change the amounts and the mixture shifts, but K holds. Heat the flask and K itself takes a new value.

What stays out of the expression

Pure solids and pure liquids are left out. Their concentration cannot change as the reaction runs. A chunk of solid has the same packing whether the chunk is large or small. So for the roasting of limestone,

CaCO3(s) ⇌ CaO(s) + CO2(g)   with   Kp = PCO2

only the gas appears. A reaction with more than one phase like this one is a heterogeneous equilibrium. When everything is in the same phase it is a homogeneous equilibrium. Water used as the solvent drops out for the same reason: it is there in such excess that its amount barely budges.

Kp, and the two tricks

Gases are easier to measure by pressure than by concentration. Kp is the same fraction built from partial pressures in atm. The two constants are tied together.

Kp = Kc(RT)Δn

Here R = 0.08206 L·atm/mol·K and T is in kelvin. The Δn is the moles of gas on the right minus the moles of gas on the left. When Δn = 0 the factor is 1, so the two constants are the same number.

Two small tricks come up again and again. Flip an equation and K becomes 1 ÷ K. Multiply an equation through by n and K becomes Kn. Add two equations together and you multiply their constants.

What the size of K tells you 10-10 10-5 10-3 1 103 105 1010 mostly reactants left barely any product forms both sides matter you can measure all of them mostly products made only traces of reactant left N₂O₄ gives K = 0.0046 at 25 °C, so it sits just left of the middle. H₂ plus I₂ gives K = 50, a little right of the middle.
The size of K is the whole story in one glance. The scale runs by powers of ten. A K below about 10-3 leaves the reactants nearly untouched. A K above about 103 all but uses them up.

Worked example 2 · From Kc to Kp

For N2O4(g) ⇌ 2 NO2(g), Kc = 4.6 × 10-3 at 25 °C. Find Kp. Then do the same for N2(g) + 3 H2(g) ⇌ 2 NH3(g), where a run at 500 K gives Kc = 0.0100.

Step 1 · count the gas molesFor the first one, 2 moles of gas on the right and 1 on the left. So Δn = 2 - 1 = +1.

Step 2 · build RTT = 25 + 273.15 = 298.15 K. RT = (0.08206 L·atm/mol·K)(298.15 K) = 24.47 L·atm/mol.

Step 3 · raise and multiplyKp = (4.6 × 10-3)(24.47)1 = 0.113.

Step 4 · now the ammonia caseGas moles: 2 on the right, 1 + 3 = 4 on the left, so Δn = 2 - 4 = -2. RT = (0.08206)(500) = 41.03 L·atm/mol, and (41.03)2 = 1683. A negative exponent means divide: Kp = 0.0100 ÷ 1683 = 5.94 × 10-6.

AnswerKp = 0.113 for the first reaction and 5.94 × 10-6 for the second. A negative Δn pushes Kp far below Kc.

Watch the sign of Δn. This sign trips people up more than anything else in the chapter. Count the gas moles on each side first, write Δn down, and only then reach for RT.
Key points
  • Kc is products over reactants, each raised to its coefficient.
  • Pure solids, pure liquids, and solvent water are left out.
  • Kp = Kc(RT)Δn, with Δn counted for gases only.
  • A big K means the products win; a small K means the reactants stay.
  • Flipping the equation inverts K, and scaling it by n raises K to the n.
  • Only a change in temperature changes the value of K.

Section 3The reaction quotient Q

The same fraction, any moment you like

K only describes the resting state. Most mixtures you meet are not resting. The reaction quotient Q fixes that. It uses the exact same fraction as K, but you plug in the amounts you have right now.

Q = [C]c[D]d ÷ [A]a[B]b   at any instant

Then you compare the two numbers. The comparison is the whole point, and it has only three outcomes.

If Q < K, the fraction is too small. There is not enough product on top. The reaction runs forward, to the right, making more product until Q climbs up to K.

If Q > K, the fraction is too large. There is too much product. The reaction runs in reverse, to the left, rebuilding reactant until Q drops to K.

If Q = K, the mixture is already at equilibrium. Nothing shifts. The net change is zero.

Two edge cases are easy once you see them. With no product at all, Q = 0, which is below any K, so the reaction must go forward. With no reactant at all, the bottom is zero, Q blows up, and the reaction must go in reverse.

Three ways Q can compare with K K Q Q < K goes right too little product, so the forward reaction wins for a while K Q sits on K Q = K no shift already at equilibrium, so both rates are matched K Q Q > K goes left too much product, so the reverse reaction wins for a while
Q always moves toward K, never away from it. The gold dot is where your mixture sits. The teal arrow is the direction of net change. The dot slides until it lands on the tick mark for K.

Why this beats guessing

Without Q you would have to guess. A flask holding plenty of product might still need to make more. Q settles it with one calculation. That is why Q comes before every ICE table in the next sections. It tells you the sign of the change.

You will meet Q again in later chapters. The same test decides whether a solid will drop out of a solution. There you compare Q with Ksp. The logic never changes, only the names.

Worked example 3 · Which way does it go?

For H2(g) + I2(g) ⇌ 2 HI(g), K = 50.0 at 731 K. A sealed flask at 731 K holds [H2] = 0.200 M, [I2] = 0.200 M, and [HI] = 0.500 M. Which way does the reaction move?

Step 1 · write Q in the same shape as KQ = [HI]2 ÷ ([H2][I2]).

Step 2 · put the numbers inTop: (0.500)2 = 0.250. Bottom: (0.200)(0.200) = 0.0400.

Step 3 · divideQ = 0.250 ÷ 0.0400 = 6.25.

Step 4 · compareQ = 6.25 and K = 50.0, so Q < K.

AnswerThe reaction runs forward. H2 and I2 keep being used up, and [HI] keeps rising, until Q reaches 50.0.

A second lookNow try [HI] = 2.00 M with [H2] = [I2] = 0.100 M. Q = (2.00)2 ÷ (0.100)(0.100) = 4.00 ÷ 0.0100 = 400. Here Q > K, so this flask runs in reverse and breaks HI back apart.

Key points
  • Q is built exactly like K but from the amounts present right now.
  • Q < K sends the reaction to the right, toward more product.
  • Q > K sends the reaction to the left, toward more reactant.
  • Q = K means the mixture is already at rest.
  • Q drifts toward K and then stops there.

Section 4Le Chatelier's principle

Push on it and it pushes back

A mixture at equilibrium is not fragile. Disturb it and it rearranges itself. Le Chatelier's principle says it this way. When you put a stress on a system at equilibrium, the system shifts in the direction that partly relieves that stress.

Partly is the honest word. It moves only far enough to make Q equal K again, and then it stops. Each kind of stress has its own rule, and all four rules come from the same Q against K test.

Adding and removing

Add more of any species and the system shifts away from it. Take some away and the system shifts toward it. Check it with Q. Add product and the top of the fraction grows, so Q > K, so the reaction runs in reverse.

Factories use this on purpose. In the ammonia process the NH3 is condensed out as fast as it forms. Removing product keeps Q below K forever, so the reaction keeps running forward.

Squeezing and stretching

Changing the volume only matters when the gas moles differ across the arrow. Squeeze a gas mixture into half the space and every concentration doubles at once. Whether Q rises or falls then depends on Δn.

The rule is short. Squeezing shifts the reaction toward the side with fewer moles of gas. Expanding shifts it toward the side with more. If the gas moles match on both sides, nothing shifts at all.

One classic trap. Pumping in an unreactive gas like argon at fixed volume changes the total pressure but changes no concentration. Q does not move, so neither does the equilibrium.

Heating and cooling

Temperature is the only stress that changes K itself. The trick is to treat heat as if it were a species in the equation. For an endothermic forward reaction, heat belongs on the left.

heat + N2O4(g) ⇌ 2 NO2(g)    ΔH = +57.2 kJ/mol

Now the rule for adding a species does the work. Heating adds heat to the left, so the reaction shifts right and K rises. Cooling removes heat from the left, so it shifts back and K falls. That is why a sealed tube of these gases darkens in hot water and pales in ice water. For an exothermic reaction heat sits on the right, and every arrow flips.

Catalysts do not count

A catalyst speeds the forward step and the reverse step by the same factor. The rates both rise, so they stay matched. K is untouched and no shift happens. What you gain is time, since a slow reaction may now settle in minutes.

Squeezing the N2O4 and NO2 equilibrium 1. at equilibrium 1.00 L, Q = K = 0.0046 N₂O₄ = 0.0500 M NO₂ = 0.0152 M halve the volume 2. the instant after 0.500 L, Q = 0.0092 every amount doubled Q is twice K, so it goes left shifts 3. settled again 0.500 L, Q = K = 0.0046 N₂O₄ = 0.104 M NO₂ = 0.0219 M, down from 0.0303 N₂O₄ NO₂
Squeezing shifts this one to the left. Big teal circles stand for N2O4, small gold ones for NO2. Two NO2 become one N2O4, so the crowding eases. The brown is paler than in panel 2, though still deeper than in panel 1.

Worked example 4 · Squeeze it and track Q

A 1.00 L flask at 25 °C holds N2O4 and NO2 at equilibrium, with Kc = 4.6 × 10-3. The N2O4 is 0.0500 M. Find [NO2]. Then push the piston to 0.500 L and show which way the reaction shifts.

Step 1 · get the starting [NO2]Kc = [NO2]2 ÷ [N2O4], so [NO2]2 = (4.6 × 10-3)(0.0500) = 2.30 × 10-4 M2. The square root gives [NO2] = 0.0152 M.

Step 2 · halve the volumeSame moles in half the space, so both concentrations double. [N2O4] = 0.1000 M and [NO2] = 0.0303 M.

Step 3 · find Q right after the squeezeQ = (0.0303)2 ÷ 0.1000 = (9.20 × 10-4) ÷ 0.1000 = 9.2 × 10-3.

Step 4 · compare with KQ = 9.2 × 10-3 is exactly twice Kc = 4.6 × 10-3. So Q > K.

AnswerThe reaction shifts left, toward N2O4. That is the side with 1 mole of gas instead of 2, so the shift lowers the crowding. Notice why Q doubled: Δn = +1, so halving the volume multiplies Q by 21.

Le Chatelier is a shortcut, not a separate law. Every one of these rules is just the Q against K test in a hurry. When a case confuses you, compute Q and trust the number instead.
Key points
  • A stressed system shifts in the direction that partly relieves the stress.
  • Adding a species shifts the reaction away from it; removing one pulls it back.
  • Squeezing shifts toward the side with fewer moles of gas.
  • An unreactive gas added at fixed volume changes nothing.
  • Heating shifts an endothermic reaction forward and raises K.
  • A catalyst changes the speed, never the position, of equilibrium.

Section 5ICE tables

Three rows that organize everything

Most equilibrium problems give you starting amounts and K, then ask for the final amounts. An ICE table keeps that bookkeeping straight. The letters stand for Initial, Change, and Equilibrium. You get one column per species and three rows.

The Initial row holds what you mixed, before any reaction. The Change row holds how much each one moves. The Equilibrium row is just the first two rows added together.

The Change row is where the thinking happens. You call the extent of reaction x. Every species changes by its coefficient times x. Reactants lose, so they get a minus sign. Products gain, so they get a plus sign. A coefficient of 2 means the change is 2x, not x.

How an ICE table is built N₂O₄ (g) ⇌ 2 NO₂ (g) row N₂O₄ NO₂ I nitial 0.0400 0 C hange − x + 2x E quil. 0.0400 − x 2x the 2 comes from the coefficient, and the sign from the side Then substitute the bottom row into K: K = (2x)² ÷ (0.0400 − x) = 0.0046, and you solve that for x.
The bottom row is what you feed into K. Add the Initial and Change rows to get the Equilibrium row, then put those expressions into the equilibrium constant and solve for x.

From the table to an answer

Once the Equilibrium row is written, the rest is algebra. Substitute those expressions into the K expression. You now have one equation in one unknown. Solve for x, then feed x back into the Equilibrium row to get every concentration.

Two sanity checks are worth the ten seconds. First, no concentration may come out negative. If it does, you probably picked the wrong root or the wrong direction. Second, put your answers back into K. The fraction should give the K you started with.

Sometimes both sides of the equation are perfect squares. Then you can take the square root of both sides and skip the quadratic formula. That shortcut shows up in the example below.

One warning about units. Put concentrations in the table, not moles, unless you convert at the end. If the problem gives moles and a volume, divide first. Doing it later is where errors sneak in.

Worked example 5 · A full ICE table

1.00 mol of H2 and 1.00 mol of I2 go into a 1.00 L flask at 731 K. For H2(g) + I2(g) ⇌ 2 HI(g), Kc = 50.0. Find all three concentrations at equilibrium.

Step 1 · starting concentrations1.00 mol ÷ 1.00 L = 1.00 M for each gas. There is no HI yet, so [HI] = 0.

Step 2 · the tableInitial: 1.00, 1.00, 0. Change: −x, −x, +2x. Equilibrium: 1.00 − x, 1.00 − x, 2x. Since Q starts at 0, the reaction must go forward, so x is positive.

Step 3 · substituteKc = (2x)2 ÷ [(1.00 − x)(1.00 − x)] = 50.0, which is (2x)2 ÷ (1.00 − x)2 = 50.0.

Step 4 · use the perfect squareBoth sides are squares, so take the square root. 2x ÷ (1.00 − x) = √50.0 = 7.0711.

Step 5 · solve for x2x = 7.0711 − 7.0711x, so 9.0711x = 7.0711 and x = 0.7795 M.

Step 6 · fill the bottom row[H2] = [I2] = 1.00 − 0.7795 = 0.220 M. [HI] = 2(0.7795) = 1.559 M, or 1.56 M.

Answer and check[H2] = [I2] = 0.220 M and [HI] = 1.56 M. Check: (1.559)2 ÷ (0.2205)2 = 2.431 ÷ 0.04862 = 50.0. It matches.

Key points
  • ICE stands for Initial, Change, Equilibrium, one column per species.
  • Each Change entry is the coefficient times x, minus for reactants and plus for products.
  • The Equilibrium row is the Initial row plus the Change row.
  • Feed the Equilibrium row into K to get one equation in x.
  • Reject any root that makes a concentration negative, then check your answer in K.

Section 6Equilibrium calculations

Four kinds of problem

Almost every equilibrium question is one of four types. Knowing which one you are holding saves a lot of flailing.

Type one gives you all the equilibrium concentrations and asks for K. Just plug them in. Type two gives initial amounts plus one equilibrium amount. Use the known change to find x, then fill the rest of the table. Type three gives K and the initial amounts and asks for the equilibrium amounts. That is the ICE table from the last section. Type four is type three with a tiny K, and it has a shortcut.

The small-x shortcut

When K is very small, almost nothing reacts. The reactant you started with barely drops. So the term (c0 − x) is nearly c0, and you can drop the x from that one spot. This is the small-x approximation. It turns an awkward quadratic into a square root.

You do not drop every x. Only the one being subtracted from a much larger number. The x values sitting alone in the numerator are the answer, so they stay.

Two tests tell you whether the shortcut was allowed. Before you start, compare c0 with K. If c0 ÷ K is larger than about 400, the shortcut is safe. After you solve, use the five percent rule. If x is under five percent of c0, keep the answer. If not, throw it out and solve the full quadratic.

x ÷ c0 × 100% < 5%   →   the shortcut holds

That same fraction has a name: the percent dissociation. It says what share of your starting material fell apart.

Deciding whether the small-x shortcut is allowed Work out c ÷ K start over the constant Is it above 400? if yes, the x you subtract hardly moves the answer yes Drop that one x take a square root no Keep every x solve the quadratic Last step: is x under five percent of c? if it is not, the shortcut answer has to go Both tests ask the same thing.
Two gates guard the shortcut. The ratio test runs before you solve. The five percent rule runs after. Failing either one sends you to the quadratic formula.

Worked example 6 · A tiny K, so the shortcut works

Phosgene breaks up a little at 100 °C: COCl2(g) ⇌ CO(g) + Cl2(g), with Kc = 2.2 × 10-10. You start with 0.100 M COCl2. Find every equilibrium concentration.

Step 1 · run the ratio testc0 ÷ K = 0.100 ÷ (2.2 × 10-10) = 4.5 × 108. That is far above 400, so the shortcut is fine.

Step 2 · the tableInitial: 0.100, 0, 0. Change: −x, +x, +x. Equilibrium: 0.100 − x, x, x.

Step 3 · drop the one xKc = x2 ÷ (0.100 − x) becomes x2 ÷ 0.100 = 2.2 × 10-10.

Step 4 · solvex2 = (2.2 × 10-10)(0.100) = 2.2 × 10-11 M2, so x = 4.69 × 10-6 M.

Step 5 · check the five percent rule(4.69 × 10-6) ÷ 0.100 × 100% = 0.0047%. Nowhere near 5%.

Answer[CO] = [Cl2] = 4.7 × 10-6 M, and [COCl2] = 0.100 M to three figures. Hardly any gas broke up, which is what a K of 10-10 promised.

When the shortcut breaks

The shortcut fails when K is not small compared with the starting amount. Then x is a real chunk of c0, and dropping it skews the answer badly. You go back to the full quadratic, in the form ax2 + bx + c = 0, and use the formula. Take the root that keeps every concentration positive.

x = [ −b + √(b2 − 4ac) ] ÷ 2a

Worked example 7 · When the shortcut fails

Start with 0.0100 M N2O4 at 25 °C. For N2O4(g) ⇌ 2 NO2(g), Kc = 4.6 × 10-3. Try the shortcut, see it fail, and then do it properly.

Step 1 · the ratio test warns youc0 ÷ K = 0.0100 ÷ (4.6 × 10-3) = 2.2. That is tiny, far under 400, so expect trouble.

Step 2 · try it anywayEquilibrium row: 0.0100 − x and 2x. Dropping the x gives (2x)2 ÷ 0.0100 = 4.6 × 10-3, so 4x2 = 4.6 × 10-5 and x = 3.39 × 10-3 M.

Step 3 · the five percent rule rejects it(3.39 × 10-3) ÷ 0.0100 × 100% = 34%. Way over 5%, so this answer is out.

Step 4 · set up the real quadratic(2x)2 ÷ (0.0100 − x) = 4.6 × 10-3. Cross multiply: 4x2 = 4.6 × 10-5 − (4.6 × 10-3)x. Rearranged: 4x2 + (4.6 × 10-3)x − 4.6 × 10-5 = 0.

Step 5 · use the formulab2 − 4ac = (4.6 × 10-3)2 + 4(4)(4.6 × 10-5) = 2.116 × 10-5 + 7.36 × 10-4 = 7.572 × 10-4. Its square root is 2.7517 × 10-2. Then x = (−4.6 × 10-3 + 2.7517 × 10-2) ÷ 8 = (2.2917 × 10-2) ÷ 8 = 2.86 × 10-3 M.

Step 6 · fill in and check[N2O4] = 0.0100 − 0.00286 = 7.14 × 10-3 M. [NO2] = 2(2.86 × 10-3) = 5.73 × 10-3 M. Check: (5.73 × 10-3)2 ÷ (7.14 × 10-3) = (3.28 × 10-5) ÷ (7.14 × 10-3) = 4.6 × 10-3. It matches Kc.

Answerx = 2.86 × 10-3 M, not 3.39 × 10-3 M. The shortcut was 18% high. The percent dissociation here is 29%, which is why no x could be dropped.

Do the ratio test first, every time. It costs one division and tells you which path to take before you waste a page of algebra on the wrong one.
Key points
  • Sort the problem first: find K, finish a table, or solve for x.
  • A tiny K lets you replace (c0 − x) with c0.
  • Never drop the x values that are your answer.
  • c0 ÷ K above about 400 means the shortcut is safe.
  • Then confirm with the five percent rule, and solve the quadratic if it fails.
  • Always put your answers back into K as a final check.

The chapter in one breath

  • Equilibrium is busy, not finished. The forward and reverse rates are equal, so the measured amounts hold still while molecules keep reacting.
  • K is one number for the resting state. Products over reactants, each raised to its coefficient, with pure solids and pure liquids left out.
  • Kp = Kc(RT)Δn. Count only gas moles for Δn, and use R = 0.08206 L·atm/mol·K with T in kelvin.
  • The size of K says where the reaction sits. Big K means the products win, small K means the reactants mostly stay.
  • Q is K's formula with today's amounts. Q < K goes right, Q > K goes left, Q = K sits still.
  • Le Chatelier is that same test in a hurry. Add, remove, squeeze, or heat, and the system shifts the way that partly relieves the push.
  • Only temperature changes K. Catalysts and unreactive gases change speed or pressure, never the position.
  • ICE tables carry the bookkeeping. Initial plus Change gives Equilibrium, and the Change entry is the coefficient times x.
  • Check every shortcut. Drop x only when c0 ÷ K is above about 400, then confirm with the five percent rule.

ReferenceWords worth knowing

These are the words this chapter leans on. Learn them in pairs where you can, since K and Q, or forward and reverse, only make sense together.

Catalyst
A substance that speeds both the forward and the reverse reaction by the same factor. It shortens the wait for equilibrium and leaves K and the final amounts alone.
Closed system
A container that lets no reactant or product escape. Equilibrium can only be reached in one.
Dynamic equilibrium
The state in which the forward and reverse rates are equal. Amounts stay constant while molecules keep reacting both ways.
Equilibrium constant (K)
The value of the products-over-reactants fraction once a reaction has settled. It depends only on temperature and is written without units.
Forward reaction
The change that runs left to right as the equation is written, turning reactants into products.
Heterogeneous equilibrium
An equilibrium with species in more than one phase, such as a solid and a gas. Pure solids and pure liquids are left out of K.
Homogeneous equilibrium
An equilibrium in which every species is in the same phase, such as all gases or all dissolved.
ICE table
A three-row grid of Initial, Change, and Equilibrium amounts. It organizes a problem so K gives one equation in one unknown.
Kc
An equilibrium constant built from concentrations in mol/L.
Kp
An equilibrium constant built from partial pressures in atm. It equals Kc(RT)Δn.
Law of mass action
The rule that gives the shape of K: products over reactants, each concentration raised to its coefficient.
Le Chatelier's principle
A stressed equilibrium shifts in the direction that partly relieves the stress.
Percent dissociation
The share of the starting material that reacted, found as x divided by c0 times 100%.
Reaction quotient (Q)
The same fraction as K, but filled in with the amounts present at this moment. Comparing Q with K gives the direction of change.
Reverse reaction
The change that runs right to left, turning products back into reactants.
Reversible reaction
A reaction that can run in either direction, written with a double arrow.
Shift
The net movement of a disturbed equilibrium toward one side until Q equals K again.
Small-x approximation
Replacing (c0 − x) with c0 when K is tiny. It is valid when c0 ÷ K is above about 400.
Stress
Any change forced on a system at equilibrium, such as adding a species, changing the volume, or changing the temperature.

Check yourself

1. Write the Kc expression for 2 SO2(g) + O2(g) ⇌ 2 SO3(g).
Kc = [SO3]2 ÷ ([SO2]2[O2]) Products go on top. Each exponent is that species' coefficient, and O2 has an unwritten coefficient of 1.
2. At equilibrium that same mixture holds [SO2] = 0.400 M, [O2] = 0.200 M, and [SO3] = 0.600 M. Find Kc.
Kc = 11.3 Top: (0.600)2 = 0.360. Bottom: (0.400)2(0.200) = (0.160)(0.200) = 0.0320. Divide: 0.360 ÷ 0.0320 = 11.25, which rounds to 11.3.
3. Convert that Kc of 11.25 to Kp at 1000 K.
Kp = 0.137 Gas moles: 2 on the right, 3 on the left, so Δn = −1. RT = (0.08206)(1000) = 82.06 L·atm/mol. A Δn of −1 means divide: 11.25 ÷ 82.06 = 0.137.
4. For PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 1.80 at 523 K. A flask holds [PCl5] = 0.100 M, [PCl3] = 0.200 M, and [Cl2] = 0.300 M. Which way does it shift?
Forward, to the right Q = (0.200)(0.300) ÷ 0.100 = 0.0600 ÷ 0.100 = 0.600. Since 0.600 is less than 1.80, Q < K, so more PCl5 must break apart.
5. Kc = 50.0 for H2 + I2 ⇌ 2 HI. What is Kc for 2 HI ⇌ H2 + I2 at the same temperature?
0.0200 Flipping an equation inverts its constant. 1 ÷ 50.0 = 0.0200. The same mixture is being described from the other side.
6. Start with 0.500 M HI at 731 K, where 2 HI ⇌ H2 + I2 has Kc = 0.0200. Find all three equilibrium concentrations.
[HI] = 0.390 M, [H2] = [I2] = 0.0551 M The table gives 0.500 − 2x, x, x. So x2 ÷ (0.500 − 2x)2 = 0.0200. Both sides are squares, so x ÷ (0.500 − 2x) = √0.0200 = 0.14142. Then x = 0.070711 − 0.28284x, so 1.28284x = 0.070711 and x = 0.05512 M. Finally [HI] = 0.500 − 2(0.05512) = 0.390 M.
7. A sealed piston holds N2(g) + 3 H2(g) ⇌ 2 NH3(g) at equilibrium. You push it to half its volume. Which way does the reaction shift?
To the right, toward NH3 The left side has 4 moles of gas and the right side has 2. Squeezing favors the side with fewer gas moles, so more ammonia forms.
8. You pump argon into that same rigid flask, raising the total pressure. What happens to the equilibrium?
Nothing shifts Argon does not react and the volume did not change, so no concentration or partial pressure of a reacting gas changed. Q is untouched, so there is no shift.
9. For N2O4(g) ⇌ 2 NO2(g), ΔH = +57.2 kJ/mol. You warm the tube. What happens to Kc and to the color?
Kc rises and the brown deepens A positive ΔH puts heat on the reactant side. Adding heat shifts the reaction right, making more brown NO2, and the constant itself takes a larger value.
10. Adding a catalyst to a reaction at equilibrium does what to K and to the time needed to get there?
K is unchanged, the wait gets shorter A catalyst lowers the barrier for both directions equally. Both rates climb by the same factor, so they stay matched at the same final amounts.
11. A weak acid HA breaks up as HA ⇌ H+ + A− with K = 1.0 × 10-5. You start with 0.250 M HA. Find x and test the shortcut.
x = 1.6 × 10-3 M, and the shortcut is fine Ratio test: 0.250 ÷ (1.0 × 10-5) = 2.5 × 104, well above 400. So x2 ÷ 0.250 = 1.0 × 10-5, giving x2 = 2.5 × 10-6 and x = 1.58 × 10-3 M. Check: 1.58 × 10-3 ÷ 0.250 = 0.63%, under 5%.
12. Why does a pure solid never appear in an equilibrium constant?
Its concentration cannot change A solid has a fixed density, so the amount packed into each liter of solid is the same whether the lump is large or small. A constant value folds into K instead of varying inside it.
Chemistry · Chapter 14

Acid-Base Equilibria

Lemon juice, soap, blood and soda all sit on one scale. This chapter shows what that scale measures. The whole story is one proton. Acids hand a proton over. Bases take one in. Water does both, which is why water sets the middle of the scale. Once you can follow that single proton, pH stops being a mystery number. You will find pH from a concentration, and a concentration from pH. You will see why some acids give up every proton and others barely let go. By the end you can mix a buffer and read a titration curve.

By the end of this chapter you can

  • Point to the acid, the base and both conjugate pairs in any proton transfer.
  • Use Kw to link [H3O+], [OH-], pH and pOH at 25 °C.
  • Tell a strong acid or base from a weak one, and say why the difference shows up in pH.
  • Find pH and percent ionization from Ka or Kb and a starting concentration.
  • Swap between Ka and Kb using Kw, and between pKa and pKb.
  • Predict whether a salt solution comes out acidic, basic or neutral.
  • Mix a buffer with the Henderson-Hasselbalch equation and read any titration curve.

Section 1Bronsted-Lowry acids, bases and conjugate pairs

One proton changes hands

A Bronsted-Lowry acid is a particle that gives a proton away. A Bronsted-Lowry base is a particle that takes one in. That is the whole rule. A proton here means H+. It is a hydrogen atom with its one electron stripped off. What is left is tiny and carries a full positive charge.

Protons never drift around loose in water. The instant one breaks free, a nearby water molecule grabs it. The result is the hydronium ion, H3O+. So when a book writes H+ in water, picture H3O+. Hydrogen chloride shows the handoff plainly.

HCl + H2O → Cl- + H3O+

HCl gave the proton away, so HCl is the acid. Water took it in, so water is the base. Nothing else moved. One proton crossed, and both partners got a new name.

Every acid leaves a base behind

Give up a proton and you turn into something new. That leftover is the conjugate base of the acid. Take a proton in and you become the conjugate acid of the base. The two halves of a conjugate acid-base pair differ by one H+. They differ in nothing else.

So every proton transfer holds two pairs. Watch hydrofluoric acid in water.

HF + H2O ⇌ F- + H3O+

HF and F- are one pair. Water and H3O+ are the other. To find a partner, take off one hydrogen and one plus of charge. HF minus H+ leaves F-. That check works every time.

The double arrow matters here. HF lets go of only a few of its protons. Fluoride takes some of them right back. The mixture settles into a balance, so you write ⇌ and not a single arrow.

One proton moves and makes two conjugate pairs one H⁺ crosses over HF + H2O ⇌ F− + H3O+ acid base conjugate base conjugate acid pair 1: HF / F− pair 2: H₂O / H₃O⁺ Each pair differs by one hydrogen and one unit of charge.
Figure 14.1 One proton leaves HF and lands on water. That single move creates two conjugate pairs at once. Subtract one H and one plus of charge to find a partner.

Water plays both parts

Water donated nothing in the HF reaction. It only accepted. Yet water hands a proton to ammonia without any fuss, making NH4+ and OH-. A species that can both give and take a proton is amphiprotic. Water is the one you will meet most.

Plenty of ions behave the same way. Bicarbonate, HCO3-, is a clear case. It can give its proton to water and become CO32-. It can also take a proton from water and become H2CO3. Dihydrogen phosphate, H2PO4-, does both too. Which path wins depends on two numbers you will meet in section 5.

Worked example · label both pairs

Bicarbonate acts as a base in this reaction. Name the acid, the base, and both conjugate pairs.

HCO3- + H2O ⇌ H2CO3 + OH-

Step 1 · track the hydrogenCompare each left side particle with its right side partner. HCO3- gained a hydrogen to become H2CO3. Water lost one to become OH-.

Step 2 · name the rolesWater gave the proton away, so water is the acid here. Bicarbonate took it in, so bicarbonate is the base.

Step 3 · write the pairsThe pairs are H2CO3 with HCO3-, and H2O with OH-. Each pair differs by one H+.

CheckCharge must balance. Left side is 1 minus. Right side is 0 plus 1 minus. Both sides match.

Key points
  • An acid donates a proton. A base accepts one. Water can do either job.
  • A conjugate pair differs by exactly one H+ and one unit of charge.
  • Every proton transfer contains two conjugate pairs, never one.
  • A loose proton in water is really a hydronium ion, H3O+.

Section 2Water autoionization and the pH scale

Water splits itself

Pure water is not just H2O molecules. Now and then one water molecule hands a proton to another. One becomes H3O+ and the other becomes OH-. This slow self-reaction is called the autoionization of water.

2 H2O ⇌ H3O+ + OH-

Its equilibrium constant gets its own name and symbol. The ion-product constant for water is Kw, and it equals the product of the two ion concentrations.

Kw = [H3O+][OH-] = 1.0 × 10-14 at 25 °C

In pure water the two ions come from the same split. So they must be equal. Each one works out to 1.0 × 10-7 mol/L, because that number squared gives 1.0 × 10-14. A solution with equal amounts of the two ions is neutral.

Here is the useful part. Kw holds in every water solution, not just pure water. Push the hydronium up and the hydroxide must fall. The product stays fixed. One ion can never go to zero.

pH shrinks the numbers

Those exponents are awkward to say out loud. So chemists take the negative base-ten logarithm. pH is −log[H3O+], and pOH is −log[OH-]. Taking the log of Kw gives a tidy partner rule.

pH + pOH = 14.00 at 25 °C

Pure water lands at pH 7.00 and pOH 7.00. Below pH 7 a solution is acidic. Above pH 7 it is basic. Each whole pH unit is a tenfold change in hydronium. So pH 3 holds a hundred times more hydronium than pH 5.

The pH scale with matching hydronium and hydroxide amounts 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH neutral 7.00 more acidic more basic [H₃O⁺] mol/L 100 10−2 10−4 10−6 10−8 10−10 10−12 10−14 [OH−] mol/L 10−14 10−12 10−10 10−8 10−6 10−4 10−2 100 stomach acid 1.0 lemon juice 2.4 cola 3.3 black coffee 5.0 blood 7.4 baking soda 8.4 ammonia spray 11.0 bleach 12.6 Multiply the two rows and you always get 1.0 × 10−¹⁴ at 25 °C.
Figure 14.2 One step along pH is a tenfold change in hydronium. The two concentration rows move in opposite directions. Their product never changes at a fixed temperature.

Worked example · pH of dilute nitric acid

You have 0.0025 M HNO3. Find [H3O+], pH, pOH and [OH-] at 25 °C.

Step 1 · hydroniumNitric acid is strong, so it ionizes fully. Each HNO3 unit gives one proton. So [H3O+] = 2.5 × 10-3 mol/L.

Step 2 · pHpH = −log(2.5 × 10-3) = 2.60. Two digits after the point match the two significant figures in 2.5.

Step 3 · pOHpOH = 14.00 − 2.60 = 11.40.

Step 4 · hydroxide[OH-] = Kw ÷ [H3O+] = (1.0 × 10-14) ÷ (2.5 × 10-3) = 4.0 × 10-12 mol/L.

CheckMultiply the two ion amounts. (2.5 × 10-3)(4.0 × 10-12) = 1.0 × 10-14. That is Kw, so the answer holds.

The trip also runs backward. Suppose a cleaner reads pH 10.50 on a meter. Raise 10 to the power of the negative pH, and [H3O+] = 10-10.50 = 3.2 × 10-11 mol/L. The pOH is then 3.50, which puts [OH-] at 3.2 × 10-4 mol/L.

Why a log scale is worth the trouble

Real solutions cover a huge range of hydronium amounts. Stomach acid holds a few million times more hydronium than blood does. Writing those two numbers side by side takes a pile of zeros, and comparing them by eye is hopeless. The log scale folds that whole range into the familiar 0 to 14 stretch, so a difference of two units always means the same thing: a factor of a hundred.

Keep one habit from this. The digits after the decimal point in a pH are the only significant figures it has. A pH of 2.60 carries two of them, which is why it came from a concentration written with two.

Worth knowing: the number 14.00 belongs to 25 °C alone. Warm the water and it splits a bit more freely. At 37 °C, Kw is about 2.4 × 10-14, so neutral water sits near pH 6.81. Your blood is still neutral by its own temperature, even though the number moved.
Key points
  • Water autoionizes, so every water solution holds both H3O+ and OH-.
  • Kw = [H3O+][OH-] = 1.0 × 10-14 at 25 °C.
  • pH = −log[H3O+] and pH + pOH = 14.00 at 25 °C.
  • One pH unit is a factor of ten in hydronium, not one small step.

Section 3Strong versus weak acids and bases

Strong means it goes all the way

A strong acid gives up its proton completely in water. Drop 0.10 mol of HCl into a litre and you get 0.10 mol of H3O+. No HCl molecules survive. The arrow points one way only. The short list is worth memorizing: HCl, HBr, HI, HNO3, HClO4, HClO3 and the first proton of H2SO4.

A strong base works the same way with hydroxide. The soluble metal hydroxides are the main group. LiOH, NaOH and KOH each release one OH-. Ca(OH)2, Sr(OH)2 and Ba(OH)2 each release two.

Worked example · two strong solutions

Find the pH of 0.10 M HCl, then the pH of 0.10 M Ba(OH)2.

Step 1 · the acidFull ionization gives [H3O+] = 0.10 mol/L. So pH = −log(0.10) = 1.00.

Step 2 · the base, per formula unitEach Ba(OH)2 unit sets two hydroxides loose. So [OH-] = 2 × 0.10 = 0.20 mol/L.

Step 3 · pOH then pHpOH = −log(0.20) = 0.70. So pH = 14.00 − 0.70 = 13.30.

CheckThe acid should land low and the base high. 1.00 and 13.30 fit that. Forgetting the factor of two would give pH 13.00, which is wrong by 0.30.

Weak means an equilibrium

A weak acid only partly gives up its proton. Most of it stays whole. Acetic acid in vinegar is the classic case. At 0.10 mol/L it sits at pH 2.87, not 1.00. Only about 1.3 percent of it has ionized. The other 98.7 percent is still CH3COOH.

A weak base behaves the same way. Ammonia takes protons from water, but only a small share of it does. Most of the ammonia stays as NH3. So you write both reactions with ⇌ and solve them as equilibria.

Same concentration, very different amounts of hydronium Each beaker holds 0.10 mol of acid per litre. Eighty acid units are drawn in each. 0.10 M HCl — strong all 80 have split into ions pH 1.00 0.10 M acetic acid — weak only 1 of the 80 has split pH 2.87 H₃O⁺conjugate baseacid molecule still whole
Figure 14.3 Strength is about how far the splitting goes, not how much acid you added. Both beakers hold 0.10 mol of acid per litre. Only the left one turns all of it into hydronium.

Strength and concentration are different ideas

Strength tells you what fraction of the acid splits apart. Concentration tells you how much acid you poured in. They are separate dials. A very dilute strong acid can have a higher pH than a concentrated weak one.

Here is a real example. A 0.0010 M solution of HCl has pH 3.00, while household vinegar at 0.10 M sits at pH 2.87. The weak acid wins on pH here, simply because it brought a hundred times more acid units to the beaker.

The same split shows up with bases. A 0.010 M NaOH solution reaches pH 12.00 because all of its hydroxide is already free. A 0.010 M ammonia solution only reaches about pH 10.6, since most of the ammonia never takes a proton at all.

FeatureStrong acidWeak acid
Arrow→ one way⇌ both ways
Fraction ionizedessentially 100%often under 5%
Intact molecules leftnone you can measuremost of what you added
pH of a 0.10 M sample1.00 for HCl2.87 for acetic acid
Kavery large, not tabulatedsmall, and tabulated
Conjugate basealmost no pull on protonsa real base you must account for
Worth knowing: water sets a ceiling on acid strength. Any acid stronger than H3O+ just converts fully and hands you H3O+ instead. So HCl and HClO4 act the same in water, even though HClO4 is the stronger acid. Chemists call this the leveling effect.
Key points
  • Strong acids and bases ionize fully, so the pH comes straight from the concentration.
  • Weak acids and bases reach an equilibrium with most of the starting material intact.
  • Count hydroxides per formula unit for bases like Ba(OH)2.
  • Strength is a fraction. Concentration is an amount. Never trade one for the other.

Section 4Ka, Kb and percent ionization

Ka puts a number on weakness

Weak acids need a measure of how far they go. That measure is the acid ionization constant, Ka. It is just the equilibrium constant for the acid giving its proton to water.

Ka = [H3O+][A-] ÷ [HA]

Liquid water is left out, as it is in any equilibrium constant. A bigger Ka means a stronger acid. Formic acid, at 1.8 × 10-4, is ten times stronger than acetic acid at 1.8 × 10-5. Bases get the matching base ionization constant, Kb, built from [BH+][OH-] over [B].

These numbers are easier to compare as logs. pKa is −log Ka. A small pKa means a strong acid. Acetic acid has pKa 4.74, and ammonium has pKa 9.25.

Acid strength ladder, from a strong acid down to very weak ones ACID Ka pKa STRENGTH BAR HCl hydrochloric strong > 1 HSO₄− hydrogen sulfate 1.2 × 10−² 1.92 H₃PO₄ phosphoric 7.5 × 10−³ 2.12 HF hydrofluoric 6.8 × 10−⁴ 3.17 HCOOH formic 1.8 × 10−⁴ 3.74 CH₃COOH acetic 1.8 × 10−⁵ 4.74 H₂CO₃ carbonic 4.3 × 10−⁷ 6.37 NH₄⁺ ammonium 5.6 × 10−¹⁰ 9.25 HCO₃− bicarbonate 4.7 × 10−¹¹ 10.33 Down the list: smaller Ka, larger pKa, weaker acid, stronger conjugate base.
Figure 14.4 A ladder of real acids ranked by Ka. Each rung down is a weaker acid and a stronger conjugate base. The pKa column just flips the exponent into a friendlier number.

Worked example · pH of 0.25 M ammonia

Ammonia has Kb = 1.8 × 10-5. Find the pH of a 0.25 M solution.

Step 1 · the reactionNH3 + H2O ⇌ NH4+ + OH-. Let x be the hydroxide made, in mol/L.

Step 2 · the tableAt balance, [NH4+] = x, [OH-] = x, and [NH3] = 0.25 − x.

Step 3 · solve1.8 × 10-5 = x2 ÷ (0.25 − x). Since Kb is tiny, try 0.25 − x ≈ 0.25. Then x2 = 4.5 × 10-6, so x = 2.1 × 10-3.

Step 4 · pHpOH = −log(2.1 × 10-3) = 2.67. So pH = 14.00 − 2.67 = 11.33.

Check the shortcutx is 0.85 percent of 0.25, well under 5 percent. The approximation was safe. The exact quadratic gives x = 2.11 × 10-3 and pH 11.32.

Ka and Kb are tied together by Kw

An acid and its conjugate base are not independent. Multiply the Ka of an acid by the Kb of its conjugate base. You always get Kw.

Ka × Kb = Kw = 1.0 × 10-14    and    pKa + pKb = 14.00

So one table covers both. Ammonia has Kb = 1.8 × 10-5. Its conjugate acid, ammonium, must have Ka = (1.0 × 10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10. Acetate runs the same way, with Kb = 5.6 × 10-10.

The rule also explains a pattern. A weaker acid has a stronger conjugate base. Their product is locked, so when one side shrinks the other must grow.

Percent ionization climbs as you dilute

Percent ionization is the ionized fraction times 100. For an acid it is [H3O+] at balance divided by the starting acid concentration. It is not a fixed property. It shifts with concentration.

Acetic acid at 0.10 M is 1.3 percent ionized, giving pH 2.87. Dilute it to 0.0010 M and the quadratic is needed, because x is no longer small next to the start. Solving x2 ÷ (0.0010 − x) = 1.8 × 10-5 gives x = 1.25 × 10-4. That is 12.5 percent ionized, and pH 3.90.

The pH still went up, because there is less acid overall. But the share that split apart grew by almost ten times. Dilution gives the ions more room, so the equilibrium shifts toward them.

Key points
  • Ka and Kb measure strength. Larger means stronger, and smaller pKa means stronger.
  • Ka × Kb = Kw for any conjugate pair, so pKa + pKb = 14.00 at 25 °C.
  • The small-x shortcut is safe when x stays under 5 percent of the starting amount.
  • Percent ionization rises on dilution, even while the pH rises too.

Section 5Polyprotic acids and salt solutions

Polyprotic acids give protons one at a time

Some acids hold more than one donatable proton. These are polyprotic acids. Carbonic acid has two. Phosphoric acid has three. They never leave all at once. Each proton goes in its own step, with its own Ka.

The steps get much harder as you go. Carbonic acid has Ka1 = 4.3 × 10-7 and Ka2 = 4.7 × 10-11. That is a drop of about 104. The reason is charge. Pulling a positive proton off a particle that is already negative takes much more energy.

Phosphoric acid gives up three protons one step at a time H₃PO₄ H₂PO₄− HPO₄²− PO₄³− − H⁺ Ka1 = 7.5 × 10−³ pKa1 = 2.12 − H⁺ Ka2 = 6.2 × 10−⁸ pKa2 = 7.21 − H⁺ Ka3 = 4.2 × 10−¹³ pKa3 = 12.38 Each step is roughly a hundred thousand times harder than the step before. So the first step sets the pH almost by itself. Pulling H⁺ off a particle that is already negative costs far more energy.
Figure 14.5 Phosphoric acid sheds protons in order, never all at once. The three Ka values drop by about 105 each step. That gap is why you can treat the first step on its own.

This gap is a gift. The second step adds so little hydronium that you can ignore it. So you treat a polyprotic acid as a simple weak acid using Ka1 alone.

Worked example · pH of carbonated water

A soda holds roughly 0.050 M H2CO3. Find its pH using Ka1 = 4.3 × 10-7.

Step 1 · first step onlyH2CO3 + H2O ⇌ HCO3- + H3O+. Let x be the hydronium made.

Step 2 · solve4.3 × 10-7 = x2 ÷ (0.050 − x) ≈ x2 ÷ 0.050. So x2 = 2.15 × 10-8 and x = 1.47 × 10-4.

Step 3 · pHpH = −log(1.47 × 10-4) = 3.83.

Check step twoThe second step would add about 4.7 × 10-11 mol/L of hydronium. Next to 1.47 × 10-4 that is nothing. Dropping it was fair.

Salt solutions are not always neutral

Dissolve a salt and its ions can react with water. That reaction is salt hydrolysis. To predict the result, look at where each ion came from. An ion left over from a strong acid or a strong base is a spectator. An ion left over from a weak partner is not.

Salt built fromExampleReacting ionResult
Strong acid + strong baseNaClneitherneutral, pH 7
Weak acid + strong baseNaCH3COOCH3COO- takes H+basic, pH above 7
Strong acid + weak baseNH4ClNH4+ gives H+acidic, pH below 7
Weak acid + weak baseNH4CH3COOboth ions reactcompare Ka with Kb
Small, high-charge metal ionAlCl3bound water gives H+acidic

Worked example · pH of 0.10 M sodium acetate

Acetic acid has Ka = 1.8 × 10-5. Find the pH of 0.10 M NaCH3COO.

Step 1 · find the active ionSodium came from NaOH, a strong base, so it does nothing. Acetate came from a weak acid, so acetate is a base.

Step 2 · get KbKb = Kw ÷ Ka = (1.0 × 10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10.

Step 3 · solve for hydroxidex2 ÷ 0.10 = 5.6 × 10-10, so x2 = 5.6 × 10-11 and x = 7.5 × 10-6 mol/L.

Step 4 · pHpOH = −log(7.5 × 10-6) = 5.13. So pH = 14.00 − 5.13 = 8.87.

CheckA salt of a weak acid should come out basic. pH 8.87 is above 7, so the sign is right. Run 0.10 M NH4Cl the same way and you get pH 5.13, the mirror image.

Amphiprotic ions need a comparison

An ion like HCO3- can act either way. So compare its two constants. As an acid it has Ka = 4.7 × 10-11, the Ka2 of carbonic acid. As a base its Kb = Kw ÷ Ka1 = 2.3 × 10-8.

Kb is about 500 times larger, so the base side wins. A baking soda solution turns out basic, which matches the taste and the feel.

Key points
  • Polyprotic acids lose protons in steps, and each Ka is far smaller than the one before.
  • Ka1 sets the pH of a polyprotic acid almost by itself.
  • Ions from strong acids and strong bases are spectators. Ions from weak partners react.
  • For an amphiprotic ion, compare its Ka with its Kb and let the larger one decide.

Section 6Buffers and titration curves

A buffer holds pH steady

A buffer mixes a weak acid with its conjugate base, both in real amounts. Add acid and the base half soaks up the extra protons. Add base and the acid half replaces them. Either way the pH barely moves.

Acetic acid with sodium acetate is a standard pair. So is H2PO4- with HPO42-. A strong acid cannot buffer anything, because it keeps no reserve of intact molecules.

The Henderson-Hasselbalch equation

Take the log of the Ka expression and rearrange. You get the Henderson-Hasselbalch equation, the fastest route to a buffer pH.

pH = pKa + log([A-] ÷ [HA])

Two readings fall out at once. Equal amounts of acid and base make the ratio 1, and log 1 is 0, so pH = pKa. And because both terms share one solution, moles work as well as concentrations.

A buffer works best within about one pH unit of its pKa, and that span is called the buffer region. Outside it the ratio gets lopsided and the reserve on one side runs thin. How much strong acid or base a buffer can soak up is its buffer capacity, which grows with the total amount of the pair.

Worked example · a buffer under attack

One litre holds 0.100 mol acetic acid and 0.100 mol sodium acetate. pKa is 4.74. Add 0.010 mol NaOH. Find the pH before and after.

Step 1 · beforeThe ratio is 0.100 ÷ 0.100 = 1. So pH = 4.74 + log(1) = 4.74.

Step 2 · react the strong baseHydroxide eats acid and makes base. Acid falls to 0.090 mol. Acetate rises to 0.110 mol.

Step 3 · afterpH = 4.74 + log(0.110 ÷ 0.090) = 4.74 + 0.09 = 4.83.

CompareThat same 0.010 mol NaOH in one litre of pure water gives [OH-] = 0.010, pOH 2.00 and pH 12.00. The buffer moved 0.09 units. Water moved 5.00.

Titration curves tell the story

A titration curve plots pH against the volume of titrant added. The equivalence point is where the added base exactly matches the acid present.

A strong acid with a strong base gives a flat run, a near vertical jump, then another flat run. Its equivalence point sits at pH 7.00, because the salt left behind is neutral. For 25.00 mL of 0.100 M HCl that jump arrives at 25.00 mL of 0.100 M NaOH.

A weak acid curve differs in three ways. It starts higher, near pH 2.87 for 0.100 M acetic acid. It has a long gentle stretch in the middle, which is the buffer region. And its equivalence point lands above pH 7, because the salt left behind is a base.

Titration curve for a weak acid with a strong base buffer region, pKa ± 1 0 2 4 6 8 10 12 14 0 10 20 30 40 50 volume of 0.100 M NaOH added (mL) pH half-equivalence: pH = pKa = 4.74 equivalence: 25.00 mL, pH 8.72 start pH 2.87
Figure 14.6 Acetic acid titrated with sodium hydroxide, computed point by point. The shaded band is the buffer region. At the half-way mark pH equals pKa, and the equivalence point lands above 7 because acetate is a base.

The half-equivalence point is the most useful spot on the curve. Half the acid has been converted, so acid and base are equal and the ratio is 1. That makes pH equal pKa, which for acetic acid reads 4.74 at 12.50 mL.

At full equivalence every acetic acid molecule has become acetate. That is 2.50 × 10-3 mol in 50.00 mL, so 0.0500 M acetate. Its Kb of 5.6 × 10-10 then gives [OH-] = 5.3 × 10-6 and pH 8.72. A polyprotic acid repeats the pattern, one jump per proton.

Worth knowing: your blood runs on a buffer. Carbonic acid and bicarbonate hold it near pH 7.4. Your lungs adjust the carbonic acid by changing how fast you breathe out CO2. Your kidneys adjust the bicarbonate. Two organs, one Henderson-Hasselbalch ratio.
Key points
  • A buffer needs a weak acid and its conjugate base together in real amounts.
  • pH = pKa + log([A-] ÷ [HA]), and moles work as well as concentrations.
  • Pick a buffer whose pKa is within one unit of the pH you want.
  • At half equivalence pH = pKa. At equivalence a weak acid titration lands above pH 7.

The chapter in one breath

  • An acid donates a proton and a base accepts one, so every transfer makes two conjugate pairs.
  • A free proton in water is really hydronium, H3O+.
  • Water autoionizes, and Kw = 1.0 × 10-14 at 25 °C ties the two ions together forever.
  • pH = −log[H3O+], and pH + pOH = 14.00 at 25 °C.
  • Strong acids and bases ionize fully. Weak ones reach an equilibrium with most material intact.
  • Ka and Kb score that weakness, and Ka × Kb = Kw for any pair.
  • Percent ionization rises as a weak acid gets more dilute.
  • Polyprotic acids lose protons in steps, and Ka1 does nearly all the work on pH.
  • Buffers follow pH = pKa + log([A-]/[HA]), and half equivalence on a curve reads pKa directly.

ReferenceWords worth knowing

Here are the terms this chapter leans on, in alphabetical order.

Acid ionization constant (Ka)
The equilibrium constant for a weak acid giving its proton to water. A larger value means a stronger acid.
Amphiprotic
Able to both donate and accept a proton. Water, HCO3- and H2PO4- all qualify.
Autoionization of water
The slow reaction in which one water molecule hands a proton to another, making H3O+ and OH-.
Base ionization constant (Kb)
The equilibrium constant for a weak base taking a proton from water. A larger value means a stronger base.
Bronsted-Lowry acid and base
An acid is a proton donor. A base is a proton acceptor. The pair of roles defines every reaction in this chapter.
Buffer
A solution of a weak acid and its conjugate base that resists pH change when acid or base is added.
Buffer capacity
How much strong acid or base a buffer can absorb before its pH shifts sharply. It grows with the total amount of the pair.
Conjugate acid-base pair
Two species that differ by one H+. The conjugate base is what an acid becomes after donating, like F- from HF. The conjugate acid is what a base becomes after accepting, like NH4+ from NH3.
Equivalence point
The point in a titration where the added titrant exactly matches the substance being titrated.
Half-equivalence point
Half way to equivalence in a weak acid titration. Acid and conjugate base are equal, so pH equals pKa.
Henderson-Hasselbalch equation
pH = pKa + log([A-]/[HA]). It gives the pH of a buffer from a ratio.
Hydronium ion
H3O+, the real form a donated proton takes in water.
Ion-product constant for water (Kw)
The product [H3O+][OH-]. It equals 1.0 × 10-14 at 25 °C.
Leveling effect
Water limits acid strength. Any acid stronger than H3O+ simply converts fully to H3O+.
Percent ionization
The ionized fraction of a weak acid or base times 100. It rises as the solution gets more dilute.
pH
The negative base-ten logarithm of [H3O+]. Each unit is a factor of ten.
pOH
The negative base-ten logarithm of [OH-]. At 25 °C it is 14.00 minus the pH.
Polyprotic acid
An acid with more than one donatable proton, released in separate steps with separate Ka values.
Salt hydrolysis
The reaction of a dissolved salt ion with water, which can leave the solution acidic or basic.
Titration curve
A plot of pH against the volume of titrant added. Its shape reveals strength and pKa.

Check yourself

1. Find the pH and pOH of 0.0050 M HCl at 25 °C.
pH 2.30 and pOH 11.70. HCl is strong, so [H3O+] = 5.0 × 10-3. pH = −log(5.0 × 10-3) = 2.30. Then pOH = 14.00 − 2.30 = 11.70.
2. Name the conjugate base and the conjugate acid of HSO4-.
SO42- and H2SO4. Remove one H+ to get the conjugate base, SO42-. Add one H+ to get the conjugate acid, H2SO4. HSO4- can do both, so it is amphiprotic.
3. A solution reads pH 3.40. Find [H3O+] and [OH-].
4.0 × 10-4 M and 2.5 × 10-11 M. [H3O+] = 10-3.40 = 4.0 × 10-4. Then [OH-] = (1.0 × 10-14) ÷ (4.0 × 10-4) = 2.5 × 10-11.
4. Find the pH and percent ionization of 0.20 M acetic acid. Ka = 1.8 × 10-5.
pH 2.72 and about 0.95 percent ionized. x2 ÷ 0.20 = 1.8 × 10-5, so x2 = 3.6 × 10-6 and x = 1.9 × 10-3. pH = −log(1.9 × 10-3) = 2.72. Percent = (1.9 × 10-3 ÷ 0.20) × 100 = 0.95.
5. HF has Ka = 6.8 × 10-4. Find Kb for F-, then the pH of 0.10 M NaF.
Kb = 1.5 × 10-11 and pH 8.08. Kb = (1.0 × 10-14) ÷ (6.8 × 10-4) = 1.5 × 10-11. Then x2 = (1.5 × 10-11)(0.10) = 1.5 × 10-12, so [OH-] = 1.2 × 10-6. pOH = 5.92, so pH = 8.08.
6. Find the pH of a buffer that is 0.30 M NH3 and 0.20 M NH4Cl. Kb for NH3 is 1.8 × 10-5.
pH 9.43. Ka for NH4+ = (1.0 × 10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10, so pKa = 9.25. Then pH = 9.25 + log(0.30 ÷ 0.20) = 9.25 + 0.18 = 9.43.
7. Why is a sodium chloride solution neutral?
Neither ion reacts with water. Na+ comes from NaOH, a strong base, and Cl- comes from HCl, a strong acid. Conjugates of strong species have no pull on protons, so both ions are spectators.
8. Which has the lower pH, 0.10 M HCl or 0.10 M HF? Why?
The HCl, at pH 1.00 against pH 2.10 for the HF. Both have the same concentration, but HCl ionizes fully while HF does not. For HF the quadratic gives [H3O+] = 7.9 × 10-3, so pH = 2.10. Strength, not concentration, separates them here.
9. Is a sodium bicarbonate solution acidic or basic? Back it with numbers.
Basic. As an acid HCO3- has Ka = 4.7 × 10-11. As a base its Kb = (1.0 × 10-14) ÷ (4.3 × 10-7) = 2.3 × 10-8. Kb is far larger, so the base behavior wins.
10. You titrate 25.00 mL of 0.100 M NaOH with 0.100 M HCl. What volume reaches equivalence, and what is the pH there?
25.00 mL, and pH 7.00. Moles of base are (0.100 mol/L)(0.02500 L) = 2.50 × 10-3. Matching that needs the same moles of acid, so 25.00 mL. The product is NaCl, a neutral salt, so pH 7.00.
11. A weak acid is titrated with NaOH. At the half-equivalence point the meter reads 3.74. What is the acid's Ka?
Ka = 1.8 × 10-4. At half equivalence [A-] equals [HA], so pH = pKa = 3.74. Then Ka = 10-3.74 = 1.8 × 10-4. That matches formic acid.
12. Why does pH + pOH = 14.00 hold only at 25 °C?
Because Kw itself changes with temperature. The sum is really −log Kw. Kw is 1.0 × 10-14 only at 25 °C. At 37 °C it is about 2.4 × 10-14, so the sum drops to about 13.62.
Chemistry · Chapter 15

Equilibria of Other Reaction Classes

Some salts pour into water and vanish. Others sit on the bottom and barely move. That gap runs water treatment, kidney stones, cave formation, film developing and the stomach scans done in a clinic. This chapter takes the equilibrium tools you already have and points them at solids. You will learn one constant, Ksp, that says how far a stubborn salt dissolves. Then you will push that balance around. Add a shared ion and solubility falls. Mix two solutions and a solid may drop out. Add ammonia or acid and a solid you could not budge goes clear in seconds.

By the end of this chapter you can

  • Write the Ksp expression for any slightly soluble salt, and say why the solid is left out.
  • Convert between Ksp and molar solubility for 1:1, 1:2 and 3:2 salts.
  • Predict how much a shared ion cuts the solubility of a salt, and work out the new value.
  • Compare the ion product Q with Ksp to say whether a solid will form on mixing.
  • Choose a reagent that drops one metal ion out of a mixture and leaves the other behind.
  • Use a formation constant Kf to find how much free metal ion is left after ligands are added.
  • Explain why aluminum and zinc hydroxides dissolve in both acid and strong base.
  • Add two equilibria, multiply their constants, and predict whether acid or a ligand will dissolve a solid.

Section 1The solubility product

What a saturated solution really is

Stir a spoon of table salt into water and it disappears. Stir in twenty spoons and some settles on the bottom. That leftover solid is not resting. Ions keep breaking away from its surface. Other ions keep landing back on it and locking in. When those two rates match, the solution is saturated. The amount dissolved stops changing, and the solid stops shrinking.

So a saturated solution is an equilibrium. For silver chloride it reads like this.

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

The equilibrium constant for a reaction of this kind has its own name. It is the solubility product constant, Ksp. You build it the usual way. Multiply the concentrations on the product side, each raised to its coefficient.

Ksp = [Ag+][Cl-] = 1.8 × 10-10

Look at what is missing. The solid is not in the expression. A pure solid has fixed crowding, so its activity is 1. A kilogram of AgCl and a single speck of AgCl give the same Ksp. Adding more solid cannot force more of it into solution. That one fact saves people a lot of wasted stirring.

A saturated solution, where ions leave the crystal surface at the same rate that ions return to it Saturation is a tie, not a stop undissolved AgCl ions leave ions leave ions return Ag+ Cl- Cl- Ag+ Cl- Cl- Ag+ The equilibrium AgCl(s) ⇌ Ag+(aq) + Cl-(aq) rate out = rate back in The constant Ksp = [Ag+][Cl-] = 1.8 × 10-10 No AgCl(s) term. A pure solid has activity 1. More solid in the beaker changes nothing.
Figure 15.1 In a saturated solution, ions leave the crystal face at the same rate they return. The dissolved amount holds steady. The solid itself never appears in the Ksp expression, because its activity is fixed at 1.

From Ksp to molar solubility

Molar solubility is the number of moles of a salt that dissolve in one liter of water to reach saturation. Call it s. The bridge between s and Ksp is the formula of the salt, so write the balanced equation first every time.

A 1:1 salt such as AgCl gives one cation and one anion per formula unit. Both ions sit at s, so Ksp = s2. A 1:2 salt such as PbI2 gives one Pb2+ and two I-. Now [Pb2+] = s and [I-] = 2s, so Ksp = (s)(2s)2 = 4s3. A 3:2 salt such as Ca3(PO4)2 gives ions at 3s and 2s, so Ksp = (3s)3(2s)2 = 108s5.

Three salt types and the solubility product formula each one gives: s squared, four s cubed, and one hundred eight s to the fifth The ion ratio writes the formula for you 1 : 1 AgCl, BaSO4 M+ X- one of each per unit [M+] = s   [X-] = s Ksp = s2 K = 1.8 × 10-10 s = 1.3 × 10-5 M 1 : 2 PbI2, CaF2 M2+ X- X- two anions per cation [M2+] = s   [X-] = 2s Ksp = (s)(2s)2 = 4s3 K = 1.4 × 10-8 s = 1.5 × 10-3 M 3 : 2 Ca3(PO4)2 M2+ M2+ M2+ X3- X3- [M2+] = 3s   [X3-] = 2s Ksp = (3s)3(2s)2 = 108s5 K = 1.3 × 10-32 s = 1.6 × 10-7 M
Figure 15.2 Three salt shapes, three formulas. The number in front of s comes from the subscripts in the salt. Lose that number and the answer is off by a lot.

Worked example · Ksp to solubility for a 1:2 salt

Lead(II) iodide has Ksp = 1.4 × 10-8 at 25 °C. Find its molar solubility, then its solubility in grams per liter. The molar mass is 461.0 g/mol.

Step 1 · balance the dissolvingPbI2(s) ⇌ Pb2+(aq) + 2 I-(aq)

Step 2 · label the ions with s[Pb2+] = s and [I-] = 2s.

Step 3 · build KspKsp = (s)(2s)2 = 4s3 = 1.4 × 10-8

Step 4 · solve for ss3 = 1.4 × 10-8 / 4 = 3.5 × 10-9, so s = (3.5 × 10-9)1/3 = 1.5 × 10-3 M

Step 5 · check it backward4(1.518 × 10-3)3 = 1.40 × 10-8. That matches.

Answer1.5 × 10-3 M. In mass terms, 1.518 × 10-3 mol/L × 461.0 g/mol = 0.70 g/L. Also note [I-] = 2s = 3.0 × 10-3 M, twice the lead.

The trip runs the other way just as well. Weigh how much salt a liter of saturated solution holds, convert to moles, then build Ksp from those numbers.

Worked example · solubility to Ksp

A liter of saturated barium sulfate holds 2.4 × 10-3 g of BaSO4 at 25 °C. The molar mass is 233.39 g/mol. Find Ksp.

Step 1 · go to moless = 2.4 × 10-3 g/L ÷ 233.39 g/mol = 1.03 × 10-5 M

Step 2 · read the ratioBaSO4(s) ⇌ Ba2+(aq) + SO42-(aq), so both ions are at 1.03 × 10-5 M.

Step 3 · multiplyKsp = (1.03 × 10-5)2 = 1.1 × 10-10

AnswerKsp = 1.1 × 10-10, which is the table value. A barium sulfate drink is safe to swallow for an X-ray because almost none of the toxic barium ever gets loose.

Why Ksp is not a ranking

Here is a trap. A smaller Ksp does not always mean a less soluble salt. Silver chloride has Ksp = 1.8 × 10-10 and dissolves to 1.3 × 10-5 M. Silver chromate has the smaller Ksp, 9.0 × 10-12, yet dissolves to 1.3 × 10-4 M. That is ten times more. The reason is the exponent. Ag2CrO4 releases three ions, so its constant is 4s3, and a cube shrinks a small number much harder than a square does. You may compare Ksp values head to head only when the salts share the same ion ratio.

Key points
  • Ksp is the equilibrium constant for a solid dissolving into its ions.
  • The solid never appears in the expression, so extra solid changes nothing.
  • Molar solubility s is moles dissolved per liter at saturation.
  • 1:1 gives s2, 1:2 gives 4s3, and 3:2 gives 108s5.
  • Compare Ksp values only for salts with the same ion ratio.
Where these numbers come from. Every Ksp and Kf in this chapter is a table value for 25 °C, matching the appendix tables in OpenStax Chemistry 2e. Tables differ in the last digit, and some list a different crystal form of the same salt. Always work from the table that is printed with the problem in front of you.

Section 2The common ion effect

A shared ion shuts the door

So far the water was pure before the salt went in. Real water is rarely pure. Suppose you try to dissolve silver chloride in a solution that already holds 0.10 M NaCl. The chloride from the salt water counts. It is the same Cl- that AgCl would release.

An ion that two solutes share is a common ion. Le Chatelier's principle tells you what happens next. Raise the amount of a product and the equilibrium shifts back toward the reactant side. Here the reactant side is solid. So less AgCl dissolves. This drop is the common ion effect.

Silver chloride dissolving in pure water compared with 0.10 molar sodium chloride, where the shared chloride ion cuts solubility by about seven thousand times A shared ion pushes the balance back to solid Ag+ Cl- Ag+ Cl- pure water s = 1.3 × 10-5 M Cl- Cl- Cl- Cl- Cl- Cl- 0.10 M NaCl s = 1.8 × 10-9 M add Cl- more solid stays undissolved Solubility on a log scale pure water, 1.3 × 10-5 M 0.10 M NaCl, 1.8 × 10-9 M each gridline is one power of ten Ksp never moved. The split did. (s)(0.10) = 1.8 × 10-10
Figure 15.3 The same salt, two different waters. Chloride already present shoves the dissolving equilibrium back toward the solid. The bar scale is logarithmic, so each step is a factor of ten.

Nothing about Ksp changes. It is a constant at a fixed temperature. What changes is how the two concentrations split the product. If one ion is already large, the other must be tiny to keep the product equal to Ksp.

Worked example · silver chloride in salt water

Find the molar solubility of AgCl in 0.10 M NaCl. Ksp = 1.8 × 10-10.

Step 1 · set up the tableLet s be the moles of AgCl that dissolve per liter. Then [Ag+] = s and [Cl-] = 0.10 + s.

Step 2 · write KspKsp = (s)(0.10 + s) = 1.8 × 10-10

Step 3 · drop the tiny termIn pure water s was only 1.3 × 10-5 M, and here it will be smaller still. So 0.10 + s ≈ 0.10.

Step 4 · solves = 1.8 × 10-10 / 0.10 = 1.8 × 10-9 M

Step 5 · test the shortcutIs 1.8 × 10-9 small next to 0.10? It is about 2 parts in 100 million. The shortcut was fair.

Answer1.8 × 10-9 M. In pure water it was 1.3 × 10-5 M, so the salt water cut solubility by a factor of about 7,500.

The effect gets stronger with more ions

When the shared ion carries an exponent, the squeeze is harder. Lead(II) iodide in 0.10 M KI is a good case. Iodide enters Ksp squared, so the added iodide gets squared too.

Worked example · lead iodide in potassium iodide

Find the molar solubility of PbI2 in 0.10 M KI. Ksp = 1.4 × 10-8.

Step 1 · label the ions[Pb2+] = s and [I-] = 0.10 + 2s ≈ 0.10

Step 2 · write and solveKsp = (s)(0.10)2 = 0.010 s = 1.4 × 10-8, so s = 1.4 × 10-6 M

Step 3 · check the shortcut2s = 2.8 × 10-6 M, far below 0.10 M. Good.

Answer1.4 × 10-6 M, down from 1.5 × 10-3 M in pure water. That is a drop of about 1,100 times.

Note that the solubility is now set almost entirely by the added ion. The salt supplies so little of its own that you can treat the added amount as fixed. That shortcut fails only when the added ion is very dilute, close to the salt's own solubility. Then you must keep the full expression and solve it properly.

Two ways people use this on purpose

Chemists lean on the common ion effect to finish a job. After a precipitate forms, you wash it with a solution holding one of its own ions instead of plain water. Pure water would redissolve some of your product. A wash made of dilute sodium chloride loses almost none of a silver chloride sample.

The same trick controls teeth. Tooth enamel is mostly hydroxyapatite, which slowly dissolves in the mouth. Fluoride in toothpaste swaps in to build fluorapatite, a salt with a far smaller Ksp. Fluoride in the water also acts as a common ion for that solid. Both effects push the enamel equilibrium back toward solid.

One warning. Adding a salt with no shared ion does not help in the same way. It can even raise solubility a little, because a crowd of ions screens the charges and makes each ion act slightly less concentrated than it is. That is the salt effect, and it is small. Ignore it unless a problem asks for it.

Key points
  • A common ion is one that the salt and some other solute both supply.
  • Adding a common ion shifts the dissolving equilibrium back toward solid, so solubility drops.
  • Ksp does not change. Only the split between the two concentrations changes.
  • If the shared ion is squared in Ksp, the squeeze is far stronger.
  • When the added ion is much larger than s, treat it as fixed and skip the small term.

Section 3Predicting precipitation

Q tells you which way the solid will go

Pour two clear solutions together and sometimes a cloud appears. A precipitate is a solid that forms inside a solution and settles out. You can predict it before you pour. Build the same product that Ksp uses, but with the concentrations you actually have. That number is the ion product, Q.

Q has the same form as Ksp. The only difference is that Q uses the moment's concentrations, which may be far from equilibrium. Then you compare.

ComparisonWhat the solution isWhat happens
Q < KspUnsaturatedNo solid forms. More salt could still dissolve.
Q = KspSaturatedAt equilibrium. Solid and ions hold steady together.
Q > KspSupersaturatedSolid forms until Q falls back to Ksp.
A number line comparing the ion product Q with the solubility product, showing unsaturated, saturated and supersaturated zones One comparison answers the question Q = Ksp Q < Ksp Q > Ksp unsaturated clear solution, no solid more salt could dissolve saturated at equilibrium supersaturated solid forms and settles out until Q falls back to Ksp Q slides left as solid drops out Example: mix 25.0 mL of 0.0020 M Pb(NO3)2 with 75.0 mL of 0.010 M KI → Q = 2.8 × 10-8, Ksp = 1.4 × 10-8
Figure 15.4 One number line decides the question. Ksp marks the saturation point. Any Q to the right of it means a solid comes out until Q slides back down to Ksp.

There is one step people skip. Mixing two solutions dilutes both of them. Before you build Q, find the new concentration of each ion in the combined volume. Use moles divided by total liters, or the shortcut of multiplying by the volume fraction.

Worked example · will lead iodide drop out?

You mix 25.0 mL of 0.0020 M Pb(NO3)2 with 75.0 mL of 0.010 M KI. Does PbI2 form? Ksp = 1.4 × 10-8.

Step 1 · total volume25.0 mL + 75.0 mL = 100.0 mL

Step 2 · dilute each ion[Pb2+] = 0.0020 M × (25.0 / 100.0) = 5.0 × 10-4 M
[I-] = 0.010 M × (75.0 / 100.0) = 7.5 × 10-3 M

Step 3 · build QQ = [Pb2+][I-]2 = (5.0 × 10-4)(7.5 × 10-3)2

Step 4 · do the arithmetic(7.5 × 10-3)2 = 5.625 × 10-5. Then (5.0 × 10-4)(5.625 × 10-5) = 2.8 × 10-8

Step 5 · compareQ = 2.8 × 10-8 and Ksp = 1.4 × 10-8. Q is about 2.0 times Ksp.

AnswerQ > Ksp, so yellow PbI2 precipitates. The margin is thin, so only a little solid appears before Q settles back to 1.4 × 10-8.

Pulling one ion out and leaving the other

Now suppose a solution holds two ions that both form solids with the same reagent. Their Ksp values are not equal, so they do not come out together. Add the reagent slowly and the less soluble salt forms first. Separating ions this way is selective precipitation. It is the backbone of the old qualitative analysis schemes, and it still drives metal recovery today.

The method is to find the reagent concentration each salt needs to just begin forming. Rearrange Ksp to solve for the reagent. Whichever salt needs less reagent starts first.

Worked example · iodide out, chloride left behind

A solution is 0.10 M in Cl- and 0.010 M in I-. You add silver nitrate drop by drop. Which solid forms first, and how much iodide is left when the second one starts? Ksp(AgCl) = 1.8 × 10-10, Ksp(AgI) = 8.3 × 10-17.

Step 1 · silver needed for AgI[Ag+] = Ksp / [I-] = 8.3 × 10-17 / 0.010 = 8.3 × 10-15 M

Step 2 · silver needed for AgCl[Ag+] = 1.8 × 10-10 / 0.10 = 1.8 × 10-9 M

Step 3 · rank themAgI needs far less silver, so pale yellow AgI forms first. White AgCl waits.

Step 4 · iodide left at the crossoverWhen [Ag+] reaches 1.8 × 10-9 M, [I-] = 8.3 × 10-17 / 1.8 × 10-9 = 4.6 × 10-8 M

AnswerAgI first. Only 4.6 × 10-8 M iodide remains, which is 4.6 × 10-6 of the original 0.010 M. Over 99.999% of the iodide is out before any chloride joins it. A clean split.

A separation works when the two Ksp values are far apart. Here they differ by seven powers of ten, so the gap is huge. Barium and calcium sulfates differ by about five powers of ten, which is still workable. Two salts with nearly equal Ksp values cannot be split this way.

Key points
  • Q has the same form as Ksp but uses the concentrations you have right now.
  • Q < Ksp means no solid. Q > Ksp means a solid forms.
  • Always dilute each ion for the combined volume before building Q.
  • Selective precipitation drops the salt with the smaller Ksp first.
  • Find the reagent amount each salt needs to start. The smaller amount goes first.
Worth knowing. Q > Ksp says a solid should form, not that it will form right away. A supersaturated solution can sit clear for hours with nothing to nucleate on. Scratch the glass or drop in one seed crystal and the solid appears at once.

Section 4Complex ion formation

Ligands wrap a metal ion

Pour ammonia into a solution of silver nitrate and a white cloud of AgOH forms. Keep pouring and the cloud clears. Something took the silver and held it in solution. That something is a complex.

A complex ion is a central metal ion bonded to several small molecules or ions around it. Each attached partner is a ligand. A ligand always has a lone pair to donate, so it acts as a Lewis base. The metal ion accepts that pair, so it acts as a Lewis acid. Water itself is a ligand, which is why a dissolved metal ion is really an aquo complex like [Al(H2O)6]3+.

The number of donor atoms attached is the coordination number. Silver likes 2 and is linear. Zinc and copper often take 4. Many ions of the first transition row take 6 in an octahedral arrangement. Common ligands include NH3, CN-, OH-, Cl-, S2O32- and H2O.

Ammonia takes a silver ion into a linear complex, which pulls the silver chloride dissolving equilibrium further to the right A ligand steals the metal ion, so the solid keeps feeding it AgCl(s) Ksp = 1.8 × 10-10 white, stubborn in water tiny Ag+ Cl- + 2 NH3 Ag NH3 NH3 linear, coordination number 2 [Ag(NH3)2]+ Kf = 1.7 × 107 free Ag+ drops to 10-11 M, so more solid dissolves Add the two steps and multiply the two constants AgCl(s) + 2 NH3 ⇌ [Ag(NH3)2]+ + Cl-    K = KspKf = 3.1 × 10-3
Figure 15.5 Two ammonia molecules replace the water around a silver ion and form a linear complex. Because the free Ag+ is swept up, the AgCl equilibrium above it is pulled to the right and more solid dissolves.

Kf measures the grip

Complex formation is an equilibrium like any other. Its constant is the formation constant, Kf. It is sometimes called a stability constant.

Ag+(aq) + 2 NH3(aq) ⇌ [Ag(NH3)2]+(aq)   Kf = 1.7 × 107 Kf = [Ag(NH3)2+] / ([Ag+][NH3]2)

Formation constants are large, often huge. Kf for [Ag(CN)2]- is about 1 × 1021. A big Kf means the complex wins. Nearly every metal ion ends up inside one, and the free ion concentration drops to almost nothing. The reverse reaction has the flipped constant, Kd = 1/Kf, and is called the dissociation constant.

Worked example · how little free silver is left

You make a solution that is 1.0 × 10-3 M in total silver and 1.00 M in NH3. Find the free [Ag+]. Kf = 1.7 × 107.

Step 1 · assume the complex winsKf is enormous and ammonia is in large excess. So treat almost all the silver as complexed: [Ag(NH3)2+] = 1.0 × 10-3 M.

Step 2 · count the ammonia usedEach complex uses 2 NH3, so 2.0 × 10-3 M is tied up. Free [NH3] = 1.00 - 0.0020 = 0.998 M.

Step 3 · rearrange Kf[Ag+] = [Ag(NH3)2+] / (Kf[NH3]2)

Step 4 · plug in[Ag+] = 1.0 × 10-3 / (1.7 × 107 × 0.996) = 5.9 × 10-11 M

Answer5.9 × 10-11 M. Of every ten million silver ions added, fewer than one is still swimming free. That is why ammonia keeps silver in solution.

Complexes dissolve solids that water cannot

Now put the two ideas together. Solid AgCl sets up a tiny supply of Ag+. Ammonia pulls that Ag+ into a complex. The solid responds by releasing more, and the dissolving keeps going. Adding the two equations gives one overall reaction, and multiplying the two constants gives its constant.

AgCl(s) + 2 NH3(aq) ⇌ [Ag(NH3)2]+(aq) + Cl-(aq) K = Ksp × Kf = (1.8 × 10-10)(1.7 × 107) = 3.1 × 10-3

Worked example · silver chloride in 1.0 M ammonia

How many moles of AgCl dissolve in one liter of 1.0 M NH3? Use K = 3.06 × 10-3.

Step 1 · set upLet s be the moles dissolved. Then the complex and the chloride are each at s, and [NH3] = 1.0 - 2s.

Step 2 · write KK = s2 / (1.0 - 2s)2 = 3.06 × 10-3

Step 3 · take the square root of both sidess / (1.0 - 2s) = (3.06 × 10-3)1/2 = 0.0553

Step 4 · solve the lines = 0.0553 - 0.1106s, so 1.1106s = 0.0553 and s = 0.0498 M

Step 5 · check[NH3] = 1.0 - 0.0996 = 0.900. Then (0.0498)2 / (0.900)2 = 3.06 × 10-3. It holds.

Answer0.050 M, against 1.3 × 10-5 M in pure water. Ammonia raised the solubility about 3,700 times. Taking the square root worked here only because both sides were perfect squares.

Key points
  • A complex ion is a metal ion surrounded by ligands that donate lone pairs.
  • Kf is the formation constant, and it is usually very large.
  • A large Kf drives free metal ion concentration down to almost zero.
  • Coordination number is how many donor atoms attach. Silver takes 2, zinc 4, and iron often 6.
  • Add a dissolving step to a complexing step and multiply Ksp by Kf.

Section 5Amphoteric hydroxides

Some hydroxides go both ways

Add a little base to a solution of Al3+ and a white gel of Al(OH)3 drops out. Split that gel between two test tubes. Pour acid into one and the gel dissolves. Pour concentrated NaOH into the other and that gel dissolves too. A solid that reacts as a base with acid and as an acid with base is amphoteric.

The two reactions are different in kind, which is the whole point. With acid, the hydroxide ions in the solid get taken by H3O+, and the metal leaves as a simple cation.

Al(OH)3(s) + 3 H3O+(aq) → Al3+(aq) + 6 H2O(l)

With extra base, the metal takes on one more hydroxide as a ligand. The product is a soluble hydroxo complex.

Al(OH)3(s) + OH-(aq) ⇌ [Al(OH)4]-(aq)

Zinc behaves the same way, with a coordination number of 4 on the acid side and 4 hydroxides on the base side. Beryllium, chromium(III), tin(II) and lead(II) hydroxides are amphoteric as well. Magnesium and iron(III) hydroxides are not. They dissolve in acid and then just sit there in base, no matter how much you add.

The V-shaped solubility curve

Plot the total dissolved metal against pH and an amphoteric hydroxide traces a V. On the left, acid eats the hydroxide and the free cation rules. Each pH unit up cuts solubility hard, because [OH-] rises tenfold and enters Ksp raised to a power. On the right, the hydroxo complex takes over and solubility climbs again. The bottom of the V is the pH where the solid is hardest to dissolve.

Total dissolved zinc against pH for zinc hydroxide, a V shaped curve with free zinc ion on the acid side and the tetrahydroxozincate complex on the base side Zn(OH)₂ solubility against pH: a V with a floor near pH 10.1 10⁻¹ 10⁻³ 10⁻⁵ 10⁻⁷ 10⁻⁹ 2 4 6 8 10 12 14 pH total dissolved Zn (M) floor: 4.8 × 10⁻⁹ M at pH 10.1 acid side [Zn2+] = Ksp / [OH-]2 solid is a base here base side [Zn(OH)42-] = 0.126[OH-]2 solid is an acid here
Figure 15.6 Total dissolved zinc against pH for Zn(OH)2. The falling branch is free Zn2+. The rising branch is [Zn(OH)4]2-. The floor sits near pH 10.1, where only about 5 × 10-9 M zinc stays in solution.

Both branches come from constants you already know how to use. The left branch is plain Ksp. For Zn(OH)2, Ksp = 4.5 × 10-17, so [Zn2+] = Ksp / [OH-]2. The right branch is a coupled reaction, Ksp times Kf for the hydroxo complex.

Zn(OH)2(s) + 2 OH- ⇌ [Zn(OH)4]2-   K = KspKf

Worked example · zinc hydroxide in 0.10 M NaOH

Ksp(Zn(OH)2) = 4.5 × 10-17 and Kf([Zn(OH)4]2-) = 2.8 × 1015. Find the dissolved zinc in 0.10 M NaOH, and compare it with pure water.

Step 1 · combine the stepsZn(OH)2(s) + 2 OH- ⇌ [Zn(OH)4]2-, so K = (4.5 × 10-17)(2.8 × 1015) = 0.126

Step 2 · write the expressionK = [Zn(OH)42-] / [OH-]2 = 0.126

Step 3 · put in 0.10 M OH-[Zn(OH)42-] = 0.126 × (0.10)2 = 0.126 × 0.010 = 1.3 × 10-3 M

Step 4 · get the pure-water valueZn(OH)2 is a 1:2 salt, so Ksp = 4s3. Then s3 = 1.125 × 10-17 and s = 2.2 × 10-6 M.

Answer1.3 × 10-3 M in 0.10 M NaOH, against 2.2 × 10-6 M in water. Strong base raised zinc solubility about 560 times. Note that the hydroxide here is both a common ion and a ligand, and the ligand role wins.

Dilute base and concentrated base do opposite things to the same solid. A trace of OH- is a common ion, so it drives zinc out of solution. A flood of OH- is a ligand, so it pulls zinc back in. That is the V in one sentence.

Why this matters in a water plant

Water treatment removes metals by raising the pH until hydroxides fall out. The operator must stop at the floor of the V. Go past it and amphoteric metals such as zinc, aluminum and lead start redissolving as hydroxo complexes. For Zn(OH)2 the floor sits near pH 10.1, and for Al(OH)3 it sits near pH 6. A plant that overshoots sends metal back out in the finished water.

Key points
  • Amphoteric hydroxides dissolve in acid and in excess strong base.
  • Acid gives the free cation. Excess base gives a hydroxo complex such as [Al(OH)4]-.
  • Al, Zn, Cr(III), Be, Sn(II) and Pb(II) hydroxides are amphoteric. Mg and Fe(III) are not.
  • Solubility against pH makes a V, and the floor is the best pH for removing the metal.
  • A little OH- is a common ion. A lot of OH- is a ligand.

Section 6Coupled equilibria

Two reactions, one constant

You have now seen the trick twice. A solid that will not dissolve gets a partner reaction that eats one of its ions. Then the solid has to keep feeding that reaction, so it dissolves. Two equilibria that share a species like this are coupled equilibria.

The bookkeeping is simple and it never changes. Add the two balanced equations and cancel the shared species. Multiply the two constants to get the constant of the sum.

If A ⇌ B has K1 and B ⇌ C has K2, then A ⇌ C has K = K1K2
Two coupled equilibria that share one ion, with the shared ion cancelling when the equations are added and the constants multiplied Add the equations, multiply the constants step 1 · the solid dissolves CaCO₃(s) ⇌ Ca2+ + CO₃2- Ksp = 8.7 × 10-9 step 2 · acid eats the anion CO₃2- + H₃O+ ⇌ HCO₃- + H₂O 1 / Ka2 = 2.1 × 1010 the shared ion CO₃2- step 1 makes it step 2 destroys it the sum, with CO₃2- cancelled CaCO₃(s) + H₃O+ ⇌ Ca2+ + HCO₃- + H₂O K = Ksp × (1 / Ka2) = (8.7 × 10-9)(2.1 × 1010) = 1.9 × 102
Figure 15.7 The shared ion is the link. One equilibrium makes it and the other destroys it. Adding the equations cancels the link, and the overall constant is the product of the two.

Watch the direction. If you have to flip an equation before adding, flip its constant to 1/K as well. If you double an equation, square its constant. These are the same rules you used for gas-phase equilibria.

Lever one: acid, for salts of weak-acid anions

Many stubborn salts carry an anion that is the conjugate base of a weak acid. Carbonate, phosphate, sulfide, fluoride, oxalate and hydroxide all qualify. Add acid and that anion gets protonated and leaves the equilibrium. Limestone is the classic case.

Worked example · acid on limestone

Find K for CaCO3(s) + H3O+ ⇌ Ca2+ + HCO3- + H2O. Use Ksp(CaCO3) = 8.7 × 10-9 and Ka2 for carbonic acid = 4.7 × 10-11.

Step 1 · the dissolving stepCaCO3(s) ⇌ Ca2+ + CO32-, Ksp = 8.7 × 10-9

Step 2 · the acid step, flippedHCO3- + H2O ⇌ CO32- + H3O+ has Ka2 = 4.7 × 10-11. Reverse it and the constant becomes 1/Ka2 = 2.13 × 1010.

Step 3 · add and cancelCO32- appears as a product in step 1 and a reactant in step 2, so it cancels. The sum is the reaction asked for.

Step 4 · multiplyK = Ksp / Ka2 = (8.7 × 10-9)(2.13 × 1010) = 1.9 × 102

AnswerK = 1.9 × 102. That is far above 1, so even dilute acid dissolves limestone easily. Rainwater carrying dissolved CO2 is weak acid, and over long spans it carves caves out of solid rock.

Push the pH lower and the bicarbonate turns into H2CO3, which falls apart into CO2 gas. The gas bubbles away and leaves the solution for good. Now a third process is pulling, and the limestone has no way to come back. That fizz is the test every geology field kit relies on.

Acid does nothing useful for a salt whose anion is a very weak base. Chloride, bromide, iodide and nitrate hold almost no protons at ordinary pH. So AgCl shrugs off nitric acid, while CaCO3 and Mg(OH)2 both give way at once.

Lever two: a ligand, for the metal side

If acid cannot touch the anion, go after the cation instead. A ligand with a large Kf removes the metal ion and the solid follows. This is how film and prints were fixed for a century. Sodium thiosulfate dissolves the unexposed silver bromide and leaves the image behind.

Worked example · thiosulfate on silver bromide

Ksp(AgBr) = 5.0 × 10-13 and Kf([Ag(S2O3)2]3-) = 2.9 × 1013. Find the molar solubility of AgBr in 1.0 M Na2S2O3.

Step 1 · add the two stepsAgBr(s) + 2 S2O32- ⇌ [Ag(S2O3)2]3- + Br-

Step 2 · multiply the constantsK = (5.0 × 10-13)(2.9 × 1013) = 14.5

Step 3 · set up with sK = s2 / (1.0 - 2s)2 = 14.5

Step 4 · square root and solves / (1.0 - 2s) = 3.808, so s = 3.808 - 7.616s. Then 8.616s = 3.808 and s = 0.442 M

Step 5 · check[S2O32-] = 1.0 - 0.884 = 0.116, and (0.442)2 / (0.116)2 = 14.5. It holds.

Answer0.44 M. In pure water AgBr reaches only 7.1 × 10-7 M, so the fixer raised solubility by a factor of about 600,000. Note that s is a large fraction of the thiosulfate, so you could not drop the 2s term here.

One caution. These overall constants are only as good as their parts. A K built from two table values carries the error of both. Treat an answer like 0.44 M as a sound estimate, not a measurement.

Key points
  • Coupled equilibria share a species. Add the equations and multiply the constants.
  • Flip an equation and its constant flips to 1/K. Double it and the constant is squared.
  • Acid dissolves salts of weak-acid anions such as carbonate, phosphate, sulfide and hydroxide.
  • Acid will not help with chloride, bromide, iodide or nitrate salts.
  • A ligand with a large Kf dissolves a solid by stripping out the metal ion.
  • Escaping gas such as CO2 adds a third pull that the system cannot undo.

The chapter in one breath

  • Ksp is the equilibrium constant for a solid dissolving, and the solid is never in it.
  • Molar solubility and Ksp convert through the salt's ion ratio: s2, 4s3 or 108s5.
  • Compare two Ksp values only when the salts have the same ion ratio.
  • A common ion pushes the equilibrium back to solid, so solubility falls, sometimes by thousands of times.
  • Build Q from the diluted concentrations and compare it with Ksp to predict a precipitate.
  • Selective precipitation drops the smaller-Ksp salt first and can split ions almost completely.
  • A ligand wraps a metal ion into a complex, and a large Kf leaves almost no free metal.
  • Amphoteric hydroxides dissolve in acid and in excess base, giving a V-shaped solubility curve.
  • Couple two equilibria by adding the equations and multiplying the constants, then read the overall K.

ReferenceWords worth knowing

These are the terms this chapter leans on. Say each one out loud and you will find the idea comes with it.

Amphoteric
Able to react as a base with acid and as an acid with base. Al(OH)3 and Zn(OH)2 both dissolve either way.
Common ion
An ion supplied by two different solutes in the same solution, such as Cl- from both AgCl and NaCl.
Common ion effect
The drop in solubility caused by adding an ion the salt already releases.
Complex ion
A central metal ion joined to several ligands, carrying an overall charge, such as [Ag(NH3)2]+.
Coordination number
How many donor atoms are bonded to the central metal ion. Silver usually takes 2 and zinc 4.
Coupled equilibria
Two equilibria that share a species, so shifting one shifts the other.
Dissociation constant (Kd)
The constant for a complex falling apart. It is 1/Kf.
Formation constant (Kf)
The equilibrium constant for a metal ion plus ligands forming a complex ion. Also called a stability constant.
Hydroxo complex
A complex in which the ligands are hydroxide ions, such as [Al(OH)4]-.
Ion product (Q)
The same product of ion concentrations as Ksp, but built from the amounts present right now.
Ligand
A molecule or ion that donates a lone pair to a metal ion. NH3, CN- and H2O are common ones.
Molar solubility
Moles of a solid that dissolve in one liter of solution to reach saturation. Its symbol here is s.
Precipitate
A solid that forms inside a solution and separates from it.
Qualitative analysis
A scheme that identifies the ions in an unknown by dropping them out in planned groups.
Salt effect
A small rise in solubility caused by unrelated ions screening charges in the solution.
Saturated solution
A solution holding all the solute it can at that temperature, with Q equal to Ksp.
Selective precipitation
Adding a reagent slowly so the least soluble salt forms first and one ion is separated from another.
Solubility product constant (Ksp)
The equilibrium constant for a slightly soluble solid breaking into its ions in water.
Supersaturated solution
A solution holding more dissolved solute than equilibrium allows, with Q above Ksp.
Unsaturated solution
A solution that could still dissolve more solute, with Q below Ksp.

Check yourself

1. Silver bromide has Ksp = 5.0 × 10-13. What is its molar solubility in pure water?
7.1 × 10-7 M. AgBr is 1:1, so Ksp = s2. Then s = (5.0 × 10-13)1/2 = 7.07 × 10-7 M. Check: (7.07 × 10-7)2 = 5.0 × 10-13.
2. Calcium fluoride has Ksp = 3.9 × 10-11. Find its molar solubility and the fluoride concentration at saturation.
s = 2.1 × 10-4 M and [F-] = 4.3 × 10-4 M. CaF2 is 1:2, so Ksp = 4s3. Then s3 = 9.75 × 10-12 and s = 2.14 × 10-4 M. Fluoride is 2s.
3. Why does the amount of undissolved solid never appear in a Ksp expression?
Because a pure solid has a fixed activity of 1. Its particles are packed at one unchanging density, so piling on more solid cannot raise anything in the expression. The surface area changes how fast equilibrium is reached, not where it lands.
4. Find the molar solubility of AgCl in 0.050 M NaCl. Ksp = 1.8 × 10-10.
3.6 × 10-9 M. [Cl-] ≈ 0.050 M, so s = Ksp / 0.050 = 1.8 × 10-10 / 0.050 = 3.6 × 10-9 M. That is roughly 3,700 times less than the 1.3 × 10-5 M it reaches in pure water.
5. Silver chromate has a smaller Ksp than silver chloride, yet it dissolves ten times more. How can both be true?
The two salts have different ion ratios. AgCl uses Ksp = s2 while Ag2CrO4 uses 4s3. A cube root of a tiny number is much larger than a square root of one, so you cannot rank them by Ksp alone.
6. You mix equal volumes of 2.0 × 10-3 M BaCl2 and 2.0 × 10-3 M Na2SO4. Does BaSO4 form? Ksp = 1.1 × 10-10.
Yes, easily. Equal volumes halve each one, so [Ba2+] = [SO42-] = 1.0 × 10-3 M. Then Q = (1.0 × 10-3)(1.0 × 10-3) = 1.0 × 10-6. That is about 9,100 times Ksp, so solid drops out.
7. A solution is 0.010 M in Ba2+ and 0.010 M in Ca2+. You add sulfate slowly. Which precipitates first, and what sulfate concentration starts it? Ksp(BaSO4) = 1.1 × 10-10, Ksp(CaSO4) = 2.4 × 10-5.
BaSO4 first, starting at 1.1 × 10-8 M sulfate. Divide each Ksp by 0.010. Barium needs 1.1 × 10-8 M and calcium needs 2.4 × 10-3 M, which is 220,000 times more. When calcium finally starts, [Ba2+] is down to 1.1 × 10-10 / 2.4 × 10-3 = 4.6 × 10-8 M.
8. Name the ligand and the coordination number in [Cu(NH3)4]2+, and give the charge on the central ion.
Ammonia is the ligand, the coordination number is 4, and copper is 2+. NH3 is neutral, so the whole 2+ charge sits on the copper. Each ammonia donates the lone pair on its nitrogen atom.
9. Work out K for AgCl(s) + 2 NH3 ⇌ [Ag(NH3)2]+ + Cl-. Use Ksp = 1.8 × 10-10 and Kf = 1.7 × 107.
K = 3.1 × 10-3. Add the dissolving step to the complexing step and multiply: (1.8 × 10-10)(1.7 × 107) = 3.06 × 10-3. K is still below 1, but with 1.0 M ammonia it is enough to dissolve about 0.050 mol per liter.
10. You have four fresh precipitates: Mg(OH)2, Al(OH)3, Fe(OH)3 and Zn(OH)2. Which dissolve when you add excess 6 M NaOH?
Al(OH)3 and Zn(OH)2. Both are amphoteric, so they pick up extra hydroxide and leave as [Al(OH)4]- and [Zn(OH)4]2-. Mg(OH)2 and Fe(OH)3 are not amphoteric and stay solid in any amount of base.
11. A buffer holds a solution at pH 10.00. How much Mg2+ can stay dissolved in contact with Mg(OH)2? Ksp = 2.1 × 10-13.
2.1 × 10-5 M. At pH 10.00, pOH = 4.00, so [OH-] = 1.0 × 10-4 M. Then [Mg2+] = Ksp / [OH-]2 = 2.1 × 10-13 / 1.0 × 10-8 = 2.1 × 10-5 M. Raise the pH two more units and it falls another factor of 10,000.
12. Nitric acid dissolves CaCO3 at once but leaves AgCl untouched. Explain the difference, and give K for the limestone reaction. Ksp(CaCO3) = 8.7 × 10-9, Ka2 = 4.7 × 10-11.
Carbonate is a strong enough base to grab protons. Chloride is not. K = 1.9 × 102. Dividing gives K = 8.7 × 10-9 / 4.7 × 10-11 = 185. Acid removes CO32- as HCO3-, then as CO2 gas, so the solid keeps feeding it. Cl- is the conjugate base of a strong acid, so H3O+ cannot take it away.
Chemistry · Chapter 16

Thermodynamics

Some changes happen all by themselves. Ice melts in a warm room. Iron rusts. A drop of dye spreads through water. None of them need a push. Other changes never start on their own, no matter how long you wait. This chapter shows you how to tell the two apart. You will meet entropy, a measure of how spread out matter and energy are. You will meet the second law, which says the entropy of the universe always grows. Then comes Gibbs free energy. Its sign tells you which way a change will go, and its size sets the value of K.

By the end of this chapter you can

  • Say what makes a change spontaneous, and why that is not the same as fast.
  • Describe entropy as the spread of matter and of energy, and use S = k ln W.
  • State the second and third laws, and add up ΔS for a system plus its surroundings.
  • Find ΔS° for a reaction from a table of standard entropies.
  • Predict the sign of ΔS° by counting moles of gas.
  • Use ΔG = ΔH − TΔS and sort out the four sign cases.
  • Link ΔG° to K, and find the crossover temperature where the sign flips.

Section 1Spontaneous change

What spontaneous means

A change is spontaneous if it keeps going on its own once it starts. Nothing outside has to drive it. A ball rolls down a hill. A hot mug cools on a desk. Salt dissolves in water. Each one runs one way and settles there.

A change that needs steady outside work is nonspontaneous. Rolling the ball back up the hill is one. Heating the mug back up is another. The reverse of a spontaneous change is always nonspontaneous. Together the pair acts like a one-way street with a sign posted on it.

Be careful with the word itself. In chemistry it does not mean sudden. It does not mean violent. It only means the change is favored in that direction.

A one-way change: gas spreads to fill both bulbs spontaneous never on its own valve shut 12 particles, one bulb valve open 12 particles, two bulbs No energy is released here. The gas spreads because spread-out is the overwhelmingly likely outcome.
A change with a direction. Open the valve and the gas fills both bulbs. Wait as long as you like and it will not crowd back into one bulb. Nothing about energy explains this, because an ideal gas releases no heat when it expands.

Spontaneous does not mean fast

Here is the trap that catches almost everyone: spontaneous says nothing at all about the speed of a change.

Diamond makes the point, because at room temperature diamond is favored to turn into graphite. Graphite is the standard form of carbon, so its free energy of formation is 0 kJ/mol. For diamond the table lists +2.9 kJ/mol.

Worked example · does diamond last forever?

The change C(diamond) → C(graphite)

Set up ΔG° = ΔG°f(graphite) − ΔG°f(diamond)

Put in numbers ΔG° = 0 kJ/mol − (+2.9 kJ/mol) = −2.9 kJ/mol

Read the sign The value is negative, so graphite is the favored side. Diamond is the one that should change.

Reality check No ring has ever turned grey. The atoms sit in strong bonds and cannot move without a huge push. The rate is near zero, so the jewel is safe.

Thermodynamics names the destination. Kinetics sets the travel time. Keep the two jobs apart and this chapter gets much easier.

Heat alone does not decide

Many spontaneous changes give off heat. Wood burns. An acid neutralizes a base. Both warm the room. So people once thought heat release was the whole rule.

It is not the whole rule, as an ice cube melting at 25 °C shows. Melting soaks heat up, so it is endothermic, and yet it still happens with no help at all. A cold pack does the same trick. Ammonium nitrate dissolves, the bag turns cold, and nobody pushes it.

So a second force has to be at work. Energy is only half the story, and the other half is how spread out the matter and the energy end up. The next section gives that idea a name and a number.

Key points
  • Spontaneous means the change goes forward with no outside drive.
  • The reverse of a spontaneous change always needs work from outside.
  • Spontaneous does not mean fast. Diamond to graphite is favored and still takes ages.
  • Giving off heat helps, but endothermic changes can be spontaneous too.
  • Spreading of matter and energy is the missing second factor.

Section 2Entropy

Entropy counts the ways

Entropy gets the symbol S. It measures how spread out the matter and the energy of a sample are. A state with many ways to arrange itself has high entropy. A state with few ways has low entropy.

Picture four gas particles in a box split down the middle. Each particle can sit left or right. That gives 2 × 2 × 2 × 2 = 16 full arrangements. Each arrangement is one microstate. Only one microstate puts all four on the left, while six of them put two on each side. So the even split is six times as likely as the lopsided pile.

Now swap in a real sample, where one mole holds 6.022 × 1023 particles. The even split still wins, but the margin becomes absurd. The one-sided case has odds so tiny that it never shows up. That is the real reason gas fills both bulbs and stays there.

Counting microstates for four particles in two halves 1 4 left, 0 right 4 3 left, 1 right 6 2 left, 2 right 4 1 left, 3 right 1 0 left, 4 right Each pair of boxes shows one example arrangement for that split. Number of arrangements behind each split, 16 in total The even split has 6 arrangements. The all-on-one-side split has only 1. Spread out wins because it has more ways to happen.
Why spread out wins. Four particles can split five ways, but the splits are not equally likely. The middle split has six arrangements behind it and the end splits have one each. Add more particles and the middle peak grows sharper.

The Boltzmann equation

Ludwig Boltzmann turned that counting into a formula.

S = k ln W

Here W is the number of microstates. The letter k is the Boltzmann constant, 1.381 × 10−23 J/K. Watch what the log does, because multiplying W by ten raises S by only k ln 10. Entropy creeps up while the number of ways explodes.

Worked example · entropy of four particles, then of a mole

Part A Four particles, two halves, so W = 24 = 16.

Put in numbers S = (1.381 × 10−23 J/K)(ln 16) = (1.381 × 10−23)(2.773) = 3.83 × 10−23 J/K

Part B Let one mole of gas double its volume. Each particle now has twice as many spots, so W doubles once per particle.

Ratio of ways Wfinal / Wstart = 2NA

Take the log ΔS = k ln(2NA) = NAk ln 2 = R ln 2

Finish ΔS = (8.314 J/mol·K)(0.6931) = +5.76 J/mol·K

Check The gas spread out, so ΔS came out positive. Note that NAk is just R, the same gas constant from Chapter 9.

Entropy is not mess. A messy room has nothing to do with it. Entropy counts real arrangements of particles and of energy, and only a sample of matter has those.

Energy spreads too

Matter is one half of the story. Energy is the other half. Particles store energy by moving, spinning and vibrating. Each of those motions has its own ladder of energy levels. The sample's energy can be shared over those levels in many ways, and each way is another microstate.

This explains why entropy climbs with temperature, since added heat lifts particles onto higher rungs. More rungs come into play, so there are more ways to park the same energy.

Four patterns follow from all this, and they answer most questions you will face.

Key points
  • Entropy S measures the spread of matter and of energy.
  • A microstate is one full arrangement of where the particles are and which levels hold the energy.
  • S = k ln W, with k = 1.381 × 10−23 J/K.
  • Doubling the volume of one mole of gas adds R ln 2 = 5.76 J/mol·K.
  • Entropy units are J/mol·K, not kJ/mol·K. That detail matters later.

Section 3The second and third laws

The second law

The second law of thermodynamics gives the real test. A change is spontaneous when the total entropy of the universe goes up.

ΔSuniv = ΔSsys + ΔSsurr > 0

Two ledgers are open at once. The system is the part you care about, such as the chemicals in a flask. The surroundings are everything else, such as the glass, the bench and the air. Either ledger may show a loss. Only the sum has to grow.

That rule is how an endothermic change can still be spontaneous. Melting ice raises the entropy of the water by a lot. It lowers the entropy of the room by a little less. The sum comes out positive, so the ice melts.

Heat and the surroundings

The surroundings are enormous, so their temperature barely budges when they take in heat. That makes their entropy change easy to write.

ΔSsurr = −qsys / T = −ΔHsys / T

Read the minus sign with care, because heat that leaves the system lands in the surroundings. An exothermic change has a negative ΔH, so −ΔH/T turns positive and the surroundings gain entropy.

Now look again at the T sitting on the bottom of that fraction. The same joule of heat buys more entropy in a cold room than in a hot one. A dollar means more to a person with almost nothing. This single fact is why temperature decides so many reactions.

Worked example · melting ice at three temperatures

One mole of ice melts. Fusion takes in 6.01 kJ/mol, which is 6010 J/mol. The melt itself happens at 273.15 K, so the system gains the same entropy each time.

System ΔSsys = 6010 J / 273.15 K = +22.0 J/K

Surroundings at 0 °C ΔSsurr = −6010 J / 273.15 K = −22.0 J/K. Total = 22.0 − 22.0 = 0 J/K, so ice and water sit together and neither one wins.

Surroundings at +10 °C ΔSsurr = −6010 J / 283.15 K = −21.2 J/K. Total = 22.0 − 21.2 = +0.8 J/K, so the ice melts.

Surroundings at −10 °C ΔSsurr = −6010 J / 263.15 K = −22.8 J/K. Total = 22.0 − 22.8 = −0.8 J/K, so the ice holds. Freezing is the spontaneous way instead.

What moved The system term never changed. Only the warmth of the room changed, and that was enough to flip the sign.

System plus surroundings entropy for melting ice +22.0 system −22.8 surroundings −10 °C total −0.8 J/K · stays frozen +22.0 system −22.0 surroundings 0 °C total 0.0 J/K · balanced +22.0 system −21.2 surroundings +10 °C total +0.8 J/K · melts ΔS for one mole of ice melting, in J/K The system bar hardly moves. The surroundings bar shrinks as the room warms, and that is what flips the total. 0
The ledger that decides. Melting always costs the surroundings entropy and always pays the system entropy. Warming the room shrinks the cost, because the same 6010 J is divided by a larger T. At 0 °C the two exactly cancel, which is what equilibrium looks like.

The third law

The third law of thermodynamics sets the zero point. A perfect crystal of a pure substance at 0 K has an entropy of exactly zero.

The reason is pure counting, because at absolute zero each particle sits in one fixed spot. All the energy rests in the lowest level. There is only one microstate, so W = 1 and S = k ln 1 = 0.

This matters more than it first looks, since enthalpy has no natural zero and tables can only list changes such as ΔH°f. Entropy, by contrast, does have a true zero of its own. So a table can list a real, absolute value for every substance. Those values are standard molar entropies, written S°, measured at 298.15 K and 1 bar. Every one of them is positive, even for elements. The next section spends them.

Key points
  • Second law: a change is spontaneous when ΔSsys + ΔSsurr is positive.
  • ΔSsurr = −ΔHsys/T at steady pressure.
  • Cold surroundings gain more entropy per joule than warm ones.
  • ΔSuniv = 0 is the fingerprint of equilibrium.
  • Third law: a perfect pure crystal at 0 K has S = 0, so S° values are absolute and positive.

Section 4Standard entropy change

Adding up the table

Because entropy has a true zero, a table can list one absolute value per substance. Those S° values are what you spend here. Every number in this chapter is taken from the standard thermodynamic table in Appendix G of OpenStax Chemistry 2e, measured at 298.15 K and 1 bar.

The standard entropy change of a reaction is built the same way as a heat of reaction.

ΔS°rxn = Σ n S°(products) − Σ n S°(reactants)

Multiply each S° by its coefficient from the balanced equation. Then subtract the reactant total from the product total. Two habits will save you here. Elements do not drop out, because their S° is not zero. And the answer arrives in J/K, since S° is tabulated in J/mol·K.

SubstanceS° (J/mol·K)SubstanceS° (J/mol·K)
H2(g)130.7H2O(l)70.0
O2(g)205.2H2O(g)188.8
N2(g)191.6NH3(g)192.8
Cl2(g)223.1HCl(g)186.9
C(graphite)5.7CO2(g)213.8
CaCO3(s)91.7CaO(s)38.1
I2(s)116.1I2(g)260.7
C2H4(g)219.3C2H6(g)229.2

Worked example · burning hydrogen to liquid water

Equation 2 H2(g) + O2(g) → 2 H2O(l)

Products 2 mol × 70.0 J/mol·K = 140.0 J/K

Reactants 2 mol × 130.7 + 1 mol × 205.2 = 261.4 + 205.2 = 466.6 J/K

Subtract ΔS° = 140.0 J/K − 466.6 J/K = −326.6 J/K

Does it make sense? Three moles of gas became zero moles of gas. A big drop was the only possible answer.

Worked example · roasting limestone

Equation CaCO3(s) → CaO(s) + CO2(g)

Products 38.1 + 213.8 = 251.9 J/K

Reactants 91.7 J/K

Subtract ΔS° = 251.9 J/K − 91.7 J/K = +160.2 J/K

Does it make sense? A solid released a mole of gas. One free gas molecule is worth far more entropy than the solid it came from. Hold on to this value, because Section 5 uses it again.

Reading the sign without a table

Often the sign is all you need, so count the moles of gas on each side of the balanced equation. Gas holds so much more entropy than liquid or solid that the gas count almost always decides.

When the gas count ties, reach for the backup clues. A solid that melts or dissolves gains entropy. Two small molecules joining into one large one lose it. More total particles beats fewer.

Try the tie case on HCl. For H2(g) + Cl2(g) → 2 HCl(g) the table gives 2(186.9) − (130.7 + 223.1) = 373.8 − 353.8 = +20.0 J/K. Two moles of gas went to two moles of gas, and the value is tiny next to the −326.6 J/K above.

Moles of gas before and after set the sign of the entropy change Count the moles of gas, then read the sign moles of gas in the equation 0 1 2 3 4 ΔS° (J/K) 2 H2​(g) + O2​(g) → 2 H2​O(l) −326.6 N2​(g) + 3 H2​(g) → 2 NH3​(g) −198.1 H2​(g) + Cl2​(g) → 2 HCl(g) +20.0 CaCO3​(s) → CaO(s) + CO2​(g) +160.2 Hollow dot = reactant side. Filled dot = product side. Equal counts give a small value, so the phases decide instead.
The gas count rules. Each reaction sits on a track of moles of gas. The arrow shows where the gas count moves, and the number on the right is the real tabulated ΔS°. Big moves give big values. The tie case gives a value near zero.
Key points
  • ΔS°rxn = sum of nS° for products minus sum of nS° for reactants.
  • S° for an element is positive, so elements never drop out of the sum.
  • ΔS° comes out in J/K, while ΔH° comes out in kJ.
  • Gas moles rising means a positive ΔS°. Gas moles falling means a negative one.
  • When the gas count ties, ΔS° is small and the phases decide.

Section 5Gibbs free energy

One number, system only

Tracking the surroundings every time is a chore, and Josiah Willard Gibbs found a way around it. Start from the second law, then swap in the heat term from Section 3.

ΔSsys − ΔHsys/T > 0

Now multiply every term by −T. Because T is always positive on the Kelvin scale, multiplying by a negative number flips the direction of the sign.

ΔHsys − TΔSsys < 0

Gibbs gave that left side a name. Gibbs free energy carries the symbol G, and its change is the whole answer.

ΔG = ΔH − TΔS

At steady temperature and pressure, one sign settles the case. A negative ΔG means the forward change is spontaneous. A positive ΔG means the reverse one is. A ΔG of zero means the system is at equilibrium. Best of all, every term belongs to the system, so the surroundings never enter the work.

You can also build ΔG° straight from a table, the same way as ΔH°. Appendix G lists a standard free energy of formation, ΔG°f, for each substance. Sum products minus reactants. That shortcut only works at 298.15 K, since T is baked into the listed values.

The unit trap

This equation is where units bite hardest. ΔH° comes out of the table in kJ. ΔS° comes out in J/K. You cannot subtract one from the other until they match.

Pick one fix and stick with it: divide ΔS° by 1000 to reach kJ/K, or multiply ΔH° by 1000 to reach J. Forget this and your answer is wrong by a factor of a thousand. The sign often comes out wrong too.

Worked example · roasting limestone, cold and hot

Equation CaCO3(s) → CaO(s) + CO2(g)

Enthalpy ΔH° = (−634.9) + (−393.5) − (−1207.6) = +179.2 kJ

Entropy ΔS° = +160.2 J/K from Section 4. Convert it: 160.2 J/K ÷ 1000 = 0.1602 kJ/K

At 298.15 K ΔG° = 179.2 kJ − (298.15 K)(0.1602 kJ/K) = 179.2 − 47.8 = +131.4 kJ

Check it another way From the free energies of formation: (−603.3) + (−394.4) − (−1129.1) = +131.4 kJ. The two routes agree.

At 1200 K ΔG° = 179.2 kJ − (1200 K)(0.1602 kJ/K) = 179.2 − 192.2 = −13.0 kJ

Meaning Limestone sits safely on a cold shelf. Push a lime kiln near 1200 K and it breaks apart on its own. Nothing about the chemicals changed. Only the size of the TΔS term changed.

A word about the name. Free energy is not energy that is free of charge. It is the share of the energy change that is free to do useful work. The rest is owed to the surroundings as spread.

The four sign cases

Since ΔH° and ΔS° each have two possible signs, there are four combinations. Two of them give the same answer at every temperature, while the other two switch over somewhere.

ΔHΔSSign of ΔGVerdict
negativepositivealways negativeSpontaneous at every temperature. Both terms help.
positivenegativealways positiveNever spontaneous. Both terms fight you.
negativenegativenegative when T is lowSpontaneous when cold. Heat release has to outweigh TΔS.
positivepositivenegative when T is highSpontaneous when hot. TΔS has to outgrow ΔH.
The four sign cases of Gibbs free energy Spontaneous at every T ΔG is negative at all temperatures. 2 H2​O2​(l) → 2 H2​O(l) + O2​(g) ΔH° = −196.0 kJ, ΔS° = +126.0 J/K Spontaneous only when hot TΔS must grow past ΔH. CaCO3​(s) → CaO(s) + CO2​(g) ΔH° = +179.2 kJ, ΔS° = +160.2 J/K, flips at 1119 K Spontaneous only when cold ΔH wins while TΔS stays small. N2​(g) + 3 H2​(g) → 2 NH3​(g) ΔH° = −91.8 kJ, ΔS° = −198.1 J/K, flips at 463 K Never spontaneous ΔG is positive at all temperatures. 3 O2​(g) → 2 O3​(g) ΔH° = +285.4 kJ, ΔS° = −137.8 J/K ΔH° negative (gives off heat) ΔH° positive (takes in heat) ΔS° positive ΔS° negative ΔG = ΔH − TΔS: four cases, four behaviours The two gold boxes are the interesting ones. Each has a crossover temperature, found from T = ΔH° / ΔS°.
Four boxes cover every reaction. Real values fill each box. Peroxide breaking down is helped by both terms. Ozone forming from oxygen is blocked by both. Ammonia synthesis and limestone roasting each own a crossover temperature, and the next section finds it.
Key points
  • ΔG = ΔH − TΔS, and the sign of ΔG decides direction.
  • Negative means forward, positive means reverse, zero means equilibrium.
  • Always convert ΔS° from J/K to kJ/K before subtracting.
  • Two sign cases hold at all temperatures and two of them switch over.
  • ΔG°f sums give ΔG° directly, but only at 298.15 K.

Section 6Free energy, K and temperature

Free energy sets the size of K

Chapter 13 gave you the equilibrium constant K. Free energy is what fixes its value.

ΔG° = −RT ln K

Here R = 8.314 J/mol·K and T is in kelvin. Since R is written in joules, put ΔG° in joules as well. Rearranging gives K = e−ΔG°/RT.

The link is easy to read off. A negative ΔG° makes K larger than 1, so products win at equilibrium. A positive ΔG° makes K smaller than 1, so reactants win. A ΔG° of exactly zero gives K = 1.

Size matters as much as sign, because at 298.15 K every 5.71 kJ of ΔG° moves K by a factor of ten. That figure is just RT ln 10. So a modest −30 kJ already buys a huge K.

One caution before the math: ΔG° describes the standard state, with each gas at 1 bar and each solute at 1 mol/L. The push you feel at some other mixture is ΔG = ΔG° + RT ln Q. When Q climbs to K, the log term cancels ΔG° and ΔG drops to zero. That is equilibrium, seen from the free energy side.

Worked example · K for making ammonia at 298.15 K

Equation N2(g) + 3 H2(g) → 2 NH3(g)

Free energy ΔG° = 2(−16.4) − 0 − 0 = −32.8 kJ = −32800 J

Set up ln K = −ΔG°/RT = −(−32800 J) / [(8.314 J/mol·K)(298.15 K)]

Denominator (8.314)(298.15) = 2478.8 J/mol

Divide ln K = 32800 / 2478.8 = 13.23

Finish K = e13.23 = 5.6 × 105

Read it Ammonia is strongly favored at room temperature. The reaction is simply far too slow to use there.

Finding the crossover temperature

In the two switching cases a single temperature marks the turn, and you find it by setting ΔG to zero in ΔG = ΔH − TΔS.

T = ΔH° / ΔS°

That value is the crossover temperature. On one side of it ΔG° is negative, on the other side it is positive, and at the crossover itself K = 1. The method assumes ΔH° and ΔS° hardly drift with temperature, which holds well enough over a few hundred kelvin.

Test the idea on boiling water, where the table gives ΔH° = +44.0 kJ and ΔS° = +118.8 J/K for H2O(l) → H2O(g). So T = 44000 J ÷ 118.8 J/K = 370 K. The real boiling point is 373 K, which is a fine result from two table entries.

Worked example · where ammonia gives up

Data ΔH° = −91.8 kJ and ΔS° = −198.1 J/K, so ΔS° = −0.1981 kJ/K

Crossover T = (−91.8 kJ) ÷ (−0.1981 kJ/K) = 463 K, which is 190 °C

Now go hotter At 700 K, ΔG° = −91.8 − (700)(−0.1981) = −91.8 + 138.7 = +46.9 kJ

Turn that into K ln K = −46900 / [(8.314)(700)] = −46900 / 5819.8 = −8.06

Finish K = e−8.06 = 3.2 × 10−4

The real squeeze K fell from 5.6 × 105 to 3.2 × 10−4. Heat is needed for speed, yet heat ruins the yield. Plants answer by running at high pressure, which Chapter 13 showed pushes a gas reaction toward the smaller side.

Free energy against temperature and the crossover point −50 +0 +50 +100 +150 +200 200 400 600 800 1000 1200 1400 ΔG° = 0, so K = 1 crossover 463 K crossover 1119 K N2​ + 3 H2​ → 2 NH3​ CaCO3​ → CaO + CO2​ ΔG° (kJ) against temperature (K) Slope of each line is −ΔS°. The line meets the axis at ΔH°, and it crosses zero at T = ΔH°/ΔS°. Below the heavy line ΔG° is negative, so K is above 1. Above it K is below 1. Ammonia loses its advantage above 463 K. Limestone gains one above 1119 K. Same equation, opposite slopes.
Two straight lines, two turning points. Plot ΔG° against T and you get a line of slope −ΔS° and intercept ΔH°. Ammonia starts favored and climbs out of favor at 463 K. Limestone starts unfavored and drops into favor at 1119 K.
Key points
  • ΔG° = −RT ln K, with R = 8.314 J/mol·K and ΔG° in joules.
  • Negative ΔG° gives K above 1. Positive gives K below 1. Zero gives K = 1.
  • At 298.15 K, 5.71 kJ of ΔG° is worth one power of ten in K.
  • ΔG = ΔG° + RT ln Q, and ΔG = 0 once Q = K.
  • The crossover temperature is T = ΔH°/ΔS°, where K = 1.

The chapter in one breath

  • Spontaneous means a change goes forward on its own. It says nothing about how fast.
  • Entropy measures the spread of matter and of energy, and S = k ln W ties it to counting.
  • The second law is the real test. A change happens when ΔSsys + ΔSsurr is positive.
  • The surroundings keep score as −ΔH/T, so cold surroundings gain more per joule.
  • The third law sets S = 0 for a perfect crystal at 0 K, which is why S° values are absolute.
  • ΔS°rxn is products minus reactants, and the moles of gas usually fix the sign.
  • ΔG = ΔH − TΔS folds both books into one number for the system alone.
  • Four sign cases: always, never, cold only, hot only. The last two have a crossover at T = ΔH°/ΔS°.
  • ΔG° = −RT ln K, so the sign of ΔG° and the size of K are the same fact twice.

ReferenceWords worth knowing

These are the terms this chapter leans on. Each one shows up in the sections above, and the list runs in alphabetical order.

Boltzmann constant (k)
The gas constant per particle, 1.381 × 10−23 J/K. It equals R/NA.
Crossover temperature
The temperature where ΔG° equals zero, found from T = ΔH°/ΔS°. On its two sides the sign of ΔG° is different.
Endothermic
Taking heat in, so ΔH is positive. Melting ice is one example.
Entropy (S)
A measure of how spread out the matter and energy of a sample are. Units are J/mol·K.
Exothermic
Giving heat off, so ΔH is negative. Burning fuel is one example.
Gibbs free energy (G)
The quantity ΔH − TΔS. Its change decides direction at steady temperature and pressure.
Kinetics
The study of how fast a change goes. It is a separate question from whether the change is favored.
Microstate
One complete arrangement of where every particle sits and which level holds each piece of energy.
Nonspontaneous change
A change that needs steady work from outside to keep going. Its reverse is spontaneous.
Reaction quotient (Q)
The same ratio as K, built from the amounts present right now instead of the equilibrium amounts.
Second law of thermodynamics
The total entropy of a system plus its surroundings rises in any spontaneous change.
Spontaneous change
A change that keeps going on its own, with no outside drive. It carries no promise of speed.
Standard entropy change (ΔS°)
The entropy change for a reaction with every substance in its standard state. Found as products minus reactants.
Standard free energy of formation (ΔG°f)
The free energy change for making one mole of a substance from its elements at 298.15 K and 1 bar. It is zero for an element in its standard form.
Standard molar entropy (S°)
The absolute entropy of one mole of a substance at 298.15 K and 1 bar. Every listed value is positive.
Surroundings
Everything outside the system, such as the flask, the bench and the air.
System
The part of the world you are tracking, such as the chemicals inside the flask.
Third law of thermodynamics
A perfect crystal of a pure substance at 0 K has an entropy of exactly zero.
Universe
In thermodynamics, the system plus its surroundings. The second law applies to that total.

Check yourself

1. A jeweller hears that graphite is favored over diamond. Should she worry about her stock?
No. ΔG° for C(diamond) → C(graphite) is 0 − 2.9 = −2.9 kJ/mol, so graphite is favored. The carbon atoms sit in strong bonds, so the rate is almost zero. Favored and fast are different questions.
2. Find ΔS° for H2O(l) → H2O(g) using S° = 70.0 and 188.8 J/mol·K.
+118.8 J/K. ΔS° = 188.8 − 70.0 = +118.8 J/K. A liquid became a gas, so a large positive value was expected.
3. Find ΔS° for I2(s) → I2(g). The table gives 116.1 and 260.7 J/mol·K.
+144.6 J/K. ΔS° = 260.7 − 116.1 = +144.6 J/K. Going straight from solid to gas is the biggest jump a substance can make.
4. Without any table, predict the sign of ΔS° for 2 NO2(g) → N2O4(g).
Negative. Two moles of gas became one mole of gas. The table agrees: 304.4 − 2(240.1) = 304.4 − 480.2 = −175.8 J/K.
5. A reaction gives off 85.0 kJ at 298 K. What is ΔSsurr?
+285 J/K. ΔSsurr = −ΔH/T = −(−85000 J)/298 K = +285 J/K. Heat left the system and raised the entropy of the room.
6. A reaction has ΔH° = +30.0 kJ and ΔS° = +80.0 J/K. Find ΔG° at 298 K, then the crossover temperature.
ΔG° = +6.2 kJ, crossover 375 K. First convert: 80.0 J/K = 0.0800 kJ/K. Then ΔG° = 30.0 − (298)(0.0800) = 30.0 − 23.8 = +6.2 kJ, so it will not go yet. Crossover: T = 30000 J ÷ 80.0 J/K = 375 K.
7. Which sign pair can never be spontaneous at any temperature?
ΔH positive with ΔS negative. Then ΔG = (positive) − T(negative), which is positive plus positive. Raising T only makes it worse.
8. A reaction has ΔG° = −20.0 kJ at 298.15 K. Find K.
K = 3.2 × 103. ln K = −ΔG°/RT = 20000 J ÷ [(8.314)(298.15)] = 20000 ÷ 2478.8 = 8.07. Then K = e8.07 = 3.2 × 103. Products are favored.
9. A classmate says ΔG° = 0 means nothing happens. Fix the statement.
It means K = 1. From ΔG° = −RT ln K, a zero gives ln K = 0 and K = 1. Plenty of reaction happens. It just stops with comparable amounts on both sides.
10. For 2 SO2(g) + O2(g) → 2 SO3(g), ΔH° = −197.8 kJ and ΔS° = −188.0 J/K. Find the crossover temperature.
1052 K. T = ΔH°/ΔS° = (−197.8 kJ) ÷ (−0.1880 kJ/K) = 1052 K. Below that the reaction is favored. Above it the yield collapses, which is why sulfuric acid plants stay well under 1052 K.
11. All tabulated S° values are positive, yet many ΔH°f values are negative. Why the difference?
Entropy has a true zero and enthalpy does not. The third law pins S = 0 at 0 K for a perfect crystal, so every warmer sample has S above zero. Enthalpy tables can only list changes from a chosen reference, and those changes may run either way.
12. Can a system lose entropy during a spontaneous change?
Yes, if the surroundings gain more. Water freezing at −10 °C loses 22.0 J/K in the system but hands the surroundings 22.8 J/K. The total is +0.8 J/K, so the second law is satisfied.
Chemistry · Chapter 17

Electrochemistry

Some reactions can push electrons through a wire. That one fact is why a battery works. In this chapter you will learn to spot those reactions and balance them in halves. Then you will build the cell that puts them to work. You will see why zinc gives up electrons and copper takes them. You will turn a voltage into a free energy and into an equilibrium constant. You will learn why a tired battery reads low. At the end you will run a cell backward on purpose, to plate metal and to keep iron from rusting.

By the end of this chapter you can

  • Split a redox reaction into two half-reactions, then balance each one in acid or in base.
  • Sketch a galvanic cell and name the anode, the cathode and the salt bridge.
  • Trace where the electrons go in the wire and where the ions go in the bridge.
  • Write any cell in short line notation.
  • Use a table of standard reduction potentials to get E°cell for any pair.
  • Move among E°cell, ΔG° and K with the right n each time.
  • Use the Nernst equation to find a cell potential away from standard conditions.
  • Use Faraday's law to link current, time and the mass of metal plated out.

Section 1Balancing redox half-reactions

Two things happen at once

A redox reaction moves electrons from one species to another. Oxidation is the loss of electrons. Reduction is the gain of electrons. One can never happen alone. The electrons have to come from somewhere and land somewhere.

You track the move with oxidation numbers. An oxidation number is a bookkeeping charge for one atom. If the number goes up, that atom lost electrons. If it goes down, that atom gained them. The reducing agent is the one that gets oxidized. The oxidizing agent is the one that gets reduced. Each name tells you what that species does to its partner.

Split the job in two

Big redox equations fight back if you balance them by eye. So split them. A half-reaction shows one side of the trade, with the electrons written in. Balance each half on its own. Then scale the halves so the electrons cancel.

Here is the recipe in acid.

  1. Write one half-reaction for each atom whose oxidation number changed.
  2. Balance every atom except oxygen and hydrogen.
  3. Add H2O to the side that needs oxygen.
  4. Add H+ to the side that needs hydrogen.
  5. Add electrons until the total charge matches on both sides.
  6. Multiply each half so the two have the same electron count.
  7. Add the halves. Cancel anything that shows up on both sides.

What changes in base

A basic solution holds almost no H+. You still balance in acid first, because the steps are easy there. Then add OH− to both sides, one for every H+ you used. On the H+ side each pair becomes water. Now cancel the water that sits on both sides. Nothing else moves.

Watch that trick on the half-reaction that eats oxygen gas. In acid it reads O2 + 4H+ + 4e− → 2H2O. Add four OH− to each side and four water molecules appear on the left. Two cancel, and you are left with the basic form. Both forms are correct. You just use the one that fits the beaker.

Balancing a half-reaction in acid, then switching it to base In acid: build the half-reaction one piece at a time STEP 1 · SKELETON MnO₄⁻ → Mn²⁺ STEP 2 · WATER FOR OXYGEN MnO₄⁻ → Mn²⁺ + 4H₂O STEP 3 · H⁺ FOR HYDROGEN MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O STEP 4 · ELECTRONS FOR CHARGE MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Charge check left: (−1) + (+8) + (−5) = +2   right: +2   they match Five electrons means manganese dropped from +7 to +2. That count of five is the n you will scale by later. Same half-reaction, basic form ACID FORM O₂ + 4H⁺ + 4e⁻ → 2H₂O ADD 4 OH⁻ TO EACH SIDE O₂ + 4H₂O + 4e⁻ → 2H₂O + 4OH⁻ cancel 2H₂O from both sides BASIC FORM O₂ + 2H₂O + 4e⁻ → 4OH⁻ Both are balanced and both are real Atoms: 4 O and 4 H on each side. Charge: −4 on each side. Still four electrons, so n stays 4. The electron count never changes when you move a half-reaction from acid into base.
One recipe, one small add-on. The acid steps build the half-reaction. Adding hydroxide to both sides moves it into base without touching the electron count.

Worked example · purple fades to clear

Permanganate ion oxidizes iron(II) in acid. Balance MnO4− + Fe2+ → Mn2+ + Fe3+.

Step 1 · two halves

Manganese goes from +7 to +2. Iron goes from +2 to +3.

MnO4− + 8H+ + 5e− → Mn2+ + 4H2O

Fe2+ → Fe3+ + e−

Step 2 · match the electrons

One half drops five electrons. The other gives up one. Multiply the iron half by 5.

5Fe2+ → 5Fe3+ + 5e−

Step 3 · add and cancel

MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O

Step 4 · check it twice

Atoms: 1 Mn, 5 Fe, 4 O and 8 H on each side.

Charge: (−1) + (+10) + (+8) = +17 on the left. (+2) + (+15) = +17 on the right. Balanced.

Key points
  • Oxidation loses electrons. Reduction gains them. They always travel together.
  • A half-reaction carries its own electrons, so you can balance it alone.
  • In acid, add water for oxygen and H+ for hydrogen, then electrons for charge.
  • For base, balance in acid and then add OH− to both sides.
  • Check atoms and charge every single time. Charge is the one people skip.
  • The electron count you end up with is the n for every equation later in this chapter.

Section 2Galvanic cells

Keep the two halves apart

Drop a strip of zinc into copper(II) nitrate and the trade happens at once. Copper coats the zinc and the solution loses its blue. You feel the beaker warm up. But the electrons jump straight across, so you get no current out of it.

A galvanic cell fixes that. Put each half-reaction in its own beaker. Each beaker holds an electrode, a solid that carries current, sitting in its own solution. One beaker plus its electrode is a half-cell. Now the electrons cannot jump. Their only road is the wire you string between the two metals. Current in a wire is current you can use.

Anode, cathode and the salt bridge

The anode is the electrode where oxidation happens. The cathode is the electrode where reduction happens. Two short phrases lock it in. An ox: anode, oxidation. Red cat: reduction, cathode. Electrons always leave the anode and travel to the cathode.

In a galvanic cell the anode is the negative terminal. It is piling up electrons. The cathode is the positive terminal, pulling them in. Zinc metal dissolves at the anode. Copper metal plates out at the cathode. You can watch the zinc strip get thin and the copper strip get fat.

There is one more problem. The anode beaker gains Zn2+ and goes positive. The cathode beaker loses Cu2+ and goes negative. Charge piles up and the current stops within seconds. A salt bridge solves it. The bridge is a tube of salty gel that lets ions walk through. Negative ions drift toward the anode. Positive ions drift toward the cathode. Both beakers stay neutral, so the current keeps flowing. A porous disc between the two solutions does the same job.

Writing a cell on one line

Drawing beakers gets old fast. Cell notation packs the same picture into one line. The anode goes on the left and the cathode on the right. A single bar is a phase boundary. A double bar is the salt bridge. Concentrations and pressures go in parentheses.

Zn(s) │ Zn2+(1 M) ││ Cu2+(1 M) │ Cu(s)

Read it left to right and you are walking the electrons. Zinc gives them up. They cross the wire. Copper ions take them. If a half-cell has no solid of its own, you add an inert electrode such as platinum or graphite. It only ferries electrons and never reacts.

A zinc and copper galvanic cell with every part labelled 1.10 volts e⁻ flow salt bridge (KNO₃ gel) NO₃⁻ K⁺ Zn Cu ANODE · oxidation · negative CATHODE · reduction · positive Zn → Zn²⁺ + 2e⁻ 1 M Zn(NO₃)₂ Cu²⁺ + 2e⁻ → Cu 1 M Cu(NO₃)₂ the strip gets thinner the strip gets heavier E°cell = 0.34 V − (−0.76 V) = 1.10 V Electrons go anode → cathode in the wire. Anions go the other way through the bridge.
The classic lab cell. Zinc dissolves on the left, copper plates on the right, and the salt bridge keeps both beakers neutral so the current does not stall.

Worked example · read a cell line

A cell is written Mg(s) │ Mg2+(1 M) ││ Ag+(1 M) │ Ag(s). Name the electrodes and write the overall reaction.

Step 1 · left is the anode

Magnesium sits on the left, so magnesium is the anode. It is oxidized.

Mg → Mg2+ + 2e−

Step 2 · right is the cathode

Silver sits on the right, so silver ion is reduced on the silver strip.

Ag+ + e− → Ag

Step 3 · match electrons and add

Double the silver half so both halves move two electrons.

Mg + 2Ag+ → Mg2+ + 2Ag

Step 4 · say what you would see

The magnesium strip pits and shrinks. Shiny silver builds up on the other strip. Nitrate ions walk toward the magnesium beaker. Here n is 2.

Key points
  • A galvanic cell keeps the two half-reactions in separate containers.
  • Oxidation happens at the anode. Reduction happens at the cathode.
  • Electrons move through the wire from anode to cathode, always.
  • In a galvanic cell the anode is negative and the cathode is positive.
  • The salt bridge passes ions so neither beaker builds up charge.
  • In cell notation the anode is on the left and the double bar is the bridge.

Section 3Standard reduction potentials

You can only measure a difference

A voltmeter needs two leads. So there is no way to measure one half-cell by itself. Every reading is a gap between two half-cells. Chemists got around this by picking one half-cell as the zero mark.

That reference is the standard hydrogen electrode, or SHE. It is a platinum strip in 1 M H+, with hydrogen gas bubbling over it at 1 bar, at 25 °C. Its potential is defined as exactly 0.000 V. Nothing in nature makes it zero. We just agreed to call it zero, the way sea level is the zero for altitude.

Now pair any half-cell with the SHE and read the meter. That reading is the standard reduction potential, written E°. The little circle means standard conditions: 1 M for every dissolved species, 1 bar for every gas, 25 °C. Every value in the table is written as a reduction, electrons on the left.

Reading the table

A positive E° means that species grabs electrons more eagerly than H+ does. A negative E° means it holds them more loosely. Copper sits at +0.34 V, so copper ions pull electrons in. Zinc sits at −0.76 V, so zinc metal would rather let them go. Fluorine tops the table at +2.87 V. Lithium sits at the bottom at −3.04 V.

Two habits follow from that. High on the table means a strong oxidizing agent. Low on the table means a strong reducing agent. And any species high on the table will oxidize anything lower down.

Half-reaction (as a reduction)E° (V)What it tells you
F2 + 2e− → 2F−+2.87Strongest common oxidizing agent
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O+1.51The purple titration workhorse
Cl2 + 2e− → 2Cl−+1.36Bleach chemistry
O2 + 4H+ + 4e− → 2H2O+1.23Air, in acid
Ag+ + e− → Ag+0.80Silver plates out easily
Fe3+ + e− → Fe2++0.77No solid metal in this one
O2 + 2H2O + 4e− → 4OH−+0.40Air, in neutral or basic water
Cu2+ + 2e− → Cu+0.34The usual cathode in the lab
2H+ + 2e− → H20.00The agreed zero, the SHE
Pb2+ + 2e− → Pb−0.13Barely below hydrogen
Ni2+ + 2e− → Ni−0.25Plating and coins
Fe2+ + 2e− → Fe−0.44Why iron rusts
Zn2+ + 2e− → Zn−0.76The usual anode in the lab
Al3+ + 3e− → Al−1.66Needs electrolysis to make
Mg2+ + 2e− → Mg−2.37A sacrificial anode metal
Li+ + e− → Li−3.04Strongest common reducing agent

Putting two half-cells together

To get the cell potential, take the cathode value minus the anode value.

E°cell = E°cathode − E°anode

Both numbers come straight out of the table as reductions. Do not flip a sign before you subtract. The subtraction already handles the flip. If the answer comes out positive, the cell runs on its own. If it comes out negative, you picked the wrong one to be the anode. Swap them and the sign turns positive.

One rule trips up almost everyone. E° never scales with coefficients. Doubling a half-reaction doubles the electrons and doubles the energy, but the potential stays put. Potential is energy per unit of charge, so the doubling cancels out. It is like density: two bricks of lead are heavier than one, but the density is the same.

A ladder of standard reduction potentials with the hydrogen electrode at zero E° in volts +3 +2 +1 0 −1 −2 −3 F₂ / F⁻   +2.87 MnO₄⁻ / Mn²⁺   +1.51 Cl₂ / Cl⁻   +1.36 O₂ / H₂O in acid   +1.23 Ag⁺ / Ag   +0.80 Cu²⁺ / Cu   +0.34 H⁺ / H₂   0.00   the SHE Fe²⁺ / Fe   −0.44 Zn²⁺ / Zn   −0.76 Al³⁺ / Al   −1.66 Mg²⁺ / Mg   −2.37 Li⁺ / Li   −3.04 stronger oxidizing agent stronger reducing agent The gap between any two rungs is the cell potential E°cell Cu above Zn 0.34 − (−0.76) = 1.10 V Each dot sits at its true height on the scale. The thin leader carries the label out to a clear row.
One ladder, every pair. Each couple sits at its own height. The vertical gap between two rungs is the standard cell potential, and the higher rung is always the cathode.

Worked example · zinc against silver

Find E°cell and n for a zinc and silver cell at standard conditions.

Step 1 · pull both values

Ag+ + e− → Ag has E° = +0.80 V.

Zn2+ + 2e− → Zn has E° = −0.76 V.

Step 2 · higher value is the cathode

Silver is higher, so silver is reduced. Zinc is the anode.

Step 3 · subtract

E°cell = (+0.80 V) − (−0.76 V) = +1.56 V

The answer is positive, so the cell runs on its own.

Step 4 · find n from the balanced equation

Zn + 2Ag+ → Zn2+ + 2Ag moves two electrons, so n = 2.

Notice what did not happen. Doubling the silver half did not change +0.80 V to +1.60 V. The potential is per charge, so it holds still.

Key points
  • No single half-cell potential can be measured. Only gaps can.
  • The hydrogen electrode is defined as exactly 0.000 V to break the tie.
  • Every tabulated E° is written as a reduction at 1 M, 1 bar and 25 °C.
  • E°cell = E°cathode − E°anode, with both values taken straight from the table.
  • A positive E°cell means the reaction as written is spontaneous.
  • Scaling a half-reaction changes n but never changes E°.

Section 4Cell potential, free energy and K

A voltage is energy per charge

One volt means one joule for every coulomb of charge moved. So if you know the charge and the voltage, you know the energy. The charge comes from the electron count. One mole of electrons carries 96,485 coulombs. That number is the Faraday constant, F.

Push n moles of electrons through a cell and the charge is nF. Multiply by the potential and you have the most useful work the cell can do. Free energy change is the mirror of that work, so it carries a minus sign.

ΔG° = −nFE°cell

Here n is moles of electrons in the balanced overall reaction. It is not a coefficient you can guess. Get it from the half-reactions after you match them. A wrong n wrecks every number that follows.

The signs all agree

Look at the minus sign and the story falls out. A positive E°cell gives a negative ΔG°. Negative ΔG° means spontaneous. So a cell with a positive potential runs on its own and can light a bulb. A negative potential gives a positive ΔG°, and that reaction needs a push from outside.

If E°cell isthen ΔG° isand K isthe reaction
positivenegativegreater than 1runs forward on its own
zerozeroequal to 1sits at equilibrium at standard state
negativepositiveless than 1needs outside energy to go forward

And K joins the family

From the chapter on free energy you already know ΔG° = −RT ln K. Now you have a second route to the same ΔG°. Set the two equal and the F and R sort themselves out.

−nFE°cell = −RT ln K  →  E°cell = (RT / nF) ln K

At 25 °C the clump RT/F times ln 10 works out to 0.0592 V. Check it yourself. Take 8.314 J/mol·K times 298.15 K, divide by 96,485 C/mol, then multiply by 2.303. That gives the shortcut form in base ten logs.

E°cell = (0.0592 V / n) log K  (at 25 °C)

This is why small voltages hide huge equilibrium constants. Each 0.0592 volts multiplies K by ten, per electron. Two electrons and a volt of potential put you past 1033. A cell you would call ordinary is a reaction that goes to completion.

The three ways to say the same thing: cell potential, free energy and K Three numbers, one fact about the reaction E°cell volts, from the table ΔG° joules per mole K no units, how far it goes ΔG° = −nFE°cell E°cell = (0.0592/n) log K ΔG° = −RT ln K The zinc and copper cell n = 2   E°cell = 1.10 V ΔG° = −(2)(96485)(1.10) = −212,000 J = −212 kJ/mol log K = (2)(1.10) / 0.0592 = 37.16 K = 1.5 × 10³⁷ A hair over one volt, and the reaction is finished. Almost no Cu²⁺ is left. Know any one corner and the right n, and you can reach the other two.
One triangle to remember. Cell potential, free energy and the equilibrium constant are three readings of the same reaction. The electron count n is the bridge between them.

Worked example · from 1.10 volts to a giant K

For Zn + Cu2+ → Zn2+ + Cu, find ΔG° and K at 25 °C.

Step 1 · get E° and n

E°cell = 0.34 V − (−0.76 V) = 1.10 V. Each half moves two electrons, so n = 2.

Step 2 · free energy

ΔG° = −(2)(96,485 C/mol)(1.10 J/C)

ΔG° = −212,267 J/mol, which rounds to −212 kJ/mol.

Step 3 · equilibrium constant

log K = nE°cell / 0.0592 V = (2)(1.10) / 0.0592 = 37.162

K = 1037.162 = 1.5 × 1037

Step 4 · check it a second way

Try ΔG° = −RT ln K. Take ln(1.5 × 1037) = 85.60.

−(8.314)(298.15)(85.60) = −212,000 J/mol. The two routes agree.

Step 5 · say what it means

A K near 1037 is enormous. Put zinc in copper sulfate and the copper ions all but vanish.

Key points
  • One volt is one joule per coulomb, so voltage times charge gives energy.
  • F = 96,485 C per mole of electrons.
  • ΔG° = −nFE°cell, and n comes from the balanced overall reaction.
  • Positive E°cell, negative ΔG° and K above 1 all say the same thing.
  • At 25 °C, E°cell = (0.0592 V / n) log K.
  • Every 0.0592 V per electron is one more power of ten in K.

Section 5The Nernst equation and batteries

Standard conditions almost never hold

Every E° in the table assumes 1 M and 1 bar. Your beaker is rarely that tidy. A battery in use is certainly not. The Nernst equation fixes the potential for whatever concentrations you actually have.

Ecell = E°cell − (RT / nF) ln Q

At 25 °C you can use the base ten form, which is easier to punch in.

Ecell = E°cell − (0.0592 V / n) log Q

The Q here is the same reaction quotient you already know. Products over reactants, each raised to its coefficient. Pure solids and pure liquids count as 1. Gases go in as pressures. Build Q from the overall cell reaction, not from one half.

Now read the minus sign. Pile up products and Q rises, log Q goes positive, and the potential drops. Pile up reactants and the potential climbs. That is the whole behavior of a battery in one line. As it runs, products build and voltage sags. When Q finally reaches K, log Q is big enough to cancel E° exactly. The cell reads zero. A dead battery is a cell that reached equilibrium.

A cell made of nothing but a difference

Here is a strange one. Put a copper strip in 0.0010 M Cu2+ and another in 1.0 M Cu2+, then join them. Both half-cells are the same couple, so E°cell is exactly zero. Yet the meter reads about 0.089 V.

That is a concentration cell. The only thing driving it is the gap in concentration. Nature wants the two solutions to even out. So copper dissolves on the dilute side and plates on the concentrated side. The dilute half-cell is always the anode. The cell dies when the two concentrations meet. Your nerve cells run on this idea, using a potassium gap instead of a copper one.

Batteries you can buy

A battery is a galvanic cell in a can. An alkaline cell runs zinc against manganese dioxide in potassium hydroxide, near 1.5 V. A lead-acid cell runs lead against lead dioxide in sulfuric acid at about 2.0 V, and six in a row make the 12 V in a car. Lithium-ion cells shuttle Li+ between a graphite layer and a metal oxide layer at roughly 3.7 V. A hydrogen fuel cell joins hydrogen with oxygen electrically and stands at 1.23 V standard.

Notice what sets them apart. In a lead-acid cell the reactants and products are mostly solids, so Q barely moves and the voltage holds steady until the acid runs thin. An alkaline cell sags as it drains. A fuel cell never discharges at all, because you keep feeding it fresh fuel.

Cell potential falls in a straight line as log Q rises, and the voltages of real batteries The zinc and copper cell, away from standard state (n = 2) 1.30 1.20 1.10 1.00 0.90 Ecell (volts) −6 −4 −2 0 2 4 6 log Q Q = 1, E = E° = 1.10 V Q = 0.01, E = 1.16 V Q = 100, E = 1.04 V slope = −0.0592 / 2 = −0.0296 V per unit of log Q Real cells, real voltages Hydrogen fuel cell 1.23 V Alkaline AA cell 1.5 V One lead-acid cell 2.0 V Lithium-ion cell 3.7 V Car battery, six cells in a row 12 V Cells in series add their voltages. Cells in parallel add current instead.
A straight line with a useful slope. Cell potential drops by 0.0592/n volts for every tenfold rise in Q. Follow the line far enough right and it crosses zero, which is where Q equals K and the cell is finished.

Worked example · a cell that is not standard

A zinc and copper cell has [Zn2+] = 1.0 M and [Cu2+] = 0.010 M. Find Ecell at 25 °C.

Step 1 · overall reaction and n

Zn(s) + Cu2+ → Zn2+ + Cu(s), with n = 2 and E°cell = 1.10 V.

Step 2 · write Q

Both metals are pure solids, so they drop out.

Q = [Zn2+] / [Cu2+] = 1.0 / 0.010 = 100

Step 3 · put it in Nernst

Ecell = 1.10 V − (0.0592 V / 2) log(100)

Ecell = 1.10 V − (0.0296 V)(2) = 1.10 V − 0.0592 V = 1.04 V

Step 4 · does it make sense?

You starved the cell of the reactant Cu2+, so the potential should fall. It did, by about 0.06 V. Small change, which is why batteries hold a steady voltage for so long.

Worked example · the concentration cell

Find Ecell for Cu │ Cu2+(0.0010 M) ││ Cu2+(1.0 M) │ Cu.

Step 1 · standard potential is zero

Both electrodes are copper, so E°cathode and E°anode are both +0.34 V. Their difference is 0.

Step 2 · the reaction just moves ions

Cu2+(1.0 M) → Cu2+(0.0010 M), and n = 2.

Q = 0.0010 / 1.0 = 1.0 × 10−3, so log Q = −3.

Step 3 · Nernst does all the work

Ecell = 0 − (0.0592 V / 2)(−3) = +0.0888 V, or 0.089 V.

Step 4 · which side is the anode?

The dilute side. Copper dissolves there to raise its own concentration. The meter stays positive until the two solutions match.

Key points
  • Nernst corrects E°cell for real concentrations and pressures.
  • At 25 °C, Ecell = E°cell − (0.0592/n) log Q.
  • Build Q from the overall cell reaction, with solids and pure liquids as 1.
  • More product means a bigger Q and a smaller potential.
  • A dead battery is a cell where Q has climbed all the way to K, so Ecell is 0.
  • A concentration cell has E°cell = 0, and the dilute half-cell is the anode.

Section 6Electrolysis and corrosion

Running a cell backward on purpose

So far the reactions have paid their own way. Now flip it. Electrolysis uses an outside power supply to force a reaction that would never go by itself. The container is an electrolytic cell. Its E°cell is negative and its ΔG° is positive, so you pay the bill in electricity.

The job names do not change. Oxidation still happens at the anode. Reduction still happens at the cathode. What flips is the sign on the terminals. The anode is now the positive post, because the supply pulls electrons out of it. The cathode is the negative post, because electrons get shoved in.

Melt sodium chloride, run current through it, and you get sodium metal at the cathode and chlorine gas at the anode. That is how sodium is made. It takes roughly four volts and a furnace above 800 °C. The table values are for water at 25 °C, so a molten salt needs its own data.

Salt water behaves differently, because water can react too. The cathode gives hydrogen gas, not sodium metal, since water is far easier to reduce than Na+. The anode gives chlorine, and sodium hydroxide builds up in the solution. That one process makes chlorine, lye and hydrogen at once.

Real cells need more voltage than the math says. That extra is called overpotential. It pays for slow steps at the electrode surface, such as gas bubbles forming. Splitting water needs 1.23 V in theory and closer to 1.8 V in practice.

Faraday's law counts the electrons

Current is charge per second, so current times time is charge. Divide the charge by F and you have moles of electrons. That is Faraday's law of electrolysis in plain words. Electrons are the reagent, and you can measure them out with an ammeter and a clock.

moles of e− = It / F   then   moles of product = moles of e− / n

The n is the electrons per formula unit in the half-reaction. Silver takes one electron per atom, copper two, aluminum three. So the same charge gives three times as many moles of silver as of aluminum.

Corrosion is a cell you did not build

Corrosion is the slow galvanic cell that eats metal in the open air. Iron is the famous victim. One patch of a damp nail acts as the anode and iron dissolves there. Another patch acts as the cathode, where oxygen from the air is reduced.

Fe → Fe2+ + 2e−   and   O2 + 2H2O + 4e− → 4OH−

Those two give 0.40 V − (−0.44 V) = 0.84 V, so rusting is spontaneous. Air and water are all it needs. The Fe2+ then oxidizes further to hydrated iron(III) oxide, which is the flaky orange solid you know as rust. Road salt makes it faster, because the ions carry current between the anode and cathode patches.

You can fight it three ways. Block the water with paint or grease. Or pick a metal that makes a tight oxide skin, the way aluminum does. Rust flakes off and bares fresh iron, but aluminum oxide clings and seals. Or set up a sacrificial anode. Bolt zinc or magnesium to the iron. Both sit lower on the table, so they oxidize instead and the iron is spared. Galvanized steel is that trick as a coating, and it keeps working even when scratched.

An electrolytic cell and the sacrificial zinc coating on galvanized steel Forced: electrolysis of molten NaCl POWER SUPPLY about 4 V − + molten NaCl, above 800 °C Na⁺ pulled in Cl⁻ pulled out CATHODE (−)  Na⁺ + e⁻ → Na ANODE (+)  2Cl⁻ → Cl₂ + 2e⁻ Reduction is still at the cathode. Only the terminal signs flip. Free: a scratch in galvanized steel iron core zinc coat zinc coat scratch e⁻ Zn²⁺ + 2e⁻ → Zn   E° = −0.76 V Fe²⁺ + 2e⁻ → Fe   E° = −0.44 V Zinc is lower, so zinc gives up its electrons first. The zinc becomes the anode and corrodes away slowly. The bare iron in the scratch becomes the cathode. A cathode does not dissolve, so the iron stays whole.
Pay or get paid. On the left a power supply forces a reaction uphill. On the right a cheaper metal is wired in so it corrodes in place of the iron, and the iron becomes the protected cathode.

Worked example · plating silver onto a spoon

You run 2.00 A through a silver plating bath for 30.0 minutes. What mass of silver lands on the spoon? Silver is 107.87 g/mol.

Step 1 · charge delivered

Change the time to seconds. 30.0 min × 60 s/min = 1800 s.

q = It = (2.00 C/s)(1800 s) = 3600 C

Step 2 · moles of electrons

3600 C ÷ 96,485 C/mol = 0.03731 mol e−

Step 3 · use n from the half-reaction

Ag+ + e− → Ag, so n = 1. One electron buys one silver atom.

moles Ag = 0.03731 mol

Step 4 · grams

(0.03731 mol)(107.87 g/mol) = 4.02 g of silver

Step 5 · sanity check

Swap in copper, where n = 2, and the same charge gives half as many moles. That comes to 1.19 g of copper. Fewer grams, and that is what the electron count predicts.

Key points
  • Electrolysis pays electricity to drive a reaction with a negative E°cell.
  • Oxidation is at the anode in every cell. Only the terminal signs swap.
  • In water, the solvent often reacts instead of the ion you expected.
  • Overpotential is the extra voltage a real cell needs above the calculated value.
  • Moles of electrons = It/F, then divide by n for moles of product.
  • Rusting is a galvanic cell worth 0.84 V, and a zinc or magnesium anode stops it.

The chapter in one breath

  • Split, then balance. Two half-reactions, water for oxygen, H+ for hydrogen, electrons for charge. Add OH− to both sides to move into base.
  • Keep the halves apart. A galvanic cell forces the electrons through a wire instead of letting them jump.
  • An ox, red cat. Oxidation at the anode, reduction at the cathode, in every cell ever built.
  • The bridge keeps the peace. Ions move through the salt bridge so neither beaker builds up charge.
  • Zero is a choice. The hydrogen electrode was set to 0.000 V so every other half-cell could be measured against it.
  • Subtract, do not flip. E°cell = E°cathode − E°anode, and E° never scales.
  • One fact, three faces. ΔG° = −nFE°cell and E°cell = (0.0592/n) log K at 25 °C.
  • Nernst explains a dying battery. As product builds, Q rises and the voltage falls. At Q = K the cell reads zero.
  • Electrons are a reagent. It/F gives moles of electrons, and that is how much metal you can plate or how much rust you can prevent.

ReferenceWords worth knowing

These are the words this chapter leans on. Come back here whenever a term stops making sense.

Anode
The electrode where oxidation happens. Negative in a galvanic cell, positive in an electrolytic one.
Cathode
The electrode where reduction happens. Positive in a galvanic cell, negative in an electrolytic one.
Cell notation
A one-line shorthand for a cell. Anode on the left, single bar for a phase boundary, double bar for the salt bridge.
Concentration cell
A cell whose two half-cells use the same couple at different concentrations. Its E°cell is zero, so all of its voltage comes from the Nernst term.
Corrosion
Unwanted oxidation of a metal by its surroundings. Rusting iron is a galvanic cell with air as the oxidizing agent.
Electrolysis
Using an outside power supply to drive a reaction that has a negative cell potential. The container it happens in is an electrolytic cell.
Faraday constant
The charge on one mole of electrons, 96,485 C/mol, written F.
Faraday's law of electrolysis
The amount of substance changed at an electrode is set by the charge passed. Moles of electrons equal It/F.
Galvanic cell
A cell that uses a spontaneous redox reaction to push current through an outside circuit. Also called a voltaic cell.
Half-reaction
One side of a redox trade, written with its own electrons so it can be balanced alone.
Nernst equation
The rule that corrects cell potential for real conditions. Ecell = E°cell − (0.0592/n) log Q at 25 °C.
Overpotential
Extra voltage a real cell needs beyond the calculated minimum, spent on slow steps at the electrode surface.
Oxidation
Loss of electrons, which raises the oxidation number.
Oxidizing agent
The species that takes the electrons and is itself reduced.
Reducing agent
The species that hands over the electrons and is itself oxidized.
Reduction
Gain of electrons, which lowers the oxidation number.
Sacrificial anode
A block or coating of a metal with a lower E° that is wired to a structure so it corrodes in its place.
Salt bridge
A path that lets ions move between the two half-cells so neither one builds up charge.
Standard hydrogen electrode
Platinum in 1 M H+ with H2 at 1 bar and 25 °C, defined as exactly 0.000 V.
Standard reduction potential
The potential of a half-reaction written as a reduction, measured against the hydrogen electrode at standard conditions.

Check yourself

1. In Zn(s) + Cu2+ → Zn2+ + Cu(s), which species is the reducing agent, and which is reduced?
Zinc metal is the reducing agent. Cu2+ is reduced. Zinc goes from 0 to +2, so it lost electrons and was oxidized. The thing that gets oxidized is the reducing agent. Copper goes from +2 to 0, so it gained them.
2. Balance this in acid: Cr2O72− + Fe2+ → Cr3+ + Fe3+. What is n?
Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O, and n = 6. The halves are Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O and Fe2+ → Fe3+ + e−, so the iron half is multiplied by 6. Charge check: (−2) + (+12) + (+14) = +24 on the left, and (+6) + (+18) = +24 on the right.
3. Write the cell notation for a cell made from an iron electrode in 1 M Fe2+ and a silver electrode in 1 M Ag+.
Fe(s) │ Fe2+(1 M) ││ Ag+(1 M) │ Ag(s) Iron sits at −0.44 V and silver at +0.80 V. Silver is higher, so silver is the cathode and goes on the right. Iron is the anode and goes on the left.
4. Find E°cell for Mg(s) │ Mg2+ ││ Fe2+ │ Fe(s), and give n.
+1.93 V, with n = 2. The cathode is iron at −0.44 V and the anode is magnesium at −2.37 V. So E°cell = (−0.44 V) − (−2.37 V) = +1.93 V. The overall reaction Mg + Fe2+ → Mg2+ + Fe moves two electrons.
5. For that magnesium and iron cell, what is ΔG°?
−372 kJ/mol ΔG° = −nFE°cell = −(2)(96,485 C/mol)(1.93 J/C) = −372,432 J/mol. That rounds to −372 kJ/mol. The sign is negative, which matches the positive cell potential.
6. Find K at 25 °C for Cu(s) + 2Ag+ → Cu2+ + 2Ag(s). Watch the value of n.
K = 3.5 × 1015 First E°cell = (+0.80 V) − (+0.34 V) = +0.46 V. The silver half had to be doubled to match copper's two electrons, so n = 2, not 1. Then log K = nE°/0.0592 = (2)(0.46)/0.0592 = 15.54, and K = 1015.54 = 3.5 × 1015.
7. A zinc and copper cell has [Zn2+] = 0.010 M and [Cu2+] = 1.0 M. What does the meter read?
1.16 V Q = [Zn2+]/[Cu2+] = 0.010/1.0 = 0.010, so log Q = −2. Then Ecell = 1.10 V − (0.0592/2)(−2) = 1.10 V + 0.0592 V = 1.16 V. Less product means a higher potential than standard.
8. Find Ecell for Ag │ Ag+(0.010 M) ││ Ag+(1.0 M) │ Ag.
+0.118 V Both sides are the same couple, so E°cell = 0. Here n = 1, and Q = 0.010/1.0 = 0.010, so log Q = −2. Then Ecell = 0 − (0.0592/1)(−2) = +0.118 V. The dilute side is the anode.
9. You electrolyze salt water. What comes off each electrode, and what builds up in the solution?
Hydrogen gas at the cathode, chlorine gas at the anode, and sodium hydroxide in the solution. Water is much easier to reduce than Na+, so you never get sodium metal from a water solution. Chloride is oxidized at the anode. The leftover Na+ and OH− make lye.
10. How long must you run 5.00 A to plate 10.0 g of copper? Copper is 63.55 g/mol.
About 6.07 × 103 s, which is 1.69 hours. First, 10.0 g ÷ 63.55 g/mol = 0.1574 mol Cu. Since Cu2+ + 2e− → Cu, that needs 0.3147 mol of electrons. Charge: (0.3147 mol)(96,485 C/mol) = 30,365 C. Time: 30,365 C ÷ 5.00 C/s = 6073 s, or 101 minutes.
11. Run 10.0 A for exactly 1.00 hour through molten aluminum oxide. How much aluminum do you get? Aluminum is 26.98 g/mol.
3.36 g Charge: (10.0 C/s)(3600 s) = 36,000 C. Moles of electrons: 36,000 ÷ 96,485 = 0.3731 mol. Since Al3+ + 3e− → Al, divide by 3 to get 0.1244 mol Al. Mass: (0.1244)(26.98) = 3.36 g. Three electrons per atom is why aluminum is so costly to make.
12. A galvanized bucket gets a deep scratch down to bare iron. Why does the iron still not rust?
The zinc coating acts as a sacrificial anode. Zinc sits at −0.76 V and iron at −0.44 V, so zinc gives up electrons more readily. The zinc corrodes and feeds electrons to the iron, which makes the iron the cathode. A cathode gains electrons instead of dissolving, so it survives.
Chemistry · Chapter 18

Representative Metals, Metalloids and Nonmetals

The main groups are the two tall columns on the left of the periodic table and the six on the right. They hold the elements you meet every day. Table salt, lime, sand, glass, rust proofing, fertilizer, bleach and battery acid all come from them. This chapter walks group by group through those elements. You will see why sodium explodes in water while magnesium barely fizzes. You will see why carbon can be the hardest solid or the slipperiest one. And you will see why the noble gases sit out almost every reaction. The rules behind all of it come from two trends you already know.

By the end of this chapter you can

  • Explain how metallic character shifts across a row and down a group, and link it to size and ionization energy.
  • Sort main group oxides as basic, acidic or amphoteric, and write the water reaction for each kind.
  • Write balanced equations for group 1 and group 2 metals reacting with water, oxygen and nitrogen.
  • Compare the allotropes of carbon and the silicate structures, and tie each one to a real use.
  • Trace nitrogen through the Haber and Ostwald processes, and rank the phosphorus acids by how many protons they give up.
  • Describe the contact process for sulfuric acid, and tell oxides, peroxides and superoxides apart.
  • Rank the halogens by oxidizing strength, predict a displacement reaction, and name the xenon fluorides.

Section 1Periodicity of the main groups

One outer shell, one personality

The main group elements fill groups 1, 2 and 13 through 18. They are also called the representative elements. What ties a group together is its outer shell. Every group 1 atom has one electron in its outer s orbital. Every group 17 atom has seven outer electrons. That count is the valence electron count. It sets the charges an element forms. It also sets how many bonds the atom makes.

Two trends explain most of this chapter. The first is size. The second is how tightly the nucleus grips the outer electrons. Chemists measure that grip with effective nuclear charge, written Zeff. It is the charge an outer electron really feels. Inner electrons screen part of the true nuclear charge.

Metallic character across and down the main groups 1 2 13 14 15 16 17 18 2 Li Be B C N O F Ne 3 Na Mg Al Si P S Cl Ar 4 K Ca Ga Ge As Se Br Kr 5 Rb Sr In Sn Sb Te I Xe Group Period across a period: more nonmetal, oxides turn acidic down a group: more metallic metal metalloid nonmetal noble gas
The main group map. Metals (teal) crowd the left and the bottom. Nonmetals (grey) crowd the top right. The metalloids (gold) run as a staircase between them. Read across for rising nonmetal character and down for rising metallic character.

Across a period the grip tightens

Walk from sodium to chlorine. Each step adds one proton and one outer electron. The new electron joins a shell that is already there, so it screens badly. Zeff climbs at every step. Atoms shrink and hold their electrons harder. Ionization energy rises from 496 kJ/mol for sodium to 1251 kJ/mol for chlorine. Electronegativity rises as well, from 0.93 to 3.16.

So the left side hands electrons out and the right side takes them in. That shift is a loss of metallic character. Sodium is a soft shiny metal. Silicon is a metalloid, shiny but a weak conductor. Chlorine is a gas that grabs electrons from almost anything.

Worked example · a quick Zeff estimate

Compare the pull on an outer electron in sodium and in chlorine. Treat the filled inner shells as perfect screens.

Step 1 · count the coreSodium has 11 protons. Ten of its electrons sit in filled inner shells.

Step 2 · subtractZeff ≈ 11 − 10 = +1.

Step 3 · repeat for chlorineChlorine has 17 protons and the same 10 core electrons. Zeff ≈ 17 − 10 = +7.

Step 4 · read the answerThe outer electron in chlorine feels about seven times the pull. That is why chlorine is far smaller than sodium. It is also why it takes 1251 kJ/mol to strip an electron from chlorine and only 496 kJ/mol from sodium. Real screening is not perfect, so these numbers are a ceiling, not an exact value.

Down a group the grip loosens

Now move from lithium to cesium. Each step adds a whole new shell. The outer electron sits farther out, and more core electrons screen it. The atomic radius grows from 152 pm to 265 pm. Ionization energy drops from 520 kJ/mol to 376 kJ/mol. A looser grip means the atom gives an electron up more easily. So metallic character grows down every main group.

Group 14 shows this best. Carbon at the top is a nonmetal. Silicon and germanium are metalloids. Tin and lead at the bottom are metals soft enough to hammer flat. One group, three kinds of element, and the only change is shell count.

Oxides give the game away

You can place an element by how its oxide treats water. A basic oxide comes from a metal and makes a base:

Na2O(s) + H2O(l) → 2 NaOH(aq)

An acidic oxide comes from a nonmetal and makes an acid:

SO3(g) + H2O(l) → H2SO4(aq)

Between them sits the amphoteric oxide, which reacts with acid and with base. Aluminum oxide is the textbook case. It dissolves in strong acid and in strong base, and it shrugs off water:

Al2O3(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2O(l) Al2O3(s) + 2 NaOH(aq) + 3 H2O(l) → 2 Na[Al(OH)4](aq)
Period 3 oxideNa2OMgOAl2O3SiO2P4O10SO3Cl2O7
Acid or basestrong basebaseamphotericweak acidacidstrong acidstrongest acid
Bondingionicionicionic, part covalentcovalent networkmolecularmolecularmolecular
Melting point (°C)113228522072171334017−92
Worth knowing. Ionization energy does not rise in a perfect line. Aluminum (578 kJ/mol) sits below magnesium (738 kJ/mol), and sulfur (1000 kJ/mol) sits below phosphorus (1012 kJ/mol). Each dip marks the start of a new orbital or a new electron pairing. The big picture still holds.
Key points
  • Main group means groups 1, 2 and 13 through 18. Outer s and p electrons run the show.
  • Across a period, Zeff rises, atoms shrink, and ionization energy and electronegativity rise.
  • Down a group, a new shell is added. Atoms grow, ionization energy falls, metallic character rises.
  • Oxides shift from basic to amphoteric to acidic as you move left to right.
  • Aluminum oxide is amphoteric. That one fact places aluminum right on the metal border.

Section 2Alkali and alkaline earth metals

Group 1: one electron to lose

The alkali metals are lithium, sodium, potassium, rubidium, cesium and francium. Each one has a single outer s electron. Each one loses it to make a +1 ion. None of them is ever found free in nature. All are soft enough to cut with a knife, and all are stored under oil. Air and water attack them on contact.

In water they all do the same thing. They make hydrogen gas and a hydroxide:

2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)

The equation is the same for every member. The speed is not. Lithium fizzes steadily. Sodium melts into a ball and darts across the surface. Potassium lights its own hydrogen with a lilac flame. Cesium reacts so fast it can crack the glass. Ionization energy explains the ladder. It takes 520 kJ/mol to strip lithium and only 376 kJ/mol to strip cesium.

First ionization energy falls down group 1 and group 2 First ionization energy (kJ/mol) and what the metal does in water 0 200 400 600 800 Li Be 520 · fizzes steadily 899 · no reaction Na Mg 496 · melts, darts about 738 · needs steam K Ca 419 · lilac flame 590 · steady fizz Rb Sr 403 · violent 549 · brisk Cs Ba 376 · explosive 503 · fast reactivity with water rises group 1 (alkali) group 2 (alkaline earth)
Down both groups the first electron gets cheaper. Group 1 (teal) always costs less than the group 2 metal beside it (gold). Water reactions track those numbers, with one famous exception noted below.

Oxygen gives three different products, and which one you get depends on the metal. Lithium makes a plain oxide. Sodium makes a peroxide, with an O–O bond. Potassium and the heavier metals make a superoxide:

4 Li(s) + O2(g) → 2 Li2O(s) 2 Na(s) + O2(g) → Na2O2(s) K(s) + O2(g) → KO2(s)

Lithium has one more trick. It is the only group 1 metal that attacks nitrogen gas at room temperature:

6 Li(s) + N2(g) → 2 Li3N(s)

Sodium metal itself is made by electrolysis. Molten sodium chloride runs in a Downs cell at about 600 °C. Calcium chloride is mixed in to drop the melting point. Sodium forms at the cathode and chlorine at the anode:

Na+ + e− → Na(l)     2 Cl− → Cl2(g) + 2 e−
Worth knowing. Lithium has the most negative standard reduction potential of the lot, −3.04 V. On paper it should be the most eager to react. In water it is the slowest. Its high melting point keeps it solid, and its hydroxide coats the surface. Speed and driving force are two different things.

Group 2: two electrons, a tighter hold

The alkaline earth metals are beryllium, magnesium, calcium, strontium, barium and radium. Each has two outer s electrons and forms a +2 ion. Pulling two electrons off costs much more than pulling one off. So group 2 metals are harder, denser and far less reactive than their group 1 neighbours.

Water shows the gap clearly. Beryllium does not react at all, even in steam. Magnesium needs steam to get going. Calcium works away in cold water at a steady pace:

Mg(s) + H2O(g) → MgO(s) + H2(g) Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g)

Magnesium burns in air with a white glare hot enough to melt steel. That is why it once filled flash bulbs and still fills emergency flares:

2 Mg(s) + O2(g) → 2 MgO(s)

The lime cycle

Limestone is calcium carbonate. Heat it past about 900 °C in a kiln and it gives up carbon dioxide. The product is quicklime, calcium oxide:

CaCO3(s) → CaO(s) + CO2(g)    ΔH = +179.2 kJ

Add water to quicklime and it hisses, swells and gets hot. That gives slaked lime, calcium hydroxide:

CaO(s) + H2O(l) → Ca(OH)2(s)    ΔH = −65 kJ

Lime mortar then closes the loop. Over months it pulls carbon dioxide back out of the air and turns to stone again. Farmers spread lime on sour soil, and power plants use it to trap sulfur dioxide from flue gas. Hard water is the same chemistry working against you. Dissolved Ca2+ and Mg2+ ions leave scale in kettles and pipes.

Worked example · hydrogen from sodium and water

Drop 2.30 g of sodium into excess water. How much hydrogen gas forms at 0 °C and 1.00 atm? How much sodium hydroxide do you end up with?

Step 1 · moles of sodium2.30 g ÷ 22.99 g/mol = 0.1000 mol Na.

Step 2 · use the ratioThe equation gives 1 mol H2 for every 2 mol Na. So 0.1000 mol ÷ 2 = 0.0500 mol H2.

Step 3 · find the volumeV = nRT/P = (0.0500 mol)(0.08206 L·atm/mol·K)(273.15 K) ÷ 1.00 atm = 1.12 L.

Step 4 · find the baseSodium and hydroxide come out 1 to 1. So 0.1000 mol NaOH × 40.00 g/mol = 4.00 g NaOH. Check it a second way. 2.30 g Na plus the oxygen and hydrogen it picks up has to weigh more, and 4.00 g does.

PropertyGroup 1Group 2
Outer electronsns1ns2
Ion formedM+M2+
First ionization energy376–520 kJ/mol503–899 kJ/mol
Cold waterall react, fast to violentBe no, Mg slow, Ca–Ba yes
Hydroxide solubilityall very solublerises down the group
Flame colourLi crimson, Na yellow, K lilacCa orange, Sr red, Ba green
Oxide in waterstrong basebase, BeO amphoteric
Key points
  • Group 1 metals lose one electron, group 2 metals lose two. Both get easier to strip down the group.
  • Every alkali metal gives hydrogen plus a hydroxide in water. Only the speed changes.
  • Lithium alone makes a nitride with N2 at room temperature. Sodium favours the peroxide with oxygen, potassium the superoxide.
  • Beryllium oxide and hydroxide are amphoteric. The rest of group 2 gives plain bases.
  • The lime cycle runs limestone → quicklime → slaked lime → limestone, and it drives mortar, soil treatment and flue gas scrubbing.

Section 3Boron, aluminum, carbon and silicon

Boron is short of electrons

Boron is the only member of group 13 that is not a metal. It is a hard, dark metalloid that melts near 2075 °C. It has three valence electrons, which is one short of a filled set of four bonds. That shortage runs all of its chemistry. Boron compounds hunt for electron pairs, so they act as Lewis acids. Boron trifluoride, BF3, grabs a pair from almost any donor.

Diborane, B2H6, shows the shortage best. It has 12 valence electrons but needs 16 for eight normal bonds. Two hydrogens solve this by bridging. Each bridge is a three-center two-electron bond, with one pair shared across B–H–B. Boron is electron deficient, not broken.

Boric acid, H3BO3, is a mild antiseptic and a very weak acid. It does not hand over its own proton. It takes a hydroxide from water instead, which leaves an H+ behind:

B(OH)3(aq) + 2 H2O(l) ⇌ B(OH)4−(aq) + H3O+(aq)    Ka = 5.8 × 10−10

Borax, Na2B4O7·10H2O, is the mined ore. Boron oxide, B2O3, goes into borosilicate glass, which barely expands when heated.

Aluminum is common, armoured and two-faced

Aluminum is the most plentiful metal in the crust, near 8% by mass. It is never free, because it holds oxygen too tightly. Pulling it out needs electricity. In the Hall–Héroult process, alumina is dissolved in molten cryolite, Na3AlF6, near 960 °C. Carbon anodes burn away as the oxide gives up its oxygen:

2 Al2O3(l) + 3 C(s) → 4 Al(l) + 3 CO2(g)

Fresh aluminum looks like it should corrode away. It does not, because the surface grows a thin tight film of Al2O3 that seals the metal below. Anodizing thickens that film on purpose. Aluminum is also an amphoteric metal. It dissolves in strong acid and in strong base, giving hydrogen both times:

2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 Na[Al(OH)4](aq) + 3 H2(g)

Aluminum wants oxygen so badly that it will steal it from other metal oxides. The thermite reaction uses that hunger to weld rail track in the field:

2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(l)    ΔH = −852 kJ

Worked example · how much aluminum a current makes

A smelting cell passes 1.00 × 106 C of charge. What mass of aluminum forms? Use F = 96,485 C/mol.

Step 1 · moles of electrons1.00 × 106 C ÷ 96,485 C/mol = 10.36 mol e−.

Step 2 · the half reactionAl3+ + 3 e− → Al. Each atom costs three electrons.

Step 3 · moles of metal10.36 mol ÷ 3 = 3.455 mol Al.

Step 4 · mass3.455 mol × 26.98 g/mol = 93.2 g. That is why a smelter is built next to a power station.

Carbon: one atom, many solids

Carbon forms more allotropes than any other element. An allotrope is a different solid form of the same element. In diamond every carbon bonds to four others at 109.5°, with bonds 154 pm long. The whole crystal is one giant molecule. Nothing can slide and no electron is free, so diamond is the hardest natural solid and a poor conductor.

In graphite each carbon bonds to only three others in a flat sheet. Those bonds are 142 pm long and very strong. The fourth electron sits in a p orbital above and below the sheet, and it roams. Sheets lie 335 pm apart and are held by weak forces only. So graphite conducts along its sheets and slides apart in your pencil.

Diamond and graphite: same atoms, different bonding Two carbons: a rigid 3D network and a stack of flat sheets Diamond · sp³, 109.5° 154 pm Every carbon holds four neighbours. No loose electrons, so it does not conduct. Hardest natural solid. Melts near 3550 °C. Used in drill bits and heat spreaders. Graphite · sp², flat sheets 335 pm apart 142 pm in sheet Three strong bonds per atom, one spare p electron. Those spare electrons roam, so graphite conducts. Weak forces between sheets let them slide.
Hard versus slippery. Diamond locks every carbon into a rigid three dimensional net. Graphite spends three bonds in the sheet and leaves one electron loose. The gap between the sheets is more than twice the bond length inside one.

Two newer allotropes matter too. A single graphite sheet on its own is graphene. Roll it up and you get a nanotube. Buckminsterfullerene, C60, is a hollow ball of 20 hexagons and 12 pentagons.

Silicon builds the ground you stand on

Silicon is second only to oxygen in the crust, near 28% by mass. You never find it free. Every silicon atom in rock is tied to oxygen in a silicate. The unit is a tetrahedron, one silicon in the middle with four oxygens at the corners. Those tetrahedra share corners, and how many they share decides the mineral.

Carbon and silicon sit in the same group, so why is CO2 a gas and SiO2 a rock? Carbon is small, so its p orbitals overlap sideways well and give strong double bonds. One carbon with two double bonds is a complete molecule. Silicon is bigger and that sideways overlap is poor. So silicon makes four single bonds to oxygen instead, and the solid never ends.

Silicate structures from single tetrahedra to a full network One building block, four different minerals Single tetrahedron SiO44− corners shared: 0 olivine, zircon separate ions Chain (SiO32−)n corners shared: 2 pyroxene, asbestos fibres Sheet (Si2O52−)n corners shared: 3 mica, clay, talc flakes Full network SiO2 corners shared: 4 quartz, sand, glass hard 3D solid
Sharing corners. Share none and you get separate SiO44− ions, as in olivine. Share two and you get chains, which is why asbestos comes as fibres. Share three and you get sheets, so mica peels. Share all four and you get quartz.

Pure silicon starts as sand reduced with coke in an arc furnace. Chip grade silicon needs a second pass through a liquid:

SiO2(s) + 2 C(s) → Si(l) + 2 CO(g) Si(s) + 3 HCl(g) → SiHCl3(g) + H2(g)

The trichlorosilane is distilled, then broken back down to deposit pure silicon. Chips need fewer than one stray atom per billion.

Worth knowing. Structure beats composition. Asbestos and talc are both magnesium silicates. Asbestos is built from chains, so it splits into sharp needles that lodge in the lung. Talc is built from sheets, so it flakes into soft powder. Same elements, very different risk.
Key points
  • Boron is electron deficient. Its compounds act as Lewis acids, and diborane bridges with three-center two-electron bonds.
  • Boric acid takes OH− from water instead of donating its own proton.
  • Aluminum is won by electrolysis in molten cryolite. Its oxide film protects it, and the metal is amphoteric.
  • Diamond is a four bond network and an insulator. Graphite is a three bond sheet with a loose electron, so it conducts and lubricates.
  • Silicates are SiO4 tetrahedra sharing corners. More sharing means a tougher, less soluble mineral.
  • CO2 is a gas and SiO2 is a solid because strong double bonds need small atoms.

Section 4Nitrogen and phosphorus

Nitrogen is locked in a triple bond

Nitrogen gas is 78% of the air you breathe, yet almost nothing reacts with it. The reason is one number. The N≡N triple bond is worth 945 kJ/mol and is only 110 pm long. It is one of the strongest bonds known. Breaking it is the whole problem of nitrogen fixation, which means turning N2 into a form life can use.

Industry solves it with the Haber process. Nitrogen from air meets hydrogen from steam reforming over an iron catalyst:

N2(g) + 3 H2(g) ⇌ 2 NH3(g)    ΔH = −92.2 kJ

The forward reaction gives out heat and shrinks the gas volume. So cold and squeezed should give the best yield. But cold is also slow. Real plants run near 450 °C and 200 atm, which is a deal struck between yield and speed. Ammonia is pulled out as a liquid as it forms, which drags the balance forward.

Ammonia itself is a trigonal pyramid with H–N–H angles near 107°. It boils at −33 °C, so it is easy to store under mild pressure. In water it is a weak base:

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)    Kb = 1.8 × 10−5

The nitrogen oxides

Nitrogen forms an oxide at every oxidation state from +1 to +5. Dinitrogen oxide, N2O, is the dental gas. Nitrogen monoxide, NO, is a colourless gas with an odd number of electrons, which makes it a radical. Your own blood vessels use it as a signal. Nitrogen dioxide, NO2, is the brown haze over a busy city. It is also a radical, and it pairs up when cooled:

2 NO2(g) ⇌ N2O4(g)    ΔH = −57.2 kJ

Brown NO2 fades to colourless N2O4 in an ice bath. Warm the tube and the colour floods back. It is one of the clearest equilibrium demonstrations there is.

Nitric acid from ammonia

The Ostwald process turns ammonia into nitric acid in three steps. A platinum and rhodium gauze does the first one near 850 °C:

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) 2 NO(g) + O2(g) → 2 NO2(g) 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

The NO from the last step is piped back and used again. Nitric acid is a strong acid, but it is also a strong oxidizer, and that second job matters more. Copper will not touch hydrochloric acid. It dissolves in nitric acid at once, because the nitrate ion does the work. The product depends on how concentrated the acid is:

Cu(s) + 4 HNO3(conc) → Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l) 3 Cu(s) + 8 HNO3(dilute) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
From air to nitric acid: the Haber and Ostwald processes Air in, fertilizer and explosives out N₂ + 3 H₂78% of air + steam reformed H₂ Haber450 °C, 200 atm, Fe 2 NH₃ΔH = −92.2 kJ + 5 O₂Pt/Rh gauze, 850 °C 4 NOcolourless radical + O₂cool to 25 °C NO₂brown 2 HNO₃strong acid, strong oxidant 3 NO₂ + H₂O → 2 HNO₃ + NO the leftover NO is recycled NH₄NO₃ and fertilizerabout 80% of all output The triple bond in N₂ is worth 945 kJ/mol, so step one needs pressure, heat and a catalyst. Every later step is downhill.
Air to acid. The Haber step is the hard one, because it has to break the triple bond. Ostwald then walks nitrogen up through NO and NO2 to nitric acid. Most of the output ends as ammonium nitrate fertilizer.

Phosphorus comes in three solids

White phosphorus is built from P4 molecules shaped like a tiny tetrahedron. The bond angles are forced to 60°, far from the 109.5° the atom wants. That strain makes it wildly reactive. White phosphorus catches fire in air near 34 °C, so it is kept under water. Heat it gently and it turns into red phosphorus, a chain polymer safe enough for match strikers. Heat it under pressure and you get black phosphorus, a layered solid much like graphite.

Phosphorus burns in air to give one of two oxides. Starve it of oxygen and you get P4O6. Give it plenty and you get P4O10, a white powder so thirsty that chemists use it to dry other gases:

P4(s) + 3 O2(g) → P4O6(s) P4(s) + 5 O2(g) → P4O10(s) P4O10(s) + 6 H2O(l) → 4 H3PO4(aq)

The phosphorus acids

Phosphoric acid, H3PO4, gives up three protons, one at a time, each much harder than the last. Its three acid constants are 7.5 × 10−3, 6.2 × 10−8 and 4.2 × 10−13. The middle one sits close to blood pH, which is why phosphate is a body buffer. Most of it is made by pouring sulfuric acid on phosphate rock:

Ca3(PO4)2(s) + 3 H2SO4(aq) → 2 H3PO4(aq) + 3 CaSO4(s)

Now the trap. Phosphorous acid is written H3PO3, so it looks triprotic. It is not. One of its three hydrogens is bonded straight to phosphorus, and a P–H hydrogen never leaves in water. Only the two O–H hydrogens are acidic, so H3PO3 is diprotic. Hypophosphorous acid, H3PO2, has two P–H bonds and gives up just one proton. The rule to carry away is simple. Count the O–H groups.

Only a hydrogen bonded to oxygen can leave Count the O–H groups, not the hydrogens Phosphoric acid H₃PO₄ P O O–H O–H O–H triprotic pKₐ₁ = 2.12 Phosphorous acid H₃PO₃ P O O–H O–H H diprotic pKₐ₁ = 1.3 Hypophosphorous acid H₃PO₂ P O O–H H H monoprotic pKₐ₁ = 1.2
Count the O–H groups. Gold boxes mark hydrogens bonded to oxygen, which can leave. Grey boxes mark hydrogens bonded straight to phosphorus, which cannot. The formula alone would fool you all three times.

Worked example · pH of 0.10 M phosphoric acid

Find the pH of 0.10 M H3PO4. Use only Ka1 = 7.5 × 10−3, since the later steps give far less H+.

Step 1 · set upLet x = [H+] at equilibrium. Then x2 / (0.10 − x) = 7.5 × 10−3.

Step 2 · check the shortcutKa1 is only about 13 times smaller than 0.10. So you cannot drop the x in the bottom. Use the full quadratic.

Step 3 · solvex2 + 7.5 × 10−3x − 7.5 × 10−4 = 0. The quadratic formula gives x = (−0.0075 + 0.05528) / 2 = 0.0239 M.

Step 4 · take the logpH = −log(0.0239) = 1.62.

Step 5 · check itDrop the shortcut back in and you would get x = 0.0274 M and pH 1.56. The quadratic answer is the honest one. Either way the acid is strong enough to sting, which fits a soft drink ingredient used in small amounts.

Key points
  • The N2 triple bond at 945 kJ/mol is why nitrogen is so slow to react.
  • Haber makes ammonia at about 450 °C and 200 atm over iron. The conditions trade yield for speed.
  • Ostwald oxidizes ammonia to NO, then NO2, then nitric acid, and recycles the leftover NO.
  • Nitric acid attacks metals as an oxidizer. Concentrated acid gives NO2, dilute acid gives NO.
  • White P4 is strained and catches fire in air. Red and black phosphorus are tamer.
  • A hydrogen bonded to phosphorus is not acidic. H3PO4 is triprotic, H3PO3 diprotic, H3PO2 monoprotic.

Section 5Oxygen and sulfur

Oxygen has a second form

Oxygen gas is 21% of the air by volume. Its double bond is worth 498 kJ/mol, just over half of the nitrogen triple bond, which is why oxygen reacts so readily. O2 also has two unpaired electrons, so a pool of liquid oxygen clings to a magnet. Industry gets it by chilling air until it liquefies, then boiling the nitrogen off at −196 °C and the oxygen at −183 °C.

Oxygen also comes as ozone, O3, an allotrope with a bent shape and a 117° angle. Both O–O bonds are 128 pm, which sits between a single bond and a double bond. Neither bond is whole, because the real structure is an average of two drawings. Making ozone costs energy:

3 O2(g) → 2 O3(g)    ΔH = +285.4 kJ

High in the air, ultraviolet light pays that bill and ozone shields the ground. At ground level the same gas is a lung irritant. Position, not chemistry, decides whether ozone helps you.

Oxide, peroxide, superoxide

Oxygen is almost always −2, but not always. In a peroxide the ion is O22− and each oxygen is −1. In a superoxide the ion is O2− and each oxygen is −½. Adding electrons fills antibonding levels, so the bond order drops and the O–O distance grows.

Adding electrons to O2 weakens the oxygen oxygen bond Each electron you add stretches the O–O bond SPECIES BOND ORDER AND LENGTH O STATE O₂ dioxygen bond order 2.0 · 121 pm two unpaired electrons, so it is drawn to a magnet 0 O₂⁻ superoxide bond order 1.5 · 133 pm one spare electron, found in KO₂ −½ O₂²⁻ peroxide bond order 1.0 · 149 pm the O–O bond in H₂O₂ and Na₂O₂ −1 O²⁻ oxide single ion, no O–O bond no O–O bond left at all −2 Four electrons in, and the bond that started at 498 kJ/mol is gone.
One family, four rungs. Going down the list you add one electron at a time. Bond order falls from 2 to 1 to 0 and the bond stretches from 121 pm to 149 pm before it breaks for good.

These ions are useful. Sodium peroxide gives hydrogen peroxide with water. Potassium superoxide gives oxygen, which is why it fills emergency breathing canisters in mines and submarines:

Na2O2(s) + 2 H2O(l) → 2 NaOH(aq) + H2O2(aq) 4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)

Hydrogen peroxide is caught in the middle of its own range, so it can be oxidized or reduced. Left alone it does both at once and tears itself apart. A speck of manganese dioxide, or the enzyme in a cut finger, makes the split instant:

2 H2O2(aq) → 2 H2O(l) + O2(g)

Sulfur: rings that open

Sulfur sits under oxygen but behaves very differently. It is too big for strong double bonds, so instead of S2 it builds eight atom rings shaped like a crown. Solid S8 is rhombic below 95.3 °C and monoclinic above it, and it melts near 115 °C. Keep heating and the rings snap open into long chains. The liquid gets darker and thicker, which is odd for a hot liquid. Pour it into cold water and the tangle freezes as rubbery plastic sulfur.

Most sulfur now comes out of oil and gas as hydrogen sulfide. The Claus process turns that waste into solid sulfur:

2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g)

Hydrogen sulfide is a weak diprotic acid with Ka1 near 1.0 × 10−7. It smells of rotten eggs at tiny amounts. At higher amounts it numbs the nerve that detects it, so the warning disappears while the danger grows.

The contact process

Sulfuric acid is made in larger amounts than any other chemical. The route is the contact process, and it has four steps. Burn sulfur, oxidize the dioxide over a catalyst, absorb the trioxide into strong acid, then add water:

S(s) + O2(g) → SO2(g) 2 SO2(g) + O2(g) ⇌ 2 SO3(g)    V2O5, 430 °C, ΔH = −198 kJ SO3(g) + H2SO4(l) → H2S2O7(l) H2S2O7(l) + H2O(l) → 2 H2SO4(l)

Only the second step is reversible, and it is the one that needs care. Cool gas favours SO3 but reacts too slowly, so the plant settles near 430 °C with a vanadium oxide catalyst. The third step looks like a detour. It is there because SO3 met with water throws a fine acid mist that no tower can trap. The oleum route keeps the product in the liquid.

The contact process for sulfuric acid Four steps to the world’s most used chemical S(s)mined or from sour gas burn in airS + O₂ → SO₂ SO₂(g)sharp smell V₂O₅, 430 °C2 SO₂ + O₂ ⇌ 2 SO₃ SO₃(g)ΔH = −198 kJ into 98% acidSO₃ + H₂SO₄ → H₂S₂O₇ oleumH₂S₂O₇ + H₂O 2 H₂SO₄98% is about 18 M Why the detour through oleum? SO₃ poured straight into water throws a fine acid mist that is very hard to collect. Oleum avoids it. Step two is the only reversible step. Low heat favours SO₃ but runs slowly, so the plant settles near 430 °C. Sulfuric acid then goes to fertilizer, batteries, metal pickling and dozens of other jobs.
The contact process. Sulfur in, sulfuric acid out. The reversible middle step sets the conditions, and the oleum step exists only to avoid acid mist.

Concentrated sulfuric acid is about 98% by mass, which works out near 18 M. It is a strong acid on the first proton and a weak one on the second, with Ka2 = 1.2 × 10−2. It is also a powerful drying agent. Pour it on table sugar and it strips out the water, leaving a black column of carbon. Sulfur dioxide gives the weaker sulfurous acid, which bleaches paper, preserves dried fruit and falls as acid rain:

SO2(g) + H2O(l) ⇌ H2SO3(aq)    Ka1 = 1.4 × 10−2

Worked example · acid from a ton of sulfur

A plant feeds 1.00 metric ton of sulfur into the contact process. If every step ran perfectly, how much sulfuric acid comes out?

Step 1 · follow the atomsEach step carries one sulfur atom forward. So 1 mol S gives 1 mol SO2, then 1 mol SO3, then 1 mol H2SO4.

Step 2 · moles of sulfur1.00 × 106 g ÷ 32.06 g/mol = 3.119 × 104 mol S.

Step 3 · mass of acid3.119 × 104 mol × 98.08 g/mol = 3.06 × 106 g = 3.06 metric tons.

Step 4 · sanity checkThe acid weighs about three times the sulfur. That fits, since 98.08 ÷ 32.06 = 3.06. The extra mass is the oxygen and water the sulfur picks up.

Worth knowing. Mixing concentrated sulfuric acid with water gives out a lot of heat. Always add the acid to the water in a slow stream, never water to the acid. Water poured on acid can boil on contact and spit the acid back out.
Key points
  • O2 has two unpaired electrons and a 498 kJ/mol double bond, so it reacts far more easily than N2.
  • Ozone is bent, costs energy to make, and shields the ground from above while irritating lungs at ground level.
  • Oxide, peroxide and superoxide run −2, −1 and −½. More electrons means a longer, weaker O–O bond.
  • Sulfur rings open on heating, which is why molten sulfur thickens instead of thinning.
  • The contact process is burn, catalyse, absorb into oleum, then dilute. Only the SO3 step is reversible.
  • Sulfuric acid is strong on the first proton, weak on the second, and a fierce drying agent besides.

Section 6Halogens and noble gases

The halogens are one electron short

The halogens are fluorine, chlorine, bromine, iodine and astatine. Each has seven outer electrons, written ns2np5. Each needs one more to fill the shell, so each is a hungry oxidizer. All five exist as X2 molecules, and the way they look changes right down the group. Fluorine is a pale yellow gas, chlorine a yellow green gas, bromine a red brown liquid and iodine a grey violet solid.

Their appetite for electrons is measured by the standard reduction potential. Fluorine leads at +2.87 V, then chlorine at +1.36 V, bromine at +1.07 V and iodine at +0.54 V. That order lets you predict every displacement reaction. Any halogen pushes a halide ion from lower in the list out of its compound:

Cl2(aq) + 2 Br−(aq) → 2 Cl−(aq) + Br2(aq)

Fluorine does not stop at halides. It takes electrons from water itself:

2 F2(g) + 2 H2O(l) → 4 HF(aq) + O2(g)

Chlorine is gentler. In water it splits its own oxidation states, a move called disproportionation. In base the same reaction runs further and gives household bleach:

Cl2(g) + H2O(l) ⇌ HOCl(aq) + HCl(aq) Cl2(g) + 2 NaOH(aq) → NaCl(aq) + NaOCl(aq) + H2O(l)
Halogen oxidizing power falls down the group but bond strength peaks at chlorine Two trends that do not match Oxidizing power falls in a straight line. Bond strength does not, and that is why fluorine is so savage. E° for X₂ + 2e⁻ → 2X⁻ (volts) F₂ +2.87 V pale yellow gas Cl₂ +1.36 V yellow green gas Br₂ +1.07 V red brown liquid I₂ +0.54 V grey violet solid X–X bond energy (kJ/mol) F₂ 159 Cl₂ 243 Br₂ 193 I₂ 151 F₂ is weakest. Three lone pairs on each tiny atom crowd together and push the bond apart. A halogen displaces any halide below it in this list.
Strength and weakness together. Oxidizing power drops steadily from fluorine to iodine. Bond energy peaks at chlorine instead. The weak F–F bond is easy to break, which adds to fluorine's bite.

That bond energy chart holds a surprise. The F–F bond is only 159 kJ/mol, weaker than the Cl–Cl bond at 242 kJ/mol. Fluorine atoms are so small that their lone pairs crowd each other and push the bond apart. A bond that cheap to break, plus a huge gain from forming F−, is why fluorine attacks almost everything.

Halogen acids and oxyacids

Three of the four hydrogen halides are strong acids. Hydrofluoric acid is the odd one out. Its Ka is 6.8 × 10−4, because the H–F bond is worth 569 kJ/mol and does not want to break. Weak does not mean safe. HF soaks through skin and attacks bone, and it is the only common acid that eats glass:

SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l)

Chlorine also makes a set of oxyacids, and each extra oxygen makes the acid stronger. Hypochlorous acid is very weak, with Ka = 2.9 × 10−8. Chlorous acid is far stronger at 1.1 × 10−2. Chloric and perchloric acids are both strong, and perchloric is one of the strongest acids known. The reason is the leftover ion. Extra oxygens pull electron density away and spread the negative charge, so the ion holds together without the proton.

Interhalogens

Halogens also bond to each other. An interhalogen has the formula AB, AB3, AB5 or AB7. The larger, less electronegative halogen always sits in the middle, because only a big atom has room for several neighbours. Chlorine monofluoride, ClF, is linear. Bromine trifluoride, BrF3, is T shaped. Iodine pentafluoride, IF5, is a square pyramid. Only iodine is wide enough to hold seven fluorines, so IF7 is the one AB7 that exists.

These compounds are savage fluorinating agents. Chlorine trifluoride sets fire to glass wool, sand and concrete. It was tested as a rocket oxidizer and then quietly dropped.

Worked example · will bromine displace iodide?

Bromine water is poured on a potassium iodide solution. Does the reaction go, and how much free energy does it release?

Step 1 · write the halvesBr2 + 2 e− → 2 Br− has E° = +1.07 V. I2 + 2 e− → 2 I− has E° = +0.54 V.

Step 2 · build the cellBromine is the stronger oxidizer, so it takes the electrons and iodide gives them up. E°cell = 1.07 V − 0.54 V = +0.53 V. A positive value means it goes.

Step 3 · the equationBr2(aq) + 2 I−(aq) → 2 Br−(aq) + I2(aq). The solution turns brown as iodine appears.

Step 4 · free energyΔG° = −nFE° = −(2)(96,485 C/mol)(0.53 V) = −1.02 × 105 J = −102 kJ.

Step 5 · turn it aroundTry iodine on bromide and E°cell = 0.54 − 1.07 = −0.53 V. Negative, so nothing happens. The ranking only works one way.

The noble gases hold back

The noble gases are helium, neon, argon, krypton, xenon and radon. Each has a filled outer shell, so each has nothing to gain by bonding. That shows up in their ionization energies, which are the highest in every period. Helium needs 2372 kJ/mol. Argon needs 1521 kJ/mol. All six are gases that boil far below room temperature, and all are single atoms rather than pairs.

Still, the heavier ones do react. Xenon needs only 1170 kJ/mol to lose an electron, which is about the same as an oxygen molecule at 1175 kJ/mol. Since oxygen can be forced to give up an electron, xenon should be able to as well. That single comparison led to the first noble gas compound. Fluorine is the partner, and the amount you use sets the product:

Xe(g) + F2(g) → XeF2(s) Xe(g) + 2 F2(g) → XeF4(s) Xe(g) + 3 F2(g) → XeF6(s)

XeF2 is linear and XeF4 is square planar, both shaped by extra lone pairs on xenon. Water turns the fluorides into xenon trioxide, which is a violent explosive:

XeF6(s) + 3 H2O(l) → XeO3(aq) + 6 HF(aq)

Their everyday jobs come from not reacting. Helium lifts balloons and cools magnets. Argon fills the space around a weld so the hot metal cannot meet oxygen. Neon glows red orange in a tube, and krypton and xenon run bright lamps and camera flashes.

Group 17 factFClBrI
State at 25 °Cgasgasliquidsolid
E° for X2/X− (V)+2.87+1.36+1.07+0.54
X–X bond (kJ/mol)159242193151
H–X bond (kJ/mol)569431366298
HX in waterweakstrongstrongstrongest
Electronegativity3.983.162.962.66
Key points
  • Halogen oxidizing power falls from fluorine to iodine, and that order predicts every displacement.
  • The F–F bond is weaker than the Cl–Cl bond because lone pairs crowd on a tiny atom.
  • HF is a weak acid because the H–F bond is strong. It is still the acid that etches glass.
  • More oxygens make a halogen oxyacid stronger, so HClO4 beats HClO by a huge margin.
  • In an interhalogen the bigger halogen goes in the middle. Only iodine can hold seven fluorines.
  • Xenon reacts because its ionization energy nearly matches that of O2. Fluorine gives XeF2, XeF4 and XeF6.

The chapter in one breath

  • Two trends run everything. Across a period the nuclear grip tightens. Down a group a new shell loosens it.
  • Oxides mark the map. Basic on the left, amphoteric at the border, acidic on the right.
  • Groups 1 and 2 give electrons away. Every alkali metal makes hydrogen and a hydroxide in water, faster as you go down.
  • Boron is short of electrons and aluminum is armoured. One acts as a Lewis acid, the other hides behind its own oxide film.
  • Structure decides properties. Diamond, graphite, quartz and asbestos are all networks of the same few atoms.
  • Nitrogen is hard to start and easy to finish. Haber breaks the triple bond, then Ostwald walks nitrogen up to nitric acid.
  • Count O–H groups, not hydrogens. That is how H3PO3 ends up diprotic.
  • Oxygen stretches as it gains electrons. Oxide, peroxide and superoxide are the same pair of atoms with different loads.
  • Halogens rank by appetite. Fluorine first, iodine last, and the ranking predicts every displacement and the noble gas compounds too.

ReferenceWords worth knowing

These are the words this chapter leans on. Come back to them before a quiz.

alkali metal
A group 1 element. One outer s electron, a +1 ion, and a fast reaction with water.
alkaline earth metal
A group 2 element. Two outer s electrons and a +2 ion. Harder and slower to react than group 1.
allotrope
A different solid form of the same element. Diamond and graphite are allotropes of carbon.
amphoteric
Able to react as an acid and as a base. Al2O3 and BeO both qualify.
contact process
The four step industrial route to sulfuric acid: burn sulfur, oxidize SO2 over V2O5, absorb SO3 into acid, then add water.
cryolite
Na3AlF6. Molten cryolite dissolves alumina so aluminum can be won at about 960 °C instead of 2072 °C.
disproportionation
One substance is oxidized and reduced in the same reaction. Chlorine in water gives HOCl and HCl.
effective nuclear charge
Written Zeff. The share of the nuclear charge an outer electron actually feels after inner electrons screen it.
Haber process
Making ammonia from N2 and H2 over iron, near 450 °C and 200 atm.
halogen
A group 17 element. Seven outer electrons, an X2 molecule, and a strong pull for one more electron.
interhalogen
A compound of two different halogens, such as BrF3 or IF7. The larger halogen sits in the middle.
metallic character
How readily an element gives electrons up and behaves as a metal. It rises down a group and falls across a period.
metalloid
An element on the staircase border with some metal and some nonmetal behaviour. Boron, silicon, germanium, arsenic, antimony and tellurium.
nitrogen fixation
Turning unreactive N2 into usable compounds such as ammonia or nitrate.
noble gas
A group 18 element. A filled outer shell, a very high ionization energy, and single atoms rather than molecules.
oleum
H2S2O7, made by feeding SO3 into concentrated sulfuric acid. It avoids the acid mist that water would cause.
Ostwald process
Making nitric acid from ammonia in three steps, through NO and then NO2.
peroxide
A compound holding the O22− ion, with oxygen at −1 and a single O–O bond.
silicate
A mineral built from SiO4 tetrahedra sharing corners. Sharing none gives separate ions, sharing all four gives quartz.
superoxide
A compound holding the O2− ion, with oxygen at −½. KO2 is the common example.

Check yourself

1. Sodium oxide, aluminum oxide and sulfur trioxide are all period 3 oxides. Sort them as basic, amphoteric or acidic, and say what fixes the order.
Na2O basic, Al2O3 amphoteric, SO3 acidic. Metallic character falls from left to right, so the oxide bonding shifts from ionic to covalent. Aluminum sits on the border, so its oxide reacts with acid and with base.
2. Drop 4.00 g of calcium into excess water. What volume of hydrogen forms at 0 °C and 1.00 atm?
2.24 L. Ca + 2 H2O → Ca(OH)2 + H2. Moles of Ca = 4.00 g ÷ 40.08 g/mol = 0.0998 mol. The ratio is 1 to 1, so 0.0998 mol H2. Then V = nRT/P = (0.0998)(0.08206)(273.15) ÷ 1.00 = 2.24 L.
3. A smelter must deliver 1.00 kg of aluminum. How much charge does that take? Use F = 96,485 C/mol.
1.07 × 107 C. Moles of Al = 1000 g ÷ 26.98 g/mol = 37.06 mol. Al3+ needs 3 electrons, so 37.06 × 3 = 111.2 mol e−. Charge = 111.2 mol × 96,485 C/mol = 1.07 × 107 C.
4. Carbon dioxide is a gas at room temperature. Silicon dioxide is sand. Both are group 14 dioxides, so why the huge difference?
Carbon can form strong double bonds and silicon cannot. Carbon is small, so its p orbitals overlap sideways well. Two double bonds finish the CO2 molecule, and separate molecules drift apart as a gas. Silicon is larger and that sideways overlap is weak, so it makes four single bonds to oxygen and builds an endless network solid.
5. A kiln is fed 1.00 kg of pure calcium carbonate. What mass of quicklime comes out?
560 g. CaCO3 → CaO + CO2. Moles of CaCO3 = 1000 g ÷ 100.09 g/mol = 9.991 mol. The ratio is 1 to 1, so 9.991 mol CaO × 56.08 g/mol = 560 g. The missing 440 g left as carbon dioxide, and 560 + 440 = 1000 g checks out.
6. Phosphorous acid is written H3PO3. How many protons can it actually give up, and how do you know?
Two. It is diprotic. Only a hydrogen bonded to oxygen is acidic. H3PO3 has two O–H groups and one hydrogen bonded straight to phosphorus. The P–H hydrogen does not leave in water. By the same rule H3PO2 is monoprotic.
7. Find the pH of a 0.050 M solution of hydrofluoric acid. Ka = 6.8 × 10−4.
pH = 2.26. Set x2/(0.050 − x) = 6.8 × 10−4, so x2 + 6.8 × 10−4x − 3.4 × 10−5 = 0. The quadratic formula gives x = (−0.00068 + 0.011682) ÷ 2 = 0.00550 M. pH = −log(0.00550) = 2.26.
8. An ammonia plant must make 1.00 kg of NH3. What volume of nitrogen gas does that need at 0 °C and 1.00 atm?
658 L. N2 + 3 H2 → 2 NH3. Moles of NH3 = 1000 g ÷ 17.03 g/mol = 58.72 mol. Nitrogen needed = 58.72 ÷ 2 = 29.36 mol. Then V = (29.36)(0.08206)(273.15) ÷ 1.00 = 658 L.
9. Why does the contact process feed SO3 into concentrated sulfuric acid instead of straight into water?
To stop an acid mist forming. SO3 met with water gives off so much heat that the acid leaves as a fine fog, and a fog is almost impossible to collect. Absorbing it into 98% acid gives liquid oleum, H2S2O7, which is then diluted to twice the acid you started with.
10. Chlorine water is added to sodium bromide solution. Will anything happen? Work out E°cell and ΔG°. Use E° = +1.36 V for Cl2 and +1.07 V for Br2.
Yes. E°cell = +0.29 V and ΔG° = −56.0 kJ. Chlorine is the stronger oxidizer, so Cl2 + 2 Br− → 2 Cl− + Br2. E°cell = 1.36 − 1.07 = +0.29 V. Then ΔG° = −nFE° = −(2)(96,485)(0.29) = −5.60 × 104 J = −56.0 kJ. The solution turns orange as bromine forms.
11. Hydrochloric acid is strong but hydrofluoric acid is weak. Fluorine is the more electronegative element, so explain the result.
The H–F bond is too strong to break. Acid strength here depends on losing the proton, not on electronegativity. H–F is worth 569 kJ/mol while H–Cl is only 431 kJ/mol. Fluoride is also small, so it holds the proton tightly in water. Ka for HF is 6.8 × 10−4, which is weak but far from harmless.
12. Xenon forms XeF2, XeF4 and XeF6, but neon forms nothing. What makes xenon different?
Xenon holds its outer electrons far more loosely. Xenon's first ionization energy is 1170 kJ/mol, close to the 1175 kJ/mol of an O2 molecule, and oxygen can be forced to give an electron up. Neon needs 2081 kJ/mol, and no partner can pay that. Xenon is also large enough to fit several fluorine atoms around it.
Chemistry · Chapter 19

Transition Metals and Coordination Chemistry

Why is ruby red and emerald green? Why does blood carry oxygen while rust does nothing of the kind? Both answers live in the middle of the periodic table. The metals there hold electrons in d orbitals, and those orbitals are close enough in energy to shift under a nudge. A ring of molecules around the metal gives that nudge. The result is color, magnetism and a huge family of compounds called complexes. This chapter shows you how to build one, name one, count its shapes and predict its color from the ligands alone.

By the end of this chapter you can

  • Write the electron configuration of a d-block atom and of its ions, taking the 4s electrons away first.
  • Pick out the metal, the ligands and the counter ions in a coordination compound, and find the metal's oxidation number.
  • Count donor atoms to get a coordination number, and sort ligands by denticity.
  • Name a coordination compound from its formula, and write the formula back from a name.
  • Spot structural, geometric and optical isomers, and say which ones are chiral.
  • Split the d orbitals with a crystal field, fill them high spin or low spin, and read the splitting off the spectrochemical series.
  • Turn a splitting energy into an absorbed wavelength, then name the color you would see.
  • Count unpaired electrons to get a magnetic moment, and explain why chelating ligands bind so tightly.

Section 1Transition metal properties and configurations

Where the d-block sits

Walk across the middle of the periodic table and you cross the d-block. It holds groups 3 through 12. Scandium through zinc make up its first row. A transition metal is a metal whose atoms or common ions carry a partly filled d subshell. That one trait drives everything else in this chapter.

These metals act alike in handy ways. They are hard and shiny. They melt at high temperatures. Tungsten melts at 3422 °C, the highest of any metal. They carry current well. Most of them show more than one charge. Iron gives you both Fe2+ and Fe3+. Manganese runs from +2 up to +7. Their compounds are often brightly colored, and many are tugged by a magnet.

Zinc, cadmium and mercury sit in the block but hold a full d10 set in every normal ion. Strict rules leave them out of the transition metal club. Most people still group them with the block for convenience.

Filling up: 4s goes in first

Energy order sets the filling order. In a neutral potassium or calcium atom, the 4s orbital lies a little below 3d. So 4s takes electrons first. Scandium is the first element with a 3d electron. Its configuration is [Ar]3d14s2.

Add protons one at a time and the pattern rolls on. Titanium is [Ar]3d24s2. Vanadium is [Ar]3d34s2. Notice that we write 3d before 4s, because 3 comes before 4. We fill 4s first, yet we list it second.

Two metals in the first row break the tidy pattern. Chromium is [Ar]3d54s1, not 3d44s2. Copper is [Ar]3d104s1, not 3d94s2. The two levels sit so close that a half filled or a filled d set is worth the shift. Molybdenum and silver do the same thing one row down.

Making ions: 4s leaves first

Here is the fact that trips people up most. When a transition metal forms a cation, the 4s electrons leave first. Every time.

Iron is [Ar]3d64s2. Pull two electrons off and you have Fe2+. The answer is [Ar]3d6, not [Ar]3d44s2. Pull a third and Fe3+ is [Ar]3d5.

Why the flip? The gap between 4s and 3d is tiny, and its size shifts with the number of d electrons. Once the 3d set holds electrons, 3d drops below 4s. The 4s orbital is also fatter and reaches further out, so its electrons are easier to strip. Both facts point the same way.

The 4s and 3d levels almost tie. In a neutral atom 4s lies lower and fills first. In a cation 3d lies lower, so the 4s electrons leave first.1 · The 4s and 3d near tie, and which way it tipsenergyNeutral Fe atom3d (five orbitals)4s (one orbital)tiny gap4s lies a little lower, so 4s fills firstenergyFe2+​ ion3d (five orbitals)4s (one orbital)tiny gap3d has dropped lower, so 4s empties firstThe same idea in boxes: strip the 4s pair, keep all six 3d electronsFe3d4s[Ar]3d6​4s2​Fe2+​3d4s[Ar]3d6​Fe3+​3d4s[Ar]3d5​
Figure 19.1 The 4s and 3d levels almost tie. In the neutral atom 4s wins by a hair, so it fills first. In the ion 3d has slipped below 4s, so the 4s pair is the first to go. Iron keeps all six of its 3d electrons through both steps.

Counting d electrons is then easy. Take the group number and subtract the charge. Iron sits in group 8, so Fe3+ has 8 − 3 = 5 d electrons. Copper sits in group 11, so Cu2+ has 9. You will lean on this count in every later section.

Key points
  • The d-block spans groups 3 to 12. A transition metal has a partly filled d subshell.
  • In neutral atoms 4s fills before 3d, but we always write 3d first.
  • Chromium and copper shift one electron to get a half filled or filled d set.
  • In cations the 4s electrons leave first, so Fe2+ is [Ar]3d6.
  • d electrons in an ion = group number − charge.

Worked example · configurations for cobalt and its ions

Write the electron configuration of Co, Co2+ and Co3+. Then count the unpaired electrons in each one.

Step 1 · place the neutral atomCobalt is element 27, in group 9. Argon covers 18 electrons. That leaves 9 to place. Fill 4s with 2, then put 7 in 3d. So Co is [Ar]3d74s2.

Step 2 · check the count18 + 7 + 2 = 27. The electrons are all accounted for.

Step 3 · take the 4s pair off firstCo2+ loses both 4s electrons. That gives [Ar]3d7. Co3+ loses the 4s pair plus one 3d electron, giving [Ar]3d6.

Step 4 · spread the d electrons by Hund's ruleFive d orbitals, filled singly first. With 7 electrons: 5 single, then 2 of them get a partner. That leaves 3 unpaired. With 6 electrons: 5 single, then 1 gets a partner, leaving 4 unpaired.

AnswerCo is [Ar]3d74s2. Co2+ is [Ar]3d7 with 3 unpaired. Co3+ is [Ar]3d6 with 4 unpaired. Those unpaired counts hold for a free ion. Section 5 shows that ligands can change them.

Worth knowing. The 4s and 3d levels are so close that a tiny change in charge reorders them. This is why d-block chemistry is rich and why main group chemistry is tidy.

Section 2Coordination compounds and ligands

A metal with a ring of partners

Drop a crystal of CuSO4 into water and the blue you see is not bare Cu2+. Six water molecules have latched onto the ion. The real species is [Cu(H2O)6]2+. Any species like that, a central metal ion with a ring of attached partners, is a coordination compound. If the whole unit carries a charge, we call it a complex ion.

Each attached partner is a ligand. A ligand always has at least one lone pair to share. It hands that pair to the metal, so the ligand is a Lewis base and the metal ion is a Lewis acid. The bond that forms is a coordinate covalent bond. Both electrons in it came from the ligand.

The atom on the ligand that actually holds the lone pair is the donor atom. In water it is the oxygen. In ammonia it is the nitrogen. In the cyanide ion it is the carbon. The metal plus its ligands make up the coordination sphere, and we wrap that in square brackets. Ions that just balance the charge from outside the brackets are counter ions.

Denticity: how many teeth does a ligand have?

Some ligands grab once. Some grab several times. The number of donor atoms one ligand uses is its denticity. The word comes from teeth, and the picture is a bite.

A monodentate ligand bites once. Water, ammonia, chloride, cyanide and hydroxide all work this way. A bidentate ligand bites twice with one body. Ethylenediamine, written en, has a nitrogen at each end and a short carbon chain between. The oxalate ion, C2O42-, bites with two oxygens.

A ligand with more teeth is polydentate. The champion in everyday use is EDTA, which bites six times with two nitrogens and four oxygens. One EDTA can take over a metal ion all by itself.

An octahedral complex with six donor atoms around one cobalt, and three ligands that bite with one, two or six donor atoms.2 · Six bonds to the metal, however many ligands that takesNH3​ClNH3​NH3​NH3​NH3​Cothe coordination sphere, written inside [ ][Co(NH3​)5​Cl]2+​coordination number 6: five ammonia plus one chlorideone donor atom (monodentate)Mammoniagrabs the metal oncetakes 1 of the six slotstwo donor atoms (bidentate)Methylenediaminetwo grabs make one ringtakes 2 of the six slotssix donor atoms (hexadentate)MEDTAwraps the ion completelytakes 6 of the six slotsThe dots are lone pairs. Every lone pair that lands on the metal counts as one bond.
Figure 19.2 Left: six donor atoms around one cobalt, boxed off as the coordination sphere. Right: the same six slots filled by six monodentate ligands, by three bidentate ligands, or by a single hexadentate EDTA.

Coordination number counts bonds, not ligands

The coordination number is the number of donor atoms bonded to the metal. It is a bond count, not a ligand count. In [Co(NH3)6]3+ both counts are 6, since each ammonia bites once. In [Co(en)3]3+ there are only 3 ligands, yet the coordination number is still 6. Three bidentate ligands supply six donor atoms.

Coordination number sets the shape. A count of 2 gives a line, as in [Ag(NH3)2]+. A count of 4 gives either a tetrahedron, as in [Zn(NH3)4]2+, or a flat square, as in [Ni(CN)4]2-. A count of 6 gives an octahedral shape, which is by far the most common. Think of the metal at the center of a cube with one ligand poking out through each face.

Coordination numberShapeExampleBond angles
2linear[Ag(NH3)2]+180°
4tetrahedral[CoCl4]2-109.5°
4square planar[Pt(NH3)2Cl2]90°
6octahedral[Fe(CN)6]3-90° and 180°
Key points
  • A coordination compound is a metal ion plus ligands. Charged, it is a complex ion.
  • Ligands donate a lone pair, so they are Lewis bases. The metal is the Lewis acid.
  • Denticity is the number of donor atoms on one ligand: mono, bi, or poly.
  • Coordination number counts bonds to the metal, not ligands.
  • Count 6 means octahedral, count 4 means tetrahedral or square planar, count 2 means linear.

Worked example · pull apart [Cr(en)2Cl2]Cl

Find the coordination number, the charge on the complex ion, and the oxidation number of chromium.

Step 1 · split inside from outsideInside the brackets: [Cr(en)2Cl2]. Outside: one Cl-. That lone chloride is a counter ion.

Step 2 · count donor atomsEach en is bidentate, so 2 en supply 2 × 2 = 4 donor atoms. The two chloride ligands add 2 more. Total = 4 + 2 = 6, so the shape is octahedral.

Step 3 · balance the outsideOne Cl- sits outside. The whole compound is neutral, so the complex ion must be 1+.

Step 4 · solve for the metalLet the oxidation number of Cr be x. en is neutral, so it adds 0. Each chloride ligand adds −1.

x + 2(0) + 2(−1) = +1, so x − 2 = +1, so x = +3.

AnswerCoordination number 6, complex ion charge 1+, chromium is +3. Chromium has 6 electrons past argon, so Cr3+ is [Ar]3d3.

Section 3Naming coordination compounds

Five moves, in order

A name for a coordination compound looks long. It is built from small pieces, always in the same order. Learn the five moves and you can name any of them.

First, name the cation, then the anion. That is the rule for every ionic compound, and it does not change here. Second, inside the brackets, list the ligands in alphabetical order, then the metal. Third, show how many of each ligand there are. Fourth, add an ending to the metal if the complex is negative. Fifth, give the metal's oxidation number in Roman numerals inside parentheses.

The whole coordination sphere becomes one single word. No spaces inside it.

Ligand names and ligand counts

A ligand that carries a negative charge gets an -o ending. Chloride becomes chlorido, cyanide becomes cyanido, hydroxide becomes hydroxido, oxalate becomes oxalato. Older books drop the id and write chloro and cyano. Both forms are still in print, so you should know each one.

Neutral ligands usually keep their own name, but three of them have special names you simply have to learn. Water becomes aqua. Ammonia becomes ammine, with two letters m. Carbon monoxide becomes carbonyl.

For counts, use the Greek prefixes di, tri, tetra, penta and hexa. If the ligand name already contains a prefix, switch to bis, tris or tetrakis and wrap the ligand in parentheses. Two ethylenediamines become bis(ethylenediamine), never diethylenediamine. The prefix never affects alphabetical order. Sort by the ligand name itself, so tetraammine files under a, not under t.

The formula for potassium hexacyanidoferrate three, split into its parts, with five naming moves listed beside the name piece each one adds.3 · One formula, five naming movesK3​[Fe(CN)6​]3−​outside the bracketsthree K+​ ionsso the complex carries a 3− chargeinside the bracketsone Fe and six CN−​so it is a complex anion1Cation firstPotassium is the counter ion, so it leads.potassium2Ligands, alphabeticalSix cyanide ligands give hexacyanido.hexacyanido3Metal nextIron follows, inside the same word....ferr...4An anion ends in -ateThe complex is negative, so use ferrate....ferrate5Oxidation number lastx + 6(−1) = −3, so x = +3. Write (III).ferrate(III)Finished name: potassium hexacyanidoferrate(III)
Figure 19.3 Every piece of the name traces back to a piece of the formula. The counter ion leads, the ligands come next in alphabetical order, and the metal closes the word with its oxidation number.

When the complex is an anion

If the coordination sphere carries a negative charge, the metal name ends in -ate. Several metals switch to a Latin stem at the same time. Iron becomes ferrate. Copper becomes cuprate. Silver becomes argentate. Gold becomes aurate. Lead becomes plumbate and tin becomes stannate. Most other metals just take the ending, as in nickelate, cobaltate and chromate.

Three worked names show the pattern. [Co(NH3)6]Cl3 is hexaamminecobalt(III) chloride. [Cu(NH3)4]SO4 is tetraamminecopper(II) sulfate. Na2[Zn(OH)4] is sodium tetrahydroxidozincate(II). Only the third one is an anion, so only the third one picks up -ate.

Key points
  • Cation first, then anion. The coordination sphere is one word.
  • Ligands come before the metal, sorted alphabetically by ligand name.
  • Negative ligands end in -o. Water is aqua, ammonia is ammine, CO is carbonyl.
  • Use bis, tris and tetrakis with parentheses for ligands that hold their own prefix.
  • A negative complex gives the metal an -ate ending, often on a Latin stem.
  • The oxidation number goes last, in Roman numerals, in parentheses.

Worked example · name [Cr(H2O)4Cl2]Br

Step 1 · find the counter ionOne Br- sits outside the brackets. It is the anion, so it is named last as bromide. The complex is the cation, so it leads.

Step 2 · charge on the complexThe compound is neutral and the bromide is −1. So the complex ion is 1+.

Step 3 · oxidation number of chromiumx + 4(0) + 2(−1) = +1. So x = +3, and we write (III).

Step 4 · name and sort the ligandsFour waters become tetraaqua. Two chlorides become dichlorido. Sort by ligand name: aqua before chlorido, since a beats c.

Step 5 · build one wordtetraaqua + dichlorido + chromium(III). The complex is positive, so chromium keeps its plain name.

Answertetraaquadichloridochromium(III) bromide. In the older style it is tetraaquadichlorochromium(III) bromide.

Worked example · go the other way

Write the formula for potassium hexafluoridocobaltate(III).

Step 1 · read the piecesPotassium is the counter cation. Hexafluorido means six F- ligands. Cobaltate means the complex is an anion. The (III) sets cobalt at +3.

Step 2 · add up the charge inside+3 + 6(−1) = −3. So the complex ion is [CoF6]3-.

Step 3 · balance with potassiumEach K is +1, so you need three of them.

AnswerK3[CoF6]

Worth knowing. Ammine has two letters m and means NH3. Amine has one and means an organic nitrogen group. One letter changes the whole meaning, so write it with care.

Section 4Isomerism in complexes

Same formula, different compound

Isomers share a formula but differ in how the atoms are joined or placed. Complexes offer more ways to do this than almost any other family of compounds. The isomers split into two camps. Structural isomers differ in which atoms are bonded to what. Stereoisomers have the same bonds but a different arrangement in space.

Four kinds of structural isomer

In an ionization isomer, a ligand and a counter ion trade places. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br hold the same atoms. Only one of them drops free sulfate into water, so a simple test tube check tells them apart.

A hydrate isomer moves a water molecule in or out of the sphere. [Cr(H2O)6]Cl3 is violet. [Cr(H2O)5Cl]Cl2·H2O is pale green. Same formula, different color.

A coordination isomer swaps ligands between two metals. [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6] are a matched pair.

A linkage isomer uses a different donor atom on the same ligand. Thiocyanate, SCN-, can bond through its sulfur or through its nitrogen. Nitrite can bond through nitrogen or through oxygen. A ligand with that choice is called ambidentate.

Cis and trans

Geometric isomers differ only in which positions the ligands take. The classic case is a flat square with two ligands of each kind. If the two chlorides touch at a corner, that is cis. If they sit straight across from each other, that is trans.

This difference is not a detail. cis-[Pt(NH3)2Cl2] is cisplatin, a drug that treats several cancers. The trans form has no such use. The two chlorides in the cis form are close enough to bind two neighboring sites on a strand of DNA. In the trans form they point the wrong way.

Octahedral complexes do the same trick. In [Co(NH3)4Cl2]+ the two chlorides are either 90° apart, which is cis, or 180° apart, which is trans. A tetrahedron cannot do this at all. Every corner of a tetrahedron touches every other corner, so there is no far side.

Cis and trans square planar isomers, cis and trans octahedral isomers, and a pair of mirror image complexes that cannot be stacked on each other.4 · Same formula, different arrangementcis, square planarClClNH3​NH3​Ptcisplatin: it fights tumourstrans, square planarClNH3​ClNH3​Pttransplatin: no use in therapycis, octahedralClNClNNNCothe two Cl sit 90° aparttrans, octahedralClClNNNNCothe two Cl sit 180° apartA mirror image you cannot stack back on the originalNNNNNNCoNNNNNNComirrorleft handed formright handed formthree bidentate rings, twisted one waythe same rings, twisted the other way
Figure 19.4 Top: cis and trans forms in a square plane and in an octahedron. Bottom: two complexes built from three bidentate ligands. They are mirror images, and no amount of turning makes one match the other.

Left handed and right handed

Some complexes come in two forms that are mirror images and cannot be laid on top of each other. Your hands work the same way. Such a molecule is chiral, and the pair are enantiomers.

[Co(en)3]3+ is the standard example. Three bidentate rings wind around the metal like the blades of a propeller, and a propeller twists one way or the other. The two forms are chemically alike in almost every test. They part ways when they meet polarized light. One form turns the plane of the light to the right, the other turns it left by the same amount. That is optical isomerism.

An equal mixture of the two shows no net turn, because the two effects cancel. Chemists call that a racemic mixture. The quick test for chirality is to hunt for a mirror plane in the molecule. If one exists, the molecule is not chiral.

Key points
  • Structural isomers differ in bonding. Stereoisomers differ only in placement.
  • Ionization, hydrate, coordination and linkage isomers are the four structural types.
  • cis means neighboring, trans means across. Tetrahedral complexes have neither.
  • Cisplatin works and transplatin does not, because of the angle alone.
  • Enantiomers are mirror image pairs. They rotate polarized light in opposite directions.
  • If a molecule has a mirror plane, it is not chiral.

Worked example · count the stereoisomers of [Co(en)2Cl2]+

Step 1 · get the shapeTwo en give 4 donor atoms. Two chlorides add 2. The coordination number is 6, so the complex is octahedral.

Step 2 · place the two chloridesThey can be 90° apart or 180° apart. That is one cis form and one trans form.

Step 3 · test the trans form for a mirror planeWith the chlorides opposite, the two en rings lie in one plane through the metal. That plane is a mirror. So trans is not chiral.

Step 4 · test the cis formWith the chlorides at 90°, no mirror plane survives. The cis form is chiral, so it exists as two enantiomers.

AnswerThree stereoisomers in all: trans, plus the two mirror forms of cis. Only the two cis forms turn polarized light, and they turn it in opposite directions.

Section 5Crystal field theory and color

Ligands push the d orbitals apart

The five d orbitals in a free ion all share one energy. Bring six ligands up along the x, y and z axes and that changes. Each ligand carries a lone pair, and lone pairs repel the electrons already on the metal. The whole d set is pushed up, but not evenly. Crystal field theory treats each ligand as a small point of negative charge and tracks that uneven push.

Two of the five d orbitals point straight down the axes, right at the incoming ligands. Those two are dz2 and dx2-y2, and they get pushed up hard. We call the pair the eg set. The other three point into the gaps between the axes, so they feel less. Those three are dxy, dxz and dyz, and they form the t2g set.

The energy gap between the two sets is the crystal field splitting energy, written Δ. The average energy does not move, so the three lower orbitals drop by 0.4Δ while the two upper ones rise by 0.6Δ.

A tetrahedral field flips the picture. Now no orbital points straight at a ligand, and the set of three ends up above the set of two. The gap is also smaller, roughly four ninths of the octahedral value.

High spin or low spin

Picture filling the split levels with 6 d electrons. You have two choices. Spread them four below and two above, leaving four unpaired. Or cram all six into the lower set, with none unpaired.

Each choice has a price. Climbing to the upper set costs Δ. Pairing two electrons in one orbital costs the pairing energy, P. Whichever is cheaper wins. If Δ is small, climbing is cheap, so electrons spread out. That is a high-spin complex. If Δ is large, pairing is cheaper, so electrons stack in the lower set. That is a low-spin complex.

The choice only exists for d4 through d7. With fewer or more electrons, both routes give the same answer. Tetrahedral complexes have a small gap, so they are high spin in practice.

The spectrochemical series

So what sets Δ? Mostly the ligands. Rank them by the size of the split and you get the spectrochemical series. Weak field ligands sit on the left, strong field ligands on the right.

I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < NO2- < CN- < CO

Two other rules help. A higher charge on the metal pulls the ligands closer, so Δ grows. And Δ grows going down a group, so 4d and 5d metals are nearly always low spin.

Octahedral d orbital splitting drawn twice, once with a weak field and high spin filling and once with a strong field and low spin filling, above a chart pairing each absorbed colour with the colour you see.5 · The split, the two fillings, and the colour you seeWeak field, high spin: d6​averageeg​ points straight at the ligandst2g​ points between the ligandsΔSmall Δ. Climbing costs less than pairing. Four unpaired.Strong field, low spin: d6​averageeg​ points straight at the ligandst2g​ points between the ligandsΔLarge Δ. Pairing costs less than climbing. None unpaired.Absorb one band and you see the opposite colour→violet410 nmyellowyou see→blue470 nmorangeyou see→green530 nmpurpleyou see→yellow580 nmvioletyou see→orange620 nmblueyou see→red700 nmgreenyou see
Figure 19.5 The same d6 ion in a weak field and in a strong field. A small gap leaves four unpaired electrons. A large gap pairs them all. Below, each absorbed band and the color that gets through.

Why complexes are colored

A d electron in the lower set can jump to the upper set if a photon hands it exactly Δ of energy. That jump is a d-d transition. The photon is swallowed, so that part of the light never reaches your eye. The color you see is what is left over.

So the color you see is the complementary color of the color absorbed. Absorb green and the solution looks purple. Absorb orange and it looks blue. Absorb violet and it looks yellow. This is the step most people get backwards. A green solution is green because it is not absorbing green.

Link the gap and the wavelength with one equation. Here h is 6.626 × 10-34 J·s and c is 2.998 × 108 m/s.

Δ = hc / λ

A big gap means a short wavelength. Swap in a stronger ligand and the band slides toward the blue end, so the color you see slides too.

Key points
  • Six ligands split the d orbitals into a lower t2g set of three and an upper eg set of two.
  • The lower set drops 0.4Δ and the upper set rises 0.6Δ.
  • Δ smaller than the pairing energy gives high spin. Δ larger gives low spin.
  • The spectrochemical series ranks ligands from weak field to strong field.
  • Δ = hc/λ, so a bigger gap absorbs a shorter wavelength.
  • You see the complementary color of the light that gets absorbed.

Worked example · from wavelength to Δ

[Ti(H2O)6]3+ absorbs most strongly at 493 nm. Find Δ in kJ/mol and name the color of the solution.

Step 1 · put the wavelength in meters493 nm × (1 m / 109 nm) = 4.93 × 10-7 m

Step 2 · energy of one photonΔ = hc/λ = (6.626 × 10-34 J·s)(2.998 × 108 m/s) / (4.93 × 10-7 m)

The top is 1.987 × 10-25 J·m. Divide: 1.987 × 10-25 / 4.93 × 10-7 = 4.03 × 10-19 J per photon.

Step 3 · scale to a mole4.03 × 10-19 J × 6.022 × 1023 /mol = 2.43 × 105 J/mol = 243 kJ/mol

Step 4 · name the color493 nm is blue-green light. Take blue-green out of white light and red plus violet are left. Your eye blends those into purple.

AnswerΔ = 243 kJ/mol, and the solution looks purple. Titanium(III) has one d electron, so it makes one clean jump. Ions with more d electrons absorb at several wavelengths, and the strongest band sets the color.

Worked example · from Δ back to a color

For [Cr(H2O)6]3+, Δ = 208 kJ/mol. What wavelength does it absorb, and what color is the solution?

Step 1 · get down to one photon208 kJ/mol = 2.08 × 105 J/mol. Divide by Avogadro's number: 2.08 × 105 / 6.022 × 1023 = 3.454 × 10-19 J.

Step 2 · rearrange for wavelengthλ = hc/Δ = 1.987 × 10-25 J·m / 3.454 × 10-19 J = 5.75 × 10-7 m

Step 3 · convert to nanometers5.75 × 10-7 m × (109 nm / 1 m) = 575 nm

Step 4 · read the color wheel575 nm is yellow. The complement of yellow is violet.

AnswerIt absorbs near 575 nm and looks violet. Swap water for ammonia and Δ grows to 258 kJ/mol. The band moves to 464 nm, which is blue, so [Cr(NH3)6]3+ looks yellow instead.

Section 6Magnetism, chelation and uses

Counting unpaired electrons with a magnet

An electron spins, and a spinning charge acts like a tiny magnet. Pair two electrons in one orbital and their magnets cancel. Leave one alone and its magnet shows. So a sample with unpaired electrons gets pulled into a magnetic field. We call that paramagnetic. A sample with every electron paired is pushed out a little instead, and we call it diamagnetic.

The pull can be weighed. Hang a sample between the poles of a magnet and record the change in apparent mass. From that you get the magnetic moment, μ, measured in Bohr magnetons. For most first row complexes a simple formula gets you close. Here n is the number of unpaired electrons.

μ = √ n(n + 2) Bohr magnetons

Run the numbers and you get 1.73 for one unpaired electron, 2.83 for two, 3.87 for three, 4.90 for four and 5.92 for five. Those values are far enough apart that a measurement tells you n with no guessing. That is how chemists prove a complex is high spin or low spin.

A few metals do something stronger. In iron, cobalt and nickel metal, whole blocks of atoms line their spins up and stay lined up. That is ferromagnetism, and it is what makes a permanent magnet.

Why a claw holds tighter

A chelate is a complex in which one ligand bites more than once and closes a ring. The name comes from the Greek word for a crab's claw. Chelating ligands bind far more tightly than the same number of single bite ligands.

Compare two nickel complexes. Both give the metal six nitrogen donors. [Ni(NH3)6]2+ uses six ammonia molecules. [Ni(en)3]2+ uses three ethylenediamines. The en version is nearly ten billion times more stable. The bonds themselves are nearly the same, so something else is at work.

That something is particle count. Replacing six waters with six ammonias trades six free molecules for six. Replacing six waters with three en molecules frees six waters while using only three ligands. The number of loose particles goes up, and rising disorder favors the change. This bonus is the chelate effect.

Heme holds iron inside a flat ring of four nitrogen donors with an oxygen molecule bound above it, and EDTA wraps a lead ion with two nitrogen and four oxygen donors.6 · Two claws that matter: heme and EDTAHeme: iron inside a flat ring of four nitrogensNNNNFeO2​rides on the sixth slotNthe protein holds on belowFour ring nitrogens and one protein nitrogen fill five slots. The sixth takes O2​ or CO.EDTA: one ligand, six donor atoms, five ringsNNOOOOPbTwo nitrogens and four oxygens take all six slots at once.Six waters leave and one EDTA arrives, so the particle count rises.
Figure 19.6 Left: heme holds iron in a flat ring of four nitrogens, with one slot left open on top for O2. Right: EDTA closes five rings around one metal ion, using all six slots by itself.

Where this shows up

Your own blood runs on a chelate. Heme wraps Fe2+ in a flat ring of four nitrogens. A nitrogen from the protein holds the fifth slot, and O2 rides on the sixth. Carbon monoxide fits that same slot and binds about 200 times tighter, which is exactly why it is so dangerous. Chlorophyll uses the same ring with Mg2+ in the middle. Vitamin B12 uses a cousin of it with cobalt.

EDTA is the workhorse of the lab and the grocery store. It takes all six slots on a metal ion at once. Salad dressing lists it because trace iron and copper ions speed up spoiling, and EDTA locks them away. Shampoo uses it to tame hard water. Doctors use a calcium form of it to pull lead out of a poisoned patient, a treatment called chelation therapy.

Complexes do heavy industrial work too. Gold is pulled from crushed rock as [Au(CN)2]-. Film photography fixed images by washing silver salts away as [Ag(S2O3)2]3-. Nickel is purified by turning it into Ni(CO)4, a gas, then heating it back to metal. Cisplatin treats cancer. And gadolinium chelates light up an MRI scan, because Gd3+ carries seven unpaired electrons.

Key points
  • Unpaired electrons make a complex paramagnetic. All paired makes it diamagnetic.
  • μ = √n(n + 2) in Bohr magnetons, so a measured μ gives n.
  • A chelate ligand bites twice or more and closes a ring.
  • The chelate effect is mostly about particle count, not stronger bonds.
  • Heme, chlorophyll and vitamin B12 are all chelates of a flat nitrogen ring.
  • EDTA, cisplatin, gold mining and MRI dyes all run on coordination chemistry.

Worked example · read the spin state off a magnet

Two iron(II) complexes are measured. [Fe(H2O)6]2+ gives μ = 5.3 Bohr magnetons. [Fe(CN)6]4- gives μ = 0. Explain both results.

Step 1 · count the d electronsIron is group 8, so Fe2+ has 8 − 2 = 6 d electrons. Both complexes are d6 and octahedral.

Step 2 · test μ against the formulaFor n = 4, μ = √(4 × 6) = √24 = 4.90. For n = 0, μ = 0. The measured 5.3 is a little above 4.90, which is normal for iron(II), so n = 4.

Step 3 · match each to a spin stateFour unpaired means t2g4eg2, which is high spin. Zero unpaired means t2g6, which is low spin.

Step 4 · check against the seriesWater sits low in the spectrochemical series, so Δ is small and the electrons spread out. Cyanide sits near the top, so Δ is large and they pair up in the lower set.

AnswerThe water complex is high spin and paramagnetic. The cyanide complex is low spin and diamagnetic. Same metal, same charge, same d count. Only the ligand changed.

Worth knowing. The spin only formula ignores any help from orbital motion. For first row metals the match is usually good. Measured values for cobalt(II) and iron(II) often run a bit high, which is why 5.3 still points to four unpaired electrons.

The chapter in one breath

  • The d-block runs from group 3 to group 12. Its metals hold a partly filled d subshell, and that is the source of their color, their many charges and their magnetism.
  • In a neutral atom 4s fills before 3d, but we write 3d first. Chromium and copper shift one electron to reach a half filled or filled d set.
  • When a cation forms, the 4s electrons leave first. So Fe2+ is [Ar]3d6, and d electrons = group number − charge.
  • A coordination compound is a metal ion plus ligands. Ligands donate lone pairs, so they are Lewis bases.
  • Coordination number counts donor atoms, not ligands. Six donors give an octahedron, four give a tetrahedron or a square, two give a line.
  • Names run cation first, then ligands in alphabetical order, then the metal with its oxidation number. A negative complex ends in -ate.
  • Complexes come as structural isomers, which differ in bonding, and stereoisomers, which differ only in placement. cis and trans change what a complex can do, as cisplatin shows.
  • Ligands split the d orbitals by Δ. A small Δ gives a high-spin complex, a large Δ gives a low-spin one, and the spectrochemical series ranks the ligands.
  • Δ = hc/λ, and you see the complementary color of the light absorbed.
  • Unpaired electrons set μ = √n(n + 2), and chelating ligands such as EDTA, heme and chlorophyll hold on by closing rings.

ReferenceWords worth knowing

These are the terms this chapter leans on most. Come back to the list whenever a formula or a name stops making sense.

bidentate
A ligand that uses two donor atoms and closes one ring, such as ethylenediamine or oxalate.
chelate
A complex in which one ligand bites twice or more, forming a ring with the metal. Named for a crab's claw.
complementary color
The color you see when one band is taken out of white light. Absorb green and you see purple.
coordination compound
A central metal ion surrounded by bonded ligands, with the sphere written inside square brackets.
coordination number
The number of donor atoms bonded to the metal. It counts bonds, not ligands.
crystal field splitting energy
Δ, the energy gap that ligands open between the t2g and eg sets of d orbitals.
d-block
Groups 3 through 12 of the periodic table, where added electrons go into d orbitals.
denticity
How many donor atoms one ligand uses. Monodentate is one, bidentate is two, hexadentate is six.
diamagnetic
Having every electron paired, so the sample is pushed slightly out of a magnetic field.
donor atom
The atom on a ligand that actually holds the lone pair given to the metal.
EDTA
A hexadentate ligand with two nitrogen and four oxygen donors. It wraps a metal ion on its own.
eg orbitals
The upper pair in an octahedral field. They point straight at the ligands, so they rise 0.6Δ.
enantiomers
A pair of complexes that are mirror images and cannot be stacked. They rotate polarized light in opposite directions.
high-spin complex
A complex with a small Δ, so electrons spread into the upper set rather than pair up.
ligand
A molecule or ion that donates at least one lone pair to a metal ion. Every ligand is a Lewis base.
low-spin complex
A complex with a large Δ, so electrons pair in the lower set rather than climb.
paramagnetic
Having one or more unpaired electrons, so the sample is pulled into a magnetic field.
spectrochemical series
Ligands ranked by the size of the split they cause, from weak field iodide up to strong field carbon monoxide.
t2g orbitals
The lower three in an octahedral field. They point between the ligands, so they drop 0.4Δ.
transition metal
A metal whose atoms or common ions carry a partly filled d subshell.

Check yourself

1. Write the electron configuration of Mn and of Mn2+, then count the unpaired electrons in the ion.
Mn is [Ar]3d54s2, Mn2+ is [Ar]3d5, with 5 unpaired. Manganese is element 25 in group 7, so argon's 18 plus 7 more. Put 2 in 4s and 5 in 3d: 18 + 5 + 2 = 25. The ion loses the 4s pair first, leaving 5 d electrons. Five electrons in five orbitals sit one apiece, so all 5 are unpaired.
2. Find the oxidation number of platinum in [PtCl6]2-.
+4 Each chloride ligand is −1, and there are six: 6 × (−1) = −6. Set up the balance: x + (−6) = −2. So x = −2 + 6 = +4. Platinum(IV) with six donor atoms is octahedral.
3. For [Co(en)2(NO2)2]Cl, give the coordination number, the charge on the complex ion and the oxidation number of cobalt.
Coordination number 6, complex charge 1+, cobalt +3. Each en is bidentate: 2 × 2 = 4 donor atoms. The two nitrite ligands add 2, so 4 + 2 = 6. One Cl- sits outside, so the complex must be 1+. Then x + 2(0) + 2(−1) = +1, which gives x = +3.
4. Why is Fe2+ written [Ar]3d6 and not [Ar]3d44s2?
Because the 4s electrons leave first. Once 3d holds electrons, the 3d level drops below 4s, and the 4s orbital reaches further from the nucleus anyway. Both effects make the 4s pair the easiest to remove. Filling order and removal order are not the same thing.
5. Name [Cr(NH3)4Cl2]Cl.
tetraamminedichloridochromium(III) chloride One chloride sits outside, so the complex is 1+ and chromium is +3, since x − 2 = +1. Ammonia becomes ammine and chloride becomes chlorido. Sort by ligand name, so ammine comes before chlorido. The older spelling is tetraamminedichlorochromium(III) chloride.
6. A complex absorbs light at 600 nm. Find Δ in kJ/mol, and say what color the solution looks.
Δ = 199 kJ/mol, and the solution looks blue. Δ = hc/λ. The product hc is (6.626 × 10-34)(2.998 × 108) = 1.987 × 10-25 J·m. Divide by 6.00 × 10-7 m to get 3.31 × 10-19 J per photon. Times 6.022 × 1023 gives 1.99 × 105 J/mol, or 199 kJ/mol. Light at 600 nm is orange, and the complement of orange is blue.
7. A complex has Δ = 290 kJ/mol. What wavelength does it absorb, and what color do you see?
412 nm, and it looks yellow. One photon carries 2.90 × 105 J/mol ÷ 6.022 × 1023 /mol = 4.816 × 10-19 J. Then λ = hc/Δ = 1.987 × 10-25 ÷ 4.816 × 10-19 = 4.12 × 10-7 m = 412 nm. That is violet light, and the complement of violet is yellow.
8. Which is low spin, [CoF6]3- or [Co(CN)6]3-? How many unpaired electrons does each have?
[Co(CN)6]3- is low spin with 0 unpaired. [CoF6]3- is high spin with 4. Both are Co3+, which is d6. Cyanide sits near the top of the spectrochemical series, so Δ beats the pairing energy and all six electrons crowd into t2g. Fluoride sits low, so Δ loses and the filling is t2g4eg2.
9. A complex is measured at μ = 3.9 Bohr magnetons. How many unpaired electrons does it have?
Three. Test values with μ = √n(n + 2). For n = 2: √(2 × 4) = √8 = 2.83. For n = 3: √(3 × 5) = √15 = 3.87. For n = 4: √(4 × 6) = √24 = 4.90. The measured 3.9 matches n = 3.
10. How many stereoisomers does square planar [Pt(NH3)2Cl2] have, and is either one chiral?
Two, cis and trans, and neither is chiral. The two chlorides either share a corner, which is cis, or sit opposite, which is trans. A flat square always has the molecular plane itself as a mirror plane, so neither form is chiral. The cis form is the drug cisplatin.
11. Why does one EDTA ion hold a metal far more tightly than six water molecules do?
The chelate effect, which is mostly about particle count. EDTA takes all six slots at once, so six waters are set free while only one ligand is used up. Two particles become seven. Rising disorder favors that trade, and once one arm of EDTA is attached the others are already nearby.
12. [CoCl4]2- is tetrahedral. Is it high spin or low spin, and how many unpaired electrons does it carry?
High spin, with 3 unpaired electrons. Cobalt is group 9, so Co2+ is 9 − 2 = 7 d electrons. A tetrahedral gap is only about four ninths of the octahedral one, and chloride is a weak field ligand on top of that, so Δ never beats the pairing energy. The lower pair of orbitals takes 4 electrons and the upper three take 1 each, leaving 3 unpaired.
Chemistry · Chapter 20

Organic Chemistry

Carbon is a quiet element with an unusual talent. It makes four bonds, and it bonds to itself without complaint. From that one habit comes a library of millions of compounds. Fuels, plastics, medicines, perfumes and your own cells are all built on carbon frames. This chapter hands you the map. You will meet the families, called functional groups. You will name them with the rules chemists share. You will draw one molecule four different ways. And you will watch a tiny change, like one oxygen moved one spot, turn a gas into a solvent or a sweet smell into vinegar.

By the end of this chapter you can

  • Draw alkanes as molecular, condensed and skeletal structures.
  • Name branched alkanes by the IUPAC rules and count structural isomers.
  • Tell alkanes, alkenes and alkynes apart by formula, shape and reaction.
  • Explain cis and trans isomers, and say when a molecule can have them.
  • Show why benzene substitutes instead of adding.
  • Match each functional group to its family, its name ending and its behavior.
  • Build addition and condensation polymers from the right monomers.

Section 1Alkanes and isomers

Four bonds, and a talent for chains

Carbon sits in group 14 of the table. It holds four valence electrons. So it forms four bonds, every time. Those four bonds spread to the corners of a tetrahedron. The angle between them is about 109.5°. Carbon also bonds to other carbon atoms with ease. Chains, rings and branches all follow from that. One element earns a whole chapter.

Organic chemistry is the chemistry of carbon compounds. The plainest ones hold carbon and hydrogen only. Those are the hydrocarbons. An alkane is a hydrocarbon with single bonds only. Each carbon carries all the hydrogen it can hold. Chemists call that saturated. The formula follows one pattern, CnH2n+2. Six carbons give C6H14.

The alkanes form a homologous series. Each member adds one CH2 unit to the one before it. The first four names are old ones: methane, ethane, propane, butane. From five on, a Greek number leads. Boiling points climb as the chain grows, because longer chains cling to each other harder. Methane is a gas you burn for heat. Octane is a liquid you pour into a tank.

CarbonsNameFormulaBoiling pointWhere you meet it
1methaneCH4−161.5 °Cnatural gas
2ethaneC2H6−88.6 °Cnatural gas
3propaneC3H8−42.1 °Cgrill tanks
4butaneC4H10−0.5 °Clighters
5pentaneC5H1236.1 °Clab solvent
6hexaneC6H1468.7 °Cglue, extraction
8octaneC8H18125.7 °Cgasoline
10decaneC10H22174.1 °Ckerosene

Three ways to draw one molecule

A molecular formula counts atoms and stops there. C4H10 tells you nothing about shape. A condensed structural formula groups the atoms carbon by carbon. Butane becomes CH3CH2CH2CH3. A skeletal structure is faster still. Each line is a bond. Each corner and each loose end is a carbon. The hydrogens are left out, because you can work them out. A corner with two lines must hold two hydrogens.

Skeletal drawings look lazy. They are not. Hiding the hydrogens lets the frame show, and the frame is where the chemistry happens.
Two isomers of C4H10, and how a chain is numbered BUTANE 2-METHYLPROPANE CH₃ CH₂ CH₂ CH₃ 12 34 CH₃CH₂CH₂CH₃ C₄H₁₀ boils at −0.5 °C CH₃ CH CH₃ CH₃ 12 3 CH₃CH(CH₃)CH₃ C₄H₁₀ boils at −11.7 °C WHICH END DO YOU NUMBER FROM? methyl branch 12 34 5 54 32 1 from the left from the right The longest chain holds five carbons, so the base is pentane. From the left the branch lands on carbon 2. From the right the same branch lands on carbon 4. The lower number wins, so the name is 2-methylpentane. Never 4-methylpentane. Same molecule, wrong name.
One formula, two molecules. Butane and 2-methylpropane share C4H10 and boil 11 degrees apart. Below, the numbering rule picks the end nearest the first branch.

Same formula, different molecule

Butane can be built two ways. Four carbons in a row give butane. Three in a row with one hanging off the middle give 2-methylpropane. Both are C4H10. Both sit in real bottles with real labels. They boil eleven degrees apart. Molecules like this are structural isomers. They share a formula but wire their atoms up differently.

Isomer counts grow fast. C4H10 has 2 ways to connect. C5H12 has 3. C6H14 has 5. C8H18 has 18. C10H22 has 75. Each new carbon opens more places to put a branch.

Chains can also close into rings. A cycloalkane has the formula CnH2n, two hydrogens short of the open chain. Cyclohexane, C6H12, is the one you will see most. Its carbons still want 109.5°, so the ring folds into a shape called a chair.

Naming by the rules

Seventy-five shapes under one formula need a naming system. The IUPAC rules do that work. Find the longest continuous carbon chain first, and let it set the base name. Number that chain from the end nearest the first branch. Name each branch as an alkyl group, so CH3 is methyl and C2H5 is ethyl. List the branches in alphabetical order with their numbers. Use di, tri and tetra for repeats. Those numbers are called locants, and the right set is the lowest one.

Common names still float around. 2-methylpropane is often called isobutane. 2,2-dimethylpropane is called neopentane. Those names are fine in a shop. The IUPAC name is the one that works everywhere.

Worked example · Naming a branched alkane

Name CH3CH2CH(CH3)CH2CH2CH3.

Step 1 · Find the longest chainCount the carbons written in a row, skipping the one in brackets. That gives 6, so the base name is hexane.

Step 2 · Number it both waysThe branch sits on the third carbon from the left. From the right it is the fourth. Pick 3.

Step 3 · Name the branchThe branch is one carbon, CH3, so it is a methyl group.

Step 4 · Write the name3-methylhexane.

CheckCarbons: 6 in the chain plus 1 in the branch is 7. For an alkane, H = 2(7) + 2 = 16. Adding the written hydrogens gives 3 + 2 + 1 + 3 + 2 + 2 + 3 = 16. The formula C7H16 agrees.

Worked example · Burning an alkane

Balance the combustion of octane, C8H18, in oxygen.

Step 1 · Write the productsA clean burn gives CO2 and H2O only.

Step 2 · Carbon first8 carbon atoms need 8 CO2.

Step 3 · Hydrogen next18 hydrogen atoms need 9 H2O.

Step 4 · Count oxygen on the right8 × 2 from CO2 plus 9 × 1 from water is 25 oxygen atoms. That is 12.5 O2.

Step 5 · Clear the halfDouble every coefficient: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O.

CheckCarbon 16 and 16. Hydrogen 36 and 36. Oxygen 50 on the left, and 32 + 18 = 50 on the right. Balanced.

Key points
  • Carbon forms four bonds at about 109.5°, so chains, rings and branches are all easy.
  • Alkanes are saturated. Their formula is CnH2n+2, and rings are CnH2n.
  • In a skeletal drawing every corner and every free end is a carbon atom.
  • Structural isomers share a formula but not a skeleton, so they are different substances.
  • Name from the longest chain, and number toward the nearer branch.
  • The correct locant set is the lowest one the structure allows.

Section 2Alkenes and alkynes

One bond more, two hydrogens fewer

An alkene holds at least one double bond between carbons. Its formula is CnH2n. An alkyne holds a triple bond, and its formula is CnH2n-2. Both are unsaturated, which means they carry less hydrogen than the matching alkane. That missing hydrogen is the whole story. The extra bonds are loose electron density, and reagents go straight for it.

Shape follows the bonding. A doubly bonded carbon is flat, with angles near 120°. A triply bonded carbon is a straight line at 180°. A single bond can spin like a wheel on an axle. A double bond cannot spin at all. The two ends are locked into one plane.

Names change at the ending. Alkenes end in -ene and alkynes end in -yne. Number the chain so the first carbon of the multiple bond gets the lowest number. CH2=CHCH2CH3 is 1-butene. CH3CH=CHCH3 is 2-butene. Two old names stuck hard. Ethene is still sold as ethylene, and ethyne is nearly always called acetylene.

Those extra bonds make these molecules worth money. Ethene is the most produced organic chemical in the world. It is the feedstock for polyethylene, and it doubles as a plant hormone that ripens fruit. Propene feeds polypropylene the same way. Ethyne burns in oxygen at around 3100 °C, which is hot enough to cut steel. A triple bond packs a lot of energy into one short, strong link.

Bond order sets the shape, and the double bond locks it BOND ORDER SETS THE SHAPE ethane, C₂H₆ ethene, C₂H₄ ethyne, C₂H₂ 109.5° · 154 pm · spins freely 120° · 134 pm · flat and locked 180° · 120 pm · straight line LOCKED MEANS TWO DIFFERENT MOLECULES CH₃CH₃ HH CH₃H HCH₃ cis-2-butenetrans-2-butene methyls on one side · boils at 3.7 °C methyls on opposite sides · boils at 0.9 °C Both are C₄H₈. Both join the same atoms the same way. The double bond will not twist, so the two forms never trade places at room temperature. 1-butene has no cis or trans form: one of its carbons holds two H atoms. ADDITION CH₃CH=CHCH₃ + Br₂ → CH₃CHBrCHBrCH₃ every atom lands in the product
Bond order, shape and locking. More bonds between two carbons means a shorter, wider angle and no twisting. That lock is what makes cis and trans real. At the bottom, bromine adds across the double bond and nothing is thrown away.

Counting the missing hydrogen

A formula alone can tell you how unsaturated a compound is. Compare its hydrogen count with the saturated count for the same carbons. Each missing pair of hydrogens means one degree of unsaturation. One degree is one double bond or one ring. A triple bond counts as two.

Worked example · Degrees of unsaturation

A hydrocarbon is C6H10. How unsaturated is it?

Step 1 · Find the saturated countFor 6 carbons, a saturated alkane holds 2(6) + 2 = 14 hydrogens.

Step 2 · Subtract14 − 10 = 4 hydrogen atoms are missing.

Step 3 · Divide by two4 ÷ 2 = 2 degrees of unsaturation.

Step 4 · Read itTwo degrees could be two double bonds, one triple bond, or one ring plus one double bond. The formula cannot tell you which.

CheckCyclohexene is C6H10, and it has one ring and one double bond. That is 2 degrees, as expected.

Addition keeps every atom

An addition reaction opens the multiple bond and attaches a piece to each carbon. Nothing is thrown away. Hydrogenation adds H2 over a platinum or nickel catalyst. Halogenation adds Br2, and the orange color fades as it goes, which makes a fast test for unsaturation. Hydration adds water with an acid catalyst and gives an alcohol. Hydrogen halides add as well. Markovnikov's rule says the hydrogen goes to the carbon that already carries more hydrogen.

Alkanes cannot add, because they have no loose bond to open. They substitute instead. Substitution swaps one atom for another and spits out a second product. Chlorine and butane under light give chlorobutane plus HCl. That HCl is waste. Addition routes every atom into the product you wanted, so they waste far less. Chemists score this with atom economy.

Worked example · Atom economy, addition against substitution

Compare Br2 adding to 2-butene with Cl2 substituting on butane.

Step 1 · Addition massesC4H8 is 56.11 g/mol and Br2 is 159.81 g/mol. Together that is 215.92 g/mol.

Step 2 · Addition productC4H8Br2 is 56.11 + 159.81 = 215.92 g/mol. Atom economy = 215.92 ÷ 215.92 = 100%.

Step 3 · Substitution massesC4H10 is 58.12 g/mol and Cl2 is 70.90 g/mol. Together that is 129.02 g/mol.

Step 4 · Substitution productC4H9Cl is 92.57 g/mol. Atom economy = 92.57 ÷ 129.02 = 0.717, or 71.7%.

CheckThe missing 28.3% leaves as HCl, which is 36.46 g/mol. 36.46 ÷ 129.02 = 0.283. The two shares add to 100%.

Key points
  • Alkenes are CnH2n and alkynes are CnH2n-2.
  • Double bonds give 120° and a flat, locked unit. Triple bonds give 180° and a straight line.
  • Degrees of unsaturation = (2C + 2 − H) ÷ 2, where C and H are the atom counts.
  • Cis and trans need two different groups on each carbon of the double bond.
  • Addition opens a multiple bond and keeps all atoms. Substitution swaps and makes waste.
  • Markovnikov's rule sends the added hydrogen to the carbon with more hydrogen already.

Section 3Aromatic compounds

A ring that refused to behave

Benzene is C6H6. Six carbons and six hydrogens, no more. Run the unsaturation count and you get (14 − 6) ÷ 2 = 4 degrees. One ring plus three double bonds fits that number. In 1865 Kekulé drew exactly that, a six-sided ring with single and double bonds taking turns.

Then the tests came back strange. An alkene wipes the color out of bromine water in seconds. Benzene sits there and does nothing. An alkene takes up hydrogen over a catalyst at room temperature. Benzene needs heat, pressure and a hard push. Something was missing from the drawing.

Measurement found it. Every carbon to carbon bond in benzene is the same length, 139 pm. A plain single bond is 154 pm. A plain double bond is 134 pm. Benzene lands between the two, at every bond in the ring. The molecule is a flat, regular hexagon with 120° angles all the way round.

Electrons that belong to the whole ring

The true structure is neither Kekulé drawing. Each ring carbon keeps one electron in a p orbital that points above and below the ring. Those six electrons spread out over all six carbons at once. That spreading is delocalization. The two Kekulé drawings are resonance structures, and the real molecule is a single blend of them. Chemists draw the blend as a hexagon with a circle inside.

Spreading electrons out lowers the energy, and the energy drop is measurable. Adding hydrogen to one double bond in cyclohexene releases 120 kJ/mol. Three separate double bonds should therefore release about 360 kJ/mol. Benzene releases only 208 kJ/mol. It began 152 kJ/mol lower in energy than the drawing predicted. That bonus is the resonance energy. A flat ring that earns a bonus like this is called aromatic. The ring needs 4n + 2 delocalized electrons to qualify, and benzene has 6, with n = 1.

Benzene's delocalization and the stability it buys THE BLEND, NOT EITHER DRAWING one Kekulé form the other Kekulé form what benzene really is ↔ ⇒ single C–C bond: 154 pm double C=C bond: 134 pm benzene, every bond: 139 pm 6 shared electrons 4n + 2 with n = 1 HEAT RELEASED ON ADDING HYDROGEN three loose C=C bonds 360 kJ/mol predicted benzene, measured 208 kJ/mol measured resonance energy = 152 kJ/mol Benzene starts lower, so less heat is left to give out.
Why benzene is calm. Neither alternating drawing is right. The six electrons sit over the whole ring, every bond comes out at 139 pm, and the ring holds 152 kJ/mol of extra stability.

Worked example · Finding the resonance energy

Use hydrogenation heats to measure benzene's extra stability.

Step 1 · Measure one double bondCyclohexene plus H2 gives cyclohexane and releases 120 kJ/mol.

Step 2 · Predict three of them3 × 120 kJ/mol = 360 kJ/mol, if the three bonds in benzene were ordinary and separate.

Step 3 · Measure benzeneBenzene plus 3 H2 gives cyclohexane and releases only 208 kJ/mol.

Step 4 · Take the difference360 − 208 = 152 kJ/mol. Benzene sat 152 kJ/mol lower in energy before the reaction started.

Check1,3-cyclohexadiene releases 232 kJ/mol, just under the 240 kJ/mol predicted for two bonds. Two neighboring bonds give a tiny bonus. A full ring gives a huge one.

Substitution keeps the prize

Addition would break the delocalized ring apart. The product would hand back that 152 kJ/mol, so addition is a bad trade. Substitution is not. Swap one hydrogen for a new group and the ring stays whole and aromatic. So benzene reacts by electrophilic aromatic substitution. Bromine with an FeBr3 catalyst gives bromobenzene plus HBr. Nitric acid with sulfuric acid gives nitrobenzene plus water. Hot concentrated sulfuric acid gives benzenesulfonic acid. In each case one hydrogen walks out and the ring survives.

Names for rings mix old and new. Methylbenzene is almost always called toluene. A ring with OH is phenol, and a ring with NH2 is aniline. When the ring itself is the branch, it is a phenyl group. Two groups on one ring can sit in three patterns. The 1,2 pattern is ortho, 1,3 is meta and 1,4 is para. The three dimethylbenzenes are known as the xylenes.

Fuse two rings and you get naphthalene, the old mothball smell. Fuse more and you reach the ring systems in soot and tar. Those big flat rings are stable, which is exactly why they stick around in the environment for so long.
Key points
  • Benzene is C6H6, flat, with 120° angles and six equal 139 pm bonds.
  • Six p electrons spread over all six carbons. That is delocalization.
  • The two Kekulé drawings are resonance forms, not two real substances.
  • Delocalization buys 152 kJ/mol of extra stability, called the resonance energy.
  • Aromatic rings are flat and hold 4n + 2 delocalized electrons.
  • Benzene substitutes rather than adds, because substitution keeps the ring aromatic.

Section 4Alcohols and ethers

The group makes the family

A hydrocarbon is a frame and little more. Swap one hydrogen for something else and the molecule gains a personality. A functional group is that reactive piece. It sets the boiling point, the solubility and the reactions. The carbon frame mostly just comes along for the ride. Chemists write R for the frame so they can watch the group alone.

Alcohols carry a hydroxyl group

An alcohol holds a hydroxyl group, written –OH, on a carbon. To name one, drop the final -e from the alkane name and add -ol. Methane becomes methanol. Propane becomes 1-propanol or 2-propanol, depending on where the OH sits. 2-propanol is the rubbing alcohol in the cabinet. You can also class an alcohol by the carbon that holds the OH. One other carbon on it makes the alcohol primary, two makes it secondary and three makes it tertiary.

That small group changes everything. Oxygen pulls electrons much harder than hydrogen, so the O–H bond is very polar. The hydrogen on one molecule is then drawn to the oxygen on the next. That attraction is a hydrogen bond, and it takes real heat to pull apart. Watch three molecules of nearly equal mass. Propane boils at −42.1 °C. Dimethyl ether boils at −24.8 °C. Ethanol boils at 78.3 °C. Only ethanol has an O–H bond, and it boils over a hundred degrees higher.

One oxygen, then one O-H bond, and the boiling point jumps SAME MASS, THREE DIFFERENT BOILING POINTS O O H propanedimethyl etherethanol C₃H₈ · 44.10 g/mol C₂H₆O · 46.07 g/mol C₂H₆O · 46.07 g/mol no polar group polar, but no O–H polar, with an O–H boils at −42.1 °C boils at −24.8 °C boils at 78.3 °C WHO CAN HOLD HANDS WITH WATER O H O H H ethanol: gives one, takes one O O H H ether: takes one, gives none Gold dashes are hydrogen bonds.
Same mass, different grip. Propane has nothing polar to offer. The ether can accept a hydrogen bond from water but cannot donate one. Ethanol does both, so it boils far higher and mixes with water freely.

Worked example · Two molecules, one molar mass

Show that ethanol and dimethyl ether have the same molar mass, then explain the boiling gap.

Step 1 · Write the shared formulaEthanol is CH3CH2OH and dimethyl ether is CH3OCH3. Both are C2H6O.

Step 2 · Add the massesC: 2 × 12.011 = 24.022 g/mol. H: 6 × 1.008 = 6.048 g/mol. O: 1 × 15.999 = 15.999 g/mol.

Step 3 · Total24.022 + 6.048 + 15.999 = 46.069, so 46.07 g/mol for both.

Step 4 · Compare boiling points78.3 − (−24.8) = 103.1 °C apart. Mass cannot explain that, because the masses match.

CheckOnly ethanol has an O–H bond, so only ethanol can donate a hydrogen bond to its neighbor. Propane, at 44.10 g/mol, boils lower still at −42.1 °C. The group is doing the work, not the mass.

Mixing with water, and running out of patience

Solubility follows the same logic. The OH end is at home in water. The carbon chain is not. Short alcohols win that argument. Methanol, ethanol and 1-propanol mix with water in any amount at all. 1-butanol only dissolves to 7.9 g per 100 mL. 1-hexanol manages 0.6 g. 1-octanol is down to 0.05 g. Add more OH groups and the trend reverses. Ethylene glycol and glycerol each hold several hydroxyls, and both are syrups that mix with water without limit.

Ethers hide the oxygen in the middle

An ether has oxygen sitting between two carbon frames, R–O–R'. Diethyl ether is the classic one. There is no O–H bond here, so an ether cannot donate a hydrogen bond. It can still accept one from water, so small ethers dissolve a little. Diethyl ether holds 6.9 g per 100 mL of water. Boiling points sit far below the matching alcohol. Diethyl ether boils at 34.6 °C and 1-butanol boils at 117.7 °C. Both are C4H10O.

Alcohols also do reactions an ether cannot. Heat an alcohol with sulfuric acid and water is pulled out, leaving an alkene. That is a dehydration. Two alcohol molecules can instead join and lose one water to make an ether. Oxidation works on alcohols as well. A primary alcohol gives an aldehyde, then a carboxylic acid. A secondary alcohol gives a ketone. A tertiary alcohol has no hydrogen left on that carbon, so it simply refuses.

Key points
  • A functional group, not the carbon frame, decides how a molecule behaves.
  • Alcohols are R–OH and take the -ol ending. Count carbons on that one carbon to call it 1°, 2° or 3°.
  • An O–H bond lets a molecule donate a hydrogen bond, which raises the boiling point sharply.
  • Ethers are R–O–R'. They accept hydrogen bonds but cannot donate any.
  • Short alcohols mix with water. Longer chains and fewer OH groups mean less mixing.
  • Primary alcohols oxidize twice, secondary alcohols once, and tertiary alcohols not at all.

Section 5Aldehydes, ketones, acids and esters

One group, four families

A carbonyl group is a carbon double bonded to oxygen, written C=O. It is strongly polar, because oxygen drags the shared electrons toward itself. That leaves the carbon a little positive and open to attack. What sits next to the carbonyl carbon decides which family you are in.

In an aldehyde the carbonyl carbon also holds a hydrogen, and it sits at the end of the chain. Names end in -al. Methanal is sold as formaldehyde. Ethanal is better known as acetaldehyde. In a ketone the carbonyl carbon holds carbon on both sides. Names end in -one, with a number when the chain is long enough to need one. Propanone is acetone, the nail polish remover. 2-butanone is a paint solvent.

A carboxylic acid puts an OH right on the carbonyl carbon, written –COOH. Names end in -oic acid. Methanoic acid is formic acid, the sting in an ant bite. Ethanoic acid is acetic acid, which is what makes vinegar sour. These are weak acids. Acetic acid has Ka = 1.8 × 10-5, so its pKa is 4.74. In a 0.10 M solution only about one molecule in eighty gives up its proton.

An ester replaces the acid's OH hydrogen with a carbon group, giving R–COO–R'. Its name takes two words. The alcohol part comes first as an alkyl name. The acid part follows with an -oate ending. Ethanol plus ethanoic acid gives ethyl ethanoate, usually called ethyl acetate. Esters smell good, which is why flavor labs love them. Methyl butanoate smells like apple. Pentyl ethanoate smells like banana.

The four carbonyl families side by side O H on the carbonyl carbon ALDEHYDE · ends in -al propanal, CH₃CH₂CHO oxidizes easily to the acid O carbon on both sides KETONE · ends in -one propanone, CH₃COCH₃ resists further oxidation O OH the acidic hydrogen CARBOXYLIC ACID · ends in -oic acid ethanoic acid, CH₃COOH weak acid, pKₐ = 4.74 O O carbon group in place of that H ESTER · alkyl + -oate ethyl ethanoate, CH₃COOCH₂CH₃ not acidic, and it smells sweet The dashed teal circle marks the carbonyl carbon in all four.
Four families, one group. Each structure holds the same C=O unit. Only the neighbors change, and that is enough to change the name, the smell and the chemistry.

Worked example · The pH of vinegar-strength acid

Find the pH of 0.10 M ethanoic acid. Ka = 1.8 × 10-5.

Step 1 · Write the equilibriumCH3COOH ⇌ H+ + CH3COO−.

Step 2 · Set up the expressionLet x be the H+ made. Then Ka = x2 ÷ (0.10 − x) = 1.8 × 10-5.

Step 3 · Assume x is smallx2 = (1.8 × 10-5)(0.10) = 1.8 × 10-6. So x = 1.3 × 10-3 M.

Step 4 · Take the logarithmpH = −log(1.34 × 10-3) = 2.87.

Checkx ÷ 0.10 = 1.3%, which is under 5%, so dropping the x was fair. A pH near 3 matches a bottle of vinegar, which is about this strength.

Making and breaking an ester

Esterification joins an acid and an alcohol with an acid catalyst. Water is the second product, so this is a condensation reaction. The reaction is reversible and settles at equilibrium, so chemists remove the water to push it forward. Run it backward with extra water and the ester splits apart, which is hydrolysis. Boil an ester with sodium hydroxide instead and you get a salt plus the alcohol. Do that to a fat and the salt is soap, a reaction called saponification.

The oxidation ladder

These families are rungs on one ladder. Start at a primary alcohol. Oxidize once and you reach the aldehyde. Oxidize again and you reach the carboxylic acid. 1-propanol becomes propanal, then propanoic acid. A secondary alcohol climbs one rung and stops at the ketone. The ketone carbon has no hydrogen left to lose, so the ladder ends there. Aldehydes sit in the middle and oxidize so readily that an open bottle slowly turns acidic.

FamilyGroupEndingExampleBehavior to remember
Aldehyde–CHO-alethanaloxidizes easily to the acid
Ketone–CO–-onepropanoneend of the oxidation ladder
Carboxylic acid–COOH-oic acidethanoic acidweak acid, pKa near 4 to 5
Ester–COO–alkyl -oateethyl ethanoatemade by condensation, split by hydrolysis
Key points
  • A carbonyl is C=O. Its neighbors decide the family.
  • Aldehydes end in -al and sit at a chain end. Ketones end in -one and sit inside the chain.
  • Carboxylic acids are –COOH and are weak acids, with pKa values near 4 to 5.
  • Esters are named alcohol part first, then the acid part with -oate.
  • Acid plus alcohol gives ester plus water. Adding water back reverses it.
  • Primary alcohol → aldehyde → acid. A secondary alcohol stops at the ketone.

Section 6Amines, amides and polymers

Nitrogen joins the frame

An amine is ammonia with carbon groups in place of hydrogen. Count the carbons bonded to the nitrogen to class it. One carbon makes it primary, two makes it secondary, three makes it tertiary. The simplest one is CH3NH2, named methanamine and usually called methylamine. Small amines smell like old fish, and that is not a coincidence. Fish spoilage releases them.

The nitrogen keeps a lone pair of electrons, so an amine is a weak base. It pulls a proton off water and leaves OH− behind. Methylamine has Kb = 4.4 × 10-4. Ammonia is weaker at 1.8 × 10-5. Aniline, which is an amine on a benzene ring, is far weaker at 4.3 × 10-10. The ring drags that lone pair into its delocalized cloud, so the pair is no longer free to grab a proton.

An amide sets nitrogen right beside a carbonyl, giving the C(=O)N unit. An acid and an amine join and lose water to build it. Ethanamide, also called acetamide, is the simplest one. Here too the nitrogen's lone pair is shared with the carbonyl, so an amide is not a base in water. The amide link is flat, strong and hard to break. In proteins that same link is called a peptide bond.

Worked example · The pH of a weak base

Find the pH of 0.10 M methylamine. Kb = 4.4 × 10-4.

Step 1 · Write the equilibriumCH3NH2 + H2O ⇌ CH3NH3+ + OH−.

Step 2 · Try the shortcutx2 = (4.4 × 10-4)(0.10) = 4.4 × 10-5, giving x = 6.6 × 10-3 M. That is 6.6% of 0.10 M, over the 5% limit, so solve it properly.

Step 3 · Solve the quadraticx2 + (4.4 × 10-4)x − 4.4 × 10-5 = 0 gives x = 6.42 × 10-3 M of OH−.

Step 4 · Convert to pHpOH = −log(6.42 × 10-3) = 2.19. pH = 14.00 − 2.19 = 11.81.

CheckA pH well above 7 is what a base should give. Ammonia at the same concentration reaches about pH 11.1, a little lower, because its Kb is smaller.

Monomers into chains

A polymer is one very long molecule built from small repeating pieces. The small starting molecule is a monomer. The piece that repeats along the chain is the repeat unit. The number of repeats is the degree of polymerization. Multiply the repeat unit mass by that number and you have the chain's molar mass. Real samples hold chains of many lengths, so the mass quoted is always an average.

An addition polymer grows from alkenes. The double bond opens and the carbons link up, one monomer at a time. Nothing leaves. Ethene gives polyethylene for bags and bottles. Propene gives polypropylene for carpet and caps. Chloroethene gives PVC for pipe. Tetrafluoroethene gives PTFE for nonstick pans. Because nothing is lost, the repeat unit has exactly the same formula as the monomer.

A condensation polymer drops a small molecule, usually water, at every link. Each monomer needs two reactive ends so the chain can keep going both ways. A diacid plus a diamine gives a polyamide. Hexanedioic acid with 1,6-hexanediamine gives nylon-6,6. A diacid plus a diol gives a polyester. Terephthalic acid with ethylene glycol gives PET, the plastic in drink bottles. Amino acids join by the same trick and give proteins, which is why a protein is a natural polyamide.

Addition keeps every atom, condensation drops water ADDITION · nothing leaves three ethene molecules, CH₂=CH₂ → repeat unit –CH₂CH₂– · 28.05 g/mol polyethylene atom loss: none CONDENSATION · water leaves at every link HO OC –(CH₂)₄– CO OH hexanedioic acid, 146.14 g/mol + H₂ N –(CH₂)₆– N H₂ 1,6-hexanediamine, 116.21 g/mol → – CO –(CH₂)₄– CO NH –(CH₂)₆– NH – nylon-6,6 repeat unit, 226.32 g/mol H₂O H₂O two water molecules per repeat unit Mass check on the repeat unit: 146.14 + 116.21 = 262.35 g/mol of monomers. 262.35 − 2(18.02) = 226.32 g/mol. The two lost waters are exactly the difference. An amide link forms at each end, so each link costs one water.
Two ways to grow a chain. Addition opens double bonds and keeps every atom, so the repeat unit matches the monomer. Condensation joins reactive ends and spits out water, so the repeat unit is lighter than the monomers that made it.

Worked example · How long is a PET chain?

A PET sample averages 38,400 g/mol. Its repeat unit is C10H8O4. Find the degree of polymerization.

Step 1 · Mass of the repeat unitC: 10 × 12.011 = 120.11 g/mol. H: 8 × 1.008 = 8.06 g/mol. O: 4 × 15.999 = 64.00 g/mol.

Step 2 · Add them120.11 + 8.06 + 64.00 = 192.17 g/mol per repeat unit.

Step 3 · Divide38,400 ÷ 192.17 = 199.8, so about 200 repeat units.

Step 4 · Count the waterEach repeat unit makes two ester links, so about 400 water molecules leave while one chain is built.

Check200 × 192.17 = 38,434 g/mol, which matches the stated average closely. The monomers weigh 166.13 + 62.07 = 228.20 g/mol, and 228.20 − 2(18.02) = 192.16 g/mol. The repeat unit mass agrees.

PolymerHow it growsMonomerRepeat unit massSmall molecule lost
Polyethyleneadditionethene28.05 g/molnone
Polypropyleneadditionpropene42.08 g/molnone
PVCadditionchloroethene62.50 g/molnone
PTFEadditiontetrafluoroethene100.01 g/molnone
Nylon-6,6condensationdiacid + diamine226.32 g/mol2 H2O
PETcondensationdiacid + diol192.17 g/mol2 H2O
Proteincondensationamino acid57.05 g/mol for glycine1 H2O
Key points
  • Amines are weak bases, because the nitrogen lone pair can take a proton.
  • A ring beside the nitrogen pulls that lone pair away, so aniline is a very weak base.
  • Amides hold C(=O)N, are not basic, and build the peptide links in proteins.
  • Addition polymers come from alkenes and lose no atoms at all.
  • Condensation polymers need two reactive ends per monomer and lose water at each link.
  • Degree of polymerization = chain molar mass ÷ repeat unit molar mass.

The chapter in one breath

  • Carbon's four bonds do the work. Chains, branches and rings all come from that one habit.
  • Alkanes are the baseline. CnH2n+2, saturated, and slow to react.
  • Naming has an order. Longest chain, lowest locants, branches listed alphabetically.
  • Isomers share a formula, not a structure. Different wiring means a different substance.
  • Multiple bonds mean reactivity and locked shape. That lock is where cis and trans come from.
  • Addition keeps every atom. Substitution makes waste. Atom economy puts a number on it.
  • Benzene spreads six electrons over six carbons. The 152 kJ/mol bonus is why it substitutes.
  • Functional groups set behavior. One O–H bond can move a boiling point by a hundred degrees.
  • Polymers are monomers repeated. Addition loses nothing, condensation loses water each link.

ReferenceWords worth knowing

Every term below appeared in bold somewhere above. Come back here whenever a word stops making sense.

addition polymer
A polymer built by opening double bonds and linking monomers with no atoms lost. Polyethylene is one.
alcohol
A compound with a hydroxyl group on a carbon, R–OH. Names end in -ol.
aldehyde
A carbonyl at the end of a chain, with a hydrogen on the carbonyl carbon. Names end in -al.
alkane
A hydrocarbon with single bonds only, CnH2n+2. Also called saturated.
alkene
A hydrocarbon with a carbon to carbon double bond, CnH2n. Names end in -ene.
alkyne
A hydrocarbon with a carbon to carbon triple bond, CnH2n-2. Names end in -yne.
amide
A nitrogen bonded straight to a carbonyl carbon. The peptide link in a protein is an amide.
amine
Ammonia with one or more hydrogens swapped for carbon groups. A weak base.
aromatic
A flat ring holding 4n + 2 delocalized electrons, which makes it unusually stable.
carbonyl group
A carbon double bonded to oxygen, C=O. The shared group of four families.
carboxylic acid
A carbonyl with an OH attached, –COOH. A weak acid with pKa near 4 to 5.
condensation polymer
A polymer built by joining reactive ends and releasing a small molecule, usually water.
delocalization
Electrons spread over three or more atoms instead of sitting in one bond.
ester
A carboxylic acid with its OH hydrogen replaced by a carbon group, R–COO–R'.
ether
An oxygen bridging two carbon frames, R–O–R'. No O–H bond, so no donated hydrogen bonds.
functional group
The reactive piece attached to a carbon frame. It sets most of the behavior.
hydrocarbon
A compound of carbon and hydrogen only.
ketone
A carbonyl with carbon on both sides, inside the chain. Names end in -one.
monomer
The small molecule that repeats to build a polymer.
structural isomer
One of two or more compounds that share a molecular formula but connect their atoms differently.

Check yourself

1. Give the molecular formula of the straight-chain alkane with 7 carbons, and its molar mass.
C7H16, 100.20 g/mol. H = 2(7) + 2 = 16. Mass = 7(12.011) + 16(1.008) = 84.077 + 16.128 = 100.205 g/mol, so 100.20 g/mol. The name is heptane.
2. Name CH3CH2CH(CH3)CH2CH3.
3-methylpentane. The longest chain holds five carbons, so the base is pentane. The methyl branch is on the middle carbon, which is number 3 from either end.
3. How many structural isomers does C5H12 have? Name them.
Three. Pentane is the straight chain. 2-methylbutane has a four-carbon chain with one branch. 2,2-dimethylpropane has a three-carbon chain with two branches on the middle carbon. There is no fourth way to connect five carbons.
4. Balance the combustion of propane, C3H8.
C3H8 + 5 O2 → 3 CO2 + 4 H2O. Carbon: 3 gives 3 CO2. Hydrogen: 8 gives 4 H2O. Oxygen on the right: 3(2) + 4(1) = 10 atoms, which is 5 O2. Both sides now hold 3 C, 8 H and 10 O.
5. A hydrocarbon is C7H12. How many degrees of unsaturation does it have?
Two. A saturated C7 alkane would hold 2(7) + 2 = 16 hydrogens. 16 − 12 = 4 missing, and 4 ÷ 2 = 2 degrees. That could be two double bonds, one triple bond, or one ring with one double bond.
6. Does propene, CH3CH=CH2, have cis and trans forms? Why or why not?
No. Cis and trans need two different groups on each carbon of the double bond. The CH2 end of propene carries two identical hydrogens, so flipping it changes nothing.
7. Hydrogenating cyclohexene releases 120 kJ/mol. Benzene plus 3 H2 releases 208 kJ/mol. Find benzene's resonance energy.
152 kJ/mol. Three separate double bonds would release 3 × 120 = 360 kJ/mol. The measured value is 208 kJ/mol. 360 − 208 = 152 kJ/mol, which is energy benzene never had to give back because delocalization had already lowered it.
8. 1-butanol boils at 117.7 °C and diethyl ether boils at 34.6 °C. Both are C4H10O. Explain the gap.
Only the alcohol has an O–H bond. 1-butanol can donate a hydrogen bond to the next molecule, so the liquid holds itself together. The ether has no O–H, so it can only accept one. Same formula, same mass, 83 degrees apart.
9. Find the pH of 0.050 M ethanoic acid. Ka = 1.8 × 10-5.
pH = 3.02. x2 ÷ 0.050 = 1.8 × 10-5, so x2 = 9.0 × 10-7 and x = 9.49 × 10-4 M. pH = −log(9.49 × 10-4) = 3.02. The ionized share is 1.9%, under 5%, so the shortcut was fair.
10. Ethanoic acid reacts with 1-propanol under acid catalysis. Name both products.
Propyl ethanoate and water. The alcohol part is named first as propyl. The acid part follows as ethanoate. Losing water makes this a condensation, and adding water back would reverse it.
11. You oxidize 2-butanol. What do you get, and can you go further?
2-butanone, and no. 2-butanol is a secondary alcohol, so it oxidizes to a ketone. That carbonyl carbon has no hydrogen left to remove, so the ladder stops. A primary alcohol would have carried on to the carboxylic acid.
12. A polyethylene sample averages 42,000 g/mol. How many ethene units are in one chain?
About 1,500. The repeat unit is –CH2CH2–, so its mass is 2(12.011) + 4(1.008) = 24.022 + 4.032 = 28.05 g/mol. 42,000 ÷ 28.05 = 1,497, which rounds to about 1.5 × 103 repeat units. Addition loses no atoms, so the monomer and repeat unit masses match.
Chemistry · Chapter 21

Nuclear Chemistry

Every reaction so far in this book moved electrons around. The nucleus sat in the middle and did not change. This chapter opens the nucleus up. You will see why some nuclei last for billions of years while others fall apart today. You will learn where a star gets its energy, and why that energy dwarfs the energy of a flame. You will date a bone from the carbon left inside it. You will follow neutrons through a reactor core. You will also meet the units that say how much radiation a body has taken in, and the ways doctors put that radiation to work.

By the end of this chapter you can

  • Read a nuclide symbol and count the protons, neutrons and nucleons in it.
  • Use the band of stability to predict whether a nuclide will decay.
  • Find a mass defect and turn it into a binding energy with E = mc2.
  • Tell alpha, beta, positron, electron capture and gamma decay apart.
  • Use a half-life to find what is left of a sample, or how old a find is.
  • Balance any nuclear equation for both mass number and charge.
  • Compare the energy released by fission with the energy released by fusion.
  • Match radiation units, shielding and medical uses to the right kind of ray.

Section 1The nucleus and nuclear stability

Nucleons and nuclide symbols

The nucleus is a tiny speck at the center of an atom. It holds almost all of the atom's mass. Protons and neutrons both live there, and together they are called nucleons. A nuclide is one exact kind of nucleus. Fix the proton count and the neutron count and you have named a nuclide.

A nuclide symbol carries two numbers to the left of the element symbol. The top one is the mass number A, which counts every nucleon. The bottom one is the atomic number Z, which counts only the protons. Subtract to get the neutrons, so N = A − Z.

23892U  →  A = 238 nucleons,   Z = 92 protons,   N = 238 − 92 = 146 neutrons

Isotopes share a proton count but differ in neutron count. Carbon-12 and carbon-14 each hold 6 protons. Carbon-14 carries two extra neutrons, and those two neutrons are what make it fall apart. A nuclide that falls apart on its own is radioactive.

The band of stability

Every proton carries a positive charge, so protons shove each other hard. What holds them together anyway is the strong nuclear force. It pulls any two nucleons together, but only across a very short range. Neutrons add that pull without adding any shove, so they work like nuclear glue.

Now plot every nuclide with neutrons up the page and protons across it. The stable ones crowd into one narrow stripe. That stripe is the band of stability. Light stable nuclei sit near a 1-to-1 ratio of neutrons to protons. Heavy ones need extra glue, so the band bends upward toward 1.5 to 1.

A nuclide above the band holds too many neutrons. One below the band holds too few. Either way it decays in whatever direction moves it toward the band. Nothing past bismuth is safe, because every nuclide with Z above 83 is radioactive.

Some counts are steadier than others. A nucleus with 2, 8, 20, 28, 50, 82 or 126 protons or neutrons is unusually stable. These counts are the magic numbers, and they act the way a filled electron shell does. Lead-208 holds 82 protons and 126 neutrons, so it is magic twice over.

The band of stability: neutrons plotted against protons for every stable nuclide 0 10 20 30 40 50 60 70 80 90 0 25 50 75 100 125 150 N = Z Z > 83 none stable C-12 O-16 Ca-40 Fe-56 Sn-120 Pb-208 above the band: beta decay below the band: positron or capture Protons (Z) Neutrons (N) Every stable nuclide, plotted
Figure 21.1 · Each small dot is a stable nuclide. They fall in a stripe that starts along the N = Z line and then bends above it, because heavy nuclei need extra neutrons. The red strip on the right marks Z > 83, where no stable nuclide exists.
Scale check. A nucleus is about one hundred-thousandth the width of its atom. Blow the atom up to the size of a stadium and the nucleus is a pea on the center spot. All the mass of the stadium sits in that pea.

Mass defect and binding energy

Weigh a helium-4 atom. Then weigh two hydrogen atoms and two neutrons on their own and add those up. The separate parts weigh more than the finished atom does. The missing mass is the mass defect.

That mass did not vanish. It left as energy at the moment the nucleus formed. Einstein's equation ties mass and energy together, and c is the speed of light, 2.998 × 108 m/s.

E = mc2

The energy that comes out is the binding energy. It is also the bill you would have to pay to pull that nucleus back apart. Divide it by the nucleon count to get the binding energy per nucleon. That is the fair way to compare nuclei of very different sizes, and it peaks near iron-56 at about 8.79 MeV.

Worked example · the binding energy of helium-4

A helium-4 atom weighs 4.002603 u. A hydrogen-1 atom weighs 1.007825 u and a neutron weighs 1.008665 u. Find the binding energy per nucleon.

Add up the loose parts

Two hydrogen atoms come to 2 × 1.007825 u = 2.015650 u, and two neutrons come to 2 × 1.008665 u = 2.017330 u. Added together, the loose parts total 4.032980 u. The two electrons ride along on the hydrogen atoms, and helium-4 has two of its own, so they cancel out.

Subtract to find the mass defect

Subtracting the finished atom from its loose parts gives 4.032980 u − 4.002603 u = 0.030377 u of missing mass.

Turn the lost mass into energy

One u is 1.6605 × 10−27 kg, so the defect is 5.0441 × 10−29 kg. Then E = mc2 = (5.0441 × 10−29 kg)(2.998 × 108 m/s)2 = 4.534 × 10−12 J.

Report it per nucleon

Nuclear energies are usually quoted in MeV, and 1 u is worth 931.5 MeV. So 0.030377 u × 931.5 MeV/u = 28.30 MeV. Divide by 4 nucleons and you get 7.074 MeV per nucleon.

Check it a second way

One MeV is 1.602 × 10−13 J. So 4.534 × 10−12 J ÷ 1.602 × 10−13 J/MeV = 28.30 MeV, which matches. For a whole mole of helium-4 that is 2.73 × 109 kJ, which is millions of times any bond energy.

Key points
  • A nuclide symbol puts A on top and Z below, and the neutron count is A − Z.
  • Neutrons add strong-force pull without adding charge, so heavy nuclei need extra neutrons.
  • Stable nuclides lie inside the band of stability, and no nuclide past Z = 83 is stable.
  • The loose parts of a nucleus weigh more than the nucleus itself, and that mass defect is its binding energy.
  • Binding energy per nucleon peaks near iron-56, so iron sits at the bottom of the nuclear energy well.

Section 2Types of radioactive decay

Alpha decay

An alpha particle is a helium-4 nucleus on its own. It is two protons and two neutrons, with a charge of 2+. A heavy nucleus can shrink by throwing one out. The mass number drops by 4 and the atomic number drops by 2.

23892U  →  23490Th  +  42He

Alpha decay is the main exit for nuclides above Z = 83. The particle is heavy and slow for its energy, so it travels only a few centimetres in air. A sheet of paper stops it.

Beta decay and positron emission

A beta particle is a fast electron fired out of the nucleus. There are no electrons stored in there waiting. Instead a neutron turns into a proton plus an electron, and the electron leaves. The mass number holds steady while the atomic number climbs by 1.

146C  →  147N  +  0−1e

This is the fix for a nuclide sitting above the band of stability. It swaps one of its spare neutrons for a proton, which walks the nuclide down into the band. A tiny neutral particle called an antineutrino leaves with the electron and carries off part of the energy.

Positron emission runs the trade the other way. A proton turns into a neutron plus a positron, which is an electron with a positive charge. The mass number holds steady and the atomic number falls by 1.

158O  →  157N  +  0+1e

A positron does not get far. It meets an ordinary electron and the two destroy each other. Their mass turns into two gamma photons of 0.511 MeV, fired in opposite directions. That event is called annihilation, and a PET scanner is built to catch both photons at once.

Electron capture

A nucleus short on neutrons has a second route. It reaches out and swallows one of its own inner electrons. That electron joins a proton, and the pair becomes a neutron. The atomic number falls by 1 and the mass number does not move, so the books balance exactly as they do for positron emission.

4019K  +  0−1e  →  4018Ar

An outer electron then drops into the empty inner spot, and an X-ray comes out. Heavier nuclides lean toward capture rather than positron emission. Their inner electrons sit closer in, so they are easier to grab.

Gamma emission

Alpha and beta decay often leave the new nucleus holding extra energy. Such a nucleus is excited. It dumps the extra as a gamma ray, a photon with a very short wavelength and no mass or charge. Neither the mass number nor the atomic number changes.

99m43Tc  →  9943Tc  +  γ

The letter m marks a metastable state, one that holds its extra energy long enough to be useful. Technetium-99m is the busiest tracer in any hospital, and it gives off a clean 0.1405 MeV gamma ray.

What each kind of radioactive decay changes inside the nucleus Five ways a nucleus can change 2 p + 2 n Alpha, α helium-4 nucleus leaves A − 4 Z − 2 down and left e− out Beta minus, β− a neutron becomes a proton A same Z + 1 down and right e+ out Positron, β+ a proton becomes a neutron A same Z − 1 up and left e− in Electron capture an inner electron is pulled in A same Z − 1 up and left photon out Gamma, γ the nucleus sheds extra energy A same Z same stays put MASS NUMBER ATOMIC NUMBER
Figure 21.2 · Five decay modes and the bookkeeping for each. The last column says how the nuclide shifts on the band of stability plot in Figure 21.1.

Decay series

One step is rarely enough for a heavy nucleus. Uranium-238 needs 14 steps to reach lead-206, which is stable. Eight of those steps are alpha and six are beta. Every step has its own daughter and its own half-life, and the whole chain is called a decay series. That series is why uranium ore always holds radium and leaks radon gas.

ParticleSymbolMassChargeChange in AChange in Z
Alpha42He4 u2+−4−2
Beta minus0−1e1/1836 u1−0+1
Positron0+1e1/1836 u1+0−1
Electron capture0−1e in1/1836 u1−0−1
Gammaγ0000
Neutron10n1.008665 u0−10

Worked example · naming the daughter nuclide

Radium-226 has Z = 88 and emits an alpha particle. Polonium-218 has Z = 84 and does the same. Name both daughters and write both equations.

Take 4 off A and 2 off Z

Taking 4 off the mass number gives 226 − 4 = 222, and taking 2 off the atomic number gives 88 − 2 = 86. Element 86 is radon, so the daughter nuclide is radon-222.

Repeat for polonium-218

The same two subtractions give 218 − 4 = 214 for the mass number and 84 − 2 = 82 for the atomic number. Element 82 is lead, so this daughter nuclide is lead-214.

Write them out

22688Ra → 22286Rn + 42He and 21884Po → 21482Pb + 42He.

Check both sides

In the radium line the mass numbers come to 222 + 4 = 226 and the atomic numbers come to 86 + 2 = 88. In the polonium line they come to 214 + 4 = 218 and 82 + 2 = 84. Both equations match the left side on each total, so both of them are right.

Key points
  • Alpha decay drops A by 4 and Z by 2, and it is the usual exit for very heavy nuclides.
  • Beta decay turns a neutron into a proton, so Z rises by 1 and A holds still.
  • Positron emission and electron capture both turn a proton into a neutron, so Z falls by 1.
  • Gamma emission changes no count at all, because only energy leaves.
  • Heavy nuclides run through a decay series, and uranium-238 takes 14 steps to reach lead-206.

Section 3Half-life and dating

What a half-life means

You cannot say when one single nucleus will decay. The event is random, and the nucleus keeps no memory of how long it has waited. But a real sample holds a staggering number of nuclei. Across that crowd the behaviour is smooth and predictable.

The half-life, written t1/2, is the time it takes for half the nuclei in a sample to decay. Each half-life halves whatever is left, not what you began with. Start at 100 percent and you pass 50, then 25, then 12.5, then 6.25. The curve bends toward zero but never arrives.

Half-lives cover an almost absurd range. Iodine-131 has a half-life of 8.02 days. Carbon-14 sits at 5,730 years. Uranium-238 takes 4.5 billion years, which is roughly the age of the Earth. Polonium-214 is done in 164 microseconds.

The fraction of a radioactive sample left after each half-life Half of what is left goes every half-life 0 1 2 3 4 5 6 0% 25% 50% 75% 100% 100% 50% 25% 12.5% 6.25% 3.125% 1.5625% Half-lives elapsed Fraction left Carbon-14: one step = 5,730 years Iodine-131: one step = 8.02 days Uranium-238: one step = 4.5 billion years
Figure 21.3 · The same curve fits every isotope. Only the width of one step changes, from microseconds to billions of years.

Decay kinetics

Nuclear decay is first order. The rate depends on how many unstable nuclei are present and on nothing else. Heating the sample does not speed it up. Neither does crushing it, dissolving it or bonding it into a new compound. That is a sharp break from ordinary reaction rates.

The rate constant for decay gets its own name, the decay constant k. It is tied to the half-life by the natural log of 2.

k = 0.693 ÷ t1/2     ln(N0 ÷ N) = kt     N ÷ N0 = (1/2)t/t1/2

Use the right-hand form when the elapsed time is a tidy number of half-lives. Use the log form for everything else. The activity of a sample is how many decays it produces each second, and it shrinks along the very same curve as the nuclei do.

Worked example · how much iodine-131 is left

A clinic receives 40.0 mg of iodine-131, whose half-life is 8.02 days. How much is left after 32.08 days?

Count the half-lives

Dividing the elapsed time by the half-life gives 32.08 days ÷ 8.02 days = 4.00 half-lives.

Halve it four times

Halving the amount once for every half-life runs 40.0 mg → 20.0 mg → 10.0 mg → 5.00 mg → 2.50 mg.

Check with the formula

N = 40.0 mg × (1/2)4 = 40.0 mg × 0.0625 = 2.50 mg. The two routes agree, so 2.50 mg is left.

Radiometric dating

Cosmic rays strike nitrogen high in the air and keep making carbon-14. Plants take that carbon in, and animals eat the plants. So every living thing carries the same small fraction of carbon-14 as the air does. Death shuts the intake off, and from that moment the carbon-14 only decays.

Measure the fraction left and you can solve for the time since death. This is radiometric dating. Carbon-14 is useful out to roughly 50,000 years, which is about nine half-lives. Past that, too little is left to measure well.

Rock needs a slower clock. Uranium-238 decays all the way to lead-206 with a half-life of 4.5 billion years. Potassium-40 becomes argon-40 in about 1.25 billion years. Counting the parent and daughter atoms in a mineral grain dates the moment the crystal locked them in.

Worked example · dating a piece of charcoal

Charcoal from an old hearth holds 54.0 percent of the carbon-14 that living wood holds. The half-life of carbon-14 is 5,730 years. How old is the charcoal?

Find the decay constant

Dividing 0.693 by the half-life gives k = 0.693 ÷ 5,730 yr = 1.21 × 10−4 yr−1.

Set up the log form

The starting amount divided by the present amount is N0 ÷ N = 1 ÷ 0.540 = 1.85, and the natural log of that ratio is ln(1.85) = 0.616.

Solve for the time

Dividing that log by the decay constant gives t = 0.616 ÷ (1.21 × 10−4 yr−1) = 5,090 years.

Check that it is sensible

5,090 years is 0.888 of one half-life, so a little under half should be gone. And (1/2)0.888 = 0.540, which is the fraction we measured. The date holds up.

Why the date needs a correction. The carbon-14 level in the air has drifted over the centuries. Labs fix this with a calibration curve built from tree rings of known age. A raw radiocarbon date is a first draft, not the final year.
Key points
  • One half-life halves whatever is left, so the fraction after n half-lives is (1/2)n.
  • Decay is first order, and k = 0.693 ÷ t1/2.
  • Temperature, pressure and chemical bonding have no effect on a decay rate.
  • Use ln(N0/N) = kt when the time is not a whole number of half-lives.
  • Carbon-14 dates things that once lived out to about 50,000 years, while uranium and potassium clocks date rock.

Section 4Nuclear equations and transmutation

Two rules and nothing more

A nuclear equation balances when two totals agree across the arrow. The mass numbers on the left must add to the mass numbers on the right. The atomic numbers must do the same, because charge is conserved. Notice what is missing from that list. Elements are not conserved, so atoms of one element really do become atoms of another.

The trick is to give every particle a nuclide symbol, even the small ones. An alpha particle is 42He. A beta particle is 0−1e and a positron is 0+1e. A neutron is 10n and a bare proton is 11H. A gamma ray is 00γ, so it adds nothing to either total.

Once every symbol has its pair of numbers, finding an unknown is two subtraction problems. Add the mass numbers you know and subtract to get the missing A. Do the same with the atomic numbers to get the missing Z. Then look up which element owns that Z, and you have named the particle.

Balancing a nuclear equation by adding mass numbers and atomic numbers on each side Two totals have to match: the mass numbers and the atomic numbers 235 92 U 1 0 n 137 52 Te 97 40 ? 1 0 n 1 0 n + → + + + LEFT SIDE RIGHT SIDE Mass numbers (top) left: 235 + 1 = 236 right: 137 + 97 + 1 + 1 = 236 236 = 236 Atomic numbers (bottom) left: 92 + 0 = 92 right: 52 + 40 + 0 + 0 = 92 92 = 92 The unknown has A = 97 and Z = 40. Element 40 is zirconium, so the missing product is zirconium-97.
Figure 21.4 · The ledger for one fission equation. Both totals come to the same number on each side, which is the only test a nuclear equation has to pass.

Worked example · finding a missing product

Fill the gap: 23592U + 10n → 13752Te + ? + 2 10n

Add the mass numbers on each side

The left side totals 235 + 1 = 236, while the pieces already written on the right total 137 + 1 + 1 = 139. The gap between those two totals is 236 − 139 = 97, so the missing particle has A = 97.

Add the atomic numbers on each side

The left side totals 92 + 0 = 92, while the known products on the right total 52 + 0 + 0 = 52. That leaves a gap of 92 − 52 = 40, so the missing particle has Z = 40.

Name the element

Element 40 is zirconium, so the missing product is 9740Zr.

Check the finished line

The mass numbers on the right now total 137 + 97 + 2 = 236, which matches the 236 on the left. The atomic numbers total 52 + 40 = 92, which matches the 92 on the left. Both totals agree across the arrow, so the equation is balanced.

Transmutation on purpose

Turning one element into another is transmutation. Radioactive decay does it by itself. You can also force it by firing a particle into a target nucleus. Ernest Rutherford did the first deliberate one in 1919. He fired alpha particles into nitrogen gas and got oxygen and protons back.

147N  +  42He  →  178O  +  11H

Neutrons make the best bullets. They carry no charge, so a nucleus does not push them away. A reactor core is swimming in spare neutrons, which makes it a fine factory for useful isotopes. The tracer most hospitals run on is built this way.

9842Mo  +  10n  →  9942Mo  →  99m43Tc  +  0−1e

A charged bullet needs a hard push to beat the repulsion of the target. That push comes from a particle accelerator, which uses electric and magnetic fields to drive ions to huge speeds. Accelerators and reactors together built every element past uranium. Those are the transuranium elements, and none of them occurs on Earth in any real amount.

23892U  +  10n  →  23992U  →  23993Np  +  0−1e  →  23994Pu  +  0−1e

Climbing higher takes heavier bullets still. Seaborgium came from firing oxygen-18 into californium-249. Runs like that may make only a few atoms, and those atoms live for seconds or less. Even so, the equation balances the same way any other does.

24998Cf  +  188O  →  263106Sg  +  4 10n

Worked example · naming the particle that flew out

Aluminium-27 is hit by an alpha particle and phosphorus-30 comes out. What else comes out?

Write what you know

The line so far reads 2713Al + 42He → 3015P + ?

Balance the mass numbers

The mass numbers on the left total 27 + 4 = 31, and the phosphorus accounts for 30 of them. That leaves 31 − 30 = 1, so the missing particle has A = 1.

Balance the atomic numbers

The atomic numbers on the left total 13 + 2 = 15, and the phosphorus already accounts for all 15. That leaves 15 − 15 = 0, so the missing particle has Z = 0.

Identify it

A = 1 with Z = 0 is a neutron, 10n. Both totals now read 31 and 15, so the equation is balanced.

Key points
  • Balance a nuclear equation on mass number and on atomic number, and nothing else.
  • Elements are not conserved, so one element can become another.
  • Give every particle a nuclide symbol, including the neutron, the beta particle and the gamma ray.
  • Transmutation fires a particle at a target nucleus, and neutrons work best because they carry no charge.
  • Every element past uranium was built in a reactor or an accelerator.

Section 5Fission and fusion

What the binding energy curve predicts

Section 1 left you with one shape worth memorising. Binding energy per nucleon climbs steeply from hydrogen, peaks near iron-56, then drifts back down. Any change that moves nucleons toward that peak releases energy. A very heavy nucleus gets closer to the peak by splitting. Very light nuclei get closer by joining up.

Binding energy per nucleon against mass number, with its peak near iron-56 Binding energy per nucleon climbs, peaks near iron-56, then falls 0 50 100 150 200 250 0 2 4 6 8 H-2 He-4 C-12 Fe-56 Sn-120 U-238 8.79 MeV FUSION climbs this way FISSION climbs this way Mass number (A) Binding energy per nucleon (MeV)
Figure 21.5 · Both arrows point uphill toward iron-56. That single picture is the reason fission works at the heavy end and fusion works at the light end.

Fission

Fission splits a heavy nucleus into two lighter ones. Uranium-235 fissions when it swallows a slow neutron. It does not split the same way twice, and dozens of product pairs show up across a real sample. Here is one of the common lines.

23592U  +  10n  →  14156Ba  +  9236Kr  +  3 10n

One fission event releases about 200 MeV. Burning a single carbon atom releases about 4 eV. The nuclear event is roughly 50 million times larger, and that gap is the whole story of nuclear power. A fuel pellet the size of a fingertip holds the energy of a tonne of coal.

Worked example · fission energy per mole

One uranium-235 fission gives about 200 MeV. How much does a mole of fissions give, and how does that compare with burning a mole of carbon at 393.5 kJ/mol?

Turn MeV into joules

1 MeV = 1.602 × 10−13 J, so 200 MeV × 1.602 × 10−13 J/MeV = 3.204 × 10−11 J per event.

Scale up by Avogadro's number

3.204 × 10−11 J × 6.022 × 1023 /mol = 1.93 × 1013 J/mol, which is 1.93 × 1010 kJ/mol.

Compare the two fuels

Dividing the nuclear figure by the chemical one gives 1.93 × 1010 kJ ÷ 393.5 kJ = 4.90 × 107. So one mole of uranium-235 fissions releases about 49 million times the energy of one mole of carbon burned.

Chain reactions and reactors

Every fission throws out two or three fresh neutrons. Each of those can split another uranium-235 nucleus. The reaction then feeds itself, and that self-feeding loop is a chain reaction.

Whether the loop grows or dies comes down to one number. The multiplication factor k is the number of fissions in one generation divided by the number in the generation before. If k is below 1 the chain fades out. If k is exactly 1 the rate holds steady. If k is above 1 the rate climbs with every generation.

Neutrons that leak out of the fuel are neutrons that cannot split anything. So you need enough fuel, packed closely enough, to keep k at 1 or above. The smallest amount that can hold a chain reaction going is the critical mass.

Natural uranium is only 0.72 percent uranium-235, and almost all the rest is uranium-238. Power reactor fuel is enriched to about 3 to 5 percent uranium-235. That is enough to hold a chain reaction in a large core full of water.

A reactor is a machine for parking k at exactly 1. The moderator, usually water or graphite, slows the fresh neutrons down, because uranium-235 captures slow neutrons far more readily than fast ones. Control rods of boron or cadmium slide in to soak up spare neutrons, and sliding them out lets the rate rise. A coolant carries the heat away to boil water, and the steam drives a turbine. Thick shielding and a sealed containment building keep the radiation inside.

Fusion

Fusion joins light nuclei into a heavier one. The Sun runs on it. Four hydrogen nuclei become one helium-4 nucleus over several steps, releasing about 26.7 MeV and two positrons along the way.

The reaction engineers chase on Earth pairs deuterium with tritium, the hydrogen isotopes of mass 2 and mass 3.

21H  +  31H  →  42He  +  10n    (17.6 MeV)

Spread 17.6 MeV over 5 nucleons and you get about 3.5 MeV each. Spread 200 MeV over 236 nucleons and you get about 0.85 MeV each. Per unit of mass, fusion is the bigger payout by a factor of four.

The obstacle is repulsion. Both nuclei are positive, so they fly apart unless they arrive at tremendous speed. That means temperatures near 100 million kelvin, plus some way to hold the hot plasma together long enough. Magnetic fields and laser compression are the two routes being built. A fusion plant feeding a power grid is still ahead of us.

Why fusion fuel is so cheap. Deuterium is already in seawater, about one hydrogen atom in 6,400. Tritium can be bred from lithium inside the reactor itself. The fuel is almost free, and it is the machine around it that is hard.
Key points
  • Energy is released whenever nucleons move toward the binding energy peak at iron-56.
  • Fission splits a heavy nucleus and releases about 200 MeV per event.
  • Spare neutrons keep a chain reaction alive, and the multiplication factor k says whether it grows, holds or fades.
  • A reactor holds k at 1 with a moderator, control rods and a coolant loop.
  • Fusion gives about 3.5 MeV per nucleon against 0.85 MeV for fission, but it needs temperatures near 100 million kelvin.

Section 6Radiation, detection and uses

Counting decays and counting dose

Radiation raises two separate questions. How busy is the source, and how much energy did a body take in? Each one has its own unit, and mixing the two is the commonest mistake here.

The busyness of a source is its activity, measured in becquerels, where one becquerel is one decay per second. The older unit is the curie, which is 3.7 × 1010 Bq, the activity of one gram of radium-226.

Energy taken in by tissue is the absorbed dose, measured in grays, where one gray is one joule absorbed per kilogram. A gray of alpha particles harms far more than a gray of gamma rays, because the alpha dumps its energy along a very short track. So the gray is multiplied by a quality factor to give the dose equivalent in sieverts. Beta and gamma rate a 1, neutrons about 10, and alpha a 20.

QuantitySI unitMeaningOlder unit
Activitybecquerel, Bq1 decay per second1 curie = 3.7 × 1010 Bq
Absorbed dosegray, Gy1 joule absorbed per kilogram1 rad = 0.01 Gy
Dose equivalentsievert, Svgray × quality factor1 rem = 0.01 Sv

Worked example · adding two kinds of dose

A hand absorbs 2.0 mGy from an alpha emitter and 3.0 mGy from a gamma source. What is the total dose equivalent?

Weight each dose by its quality factor

Alpha rates a 20, so 2.0 mGy × 20 = 40 mSv. Gamma rates a 1, so 3.0 mGy × 1 = 3.0 mSv.

Add the weighted doses

Adding the two together gives 40 mSv + 3.0 mSv = 43 mSv of dose equivalent.

Put it in context

Natural background runs about 3.1 mSv a year, so 43 mSv ÷ 3.1 mSv is about 14 years' worth in one exposure. Notice that the alpha dose was smaller in grays yet counted for far more in sieverts.

Detecting what you cannot see

A Geiger-Müller counter is a sealed tube of gas with a high-voltage wire down the middle. A ray passing through knocks electrons off the gas atoms, those electrons rush to the wire and free more on the way, and each pulse is counted as a click.

A scintillation counter works on light instead. Radiation strikes a crystal, the crystal flashes, and a sensor counts the flashes and measures their brightness. Brightness reveals the energy of the ray, so this instrument can name the isotope. Workers also wear a dosimeter, which keeps a running total of the dose taken.

What radiation does inside tissue

These rays are called ionizing because they strip electrons off the molecules they pass. A cell is mostly water, so most hits split water into reactive fragments called free radicals, which then attack DNA, proteins and membranes nearby. A cell can usually repair one broken DNA strand, but repairs sometimes fail or finish wrong, and that is where the long-term risk sits.

Where the source sits matters enormously. Alpha emitters outside the body are almost harmless, because dead skin stops the particles before they reach anything living. Breathe them in and they become the worst case, since all their energy lands inside living tissue. Radon seeping into a basement works this way, and it is a leading cause of lung cancer.

How far alpha, beta and gamma radiation get through paper, aluminium and lead The same source, three kinds of ray, three shields source Paper 0.1 mm Aluminium 3 mm Lead 5 cm Alpha α stopped by paper Beta β stopped by a few mm of metal Gamma γ lead cuts it, never to zero Alpha carries the most charge and mass, so it loses its energy in the shortest distance.
Figure 21.6 · Penetrating power runs opposite to charge and mass. Gamma rays are never stopped completely, only cut down, so gamma shielding is quoted as the thickness that halves the beam.
Doses worth memorising. Natural background averages about 3.1 mSv a year, mostly from radon and from the potassium-40 in your own body. A chest X-ray adds about 0.1 mSv and a chest CT about 7 mSv. The yearly limit for radiation workers is 50 mSv. A single dose near 1 Sv brings on radiation sickness, and about 4 Sv is fatal to half of those exposed without treatment.

Protection comes down to three words. Spend less time near the source, keep more distance from it, and put shielding in between. Distance is the cheapest, because intensity falls with the square of the distance.

Worked example · backing away from a source

A survey meter reads 48 µSv/h at 1.0 m from a small gamma source. What will it read at 4.0 m?

Find the distance ratio and square it

Dividing the new distance by the old one gives 4.0 m ÷ 1.0 m = 4.0, and intensity falls as 1 ÷ d2, so divide the reading by 4.02 = 16.

Do the division

Dividing the close-up reading by that factor gives 48 µSv/h ÷ 16 = 3.0 µSv/h.

Check the logic

Four times the distance gives one sixteenth the dose rate, so two steps back can do more for you than a slab of lead.

Radiation put to work

A radioactive tracer is a short-lived isotope stuck onto a molecule the body already handles. Technetium-99m has a half-life of 6.01 hours and gives a clean gamma ray, so it maps bone, heart and kidney function. Fluorine-18 lasts 110 minutes and emits positrons, and on a sugar molecule it shows a PET scanner which tissues burn the most fuel.

Treatment uses much larger doses, aimed to kill. Iodine-131 is swallowed, and the thyroid pulls iodine from the blood whatever isotope it is, so its beta particles destroy diseased thyroid tissue from within. Cobalt-60 gamma beams treat tumours from outside, and tiny iodine-125 seeds can be placed inside one.

Half-life choice is part of the design. The isotope must last long enough to reach its target and finish the job, then fade fast so the patient carries no source for months. Outside the clinic the same rays sterilise surgical supplies, gauge the thickness of rolled steel and trip the americium-241 smoke alarm on your ceiling.

Key points
  • Becquerels measure the source, grays measure energy absorbed, and sieverts measure expected harm.
  • Multiply grays by the quality factor to get sieverts, and alpha radiation carries a factor of 20.
  • Geiger-Müller tubes count rays, while scintillation counters also measure their energy.
  • Ionizing radiation mostly damages tissue by splitting water into free radicals that attack DNA.
  • Time, distance and shielding are the three defences, and intensity falls as 1 ÷ d2.
  • Tracers use short half-lives and gamma or positron emitters, while treatment uses beta and gamma emitters at higher dose.

The chapter in one breath

  • A nuclide is set by its proton count Z and its mass number A, and N = A − Z.
  • Stable nuclides lie in the band of stability, and nothing above Z = 83 is stable.
  • The parts of a nucleus weigh more than the nucleus, and E = mc2 turns that mass defect into its binding energy.
  • Binding energy per nucleon peaks near iron-56 at about 8.79 MeV.
  • Alpha decay cuts A by 4 and Z by 2, beta raises Z by 1, and positron emission and electron capture each lower Z by 1.
  • Decay is first order, so one half-life always halves what is left and k = 0.693 ÷ t1/2.
  • Carbon-14 at 5,730 years dates what once lived, and uranium-238 at 4.5 billion years dates rock.
  • Nuclear equations balance on mass number and atomic number only, which lets elements change.
  • Fission splits heavy nuclei for about 200 MeV each, and spare neutrons carry a chain reaction that k keeps in check.
  • Fusion pays about 3.5 MeV per nucleon but demands temperatures near 100 million kelvin.
  • Becquerels measure a source, grays measure absorbed energy, and sieverts measure expected harm.

ReferenceWords worth knowing

Twenty terms carry most of this chapter. Read down the list once now, then come back to it while you work the questions below.

Alpha particle
A helium-4 nucleus thrown out of a decaying nucleus, written 42He and carrying a 2+ charge.
Band of stability
The narrow stripe on a neutron-against-proton plot where every stable nuclide falls.
Becquerel
The SI unit of activity, equal to one decay per second. 1 curie = 3.7 × 1010 Bq.
Beta particle
A fast electron fired from a nucleus when a neutron turns into a proton. Written 0−1e.
Binding energy
The energy released when a nucleus forms, and the energy needed to take it apart again.
Chain reaction
A sequence in which neutrons from one fission trigger more fissions, so the reaction feeds itself.
Critical mass
The smallest amount of fuel, in a close enough shape, that can keep a chain reaction going.
Electron capture
Decay in which a nucleus pulls in one of its own inner electrons, lowering Z by 1 and leaving A alone.
Fission
The splitting of a heavy nucleus into two lighter nuclei plus spare neutrons.
Fusion
The joining of light nuclei into a heavier one, which is the energy source of every star.
Gamma ray
A very short wavelength photon released by an excited nucleus. It changes neither A nor Z.
Gray
The SI unit of absorbed dose, one joule of radiation energy taken in per kilogram of tissue.
Half-life
The time needed for half the unstable nuclei in a sample to decay. Written t1/2.
Magic number
A nucleon count of 2, 8, 20, 28, 50, 82 or 126 that makes a nucleus unusually stable.
Mass defect
The difference between the mass of a nucleus and the total mass of its separated nucleons.
Nuclide
One exact kind of nucleus, named by its proton count and its neutron count.
Positron
The antiparticle of the electron, with the same mass and a 1+ charge. Written 0+1e.
Radiometric dating
Finding an age from the fraction of a radioactive parent isotope still left in a sample.
Sievert
The unit of dose equivalent, found by multiplying the absorbed dose in grays by a quality factor.
Transmutation
Any change of one element into another, whether by decay or by firing a particle at a nucleus.

Check yourself

1. Radon-222 has Z = 86 and decays by alpha emission. Write the balanced equation and name the daughter.
22286Rn → 21884Po + 42He Mass: 222 − 4 = 218. Charge: 86 − 2 = 84, and element 84 is polonium, so the daughter is polonium-218.
2. Strontium-90 has Z = 38 and is a beta emitter. What does it become, and why does strontium-90 end up in bone?
Yttrium-90. Beta decay leaves A at 90 and raises Z from 38 to 39. Strontium sits directly below calcium in group 2, so the body treats it like calcium and files it in bone.
3. A wooden handle from a grave holds 25.0 percent of the carbon-14 of living wood. How old is it?
About 11,500 years. 25.0 percent is two halvings, since 100 → 50 → 25. Two half-lives is 2 × 5,730 yr = 11,460 yr, which rounds to 1.15 × 104 years.
4. A vial holds 100.0 mg of iodine-131, half-life 8.02 days. How much iodine-131 is left after 24.06 days?
12.5 mg. 24.06 d ÷ 8.02 d = 3.00 half-lives. So 100.0 mg × (1/2)3 = 100.0 mg × 0.125 = 12.5 mg.
5. Uranium-238 has a half-life of 4.5 × 109 years. Find its decay constant.
1.5 × 10−10 yr−1. k = 0.693 ÷ t1/2 = 0.693 ÷ (4.5 × 109 yr) = 1.54 × 10−10 yr−1. Only about one atom in six billion decays in any given year.
6. A deuterium atom weighs 2.014102 u. Hydrogen-1 weighs 1.007825 u and a neutron weighs 1.008665 u. Find the binding energy per nucleon, using 1 u = 931.5 MeV.
1.112 MeV per nucleon. Parts: 1.007825 u + 1.008665 u = 2.016490 u. Mass defect: 2.016490 − 2.014102 = 0.002388 u. Energy: 0.002388 u × 931.5 MeV/u = 2.224 MeV. Per nucleon: 2.224 ÷ 2 = 1.112 MeV, far below the 8.79 MeV of iron-56.
7. A technician absorbs 0.50 mGy of neutron radiation, quality factor 10. What is the dose equivalent, and how does it compare with 0.50 mGy of gamma?
5.0 mSv. 0.50 mGy × 10 = 5.0 mSv. The same 0.50 mGy of gamma has a quality factor of 1, so it counts as only 0.50 mSv. Equal energy, ten times the expected harm.
8. A meter reads 90 µSv/h at 2.0 m from a gamma source. What is the reading at 6.0 m?
10 µSv/h. The distance triples, since 6.0 ÷ 2.0 = 3.0. Intensity falls as 1 ÷ d2, so divide by 3.02 = 9. Then 90 µSv/h ÷ 9 = 10 µSv/h.
9. Why does a stable heavy nucleus need far more neutrons than protons, while carbon-12 gets by with an even split?
Because electrical repulsion grows faster than the strong force reaches. The strong force only pulls on close neighbours, but every proton repels every other proton across the whole nucleus. Extra neutrons add pull without adding repulsion, so the band of stability bends above the N = Z line.
10. Positron emission and electron capture both lower Z by 1 and leave A alone. How can you tell them apart?
By what leaves the atom. Positron emission throws a positron out, which soon annihilates with an electron and makes two 0.511 MeV photons. Electron capture pulls an inner electron in and gives off a characteristic X-ray as an outer electron fills the hole. Heavier nuclides favour capture.
11. Fission of uranium-235 releases about 200 MeV and fusion of deuterium with tritium releases 17.6 MeV. Why is fusion still called the richer reaction?
Because you have to divide by the nucleons involved. Fission: 200 MeV ÷ 236 nucleons = 0.85 MeV each. Fusion: 17.6 MeV ÷ 5 nucleons = 3.5 MeV each. Per unit of mass, fusion pays about four times as much.
12. An alpha emitter is far less dangerous than a gamma emitter on a bench, yet far more dangerous once inhaled. Explain both halves of that.
It comes down to where the energy lands. Alpha particles cannot get through dead skin, so an outside source does nothing. Inhaled, the same particles dump all their energy into a tiny volume of living tissue, and their quality factor of 20 reflects that. Gamma rays pass through the body either way, which is why they need lead or thick concrete.