🧪 Chemistry Final Exam · Chapters 1–21
Chemistry · Final Exam

Final Exam Study Guide

The final covers all 21 chapters. Taking them one at a time would mean 21 piles of notes, so this guide groups them into seven themes instead. Each theme pulls together the chapters that share a way of thinking. Measurement and the mole teach you to count. Structure and bonding explain shape. States and solutions handle what matter does in bulk. Change and rate ask how far and how fast. Equilibrium systems hold the balance. Energy and electron transfer set direction. The last theme walks the real elements and the nucleus itself. Work one theme a day and all 21 chapters fit into a week.

The seven themes and the chapters in each

  • Measurement and the mole: chapters 1 and 3.
  • Structure and bonding: chapters 2, 6, 7 and 8.
  • States and solutions: chapters 9, 10 and 11.
  • Change and rate: chapters 4, 5 and 12.
  • Equilibrium systems: chapters 13, 14 and 15.
  • Energy and electron transfer: chapters 16 and 17.
  • Descriptive and nuclear: chapters 18, 19, 20 and 21.
  • Every theme carries what to know, the equations with their units, worked examples, the traps, and a 10-item self-test.

Theme 1 · Chapters 1 and 3Measurement and the mole

What to know

Matter is anything with mass that takes up space. In a solid the particles sit in fixed spots, so shape and volume both hold. In a liquid they touch but slide, so volume holds and shape gives way. In a gas they fly apart and fill the container, so neither holds.

Sort every sample three ways. An element holds one kind of atom. A compound holds two or more elements bonded in a fixed ratio. A mixture holds substances that are not bonded, so the ratio can shift. A physical change leaves the substance itself alone, while a chemical change makes new substances.

Every number in chemistry carries a unit, and the unit carries meaning. The significant figures of a measurement are the digits that say something real about its size. Multiplication and division keep the fewest significant figures in the problem. Addition and subtraction keep the fewest decimal places.

Dimensional analysis is the habit that makes unit work safe. Write the number with its unit, then multiply by fractions that cancel what you do not want and leave what you do. If the units come out right, the setup is right.

The mole is a counting word, like a dozen. One mole holds 6.022 × 1023 particles, which is Avogadro's number. Molar mass is the grams in one mole, read off the periodic table in g/mol. Those two facts build one bridge that almost every calculation in this book walks across.

The mole bridge, in both directions
  • Grams to moles: divide by the molar mass.
  • Moles to particles: multiply by Avogadro's number.
  • Moles to liters of solution: divide by the molarity.
  • Moles of one substance to moles of another: use the coefficients of a balanced equation.
  • Every route runs through moles. If you are stuck, convert to moles first.

Percent composition is the share of each element by mass in a compound. An empirical formula gives the smallest whole-number ratio of atoms, and a molecular formula gives the real count. Get the empirical formula from masses or percents, then compare its mass with the measured molar mass to find the multiplier.

Molarity is moles of solute per liter of solution. Dilution keeps the solute and grows the volume, so the moles hold still while the concentration falls.

Equations, units and symbols

EquationWhat each symbol meansUnits
d = m ÷ V Density is mass per unit of volume. It is a property of the substance, not the sample size. g/mL for a liquid, g/cm3 for a solid, g/L for a gas. 1 mL = 1 cm3.
K = °C + 273.15 The kelvin scale starts at absolute zero, so it never goes negative. K and °C. A change of 1 K equals a change of 1 °C.
°F = (9 ÷ 5)(°C) + 32 Runs the other way as °C = (°F − 32)(5 ÷ 9). °F and °C.
moles = mass ÷ molar mass Molar mass is the grams in one mole, summed from the periodic table. mol = g ÷ (g/mol).
particles = moles × NA NA is Avogadro's number. The particles may be atoms, molecules or formula units. NA = 6.022 × 1023/mol.
% by mass = (mass of that element in one mole ÷ molar mass) × 100% Do it for every element and the shares add to 100%. A percent.
multiplier = molar mass ÷ empirical formula mass Round it to a whole number, then multiply every subscript in the empirical formula. A bare number.
M = moles of solute ÷ liters of solution Molarity. Note that it is per liter of the whole solution, not per liter of solvent. mol/L, written M.
M1V1 = M2V2 Dilution, with before on the left and after on the right. The moles of solute do not change. Any matching volume unit on both sides.

Worked examples

Example T1.1 · Density, significant figures and units

A metal cube measures 2.50 cm on each side and weighs 67.5 g. Find its density in g/cm3, then convert that to kg/m3.

Step 1 · find the volume

A cube's volume is the edge cubed. V = (2.50 cm)3 = 15.625 cm3.

Step 2 · divide

d = 67.5 g ÷ 15.625 cm3 = 4.32 g/cm3.

Step 3 · count the significant figures

Both measurements carry three, so the answer carries three. Writing 4.32 g/cm3 is honest. Writing 4.32000 would claim precision the ruler never gave you.

Step 4 · chain the unit conversions

4.32 g/cm3 × (1 kg ÷ 1000 g) × (106 cm3 ÷ 1 m3) = 4320 kg/m3.

Answer and the check

4.32 g/cm3, which is 4320 kg/m3. Watch the cube on the second conversion. There are 100 cm in a metre, so there are 1003 = 106 cubic centimetres in a cubic metre. Forgetting to cube that factor is the single most common unit error in this whole subject.

Example T1.2 · Molar mass, percent composition and the mole bridge

Glucose is C6H12O6. Find its molar mass and its percent composition. Then find how many molecules sit in 25.0 g of it.

Step 1 · build the molar mass

Carbon: 6 × 12.011 = 72.066 g/mol. Hydrogen: 12 × 1.008 = 12.096 g/mol. Oxygen: 6 × 15.999 = 95.994 g/mol. Add them: 180.16 g/mol.

Step 2 · each element's share

Carbon: 72.066 ÷ 180.156 × 100% = 40.00%. Hydrogen: 12.096 ÷ 180.156 × 100% = 6.71%. Oxygen: 95.994 ÷ 180.156 × 100% = 53.28%.

Step 3 · check that the shares close

40.00 + 6.71 + 53.28 = 99.99%, which is 100% to the rounding. If your shares miss by more than a hundredth, something went wrong.

Step 4 · grams to moles

25.0 g ÷ 180.16 g/mol = 0.1388 mol.

Step 5 · moles to molecules

(0.1388 mol)(6.022 × 1023/mol) = 8.36 × 1022 molecules.

Answer

180.16 g/mol, 40.00% carbon, 6.71% hydrogen, 53.28% oxygen, and 8.36 × 1022 molecules in a 25.0 g spoonful. That molecule count is bigger than the number of stars anyone has ever counted, which is what a mole really means.

Example T1.3 · Empirical formula, then a dilution

A compound is 40.00% carbon, 6.71% hydrogen and 53.28% oxygen by mass, with a molar mass of 180.16 g/mol. Find its molecular formula. Then dilute 25.0 mL of 6.00 M HCl to 500.0 mL and find the new concentration.

Step 1 · treat the percents as grams

Imagine exactly 100 g of the compound. That gives 40.00 g carbon, 6.71 g hydrogen and 53.28 g oxygen.

Step 2 · go to moles

Carbon: 40.00 ÷ 12.011 = 3.330 mol. Hydrogen: 6.71 ÷ 1.008 = 6.657 mol. Oxygen: 53.28 ÷ 15.999 = 3.330 mol.

Step 3 · divide by the smallest

3.330 ÷ 3.330 = 1.000 for carbon. 6.657 ÷ 3.330 = 1.999 for hydrogen. 3.330 ÷ 3.330 = 1.000 for oxygen. So the empirical formula is CH2O.

Step 4 · find the multiplier

The empirical formula mass is 12.011 + 2.016 + 15.999 = 30.026 g/mol. Then 180.16 ÷ 30.026 = 6.00, so every subscript gets multiplied by 6. The molecular formula is C6H12O6.

Step 5 · now the dilution

M2 = M1V1 ÷ V2 = (6.00 M)(25.0 mL) ÷ 500.0 mL = 150 ÷ 500.0 = 0.300 M.

Answer

C6H12O6, which is the glucose from the example above, and a diluted acid at 0.300 M. The volume grew by a factor of 20, so the concentration fell by the same factor of 20. That check takes two seconds and catches most arithmetic slips.

The traps people fall into

Watch for these
  • Dropping or inventing significant figures. Multiplication keeps the fewest figures. Addition keeps the fewest decimal places.
  • Forgetting to cube or square a unit factor. A cubic metre holds 106 cubic centimetres, not 100.
  • Using Celsius where kelvin is needed. Any gas law or thermodynamics problem wants kelvin.
  • Reading molarity as per liter of solvent. It is per liter of the finished solution.
  • Skipping moles. Mass never converts straight to particles or to another substance.
  • Rounding an empirical ratio too early. A ratio of 1.50 means multiply everything by 2, not round down to 1.
  • Mixing up the empirical and the molecular formula. The multiplier is the measured molar mass over the empirical formula mass.
  • Calling a solution a compound. Salt water is a homogeneous mixture, since the ratio can shift.

Self-test · Theme 1

1. How many significant figures are in 0.03080?
Four. The zeros in front only place the decimal point, so they do not count. The 3, the 0 between the digits, the 8 and the final 0 all do. A trailing zero after the decimal point is there because someone measured it.
2. Convert 1.25 kg to milligrams.
1.25 × 106 mg. Chain two factors: 1.25 kg × (1000 g ÷ 1 kg) × (1000 mg ÷ 1 g). The kilograms and grams cancel, leaving 1.25 × 106 mg. Three significant figures in, three out.
3. Convert 98.6 °F to Celsius and to kelvin.
37.0 °C and 310.2 K. First °C = (98.6 − 32)(5 ÷ 9) = (66.6)(5 ÷ 9) = 37.0 °C. Then K = 37.0 + 273.15 = 310.2 K. Subtract before you scale, never after.
4. A 15.0 mL sample of a liquid weighs 11.8 g. Find its density.
0.787 g/mL. d = 11.8 g ÷ 15.0 mL = 0.787 g/mL. Three significant figures in each measurement, so three in the answer. A density under 1 g/mL means this liquid floats on water.
5. How many moles are in 50.0 g of CaCO3? Its molar mass is 100.09 g/mol.
0.500 mol. 50.0 g ÷ 100.09 g/mol = 0.4996 mol, which rounds to 0.500 mol. Half a molar mass gives half a mole, which is a useful sanity check to run in your head.
6. How many atoms are in 0.250 mol of iron?
1.51 × 1023 atoms. (0.250 mol)(6.022 × 1023/mol) = 1.5055 × 1023, which rounds to 1.51 × 1023. A quarter of a mole gives a quarter of Avogadro's number.
7. Find the molar mass of (NH4)2SO4.
132.13 g/mol. The subscript 2 outside the bracket doubles both the nitrogen and the hydrogen. Nitrogen: 2 × 14.007 = 28.014. Hydrogen: 8 × 1.008 = 8.064. Sulfur: 32.06. Oxygen: 4 × 15.999 = 63.996. Add: 132.13 g/mol.
8. How much NaCl do you weigh out for 250.0 mL of 0.150 M solution? NaCl is 58.44 g/mol.
2.19 g. Moles = (0.2500 L)(0.150 mol/L) = 0.0375 mol. Mass = (0.0375 mol)(58.44 g/mol) = 2.19 g. Dissolve that and then add water up to the 250.0 mL mark, not to 250.0 mL of water.
9. Dilute 10.0 mL of 12.0 M HCl to 250.0 mL. What is the new molarity?
0.480 M. M2 = (12.0 M)(10.0 mL) ÷ 250.0 mL = 120 ÷ 250.0 = 0.480 M. The volume grew 25 times, so the concentration fell 25 times, and 12.0 ÷ 25 = 0.480.
10. An iron oxide is 69.9% Fe and 30.1% O by mass. Find its empirical formula. Iron is 55.845 g/mol.
Fe2O3. Take 100 g. Iron: 69.9 ÷ 55.845 = 1.252 mol. Oxygen: 30.1 ÷ 15.999 = 1.881 mol. Divide by the smaller: iron 1.000 and oxygen 1.503. A ratio of 1.5 means double both, giving 2 iron to 3 oxygen. Rounding 1.503 down to 1 would have given FeO, which is a different compound.

Theme 2 · Chapters 2, 6, 7 and 8Structure and bonding

What to know

An atom holds protons and neutrons in a tiny nucleus with electrons spread around it. The proton count is the atomic number Z and it names the element. Isotopes share a proton count and differ in neutrons. The atomic mass on the periodic table is a weighted average over the isotopes as they occur in nature, so the common isotope pulls the average toward itself.

Atoms become ions by losing or gaining electrons. Metals on the left give electrons away to make cations. Nonmetals on the right take them in to make anions. An ionic compound is a lattice of those ions in whatever ratio makes the charges cancel, so the formula is the smallest such ratio. A molecular compound shares electrons instead, and its formula counts the atoms in one molecule.

Light is a wave and a stream of packets at once. Wavelength and frequency are locked together, and each packet carries an energy set by the frequency. Bohr fixed hydrogen's electron into levels with a definite energy each, and the gap between two levels is exactly the energy of the photon that moves an electron between them. That is why a hydrogen lamp gives sharp lines instead of a smear.

Quantum numbers name an orbital. The main level n sets the size, the letter sets the shape as s, p, d or f, and within a subshell the orbitals differ only in direction. An electron configuration lists how the electrons fill those orbitals in order of rising energy. Hund's rule says a subshell fills singly before any orbital doubles up, because two electrons in one orbital repel.

Periodic trends, and the one reason behind them
  • Across a period the nuclear grip tightens, so atoms shrink.
  • Down a group a new shell is added, so atoms grow.
  • Ionization energy and electronegativity rise across a period and fall down a group.
  • A cation is always smaller than its atom. An anion is always larger.
  • Every one of those trends traces back to how far the outer electron sits and how much charge it feels.

An ionic bond is the pull between opposite charges, and its strength shows up as the lattice energy. Bigger charges and smaller ions both raise it, which is why MgO holds together far harder than NaCl. A covalent bond is a shared pair. How evenly that pair is shared depends on the electronegativity gap, so a bond runs from nonpolar through polar to fully ionic as the gap grows.

A Lewis structure counts valence electrons and spends them to give every atom a full shell, usually an octet. When two or more good structures differ only in where the double bonds sit, the real molecule is a blend of them, and those drawings are resonance forms. Formal charge is the tie-breaker when several structures look legal. Keep the formal charges small and put any negative one on the more electronegative atom.

VSEPR turns a Lewis structure into a shape. Count the electron groups around the central atom, counting each lone pair and each bond as one group whether it is single, double or triple. The groups spread as far apart as they can. Then remove the lone pairs from the picture to read off the shape, because a lone pair takes up more room than a bond and squeezes the remaining angles.

Electron groups to shape
  • 2 groups: linear, 180°.
  • 3 groups: trigonal planar, 120°. With one lone pair the shape is bent.
  • 4 groups: tetrahedral, 109.5°. One lone pair gives trigonal pyramidal, two give bent.
  • 5 groups: trigonal bipyramidal. 6 groups: octahedral, 90°.
  • A molecule is polar when the bond dipoles do not cancel. Symmetry is what cancels them.

Valence bond theory builds a bond from overlapping orbitals, and hybrid orbitals explain the measured angles. Four groups means sp3 at 109.5°, three means sp2 at 120°, and two means sp at 180°. A head-on overlap is a sigma bond and a side-by-side overlap is a pi bond. A single bond is one sigma, a double bond is one sigma plus one pi, and a triple bond is one sigma plus two pi. A sigma bond lets the ends spin and a pi bond does not, which is why a double bond locks a molecule flat.

Molecular orbital theory spreads the electrons over the whole molecule instead. Combining two atomic orbitals gives one bonding level lower in energy and one antibonding level higher. Bond order is the bonding electrons minus the antibonding ones, divided by two. A higher bond order means a shorter and stronger bond. The theory also explains magnetism, since a molecule with unpaired electrons in its orbitals is paramagnetic. Oxygen is the famous case, and liquid oxygen really does cling to a magnet.

Equations, units and symbols

EquationWhat each symbol meansUnits
average atomic mass = Σ(fraction × isotope mass) The fraction is the natural abundance written as a decimal. The fractions add to 1. Atomic mass units, u.
c = λν λ is wavelength and ν is frequency. A longer wave means a lower frequency. c = 2.998 × 108 m/s, λ in m, ν in Hz, which is s-1.
E = hν = hc ÷ λ The energy of one photon. h is Planck's constant. h = 6.626 × 10-34 J·s, so hc = 1.987 × 10-25 J·m. E in J.
En = −(2.179 × 10-18 J) ÷ n2 Bohr's levels for hydrogen. n is the level number, and the minus sign means the electron is bound. J per atom. Times NA for J/mol.
ΔE = Efinal − Einitial Negative means a photon came out. Positive means one went in. Take the size of ΔE to get the photon. J per atom.
formal charge = (valence electrons) − (lone pair electrons) − (number of bonds) Count the atom's own valence electrons from its group, then subtract what the drawing gives it. A charge in units of e.
bond order from resonance = (bonding pairs) ÷ (positions they spread over) Half-integer answers are normal, and they fit the measured bond lengths. A bare number.
bond order from MO theory = (bonding − antibonding) ÷ 2 Count the electrons in each kind of level. A bond order of 0 means the molecule does not form. A bare number.

Worked examples

Example T2.1 · Average atomic mass from two isotopes

Chlorine is 75.76% chlorine-35, at 34.96885 u, and 24.24% chlorine-37, at 36.96590 u. Find the average atomic mass, and say which isotope the answer sits nearer.

Step 1 · turn percents into fractions

75.76% becomes 0.7576 and 24.24% becomes 0.2424. Those two add to 1.0000, as they must.

Step 2 · weight each isotope

(0.7576)(34.96885 u) = 26.4889 u. (0.2424)(36.96590 u) = 8.9605 u.

Step 3 · add

26.4889 + 8.9605 = 35.4494 u, which is 35.45 u.

Step 4 · read it

The answer sits much nearer 35 than 37, because chlorine-35 makes up three quarters of the sample. A weighted average always leans toward the common isotope.

Answer and the check

35.45 u, which is the value printed on the periodic table. One good check is that the answer must land between the two isotope masses. If it lands outside that range, you multiplied by the wrong fraction.

Example T2.2 · One spectral line, two ways

Hydrogen shows a bright blue-green line at 486 nm. Find the photon's frequency and its energy, in joules and in kJ/mol. Then show that the line comes from the n = 4 to n = 2 drop.

Step 1 · put the wavelength in meters

486 nm × (1 m ÷ 109 nm) = 4.86 × 10-7 m.

Step 2 · use c = λν

ν = c ÷ λ = (2.998 × 108 m/s) ÷ (4.86 × 10-7 m) = 6.17 × 1014 Hz.

Step 3 · use E = hν

E = (6.626 × 10-34 J·s)(6.1687 × 1014 s-1) = 4.087 × 10-19 J per photon.

Step 4 · scale to a mole

(4.087 × 10-19 J)(6.022 × 1023/mol) = 2.46 × 105 J/mol, which is 246 kJ/mol.

Step 5 · now from Bohr's levels

ΔE = −(2.179 × 10-18 J)(1 ÷ 42 − 1 ÷ 22) = −(2.179 × 10-18)(0.0625 − 0.2500) = −(2.179 × 10-18)(−0.1875) = 4.086 × 10-19 J.

Answer

6.17 × 1014 Hz, 4.087 × 10-19 J per photon, and 246 kJ/mol. The two routes agree to three figures, so the line really is the n = 4 to n = 2 drop. The minus sign in Bohr's formula came out positive once the subtraction was done, which tells you energy left the atom as light.

Example T2.3 · From valence electrons to shape and polarity

Draw SO2, give its electron group count, its shape, its bond order and whether it is polar. Then do the same for CO2 and explain the difference.

Step 1 · count the valence electrons

Sulfur is group 16, so it brings 6. Each oxygen brings 6 as well. Total = 6 + 2(6) = 18 electrons, which is 9 pairs.

Step 2 · spend them

Put sulfur in the middle. One good structure has one S=O double bond, one S–O single bond, a lone pair on sulfur, and lone pairs filling both oxygens. Swapping which oxygen holds the double bond gives a second structure that is just as good, so SO2 is a resonance pair.

Step 3 · count electron groups and read the shape

Sulfur holds two bonded oxygens plus one lone pair, so three groups. Three groups spread to trigonal planar at 120°. Take the lone pair out of the picture and the shape is bent, with a measured angle near 119°.

Step 4 · bond order from the resonance

Three bonding pairs are spread over two bond positions, so the bond order is 3 ÷ 2 = 1.5. Both S–O bonds are the same length, which is what a blend has to give.

Step 5 · now CO2, for contrast

Carbon brings 4 and each oxygen 6, so 16 electrons. Two C=O double bonds use them all, with no lone pair on carbon. Two groups means linear at 180°.

Answer

SO2 is bent and polar. CO2 is linear and nonpolar. Both have polar bonds, so the bonds are not the deciding factor. In CO2 the two dipoles point in exactly opposite directions and cancel. In SO2 the lone pair bends the molecule, so they cannot cancel. Shape decides polarity, every time.

The traps people fall into

Watch for these
  • Averaging the isotope masses instead of weighting them. Multiply each one by its abundance first.
  • Confusing wavelength and frequency. They move in opposite directions, since their product is fixed at c.
  • Leaving the minus sign on a photon energy. ΔE tells you which way the energy went. The photon carries the size of it.
  • Counting a double bond as two electron groups. Each bond is one group, however many pairs it holds.
  • Ignoring lone pairs when reading a shape. They take up more room than bonds and squeeze the angles.
  • Calling a molecule polar because its bonds are. Symmetry can cancel every bond dipole, as in CO2 and CCl4.
  • Treating resonance forms as two real substances. The molecule is one blend of them, with one set of bond lengths.
  • Forgetting that only sigma bonds let the ends spin. A pi bond locks the geometry flat.
  • Expecting an ion to be the same size as its atom. Cations shrink and anions grow.

Self-test · Theme 2

1. How many protons, neutrons and electrons are in an 56Fe3+ ion? Iron is Z = 26.
26 protons, 30 neutrons, 23 electrons. Z gives the protons, so 26. Neutrons are 56 − 26 = 30. A 3+ charge means three electrons are missing, so 26 − 3 = 23. The charge never changes the nucleus.
2. Boron is 19.9% boron-10, at 10.0129 u, and 80.1% boron-11, at 11.0093 u. Find its average atomic mass.
10.81 u. (0.199)(10.0129) = 1.9926 u and (0.801)(11.0093) = 8.8185 u. Add them: 10.811 u. The answer leans toward boron-11, which is four fifths of the sample.
3. Name Fe2(SO4)3, and explain the subscripts.
Iron(III) sulfate. Sulfate is SO42-. Three of them carry 6−, so two iron ions must carry 6+, which makes each one Fe3+. Iron has more than one charge, so the Roman numeral is required. The formula is just the smallest ratio that cancels the charges.
4. Write the electron configuration of calcium, Z = 20, the long way and the short way.
1s22s22p63s23p64s2, or [Ar]4s2. Count the electrons as you go: 2 + 2 + 6 + 2 + 6 + 2 = 20. Argon covers the first 18, so the short form says the same thing in four characters. The two outer 4s electrons are the ones calcium loses to become Ca2+.
5. A photon has a frequency of 4.50 × 1014 Hz. Find its wavelength in nanometers.
666 nm. λ = c ÷ ν = (2.998 × 108 m/s) ÷ (4.50 × 1014 s-1) = 6.66 × 10-7 m. Multiply by 109 nm per metre to get 666 nm, which is red light.
6. How much energy does it take to pull the electron right off a hydrogen atom in its lowest level? Give it per mole.
1312 kJ/mol. The electron starts at n = 1 and ends free, which is n at infinity where the energy is 0. So ΔE = 0 − [−(2.179 × 10-18) ÷ 12] = 2.179 × 10-18 J per atom. Times Avogadro's number: (2.179 × 10-18)(6.022 × 1023) = 1.312 × 106 J/mol = 1312 kJ/mol.
7. Which is larger, a sodium atom or a sodium ion? Why?
The atom, by a lot. Sodium's one outer 3s electron is the only thing in that shell. Lose it and the whole shell is gone, so the ion is just the neon core. The remaining electrons also feel the same 11 protons with fewer of them to share the pull, so they are drawn in closer still.
8. In the common Lewis structure of NO3-, nitrogen makes four bonds and holds no lone pair. Find its formal charge.
+1. Nitrogen is group 15, so it brings 5 valence electrons. Formal charge = 5 − 0 lone pair electrons − 4 bonds = +1. The three oxygens then carry charges that add with this one to give the ion's total of 1−.
9. Give the electron group count, the shape, the angle and the polarity of NH3.
Four groups, trigonal pyramidal, near 107°, and polar. Nitrogen holds three bonded hydrogens plus one lone pair, so four groups spread toward a tetrahedron. Dropping the lone pair from the picture leaves a pyramid. The lone pair pushes harder than a bond, which squeezes 109.5° down to about 107°. The shape is lopsided, so the dipoles cannot cancel.
10. Give the bond order of O2, O2+ and O2-, and rank their bond lengths.
2, 2.5 and 1.5, so O2+ has the shortest bond and O2- the longest. Removing an electron from O2 takes it out of an antibonding level, which raises the bond order by half. Adding one fills an antibonding level instead and lowers the bond order by half. Higher bond order always means a shorter, stronger bond, so the length runs O2+ < O2 < O2-.

Theme 3 · Chapters 9, 10 and 11States and solutions

What to know

Gas pressure is particles drumming on a wall. More particles, faster particles or a smaller box all raise it. The units all describe the same thing, and 1 atm = 760 torr = 760 mmHg = 101.325 kPa.

The simple gas laws each hold two things still and watch two others trade. Volume rises with temperature, falls as pressure rises, and rises with the amount of gas. The combined gas law packs all of that into one relation for a fixed amount of gas. The ideal gas law then ties all four quantities together at once, and it is the only gas equation you really have to remember.

STP is 0 °C and 1 atm. Put one mole into the ideal gas law at STP and out comes the molar volume, 22.4 L/mol, whatever the gas. So at STP the density of a gas is just its molar mass divided by 22.4.

In a mixture each gas pushes on the wall as if it were alone. Dalton's law adds those partial pressures to get the total, and each gas's share of the pressure equals its share of the moles. Light gases also escape through a pinhole faster than heavy ones, which is Graham's law: the rate goes as one over the square root of the molar mass.

Real gases stray from ideal at high pressure and low temperature, because particles do take up room and do attract each other. The van der Waals constants a and b correct for those two effects.

Intermolecular forces, weakest to strongest
  • Dispersion forces act between all molecules, and they grow with the number of electrons.
  • Dipole-dipole forces act between polar molecules, lining positive ends to negative ends.
  • Hydrogen bonding is a strong special case, needing H bonded to N, O or F.
  • Ion-dipole forces act when an ion sits in a polar solvent, which is what dissolving a salt looks like.
  • Stronger forces mean higher boiling point, higher surface tension, higher viscosity and lower vapor pressure.

Heat a solid and a heating curve shows three slopes and two flats. On a slope only the temperature changes, so the heat is mass times specific heat times the temperature change. On a flat only the state changes, so the temperature holds while every joule goes into breaking the forces. The heat for a flat is the mass times the heat of fusion or of vaporization.

Vapor pressure is the push of the vapor sitting above its own liquid in a closed container. It climbs steeply with temperature, and a liquid boils when its vapor pressure reaches the pressure above it. The Clausius-Clapeyron relation puts numbers on that climb. A phase diagram maps which state is stable at each pressure and temperature, with the triple point where all three meet and the critical point where liquid and gas stop being different.

Dissolving takes three steps. Pull the solute apart, open a hole in the solvent, then let the two mix. The first two cost energy and the third pays it back, so the net can go either way. Like dissolves like, since polar solvents handle polar and ionic solutes while nonpolar solvents handle nonpolar ones.

Concentration has several measures and each has its own use. Molarity is per liter of solution and shifts a little with temperature, because volume does. Molality is per kilogram of solvent and never shifts. Mass percent and mole fraction are pure ratios. Density is the bridge between a mass-based measure and a volume-based one.

Colligative properties depend on how many solute particles there are and not on what they are. Solute lowers the vapor pressure, so the boiling point rises and the freezing point falls. The van't Hoff factor i counts the particles each formula unit gives, so NaCl counts twice and CaCl2 counts three times. Osmotic pressure follows the same logic and is the most sensitive of the four.

Equations, units and symbols

EquationWhat each symbol meansUnits
1 atm = 760 torr = 101.325 kPa = 1.01325 bar The same pressure in four unit systems. Convert before you use any gas law. atm, torr, kPa, bar.
P1V1 ÷ T1 = P2V2 ÷ T2 The combined gas law, for a fixed amount of gas. Drop whatever is held constant. Any matching P and V units. T must be in kelvin.
PV = nRT The ideal gas law. n is moles, and the other three are pressure, volume and temperature. R = 0.08206 L·atm/mol·K, so P in atm, V in L, T in K.
M = dRT ÷ P Molar mass from a gas density, from putting n = m ÷ M into the ideal gas law. d in g/L gives M in g/mol.
At STP, molar volume = 22.4 L/mol STP is 0 °C and 1 atm, so density at STP is molar mass divided by 22.4. L/mol, and g/L for the density.
Ptotal = ΣPi, and Pi = XiPtotal Dalton's law. Xi is the mole fraction, which is that gas's moles over the total moles. Any pressure unit. X is a bare fraction.
rateA ÷ rateB = √(MB ÷ MA) Graham's law for effusion. Note which molar mass goes on top: the other gas's. A bare ratio.
q = mcΔT Heat on a slope of a heating curve. ΔT is always final minus initial. c in J/g·°C. For water: ice 2.09, liquid 4.184, steam 1.84. q in J.
q = nΔHfus or nΔHvap Heat on a flat of a heating curve, where the temperature holds still. For water, ΔHfus = 6.01 kJ/mol and ΔHvap = 40.7 kJ/mol.
ln(P2 ÷ P1) = −(ΔHvap ÷ R)(1 ÷ T2 − 1 ÷ T1) Clausius-Clapeyron. Two pressures and two temperatures, so you can solve for any one of them. R = 8.314 J/mol·K, so ΔHvap in J/mol and T in K.
m = moles of solute ÷ kilograms of solvent Molality. Solvent, not solution, and kilograms, not liters. mol/kg, written m.
ΔTb = iKbm and ΔTf = iKfm Boiling point rise and freezing point drop. i is the particles per formula unit. For water, Kb = 0.512 °C/m and Kf = 1.86 °C/m.
Π = iMRT Osmotic pressure. This one uses molarity, not molality. R = 0.08206 L·atm/mol·K, T in K, Π in atm.

Worked examples

Example T3.1 · Molar mass of an unknown gas

A 0.500 L flask holds 1.25 g of a gas at 25 °C and 0.980 atm. Find the molar mass. Then find the gas density at STP.

Step 1 · fix the units

Volume is already in liters and pressure in atm. Temperature must be kelvin: T = 25 + 273.15 = 298.15 K.

Step 2 · find the moles

n = PV ÷ RT. The bottom is (0.08206 L·atm/mol·K)(298.15 K) = 24.47 L·atm/mol. The top is (0.980 atm)(0.500 L) = 0.490 L·atm. So n = 0.490 ÷ 24.47 = 0.02003 mol.

Step 3 · grams over moles

M = 1.25 g ÷ 0.02003 mol = 62.4 g/mol.

Step 4 · density at STP

At STP one mole of any gas fills 22.4 L, so d = 62.4 g/mol ÷ 22.4 L/mol = 2.79 g/L.

Answer and the check

62.4 g/mol, with a density of 2.79 g/L at STP. Check the size of the answer. Air averages about 29 g/mol and has a density near 1.3 g/L at STP, so this gas is roughly twice as heavy as air. That fits a molar mass near 62.

Example T3.2 · A whole heating curve

How much heat takes 50.0 g of ice at −15.0 °C all the way to steam at 115.0 °C? Use c = 2.09 for ice, 4.184 for water and 1.84 for steam, all in J/g·°C, with ΔHfus = 6.01 kJ/mol and ΔHvap = 40.7 kJ/mol. Water is 18.02 g/mol.

Step 1 · get the moles for the two flats

50.0 g ÷ 18.02 g/mol = 2.775 mol.

Step 2 · warm the ice to 0 °C

q = (50.0 g)(2.09 J/g·°C)(15.0 °C) = 1570 J = 1.57 kJ.

Step 3 · melt it, with no temperature change

q = (2.775 mol)(6.01 kJ/mol) = 16.68 kJ.

Step 4 · warm the water to 100 °C

q = (50.0 g)(4.184 J/g·°C)(100.0 °C) = 20,920 J = 20.92 kJ.

Step 5 · boil it, then warm the steam

Boiling: q = (2.775 mol)(40.7 kJ/mol) = 112.93 kJ. Warming the steam: q = (50.0 g)(1.84 J/g·°C)(15.0 °C) = 1380 J = 1.38 kJ.

Answer and what it tells you

Add the five pieces: 1.57 + 16.68 + 20.92 + 112.93 + 1.38 = 153 kJ. Look at the shares. Boiling alone takes 113 kJ, which is about three quarters of the whole bill. Pulling molecules out of a liquid into a gas costs far more than heating them ever does, and that is why steam burns are so bad.

Example T3.3 · Antifreeze, road salt and osmosis

Dissolve 93.0 g of ethylene glycol, 62.07 g/mol, in 500.0 g of water. Find the new freezing and boiling points. Then compare 8.77 g of NaCl, 58.44 g/mol, in the same water. Finish with the osmotic pressure of 0.300 M NaCl at 25 °C.

Step 1 · moles and molality for the glycol

93.0 g ÷ 62.07 g/mol = 1.498 mol. The solvent is 0.5000 kg, so m = 1.498 ÷ 0.5000 = 3.00 m.

Step 2 · the two shifts

Glycol is a nonelectrolyte, so i = 1. ΔTf = (1)(1.86 °C/m)(3.00 m) = 5.58 °C and ΔTb = (1)(0.512 °C/m)(3.00 m) = 1.54 °C.

Step 3 · read the new points

Freezing: 0.00 − 5.58 = −5.58 °C. Boiling: 100.00 + 1.54 = 101.54 °C. The freezing shift is much larger, because Kf is more than three times Kb.

Step 4 · now the road salt

8.77 g ÷ 58.44 g/mol = 0.150 mol, so m = 0.150 ÷ 0.5000 = 0.300 m. But NaCl splits into two ions, so i = 2. ΔTf = (2)(1.86)(0.300) = 1.12 °C, giving −1.12 °C.

Step 5 · osmotic pressure

Π = iMRT = (2)(0.300 mol/L)(0.08206 L·atm/mol·K)(298.15 K). The last two give 24.47, so Π = (2)(0.300)(24.47) = 14.7 atm.

Answer

The glycol solution freezes at −5.58 °C and boils at 101.54 °C. The salt solution freezes at only −1.12 °C, because there is ten times less of it by moles, even though each unit counts twice. And the same 0.300 M salt solution pushes 14.7 atm across a membrane, which is about 15 times normal air pressure. That is why osmotic pressure is the most sensitive of the four properties.

The traps people fall into

Watch for these
  • Leaving temperature in Celsius. Every gas law needs kelvin, and a Celsius value can even be negative.
  • Using the wrong R. Use 0.08206 L·atm/mol·K with atm and liters, and 8.314 J/mol·K with energy.
  • Using 22.4 L/mol away from STP. It is the molar volume at 0 °C and 1 atm only.
  • Flipping Graham's law. The lighter gas is faster, so the other gas's molar mass goes on top.
  • Using q = mcΔT on a flat. The temperature does not change there, so the heat comes from the phase change instead.
  • Forgetting a step in a heating curve. Ice to steam is five pieces, and the two flats are the biggest.
  • Swapping molarity and molality. Freezing and boiling shifts use molality. Osmotic pressure uses molarity.
  • Forgetting i. NaCl gives 2 particles and CaCl2 gives 3, so each shift is multiplied.
  • Thinking a colligative property cares which solute it is. It only counts particles.

Self-test · Theme 3

1. Convert 745 torr to atmospheres and to kilopascals.
0.980 atm and 99.3 kPa. 745 torr ÷ 760 torr/atm = 0.980 atm. For kPa, 745 torr × (101.325 kPa ÷ 760 torr) = 99.3 kPa. Both answers sit just under the 1 atm values, which is what a reading just under 760 torr should give.
2. A gas fills 2.00 L at 1.00 atm and 273 K. What volume does it fill at 2.00 atm and 546 K?
2.00 L, unchanged. Rearranged, V2 = V1(P1 ÷ P2)(T2 ÷ T1) = (2.00 L)(1.00 ÷ 2.00)(546 ÷ 273) = (2.00)(0.500)(2.00) = 2.00 L. Doubling the pressure halves the volume and doubling the temperature doubles it, so the two effects cancel exactly.
3. How many moles of gas fill 5.00 L at 2.00 atm and 300. K?
0.406 mol. n = PV ÷ RT. The top is (2.00 atm)(5.00 L) = 10.0 L·atm. The bottom is (0.08206)(300.) = 24.62 L·atm/mol. So n = 10.0 ÷ 24.62 = 0.406 mol.
4. Find the density of CO2 at STP. Its molar mass is 44.01 g/mol.
1.96 g/L. At STP every gas has a molar volume of 22.4 L/mol, so d = 44.01 g/mol ÷ 22.4 L/mol = 1.96 g/L. Air averages near 1.3 g/L, so carbon dioxide sinks, which is how a dry ice fog hugs the floor.
5. A flask holds nitrogen at 0.600 atm and oxygen at 0.250 atm. Find the total pressure and the mole fraction of nitrogen.
0.850 atm, and X = 0.706. Dalton's law adds the partial pressures: 0.600 + 0.250 = 0.850 atm. Each gas's share of the pressure is its share of the moles, so X = 0.600 ÷ 0.850 = 0.706. The two fractions add to 1, since 0.706 + 0.294 = 1.000.
6. How much faster does helium effuse than neon? He is 4.003 g/mol and Ne is 20.18 g/mol.
2.25 times faster. Graham's law puts the other gas's mass on top: rateHe ÷ rateNe = √(20.18 ÷ 4.003) = √5.041 = 2.25. Neon is about five times heavier, and the square root turns that into a factor of 2.25 in speed.
7. How much heat warms 125 g of water from 22.0 °C to 98.0 °C?
39.7 kJ. ΔT = 98.0 − 22.0 = 76.0 °C. Then q = (125 g)(4.184 J/g·°C)(76.0 °C) = 39,748 J, which is 39.7 kJ. Note that boiling that same water away would cost about 282 kJ, seven times more.
8. Water's vapor pressure is 760 torr at 373.15 K, and ΔHvap = 40.7 kJ/mol. Estimate it at 353.15 K.
About 360 torr. First ΔHvap ÷ R = 40,700 ÷ 8.314 = 4895 K. Next 1 ÷ 353.15 − 1 ÷ 373.15 = 0.00283244 − 0.00267989 = 1.5255 × 10-4. Then ln(P2 ÷ 760) = −(4895)(1.5255 × 10-4) = −0.7467, so P2 ÷ 760 = e-0.7467 = 0.474 and P2 = 360 torr. The measured value is 355 torr, so two table numbers landed within 2%.
9. A brine is 12.0% NaCl by mass with a density of 1.086 g/mL. Find its molarity and its molality.
2.23 M and 2.33 m. Take 1.000 L, which weighs 1086 g. The salt is 0.120 × 1086 = 130.3 g, or 130.3 ÷ 58.44 = 2.230 mol, so the molarity is 2.23 M. The water is 1086 − 130.3 = 955.7 g = 0.9557 kg, so the molality is 2.230 ÷ 0.9557 = 2.33 m. Molality is the larger number, because a kilogram of solvent is less than a liter of solution.
10. Find the freezing point of 0.200 m CaCl2 in water.
−1.12 °C. Each CaCl2 unit gives one Ca2+ and two Cl-, so i = 3. Then ΔTf = (3)(1.86 °C/m)(0.200 m) = 1.12 °C, and the freezing point is 0.00 − 1.12 = −1.12 °C. Using i = 1 would have given only −0.372 °C, which is why calcium chloride beats sugar on an icy step.

Theme 4 · Chapters 4, 5 and 12Change and rate

What to know

A chemical equation is balanced when every element has the same atom count on both sides. You balance it by changing coefficients only, never subscripts, because changing a subscript changes the substance. Balance the complicated formula first, leave free elements for last, and double everything if a half coefficient turns up.

Most reactions fall into a few families. Combination joins two things into one. Decomposition splits one into two. Single replacement swaps a free element into a compound. Double replacement trades partners, and a precipitation or a neutralization is that pattern. Combustion burns a fuel in oxygen to give carbon dioxide and water.

A redox reaction moves electrons, and you track that with oxidation numbers. A free element is 0. A monatomic ion carries its charge. Oxygen is usually −2 and hydrogen usually +1. The numbers must add to the overall charge on the species. If any number changes across the arrow, the reaction is redox.

The coefficients of a balanced equation are a mole ratio, and that ratio is the only way to cross from one substance to another. So every stoichiometry problem has the same shape. Convert what you are given into moles, cross the arrow with the coefficient ratio, then convert out to whatever the question asks for.

Limiting reactant and yield
  • The limiting reactant runs out first and sets how much product you can get.
  • Find it by taking each reactant all the way to the product. The smallest answer wins.
  • Comparing grams or even moles directly is not enough, because the coefficients differ.
  • The theoretical yield is what the limiting reactant allows. The actual yield is what you weigh.
  • Percent yield is actual over theoretical. Over 100% means a wet or impure product, not magic.

Energy comes in two forms. Kinetic energy is motion and potential energy is stored position. Heat is energy moving because of a temperature gap, and work is energy moving because something was pushed. A system gives off heat in an exothermic change and takes heat in during an endothermic one.

Specific heat is the heat it takes to raise one gram by one degree. Water's value of 4.184 J/g·°C is unusually large, which is why oceans steady the climate and why a water bath is so stable. A calorimeter uses that fact in reverse. Measure the temperature change of a known mass of water and you know the heat the reaction released, with the sign flipped.

Enthalpy change at steady pressure, written ΔH, is the heat of a reaction. It scales with the amount, so doubling the equation doubles ΔH, and reversing the equation flips its sign. Hess's law is the payoff. If several steps add up to your target reaction, their ΔH values add up too, so you can flip, scale and stack known reactions to reach a reaction nobody ever measured.

A standard enthalpy of formation is the ΔH for making one mole of a substance from its elements in their standard states, so it is zero for an element in that form. Then any reaction's ΔH° is the sum over products minus the sum over reactants, each term times its coefficient. Bond energies give a rougher second route, since breaking bonds costs and forming them pays.

Kinetics asks how fast. A rate law links the rate to concentrations through a rate constant and a set of exponents, and those exponents are found by experiment, not read off the equation. The only exception is an elementary step, where the coefficients do give the rate law. Add the exponents and you have the overall order.

An integrated rate law turns a rate into a clock. Each order has its own form, and each form is already the shape of a straight line. So plot the data three ways and only one plot comes out straight. That plot names the order and its slope gives k.

Collision theory says a reaction needs a crash that is hard enough and aimed correctly. The activation energy is the least energy a crash needs, and the top of that hill is the transition state. Note that activation energy is not ΔH. The ΔH is the gap from start to finish, while the activation energy is the height of the hill in between. The Arrhenius equation ties the rate constant to that barrier and the temperature, so two rate constants at two temperatures give you the barrier.

A mechanism is the list of elementary steps a reaction really takes. The slowest step sets the pace, so its rate law is usually the measured one, which is how a mechanism gets tested. A catalyst opens a path with a lower barrier and is remade at the end. It speeds both directions by the same factor, so it changes neither ΔH nor the position of equilibrium.

Equations, units and symbols

EquationWhat each symbol meansUnits
moles of B = moles of A × (coefficient of B ÷ coefficient of A) The only bridge across the arrow. Everything else is a unit conversion. mol to mol.
percent yield = (actual yield ÷ theoretical yield) × 100% Both yields in the same unit, usually grams. The theoretical yield comes from the limiting reactant. A percent.
q = mcΔT m is mass, c is specific heat, and ΔT is final minus initial. c in J/g·°C, so q in J. Water is 4.184 J/g·°C.
qreaction = −qsolution Calorimetry. Heat the solution gained is heat the reaction lost, so the sign flips. J, or kJ after dividing by 1000.
ΔH = qreaction ÷ moles Turns a measured heat into a per-mole value you can compare with a table. kJ/mol.
ΔH°rxn = ΣnΔH°f(products) − ΣnΔH°f(reactants) n is each coefficient. ΔH°f is zero for an element in its standard state. kJ/mol in the table, kJ in the answer.
ΔH ≈ Σ(bonds broken) − Σ(bonds formed) Breaking costs and forming pays. This is an estimate, since a tabulated bond energy is an average. kJ/mol for each bond, kJ for the answer.
rate = k[A]m[B]n m and n come from experiment. Their sum is the overall order. k depends on temperature. rate in M/s. The unit of k depends on the order.
Zero order: [A] = [A]0 − kt [A] against t is straight, with slope −k. Its half-life is [A]0 ÷ 2k. k in M/s.
First order: ln[A] = ln[A]0 − kt ln[A] against t is straight, with slope −k. Its half-life is 0.693 ÷ k and does not depend on the start. k in s-1.
Second order: 1 ÷ [A] = 1 ÷ [A]0 + kt 1 ÷ [A] against t is straight, with slope +k. Its half-life is 1 ÷ k[A]0. k in M-1s-1.
k = Ae−Ea/RT The Arrhenius equation. A bundles collision frequency with correct aim. R = 8.314 J/mol·K, so Ea in J/mol and T in K.
ln(k2 ÷ k1) = −(Ea ÷ R)(1 ÷ T2 − 1 ÷ T1) Two rate constants at two temperatures are enough to get the barrier. Ea in J/mol. Divide by 1000 for kJ/mol.

Worked examples

Example T4.1 · Limiting reactant and percent yield

You feed 28.0 g of N2 and 6.00 g of H2 into a reactor running N2 + 3 H2 → 2 NH3. You collect 28.5 g of ammonia. Find the limiting reactant, the theoretical yield and the percent yield. Use 28.01 for N2, 2.016 for H2 and 17.03 for NH3, all in g/mol.

Step 1 · moles of each reactant

Nitrogen: 28.0 ÷ 28.01 = 0.9996 mol. Hydrogen: 6.00 ÷ 2.016 = 2.976 mol.

Step 2 · take each one to the product

From nitrogen: (0.9996 mol)(2 ÷ 1) = 1.999 mol NH3. From hydrogen: (2.976 mol)(2 ÷ 3) = 1.984 mol NH3.

Step 3 · the smaller answer wins

1.984 is smaller than 1.999, so hydrogen is limiting and runs out first. It is close, which is exactly why you cannot judge this by eye.

Step 4 · the theoretical yield

(1.984 mol)(17.03 g/mol) = 33.8 g of ammonia.

Step 5 · the percent yield

(28.5 g ÷ 33.8 g) × 100% = 84.3%.

Answer and the check

Hydrogen is limiting, the theoretical yield is 33.8 g and the actual run reached 84.3% of it. Check that the percent is under 100%, because anything over means a weighing problem or a wet product. Note that nitrogen came within one percent of running out too, so almost nothing is left over.

Example T4.2 · A calorimeter, then a table

Mix 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH, both starting at 22.0 °C. The mixture reaches 28.9 °C. Find ΔH per mole of water formed. Take the density as 1.00 g/mL and c as 4.184 J/g·°C. Then get ΔH° for burning methane from a table of formation values.

Step 1 · heat the solution took in

The mixture weighs 100.0 g and warmed by 28.9 − 22.0 = 6.9 °C. So q = (100.0 g)(4.184 J/g·°C)(6.9 °C) = 2887 J.

Step 2 · flip the sign

That heat came out of the reaction, so qreaction = −2887 J. The reaction is exothermic, which is why the thermometer rose.

Step 3 · count the moles

(0.0500 L)(1.00 mol/L) = 0.0500 mol of each, and they trade one for one, so 0.0500 mol of water formed.

Step 4 · go per mole

ΔH = (−2887 J) ÷ (0.0500 mol) = −57,700 J/mol, which is −57.7 kJ/mol.

Step 5 · now the table route

For CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), use ΔH°f of −74.6 for CH4, −393.5 for CO2 and −285.8 for H2O(l), all in kJ/mol. Oxygen is an element, so it is 0. Products: −393.5 + 2(−285.8) = −965.1 kJ. Reactants: −74.6 + 0 = −74.6 kJ. Subtract: −965.1 − (−74.6) = −890.5 kJ.

Answer

Neutralization gives about −57.7 kJ per mole of water, close to the accepted value near −57 kJ/mol. Burning one mole of methane gives −890.5 kJ. Both signs are negative, so both release heat. Watch the double minus in the last subtraction, since dropping it would turn −890.5 into −1040 kJ.

Example T4.3 · A rate law, a half-life and a barrier

Three runs are measured. Doubling [A] makes the rate four times larger. Doubling [B] changes nothing. Write the rate law. Then find the half-life for a first-order decay with k = 2.50 × 10-3 s-1. Finally, find Ea from k = 2.50 × 10-3 s-1 at 300. K and k = 3.80 × 10-2 s-1 at 330. K.

Step 1 · read the orders off the data

Doubling [A] multiplied the rate by 4, and 22 = 4, so the order in A is 2. Doubling [B] multiplied it by 1, and 20 = 1, so the order in B is 0.

Step 2 · write the law

rate = k[A]2. The overall order is 2 + 0 = 2. B appears in the equation and not in the rate law, which is normal and only experiment could have told you.

Step 3 · the first-order half-life

t1/2 = 0.693 ÷ k = 0.693 ÷ (2.50 × 10-3 s-1) = 277 s. It does not depend on the starting concentration, which is what makes first order special.

Step 4 · set up the Arrhenius pair

k2 ÷ k1 = (3.80 × 10-2) ÷ (2.50 × 10-3) = 15.2, and ln(15.2) = 2.721.

Step 5 · solve for the barrier

1 ÷ 300. − 1 ÷ 330. = 0.0033333 − 0.0030303 = 3.030 × 10-4 K-1. Then Ea = R × ln(k2 ÷ k1) ÷ that difference = (8.314)(2.721) ÷ (3.030 × 10-4) = 22.62 ÷ (3.030 × 10-4) = 7.47 × 104 J/mol.

Answer

rate = k[A]2, a half-life of 277 s, and Ea = 74.7 kJ/mol. Check the size of that barrier. A 30 K rise multiplied the rate 15 times, and a barrier near 75 kJ/mol is exactly the kind that reacts this hard to a modest temperature change.

The traps people fall into

Watch for these
  • Changing a subscript to balance. That makes a different substance. Only coefficients may move.
  • Comparing grams to find the limiting reactant. Take each reactant all the way to the product and compare those answers.
  • Reporting a percent yield over 100%. Check the arithmetic, then suspect a wet or impure sample.
  • Getting ΔT backward. It is always final minus initial, so a cooling sample gives a negative value.
  • Keeping the sign in calorimetry. The heat the water gained is the heat the reaction lost, so the sign flips.
  • Losing a double minus in a formation sum. Subtracting a negative reactant total raises the answer.
  • Reading rate law exponents off the balanced equation. They come from experiment, except for an elementary step.
  • Confusing Ea with ΔH. One is the hill, the other is the drop from start to finish.
  • Using kJ with R = 8.314 J/mol·K. Put the barrier in joules per mole first.
  • Thinking a catalyst raises the yield. It lowers the barrier and leaves ΔH and the equilibrium position alone.

Self-test · Theme 4

1. Balance Fe + O2 → Fe2O3.
4 Fe + 3 O2 → 2 Fe2O3. Start with 1 Fe2O3 and you need 1.5 O2, so double everything. Then iron is 4 and 4, and oxygen is 6 and 6. Leave the free element for last, since changing its coefficient disturbs nothing else.
2. Find the oxidation number of chromium in Cr2O72-.
+6. Each oxygen is −2, and there are seven, so the oxygens add to −14. The ion's total must be −2, so the two chromiums must add to +12. Then 12 ÷ 2 = +6 each. Notice that +6 is not a charge you would ever see on a free chromium ion. An oxidation number is bookkeeping, not a real charge.
3. What kind of reaction is AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)?
A double replacement, and specifically a precipitation. The two pairs trade partners. What drives it is that silver chloride is barely soluble, so it drops out as a solid and leaves the solution for good. No oxidation number changed, so this is not a redox reaction.
4. How many moles of O2 burn 2.50 mol of propane? C3H8 + 5 O2 → 3 CO2 + 4 H2O.
12.5 mol. The ratio is 5 moles of oxygen for every 1 mole of propane, so (2.50 mol)(5 ÷ 1) = 12.5 mol. That is why a grill tank needs so much air, and why burning it indoors is dangerous.
5. You have 10.0 g of H2 and 64.0 g of O2 for 2 H2 + O2 → 2 H2O. Which runs out first, and how much water forms?
Oxygen is limiting, and 72.1 g of water forms. Moles: 10.0 ÷ 2.016 = 4.96 mol H2 and 64.0 ÷ 32.00 = 2.00 mol O2. Take each to the product: hydrogen gives (4.96)(2 ÷ 2) = 4.96 mol, oxygen gives (2.00)(2 ÷ 1) = 4.00 mol. The smaller wins, so 4.00 mol of water, which is (4.00)(18.02) = 72.1 g. Hydrogen had the larger mole count and still was not limiting.
6. A run has a theoretical yield of 45.0 g and an actual yield of 38.2 g. Find the percent yield.
84.9%. (38.2 g ÷ 45.0 g) × 100% = 84.9%. The missing 6.8 g went to side reactions, stayed stuck to the glass or never reacted at all. Any value under 100% is normal, and anything over it needs explaining.
7. How much heat warms 250. g of water from 20.0 °C to 75.0 °C?
57.5 kJ. ΔT = 75.0 − 20.0 = 55.0 °C. Then q = (250. g)(4.184 J/g·°C)(55.0 °C) = 57,530 J = 57.5 kJ. Water's large specific heat is why this takes so much more energy than the same mass of metal would.
8. Given A → B with ΔH = −50 kJ and B → C with ΔH = +20 kJ, find ΔH for A → C.
−30 kJ. The two steps add to the target, since the B cancels, so their enthalpies add too: −50 + 20 = −30 kJ. That is Hess's law. If the question had asked for C → A, you would flip the sign and answer +30 kJ.
9. Estimate ΔH for H2 + Cl2 → 2 HCl from bond energies of 436 for H–H, 242 for Cl–Cl and 431 for H–Cl, all in kJ/mol.
About −184 kJ. Bonds broken: 436 + 242 = 678 kJ, which costs energy. Bonds formed: 2 × 431 = 862 kJ, which pays it back. Then ΔH ≈ 678 − 862 = −184 kJ. The measured value is near −185 kJ, so an estimate from averages lands within 1%.
10. A first-order reaction has k = 0.0231 min-1. Find the half-life, and the fraction left after 90.0 min.
30.0 min, and 12.5% left. t1/2 = 0.693 ÷ 0.0231 min-1 = 30.0 min. Then 90.0 ÷ 30.0 = 3.00 half-lives, so halve three times: 100% to 50% to 25% to 12.5%. The log form agrees, since kt = (0.0231)(90.0) = 2.079 and e-2.079 = 0.125.

Theme 5 · Chapters 13, 14 and 15Equilibrium systems

What to know

At dynamic equilibrium the forward and reverse rates are equal, so the measured amounts hold still while molecules keep reacting both ways. Equal rates do not mean equal amounts. The container must be closed, and you reach the same resting mixture from either side at a given temperature.

The equilibrium constant K is products over reactants, each amount raised to its coefficient from the balanced equation. Pure solids, pure liquids and solvent water stay out, since their crowding cannot change. K carries no units and depends only on temperature. The reaction quotient Q is the same fraction filled with what you have right now, and comparing the two gives the direction: below K it runs forward, above K it runs in reverse.

Le Chatelier's principle is that same test in a hurry. Add a species and the system shifts away from it. Squeeze a gas mixture and it shifts toward the side with fewer moles of gas. Heating shifts an endothermic reaction forward and raises K, and temperature is the only stress that changes K at all. A catalyst and an unreactive gas change nothing about the position.

An ICE table organizes the algebra, with each Change entry equal to the coefficient times x. When K is tiny you may replace (c0 − x) with c0, as long as c0 ÷ K clears about 400 and x ends under five percent of c0. Sometimes both sides are perfect squares, and then a square root beats the quadratic formula.

Acid-base chemistry is one proton changing hands. An acid donates it, a base accepts it, and a conjugate pair differs by exactly one H+. Water does both jobs, so it sets the middle of the scale. The ion-product constant Kw locks the two water ions together at 1.0 × 10-14 at 25 °C, so pushing one up forces the other down and neither can reach zero.

Strong, weak, and the salts they leave behind
  • Strong acids and bases ionize fully, so the pH comes straight from the concentration.
  • A weak acid or base reaches an equilibrium, scored by Ka or Kb, with most material intact.
  • Ka × Kb = Kw for any conjugate pair, so a weaker acid has a stronger conjugate base.
  • Percent ionization climbs as a weak acid gets more dilute, even while the pH rises.
  • A salt is neutral only when both ions came from strong partners. Otherwise the ion from the weak partner reacts.
  • A polyprotic acid loses protons in steps, and Ka1 sets the pH almost by itself.

A buffer holds a weak acid and its conjugate base together in real amounts, so added acid or base is soaked up and the pH barely moves. The Henderson-Hasselbalch equation reads a buffer pH off a ratio, and a buffer works best within about one pH unit of its pKa. On a titration curve, the half-equivalence point reads pKa directly, and a weak acid's equivalence point lands above pH 7 because the salt left behind is a base.

Solubility is equilibrium too. Ksp is the constant for a solid breaking into its ions, and the solid is never in the expression. The bridge from Ksp to molar solubility is the salt's ion ratio, so a 1:1 salt gives s2, a 1:2 or 2:1 salt gives 4s3, and a 3:2 salt gives 108s5. Because of those exponents, two Ksp values can be compared head to head only when the ratios match.

A shared ion pushes the dissolving equilibrium back toward solid, which is the common ion effect. To predict a precipitate, dilute every ion for the combined volume, build Q, and compare it with Ksp. A ligand with a large formation constant strips the metal ion out and dissolves a solid water could not touch. Acid does the same job from the anion side, but only for salts whose anion is the conjugate base of a weak acid. For coupled equilibria, add the equations and multiply the constants, flipping a constant to 1 ÷ K whenever you flip its equation.

Equations, units and symbols

EquationWhat each symbol meansUnits
Kc = products over reactants, each raised to its coefficient Square brackets are molarity at equilibrium. Solids, pure liquids and solvent water are left out. [ ] in mol/L. K has no units.
Kp = Kc(RT)Δn Δn is gas moles on the right minus gas moles on the left. Only gases count. R = 0.08206 L·atm/mol·K, T in K.
Q, built the same way from current amounts Q < K runs forward, Q > K runs in reverse, Q = K sits still. No units.
Kw = [H3O+][OH-] = 1.0 × 10-14 Holds in any water solution at 25 °C, so pH + pOH = 14.00. [ ] in mol/L.
pH = −log[H3O+], and [H3O+] = 10-pH Only the digits after the decimal point are significant figures. pH is a bare number.
[H3O+] = √(Kac0), and [OH-] = √(Kbc0) The weak acid and weak base shortcuts, good while x stays under 5% of c0. mol/L.
Ka × Kb = Kw For a conjugate pair, so pKa + pKb = 14.00 at 25 °C. No units.
pH = pKa + log([A-] ÷ [HA]) Henderson-Hasselbalch. Moles work as well as concentrations, since both share one volume. Bare numbers.
Ksp = s2, 4s3, or 108s5 By ion ratio: 1:1, then 1:2 or 2:1, then 3:2. s is molar solubility. s in mol/L.
Kf = [MLn] ÷ ([M][L]n) Formation constant for a complex ion. Its reverse is Kd = 1 ÷ Kf. No units.
Koverall = K1 × K2 Add the two equations and multiply the constants. Flip an equation and its constant inverts. No units.

Worked examples

Example T5.1 · An ICE table with a perfect square

Hydrogen iodide breaks up: 2 HI(g) ⇌ H2(g) + I2(g), with Kc = 0.0200 at 731 K. Start with 0.100 M HI and find all three concentrations.

Step 1 · build the table

Initial: 0.100, 0, 0. Change: −2x, +x, +x. Equilibrium: 0.100 − 2x, x, x. The coefficient of 2 on HI is why its change is 2x.

Step 2 · substitute

Kc = [H2][I2] ÷ [HI]2 = x2 ÷ (0.100 − 2x)2 = 0.0200.

Step 3 · spot the perfect squares

Both the top and the bottom are squares, so take the square root of both sides. x ÷ (0.100 − 2x) = √0.0200 = 0.1414.

Step 4 · solve the line

x = 0.01414 − 0.2828x, so 1.2828x = 0.01414 and x = 0.01102 M.

Step 5 · fill the bottom row

[H2] = [I2] = 0.0110 M. [HI] = 0.100 − 2(0.01102) = 0.0780 M.

Answer and the check

Check it: (0.01102)2 ÷ (0.0780)2 = (1.215 × 10-4) ÷ (6.08 × 10-3) = 0.0200. It matches Kc. The shortcut was not available here, since 0.100 ÷ 0.0200 is only 5, nowhere near 400. The perfect square saved you from the quadratic formula instead.

Example T5.2 · Design a buffer at a target pH

You want a buffer at pH 5.00 from acetic acid and sodium acetate. Acetic acid has pKa = 4.74. What ratio do you need, and how much acetate goes with 0.100 mol of acid in 1.00 L?

Step 1 · check that the pair fits

The target is 5.00 and the pKa is 4.74, which is 0.26 units apart. A buffer works well within about one unit, so this pair is a good choice.

Step 2 · solve for the log term

5.00 = 4.74 + log([A-] ÷ [HA]), so log([A-] ÷ [HA]) = 5.00 − 4.74 = 0.26.

Step 3 · undo the log

[A-] ÷ [HA] = 100.26 = 1.8. So you need about 1.8 times as much acetate as acetic acid.

Step 4 · put real amounts on it

With 0.100 mol of acetic acid, you need (1.8)(0.100) = 0.18 mol of sodium acetate in the same 1.00 L.

Step 5 · check it forward

pH = 4.74 + log(0.18 ÷ 0.100) = 4.74 + log(1.8) = 4.74 + 0.26 = 5.00.

Answer

A ratio near 1.8 to 1 of acetate to acid, which is 0.18 mol of the salt beside 0.100 mol of the acid. Notice that the pH above pKa means more base than acid, and a pH below it would mean more acid. If the target had been 7.00, this pair would be useless and you would reach for a phosphate pair instead.

Example T5.3 · Will calcium fluoride drop out?

Mix 50.0 mL of 0.0020 M Ca(NO3)2 with 50.0 mL of 0.0040 M NaF. Does CaF2 precipitate? Its Ksp is 3.9 × 10-11.

Step 1 · total volume

50.0 mL + 50.0 mL = 100.0 mL, so each solution was cut in half.

Step 2 · dilute both ions

[Ca2+] = (0.0020 M)(50.0 ÷ 100.0) = 0.0010 M. [F-] = (0.0040 M)(50.0 ÷ 100.0) = 0.0020 M.

Step 3 · build Q with the right exponent

Q = [Ca2+][F-]2 = (0.0010)(0.0020)2. First (0.0020)2 = 4.0 × 10-6.

Step 4 · multiply

Q = (1.0 × 10-3)(4.0 × 10-6) = 4.0 × 10-9.

Step 5 · compare

Q = 4.0 × 10-9 against Ksp = 3.9 × 10-11. The ratio is about 100, so Q is far above Ksp.

Answer

Yes, CaF2 precipitates, and it keeps forming until Q falls back to 3.9 × 10-11. Two things go wrong most often here. Skipping the dilution would have given Q eight times too large. Forgetting to square the fluoride would have given 2.0 × 10-6, which is wrong by 500 times.

The traps people fall into

Watch for these
  • Calling equilibrium dead. The rates are equal, not zero.
  • Putting a solid or solvent water into K. They never appear.
  • Losing a coefficient. It is an exponent in K and a multiplier on x in the table.
  • Dropping the x that is your answer. Drop it only where it is subtracted from a much larger number, and only after the ratio test.
  • Forgetting the second hydroxide on Ba(OH)2 or Ca(OH)2. Each unit gives two.
  • Treating a weak acid as strong. A 0.10 M acetic acid solution is pH 2.87, not 1.00.
  • Counting pH significant figures from the whole number. Only the digits after the point count.
  • Calling every salt neutral. Check where each ion came from first.
  • Comparing Ksp values across different ion ratios. The exponent, not just the constant, sets the solubility.
  • Skipping the dilution before building Q. Mixing weakens both solutions before anything reacts.

Self-test · Theme 5

1. Write Kc for 2 NO(g) + O2(g) ⇌ 2 NO2(g).
Kc = [NO2]2 ÷ ([NO]2[O2]). Products on top, each raised to its coefficient. The oxygen has an unwritten coefficient of 1, so it enters to the first power.
2. For that same reaction, Kc = 6.0 × 1013 at 298 K. Find Kp.
Kp = 2.5 × 1012. Gas moles: 2 on the right against 3 on the left, so Δn = −1. RT = (0.08206)(298) = 24.45 L·atm/mol. A negative exponent means divide: Kp = (6.0 × 1013) ÷ 24.45 = 2.5 × 1012.
3. A reaction has K = 1.8 and a mixture gives Q = 0.40. Which way does it move, and what happens to Q?
Forward, and Q climbs to 1.8. Q below K means the top of the fraction is too small, so there is not enough product yet. The reaction makes more product, which raises Q, and it stops the moment Q reaches 1.8.
4. Find the pH of 0.0500 M HNO3.
pH = 1.30. Nitric acid is strong and gives one proton per unit, so [H3O+] = 0.0500 M. Then pH = −log(0.0500) = 1.30. Three significant figures in the concentration would have given three digits after the point, and here two gives two.
5. Find the pH of 0.200 M ammonia. Kb = 1.8 × 10-5.
pH = 11.28. The ratio 0.200 ÷ (1.8 × 10-5) = 1.1 × 104 clears 400, so use the shortcut. [OH-] = √[(1.8 × 10-5)(0.200)] = √(3.6 × 10-6) = 1.9 × 10-3 M. Then pOH = 2.72 and pH = 14.00 − 2.72 = 11.28.
6. Find the pH of 0.10 M NH4Cl. Ammonia has Kb = 1.8 × 10-5.
pH = 5.13. Chloride is a spectator and ammonium is an acid, so first convert: Ka = Kw ÷ Kb = (1.0 × 10-14) ÷ (1.8 × 10-5) = 5.6 × 10-10. Then [H3O+] = √[(5.6 × 10-10)(0.10)] = √(5.6 × 10-11) = 7.5 × 10-6 M, so pH = 5.13. A salt of a weak base comes out acidic, which it did.
7. A weak acid with pKa = 4.20 is titrated with NaOH. What is the pH at half equivalence?
pH = 4.20. Half the acid has become its conjugate base, so the two amounts are equal and the ratio is 1. Since log(1) = 0, Henderson-Hasselbalch collapses to pH = pKa. This is the easiest way to measure a pKa in a lab.
8. Silver bromide has Ksp = 5.0 × 10-13. Find its molar solubility in pure water.
7.1 × 10-7 M. AgBr is a 1:1 salt, so Ksp = s2 and s = √(5.0 × 10-13). Rewrite it as √(50 × 10-14) = 7.1 × 10-7 M. Shifting the exponent to an even power makes the root easy to take by hand.
9. Find the solubility of AgBr in 0.010 M NaBr, and compare it with pure water.
5.0 × 10-11 M, about 14,000 times lower. Let [Ag+] = s and [Br-] = 0.010 + s, which is near enough 0.010. Then s = (5.0 × 10-13) ÷ 0.010 = 5.0 × 10-11 M. Divide the pure water value: (7.1 × 10-7) ÷ (5.0 × 10-11) = 1.4 × 104. That is the common ion effect doing its work.
10. Find K for AgBr(s) + 2 S2O32- ⇌ [Ag(S2O3)2]3- + Br-. Use Ksp = 5.0 × 10-13 and Kf = 2.9 × 1013.
K = 14.5. Add the dissolving step to the complexing step and the free Ag+ cancels, leaving the reaction asked for. So multiply the constants: (5.0 × 10-13)(2.9 × 1013) = 14.5. A K above 1 means thiosulfate dissolves silver bromide easily, which is exactly how film was once fixed.

Theme 6 · Chapters 16 and 17Energy and electron transfer

What to know

A change is spontaneous if it keeps going on its own, and that says nothing about speed. Heat release helps but does not decide, since ice melts in a warm room while taking heat in. The missing factor is how spread out the matter and energy end up, measured as entropy. Entropy counts arrangements, and S = k ln W ties it to that count.

The second law is the real test: a change happens when the total entropy of system plus surroundings rises. Either ledger may lose, and only the sum must grow. The surroundings keep score as −ΔH ÷ T, so cold surroundings gain more entropy per joule than warm ones. The third law sets a perfect pure crystal at 0 K to exactly zero entropy, which is why tables can list absolute S° values. Every one of them is positive, including for elements.

Gibbs free energy folds both ledgers into one number for the system alone, as ΔG = ΔH − TΔS. Negative means the forward change is favored, positive means the reverse, and zero means equilibrium. The four sign cases are worth memorizing. Both terms helping means always spontaneous. Both fighting means never. One of each means a switch-over at a crossover temperature.

Free energy also fixes the size of K through ΔG° = −RT ln K. At 298.15 K every 5.71 kJ is one power of ten in K. Keep ΔG° and ΔG apart, since the circle means the standard state while ΔG = ΔG° + RT ln Q describes the mixture you have. Once Q reaches K, the two terms cancel and ΔG is zero.

Electrochemistry turns all of that into a voltage. Oxidation loses electrons and reduction gains them, and they always travel together. Split a redox reaction into half-reactions, balance each one with water for oxygen, H+ for hydrogen and electrons for charge, then scale so the electrons cancel. For a basic solution, balance in acid and then add OH- to both sides.

Cell rules that never change
  • An ox and red cat: oxidation at the anode, reduction at the cathode, in every cell ever built.
  • Electrons travel through the wire from anode to cathode, always.
  • In a galvanic cell the anode is negative. In an electrolytic cell it is positive.
  • A salt bridge passes ions so neither half-cell builds up charge.
  • In cell notation the anode is on the left and the double bar is the bridge.
  • E°cell = E°cathode − E°anode, with both values taken straight from the table.
  • E° never scales with coefficients, because potential is energy per unit of charge.

The hydrogen electrode was defined as exactly 0.000 V so every other half-cell could be measured against it. A high table value means a strong oxidizing agent, and any species high on the table will oxidize anything lower down. A positive E°cell, a negative ΔG° and a K above 1 are the same fact three times over.

The Nernst equation corrects a potential for real concentrations. Pile up product and Q rises, so the voltage sags. That is a battery's whole life, and a dead battery is simply a cell that reached Q = K. A concentration cell is the limit of the idea, with E°cell = 0 and the dilute half-cell as the anode.

Electrolysis pays electricity to force a reaction with a negative potential. Faraday's law treats electrons as a reagent measured out with an ammeter and a clock. In water the solvent often reacts instead of the ion you expected, and a real cell needs extra voltage called overpotential. Corrosion is a galvanic cell you did not build, worth 0.84 V for iron in damp air, and a sacrificial anode of zinc or magnesium stops it.

Equations, units and symbols

EquationWhat each symbol meansUnits
S = k ln W W is the number of microstates. k is the Boltzmann constant, R divided by NA. k = 1.381 × 10-23 J/K.
ΔSuniv = ΔSsys + ΔSsurr > 0 The second law. Zero marks equilibrium. J/K.
ΔSsurr = −ΔHsys ÷ T At steady pressure. An exothermic change gives the surroundings a positive value. ΔH in J, T in K.
ΔS°rxn = ΣnS°(products) − ΣnS°(reactants) Elements stay in the sum, because S° for an element is not zero. S° in J/mol·K, answer in J/K.
ΔG = ΔH − TΔS Match the units before subtracting. ΔH in kJ needs ΔS in kJ/K. kJ, with T in K.
ΔG° = −RT ln K, and ΔG = ΔG° + RT ln Q Use R in joules, so put ΔG° in joules too. R = 8.314 J/mol·K, T in K.
T = ΔH° ÷ ΔS° The crossover temperature, where ΔG° = 0 and K = 1. Same energy unit top and bottom. Answer in K.
E°cell = E°cathode − E°anode A positive answer means the cell runs by itself. A negative one means you picked the halves backward. Volts, where 1 V = 1 J/C.
ΔG° = −nFE°cell n is moles of electrons in the balanced overall reaction, not a coefficient you guess. F = 96,485 C/mol. Answer in J/mol.
E°cell = (0.0592 V ÷ n) log K Valid at 25 °C. Every 0.0592 V per electron is one power of ten in K. Volts.
Ecell = E°cell − (0.0592 V ÷ n) log Q The Nernst equation at 25 °C. Build Q from the overall reaction, with solids as 1. Volts.
moles of e- = It ÷ F, then divide by n Faraday's law. Convert the time to seconds first. I in A, t in s, F in C/mol.

Worked examples

Example T6.1 · Ammonia synthesis, from entropy to a crossover

For N2(g) + 3 H2(g) → 2 NH3(g), ΔH° = −91.8 kJ. Find ΔS° from the table values 191.6 for N2, 130.7 for H2 and 192.8 for NH3, all in J/mol·K. Then find ΔG° at 298.15 K and the crossover temperature.

Step 1 · add the products

2 × 192.8 = 385.6 J/K.

Step 2 · add the reactants

191.6 + 3(130.7) = 191.6 + 392.1 = 583.7 J/K.

Step 3 · subtract

ΔS° = 385.6 − 583.7 = −198.1 J/K. Four moles of gas became two, so a negative answer was the only possibility.

Step 4 · fix the units and find ΔG°

ΔS° = −0.1981 kJ/K. Then ΔG° = −91.8 − (298.15)(−0.1981) = −91.8 + 59.1 = −32.7 kJ.

Step 5 · the crossover

T = ΔH° ÷ ΔS° = (−91.8 kJ) ÷ (−0.1981 kJ/K) = 463 K, which is 190 °C.

Answer

ΔS° = −198.1 J/K and ΔG° = −32.7 kJ at room temperature. Both ΔH° and ΔS° are negative, so this is the cold-only sign case. Below 463 K the reaction is favored and above it the ammonia falls apart. That is the squeeze a real plant lives with, since heat is needed for speed and ruins the yield.

Example T6.2 · From ΔG° to K, and the size check

For the same synthesis, ΔG°f for NH3(g) is −16.4 kJ/mol. Find ΔG° that way, then find K at 298.15 K.

Step 1 · sum the formation values

Both elements are zero in their standard states, so ΔG° = 2(−16.4) − 0 − 0 = −32.8 kJ. That matches the −32.7 kJ from the other route, to the rounding.

Step 2 · put it in joules

−32.8 kJ = −32,800 J/mol, because R is written in joules.

Step 3 · build RT

RT = (8.314 J/mol·K)(298.15 K) = 2478.8 J/mol.

Step 4 · solve for ln K

ln K = −ΔG° ÷ RT = 32,800 ÷ 2478.8 = 13.23.

Step 5 · undo the log

K = e13.23 = 5.6 × 105.

Answer and the size check

K = 5.6 × 105, so ammonia is strongly favored at room temperature and the reaction is simply far too slow to use there. Check the size: 32.8 kJ divided by 5.71 kJ per power of ten gives 5.7 powers, and 105.7 is about 5 × 105. The two agree.

Example T6.3 · One cell, four answers

Build a cell from iron and silver. Use E° = +0.80 V for Ag+ + e- → Ag and −0.44 V for Fe2+ + 2e- → Fe. Find E°cell, n, ΔG° and K. Then find Ecell when [Fe2+] = 0.100 M and [Ag+] = 0.0100 M.

Step 1 · pick the cathode

Silver is higher on the table, so silver ion is reduced and iron is the anode. E°cell = (+0.80) − (−0.44) = +1.24 V.

Step 2 · balance and count electrons

Fe + 2 Ag+ → Fe2+ + 2 Ag moves two electrons, so n = 2. Doubling the silver half left its +0.80 V alone.

Step 3 · free energy

ΔG° = −nFE° = −(2)(96,485 C/mol)(1.24 J/C) = −239,300 J/mol, or −239 kJ/mol.

Step 4 · the constant

log K = nE° ÷ 0.0592 = (2)(1.24) ÷ 0.0592 = 2.48 ÷ 0.0592 = 41.9. So K = 1041.9 = 7.9 × 1041.

Step 5 · now Nernst

Q = [Fe2+] ÷ [Ag+]2 = 0.100 ÷ (0.0100)2 = 0.100 ÷ (1.00 × 10-4) = 1.00 × 103, so log Q = 3.00. Then Ecell = 1.24 − (0.0592 ÷ 2)(3.00) = 1.24 − 0.0888 = 1.15 V.

Answer

+1.24 V, n = 2, ΔG° = −239 kJ/mol, K near 1042, and 1.15 V at those concentrations. The potential fell because product built up and reactant was thin. Note the exponent on silver in Q, since forgetting to square it would have given log Q = 1.00 and a wrong answer of 1.21 V.

The traps people fall into

Watch for these
  • Reading spontaneous as fast. Direction and rate are different questions.
  • Subtracting J/K from kJ. Convert first, or the answer is off by a thousand.
  • Dropping elements from a ΔS° sum. Their S° is positive, unlike ΔH°f and ΔG°f.
  • Using R = 0.08206 with ln K. That R belongs to gas laws. Use 8.314 J/mol·K here.
  • Flipping the anode's sign before subtracting. The subtraction already handles it.
  • Scaling E° with the coefficients. Doubling a half-reaction changes n, never E°.
  • Guessing n. Read it off the half-reactions after the electrons are matched.
  • Building Q from one half-reaction. Q comes from the overall cell reaction, exponents and all.
  • Leaving time in minutes in q = It. Convert to seconds first.
  • Calling a dead battery empty. It reached Q = K, so the potential is zero with material still inside.

Self-test · Theme 6

1. Predict the sign of ΔS° for 2 H2O(l) → 2 H2(g) + O2(g).
Positive, and strongly so. Zero moles of gas became three. A gas holds far more spread than a liquid, so splitting water apart raises entropy a lot. Both ΔH° and ΔS° are positive here, which is the hot-only sign case.
2. A reaction has ΔH = −25.0 kJ and ΔS = −80.0 J/K. Find ΔG at 350 K and the crossover temperature.
ΔG = +3.0 kJ, and the crossover is 313 K. Convert: ΔS = −0.0800 kJ/K. Then ΔG = −25.0 − (350)(−0.0800) = −25.0 + 28.0 = +3.0 kJ, so it is not spontaneous there. The crossover is T = 25,000 J ÷ 80.0 J/K = 313 K, and 350 K is above it, which matches the positive answer.
3. A reaction takes in 50.0 kJ at 298.15 K. Find ΔSsurr.
−168 J/K. Taking heat in means ΔHsys = +50.0 kJ, or +50,000 J. Then ΔSsurr = −50,000 J ÷ 298.15 K = −168 J/K. The surroundings lost heat, so they lost spread, and the system's own gain has to beat that for the change to go.
4. Find K at 298.15 K for a reaction with ΔG° = +15.0 kJ.
K = 2.4 × 10-3. RT = (8.314)(298.15) = 2478.8 J/mol. Then ln K = −15,000 ÷ 2478.8 = −6.051, so K = e-6.051 = 2.4 × 10-3. A positive ΔG° always gives a K below 1, so the reactants win.
5. Find E°cell and n for a zinc and silver cell. Use +0.80 V for silver and −0.76 V for zinc.
+1.56 V with n = 2. Silver is higher, so it is the cathode: E°cell = (+0.80) − (−0.76) = +1.56 V. The balanced reaction Zn + 2 Ag+ → Zn2+ + 2 Ag moves two electrons. The positive potential says the cell runs on its own.
6. Find ΔG° for that zinc and silver cell.
−301 kJ/mol. ΔG° = −nFE° = −(2)(96,485)(1.56). First 2 × 96,485 = 192,970, then times 1.56 gives 301,000 J. So ΔG° = −3.01 × 105 J/mol, which is −301 kJ/mol.
7. Find K for that same cell at 25 °C.
K = 5.0 × 1052. log K = nE° ÷ 0.0592 = (2)(1.56) ÷ 0.0592 = 3.12 ÷ 0.0592 = 52.7. So K = 1052.7 = 5.0 × 1052. A cell you would call ordinary is a reaction that goes to completion.
8. A zinc and copper cell has [Zn2+] = 2.00 M and [Cu2+] = 0.0500 M. Find Ecell, given E°cell = 1.10 V.
1.05 V. Both metals are solids and drop out, so Q = [Zn2+] ÷ [Cu2+] = 2.00 ÷ 0.0500 = 40.0, and log(40.0) = 1.602. Then Ecell = 1.10 − (0.0592 ÷ 2)(1.602) = 1.10 − 0.0474 = 1.05 V. Product is high and reactant is low, so the potential fell.
9. A current of 1.50 A runs through a copper plating bath for 20.0 minutes. What mass of copper lands? Copper is 63.55 g/mol.
0.593 g. Time in seconds: 20.0 × 60 = 1200 s. Charge: q = (1.50 C/s)(1200 s) = 1800 C. Moles of electrons: 1800 ÷ 96,485 = 0.01866 mol. Copper takes two electrons per atom, so 0.01866 ÷ 2 = 0.009328 mol. Mass: (0.009328)(63.55) = 0.593 g.
10. Why does a dead battery read zero volts when there is still material inside it?
Because Q has climbed all the way to K. As the cell runs, product builds and Q rises, so the Nernst term grows and eats into E°cell. When Q finally equals K, the log term cancels E°cell exactly and the meter reads zero. A dead battery is a cell that reached equilibrium, not one that ran out of atoms.

Theme 7 · Chapters 18, 19, 20 and 21Descriptive and nuclear

What to know

The main groups run from 1 and 2 across to 13 through 18. Two trends explain nearly all of their chemistry. Across a period the nuclear grip tightens, atoms shrink, and the elements shift from giving electrons away to taking them in. Down a group a new shell is added, the grip loosens, and metallic character grows. One quick test places any element: a metal oxide makes a base, a nonmetal oxide makes an acid, and the amphoteric oxides in between react with both.

Main group facts worth knowing cold
  • Every alkali metal gives hydrogen plus a hydroxide in water. Only the speed changes.
  • With oxygen, lithium gives the oxide, sodium the peroxide, and potassium and heavier the superoxide.
  • Only lithium attacks N2 at room temperature.
  • Beryllium will not react with water, and BeO and Al2O3 are amphoteric.
  • Diamond is a four-bond network and an insulator. Graphite is a three-bond sheet with one loose electron, so it conducts.
  • CO2 is a gas and SiO2 is rock, because strong double bonds need small atoms.
  • Count O–H groups, not hydrogens: H3PO4 is triprotic, H3PO3 diprotic, H3PO2 monoprotic.
  • Halogen oxidizing power runs F > Cl > Br > I, and that order predicts every displacement.
  • More oxygens make a halogen oxyacid stronger, so HClO4 far outdoes HClO.

The d-block metals carry a partly filled d subshell, which is where their color and magnetism come from. In a neutral atom 4s fills before 3d, yet we write 3d first, and chromium and copper each shift one electron to reach a half filled or filled d set. When a cation forms, the 4s electrons always leave first, so d electrons equal the group number minus the charge.

A coordination compound is a metal ion with ligands attached, each donating a lone pair, so the ligand is a Lewis base and the metal is a Lewis acid. The coordination number counts donor atoms, not ligands, so three bidentate ligands give six. Names run cation first, then the ligands in alphabetical order, then the metal with its oxidation number in Roman numerals, and a negative complex takes an -ate ending.

Ligands split the d orbitals into a lower set of three and an upper set of two, separated by Δ. A small Δ leaves the electrons spread out, which is high spin. A large Δ stacks them low, which is low spin, and the choice only exists for d4 through d7. A d electron can absorb a photon worth exactly Δ, so you see the complementary color of the light that was taken out. Unpaired electrons make a complex paramagnetic, and the magnetic moment gives their count.

Carbon earns its own chapter because it makes four bonds and bonds to itself without complaint. Alkanes are saturated at CnH2n+2. Alkenes and rings are CnH2n, and alkynes are CnH2n-2. Structural isomers share a formula and wire up differently, so they are different substances. Naming always runs longest chain, lowest locants, branches in alphabetical order.

A multiple bond is loose electron density, so reagents attack it. An addition reaction keeps every atom, while substitution throws a second product away. Benzene is the exception that proves the rule, since its six delocalized electrons buy 152 kJ/mol of extra stability, and addition would hand that back. So an aromatic ring substitutes and stays whole.

A functional group sets the behavior and the carbon frame mostly comes along. An O–H bond lets a molecule donate a hydrogen bond, which can move a boiling point by a hundred degrees. The oxidation ladder runs primary alcohol to aldehyde to carboxylic acid, a secondary alcohol stops at the ketone, and a tertiary alcohol refuses. An addition polymer loses nothing, so its repeat unit matches the monomer. A condensation polymer drops water at each link, so each monomer needs two reactive ends.

Finally the nucleus. A nuclide symbol puts the nucleon count A on top and the proton count Z below, so N = A − Z. Stable nuclides crowd into the band of stability, and nothing past Z = 83 is stable. A nucleus weighs less than its separate parts, and that mass defect is its binding energy. Divide by the nucleon count and the value peaks near iron-56 at about 8.79 MeV, which is why heavy nuclei split and light ones fuse.

Alpha decay drops A by 4 and Z by 2. Beta decay raises Z by 1 while A holds. Positron emission and electron capture each lower Z by 1. Gamma emission changes neither. Decay is first order, so one half-life always halves what is left, and nothing you do to a sample changes the rate. A nuclear equation balances on A and Z only, which is exactly why elements can change. For radiation, becquerels measure the source, grays measure energy absorbed, and sieverts measure expected harm.

Equations, units and symbols

EquationWhat each symbol meansUnits
Zeff ≈ Z − (core electrons) A ceiling on the pull an outer electron feels, since real screening is not perfect. A charge in units of e.
d electrons = group number − charge Works for any d-block cation, because the 4s electrons leave first. A bare count.
x + Σ(ligand charges) = charge on the complex ion x is the metal's oxidation number. Counter ions outside the brackets set that charge. Charges in units of e.
Δ = hc ÷ λ, then times NA for a mole The field splitting from the wavelength absorbed. You see the complement of that color. hc = 1.987 × 10-25 J·m, NA = 6.022 × 1023/mol.
μ = √(n(n + 2)) n is the unpaired electron count. The values 0, 1.73, 2.83, 3.87, 4.90 and 5.92 cover n = 0 to 5. Bohr magnetons.
degrees of unsaturation = (2C + 2 − H) ÷ 2 C and H are the atom counts. One degree is one ring or one double bond, and a triple bond is two. A bare count.
degree of polymerization = (chain molar mass) ÷ (repeat unit molar mass) For a condensation polymer, subtract the water lost at each link before dividing. A bare count.
N = A − Z Neutrons equal nucleons minus protons. Bare counts.
binding energy = Δm × 931.5 MeV/u Δm is the mass defect. Build the parts from Z hydrogen atoms and N neutrons so the electrons cancel. 1 u = 931.5 MeV = 1.6605 × 10-27 kg. 1 MeV = 1.602 × 10-13 J.
k = 0.693 ÷ t1/2 The decay constant, since 0.693 is the natural log of 2. Reciprocal time.
N = N0(1/2)t/t1/2, or ln(N0 ÷ N) = kt The first for a whole number of half-lives, the second for anything else. Matching amount units. t in the unit k uses.
sieverts = grays × quality factor Beta and gamma rate 1, neutrons about 10, alpha 20. 1 Gy = 1 J/kg. 1 Ci = 3.7 × 1010 Bq.

Worked examples

Example T7.1 · Two cobalt complexes, same metal, different everything

Compare [Co(NH3)6]3+ with [CoF6]3-. Give the oxidation number, the d count, the coordination number and the spin state of each. Then find Δ in kJ/mol for a complex that absorbs at 430 nm, and name the color you would see.

Step 1 · the oxidation number in each

Ammonia is neutral, so x + 6(0) = +3 gives x = +3. Fluoride is 1− each, so x + 6(−1) = −3 gives x = +3 as well. Both complexes hold cobalt(III).

Step 2 · the d count and the shape

Cobalt is group 9, so 9 − 3 = 6 d electrons in both. Both ligands are monodentate and there are six, so both have a coordination number of 6 and are octahedral.

Step 3 · now the ligands part ways

Ammonia sits high in the spectrochemical series and fluoride sits low. So ammonia opens a large Δ and the six electrons pair up low, giving low spin with 0 unpaired. Fluoride opens a small Δ, so the electrons spread out, giving high spin with 4 unpaired.

Step 4 · what a magnet would show

For n = 0, μ = 0 and the ammonia complex is diamagnetic. For n = 4, μ = √(4 × 6) = √24 = 4.90 Bohr magnetons, so the fluoride complex is paramagnetic.

Step 5 · the color calculation

430 nm is 4.30 × 10-7 m. Then Δ = (1.987 × 10-25 J·m) ÷ (4.30 × 10-7 m) = 4.62 × 10-19 J per photon. Times Avogadro's number: (4.62 × 10-19)(6.022 × 1023) = 2.78 × 105 J/mol = 278 kJ/mol.

Answer

Both are cobalt(III), d6, coordination number 6. The ammonia complex is low spin and diamagnetic, the fluoride one is high spin with μ = 4.90. The 430 nm band means Δ = 278 kJ/mol, and since blue light was removed the solution looks orange. Same metal, same charge, same d count, and only the ligand changed.

Example T7.2 · One formula, two compounds, 83 degrees apart

A compound is C4H10O. How unsaturated is it? Name two compounds that fit, give their molar mass, and explain why their boiling points differ by so much. 1-butanol boils at 117.7 °C and diethyl ether boils at 34.6 °C.

Step 1 · run the unsaturation count

For 4 carbons a saturated chain holds 2(4) + 2 = 10 hydrogens. The formula has 10, so 10 − 10 = 0 and there are zero degrees of unsaturation. No rings and no double bonds.

Step 2 · name two that fit

Put the oxygen on the end as an O–H and you get 1-butanol, an alcohol. Put it in the middle between two carbon frames and you get diethyl ether. Both are C4H10O, so they are structural isomers.

Step 3 · the shared molar mass

Carbon: 4 × 12.011 = 48.044. Hydrogen: 10 × 1.008 = 10.080. Oxygen: 15.999. Total = 74.12 g/mol for both.

Step 4 · find the real difference

117.7 − 34.6 = 83.1 °C apart, and the masses are identical, so mass cannot be the reason.

Step 5 · read the functional group

Only the alcohol has an O–H bond, so only the alcohol can donate a hydrogen bond to its neighbor. The ether hides its oxygen in the middle with no hydrogen on it, so it can accept a hydrogen bond from water but never donate one.

Answer

Zero degrees of unsaturation, both at 74.12 g/mol, and 83.1 °C apart because of one O–H bond. The two also behave differently in reactions. Heat the alcohol with sulfuric acid and water is pulled out to leave an alkene. Oxidize it and you climb to butanal and then to butanoic acid. The ether does none of that.

Example T7.3 · Binding energy, then a date

A lithium-7 atom weighs 7.016003 u. A hydrogen-1 atom weighs 1.007825 u and a neutron weighs 1.008665 u. Find the binding energy per nucleon. Then date a sample holding 6.25% of its original carbon-14, where the half-life is 5,730 years.

Step 1 · count the parts

Lithium-7 has Z = 3 and N = 7 − 3 = 4. Three hydrogen atoms come to 3 × 1.007825 = 3.023475 u. Four neutrons come to 4 × 1.008665 = 4.034660 u. Together that is 7.058135 u.

Step 2 · subtract

Δm = 7.058135 − 7.016003 = 0.042132 u. The parts weigh more, which they always do.

Step 3 · turn mass into energy

(0.042132 u)(931.5 MeV/u) = 39.25 MeV. Divide by 7 nucleons: 5.61 MeV per nucleon.

Step 4 · now the date, the easy way

6.25% is 1 ÷ 16, and 16 is 24. So exactly four half-lives have passed, since halving four times runs 100% to 50% to 25% to 12.5% to 6.25%.

Step 5 · multiply

4 × 5,730 years = 22,920 years, which is 2.29 × 104 years.

Answer and the size check

5.61 MeV per nucleon, and a sample about 2.29 × 104 years old. Lithium sits well below the iron-56 peak of 8.79 MeV, which is why fusing light nuclei pays so well. And 22,900 years is inside carbon-14's useful range of roughly 50,000 years, so the date is trustworthy.

The traps people fall into

Watch for these
  • Counting hydrogens on a phosphorus acid. Count the O–H groups, so H3PO3 is diprotic.
  • Giving every alkali metal the same oxide. Lithium, sodium and potassium each give a different one.
  • Taking 3d electrons off first. The 4s pair always goes first, so Fe2+ is [Ar]3d6.
  • Counting ligands for a coordination number. Count donor atoms, and watch for counter ions outside the brackets.
  • Naming the color absorbed. You see the complement, so a green solution is not absorbing green.
  • Thinking high spin means more electrons. It means the same electrons are spread out, because Δ is small.
  • Counting a triple bond as one degree of unsaturation. It costs four hydrogens, so it is two.
  • Making benzene add like an alkene. Addition would cost the 152 kJ/mol bonus, so the ring substitutes.
  • Oxidizing a tertiary alcohol. That carbon has no hydrogen left to lose.
  • Thinking the parts of a nucleus weigh less. They weigh more, and that defect is the binding energy.
  • Mixing becquerels, grays and sieverts. They measure the source, the energy absorbed and the expected harm.

Self-test · Theme 7

1. Sort MgO, Al2O3 and P4O10 as basic, amphoteric or acidic.
Basic, amphoteric, acidic. Oxides shift that way as you move left to right across a period. Magnesium is a metal, phosphorus is a nonmetal, and aluminum sits on the border, which is exactly why its oxide reacts with both acid and base.
2. Which alkali metal reacts with nitrogen gas at room temperature, and what forms?
Lithium, giving Li3N. The N≡N bond is one of the strongest known, so almost nothing touches nitrogen without heat or a catalyst. Lithium is the one exception in group 1. Its small, high-charge ion makes a nitride lattice stable enough to pay for breaking that bond.
3. How many protons does H3PO3 give up in water, and why?
Two, so it is diprotic. It looks triprotic from the formula, but 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. H3PO2 has two P–H bonds and gives up just one.
4. For [Fe(CN)6]3-, find the iron's oxidation number, its d count and the coordination number.
+3, d5, and 6. Each cyanide is 1−, so x + 6(−1) = −3 gives x = +3. Iron is group 8, so 8 − 3 = 5 d electrons. Cyanide is monodentate, so six of them give six donor atoms and an octahedral shape.
5. Name [Pt(NH3)2Cl2], and say which isomer is the cancer drug.
Diamminedichloridoplatinum(II), and the cis isomer is cisplatin. Two neutral ammonias and two chlorides at 1− each make the complex neutral, so platinum is +2. Sort ammine before chlorido. The cis form holds its two chlorides at one corner of the flat square, close enough to bind two neighboring sites on a DNA strand. The trans form points them apart and does nothing.
6. A Cr3+ complex holds three unpaired electrons. Find μ.
3.87 Bohr magnetons. Chromium is group 6, so 6 − 3 = 3 d electrons, and three electrons in five orbitals stay unpaired by Hund's rule. Then μ = √(3 × 5) = √15 = 3.87. Chromium(III) is d3, so there is no high spin or low spin choice to make.
7. How unsaturated is C6H6, and what structure fits?
Four degrees, which fits a benzene ring. A saturated chain of 6 carbons holds 2(6) + 2 = 14 hydrogens, so 14 − 6 = 8 are missing and 8 ÷ 2 = 4 degrees. One ring plus three double bonds comes to four. The six electrons spread over all six carbons instead of sitting in three fixed bonds, which is why every bond measures the same 139 pm.
8. Which boils higher, 1-propanol or dimethyl ether, and why?
1-propanol, by a wide margin. The alcohol has an O–H bond, so one molecule's hydrogen is drawn to the next molecule's oxygen and it takes real heat to pull them apart. The ether can accept a hydrogen bond but has no hydrogen on its oxygen to donate one. A single functional group can move a boiling point by a hundred degrees.
9. A polyethylene sample averages 56,100 g/mol. Its monomer is ethene, C2H4. Find the degree of polymerization.
About 2000 repeat units. The repeat unit mass is 2(12.011) + 4(1.008) = 24.022 + 4.032 = 28.05 g/mol. Polyethylene is an addition polymer, so nothing leaves and the repeat unit matches the monomer exactly. Then 56,100 ÷ 28.05 = 2000. For a condensation polymer you would subtract the water lost at each link first.
10. Polonium-210 has Z = 84 and emits an alpha particle. Name the daughter, and then say what 1.00 mGy of neutrons works out to in sieverts.
Lead-206, and 10 mSv. Alpha decay takes 4 off the mass number, giving 210 − 4 = 206, and 2 off the atomic number, giving 84 − 2 = 82, which is lead. For the dose, neutrons carry a quality factor of about 10, so 1.00 mGy × 10 = 10 mSv. The same gray of gamma would count for only 1 mSv, and of alpha for 20 mSv.

Cram sheet

One page for the night before. Everything below is for 25 °C unless a line says otherwise.

Constants, in one place

  • NA = 6.022 × 1023/mol, and one mole of gas fills 22.4 L at STP.
  • R = 0.08206 L·atm/mol·K for gas work, and R = 8.314 J/mol·K for energy work.
  • 0 °C = 273.15 K. Standard conditions here mean 298.15 K.
  • 1 atm = 760 torr = 760 mmHg = 101.325 kPa = 1.01325 bar.
  • h = 6.626 × 10-34 J·s and c = 2.998 × 108 m/s, so hc = 1.987 × 10-25 J·m.
  • Hydrogen's levels: En = −(2.179 × 10-18 J) ÷ n2, so ionizing it costs 1312 kJ/mol.
  • Water: c = 4.184 J/g·°C liquid, 2.09 ice, 1.84 steam. ΔHfus = 6.01 kJ/mol and ΔHvap = 40.7 kJ/mol.
  • Water: Kb = 0.512 °C/m and Kf = 1.86 °C/m.
  • Kw = 1.0 × 10-14, so pH + pOH = 14.00 and pure water is pH 7.00.
  • k = 1.381 × 10-23 J/K, which is R ÷ NA.
  • F = 96,485 C/mol, 1 V = 1 J/C, and the hydrogen electrode is exactly 0.000 V.
  • RT at 298.15 K = 2478.8 J/mol. RT ln 10 = 5.71 kJ/mol, and 0.0592 V per electron, are both one power of ten in K.
  • 1 u = 1.6605 × 10-27 kg = 931.5 MeV. 1 MeV = 1.602 × 10-13 J. Binding energy peaks near iron-56 at 8.79 MeV per nucleon.
  • 1 Gy = 1 J/kg. 1 Ci = 3.7 × 1010 Bq. Quality factors: beta and gamma 1, neutrons 10, alpha 20.

Every equation, by theme

ThemeEquations
Measurement and the moled = m ÷ V; K = °C + 273.15; moles = mass ÷ molar mass; particles = moles × NA; M = mol ÷ L; M1V1 = M2V2; multiplier = molar mass ÷ empirical formula mass
Structure and bondingaverage mass = Σ(fraction × mass); c = λν; E = hν = hc ÷ λ; En = −2.179 × 10-18 ÷ n2; formal charge = valence − lone pair electrons − bonds; bond order = (bonding − antibonding) ÷ 2
States and solutionsP1V1 ÷ T1 = P2V2 ÷ T2; PV = nRT; M = dRT ÷ P; Pi = XiPtotal; rateA ÷ rateB = √(MB ÷ MA); q = mcΔT; q = nΔHfus or nΔHvap; ln(P2 ÷ P1) = −(ΔHvap ÷ R)(1 ÷ T2 − 1 ÷ T1); ΔTb = iKbm; ΔTf = iKfm; Π = iMRT
Change and ratemole ratio from the coefficients; percent yield = actual ÷ theoretical × 100%; qrxn = −qsolution; ΔH° = ΣnΔH°f(products) − ΣnΔH°f(reactants); ΔH ≈ bonds broken − bonds formed; rate = k[A]m[B]n; ln[A] = ln[A]0 − kt; 1 ÷ [A] = 1 ÷ [A]0 + kt; t1/2 = 0.693 ÷ k; ln(k2 ÷ k1) = −(Ea ÷ R)(1 ÷ T2 − 1 ÷ T1)
Equilibrium systemsK = products over reactants, each to its coefficient; Kp = Kc(RT)Δn; Q against K for direction; Kw = [H3O+][OH-]; pH = −log[H3O+]; [H3O+] = √(Kac0); KaKb = Kw; pH = pKa + log([A-] ÷ [HA]); Ksp = s2, 4s3 or 108s5; Koverall = K1K2
Energy and electron transferS = k ln W; ΔSsurr = −ΔH ÷ T; ΔS° = ΣnS°(products) − ΣnS°(reactants); ΔG = ΔH − TΔS; ΔG° = −RT ln K; T = ΔH° ÷ ΔS°; E°cell = E°cathode − E°anode; ΔG° = −nFE°cell; E°cell = (0.0592 ÷ n) log K; Ecell = E°cell − (0.0592 ÷ n) log Q; moles of e- = It ÷ F
Descriptive and nuclearZeff ≈ Z − core electrons; d electrons = group − charge; Δ = hc ÷ λ; μ = √(n(n + 2)); degrees of unsaturation = (2C + 2 − H) ÷ 2; N = A − Z; binding energy = Δm × 931.5 MeV/u; k = 0.693 ÷ t1/2; ln(N0 ÷ N) = kt; sieverts = grays × quality factor

Twelve rules that settle most questions

  • Every route between two substances runs through moles. Convert to moles first.
  • Shape decides polarity. Symmetry can cancel every bond dipole.
  • Any gas law or thermodynamics problem needs kelvin, never Celsius.
  • On a slope of a heating curve use q = mcΔT. On a flat use the phase change.
  • Colligative properties count particles, so never forget i.
  • Rate law exponents come from experiment, except for an elementary step.
  • Activation energy is the hill. ΔH is the drop from start to finish.
  • Find Q before you build an ICE table, because it gives you the sign of x.
  • Drop x only when c0 ÷ K clears 400, then confirm with the five percent rule.
  • Spontaneous is about direction, never about speed.
  • Positive E°cell, negative ΔG° and K above 1 are one fact said three ways.
  • Put every answer back into the equation it came from. If it does not reproduce the constant, something is wrong.

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