Thermodynamics
Some changes happen all by themselves. Ice melts in a warm room. Iron rusts. A drop of dye spreads through water. None of them need a push. Other changes never start on their own, no matter how long you wait. This chapter shows you how to tell the two apart. You will meet entropy, a measure of how spread out matter and energy are. You will meet the second law, which says the entropy of the universe always grows. Then comes Gibbs free energy. Its sign tells you which way a change will go, and its size sets the value of K.
By the end of this chapter you can
- Say what makes a change spontaneous, and why that is not the same as fast.
- Describe entropy as the spread of matter and of energy, and use S = k ln W.
- State the second and third laws, and add up ΔS for a system plus its surroundings.
- Find ΔS° for a reaction from a table of standard entropies.
- Predict the sign of ΔS° by counting moles of gas.
- Use ΔG = ΔH − TΔS and sort out the four sign cases.
- Link ΔG° to K, and find the crossover temperature where the sign flips.
Section 1Spontaneous change
What spontaneous means
A change is spontaneous if it keeps going on its own once it starts. Nothing outside has to drive it. A ball rolls down a hill. A hot mug cools on a desk. Salt dissolves in water. Each one runs one way and settles there.
A change that needs steady outside work is nonspontaneous. Rolling the ball back up the hill is one. Heating the mug back up is another. The reverse of a spontaneous change is always nonspontaneous. Together the pair acts like a one-way street with a sign posted on it.
Be careful with the word itself. In chemistry it does not mean sudden. It does not mean violent. It only means the change is favored in that direction.
Spontaneous does not mean fast
Here is the trap that catches almost everyone: spontaneous says nothing at all about the speed of a change.
Diamond makes the point, because at room temperature diamond is favored to turn into graphite. Graphite is the standard form of carbon, so its free energy of formation is 0 kJ/mol. For diamond the table lists +2.9 kJ/mol.
Worked example · does diamond last forever?
The change C(diamond) → C(graphite)
Set up ΔG° = ΔG°f(graphite) − ΔG°f(diamond)
Put in numbers ΔG° = 0 kJ/mol − (+2.9 kJ/mol) = −2.9 kJ/mol
Read the sign The value is negative, so graphite is the favored side. Diamond is the one that should change.
Reality check No ring has ever turned grey. The atoms sit in strong bonds and cannot move without a huge push. The rate is near zero, so the jewel is safe.
Thermodynamics names the destination. Kinetics sets the travel time. Keep the two jobs apart and this chapter gets much easier.
Heat alone does not decide
Many spontaneous changes give off heat. Wood burns. An acid neutralizes a base. Both warm the room. So people once thought heat release was the whole rule.
It is not the whole rule, as an ice cube melting at 25 °C shows. Melting soaks heat up, so it is endothermic, and yet it still happens with no help at all. A cold pack does the same trick. Ammonium nitrate dissolves, the bag turns cold, and nobody pushes it.
So a second force has to be at work. Energy is only half the story, and the other half is how spread out the matter and the energy end up. The next section gives that idea a name and a number.
- Spontaneous means the change goes forward with no outside drive.
- The reverse of a spontaneous change always needs work from outside.
- Spontaneous does not mean fast. Diamond to graphite is favored and still takes ages.
- Giving off heat helps, but endothermic changes can be spontaneous too.
- Spreading of matter and energy is the missing second factor.
Section 2Entropy
Entropy counts the ways
Entropy gets the symbol S. It measures how spread out the matter and the energy of a sample are. A state with many ways to arrange itself has high entropy. A state with few ways has low entropy.
Picture four gas particles in a box split down the middle. Each particle can sit left or right. That gives 2 × 2 × 2 × 2 = 16 full arrangements. Each arrangement is one microstate. Only one microstate puts all four on the left, while six of them put two on each side. So the even split is six times as likely as the lopsided pile.
Now swap in a real sample, where one mole holds 6.022 × 1023 particles. The even split still wins, but the margin becomes absurd. The one-sided case has odds so tiny that it never shows up. That is the real reason gas fills both bulbs and stays there.
The Boltzmann equation
Ludwig Boltzmann turned that counting into a formula.
S = k ln W
Here W is the number of microstates. The letter k is the Boltzmann constant, 1.381 × 10−23 J/K. Watch what the log does, because multiplying W by ten raises S by only k ln 10. Entropy creeps up while the number of ways explodes.
Worked example · entropy of four particles, then of a mole
Part A Four particles, two halves, so W = 24 = 16.
Put in numbers S = (1.381 × 10−23 J/K)(ln 16) = (1.381 × 10−23)(2.773) = 3.83 × 10−23 J/K
Part B Let one mole of gas double its volume. Each particle now has twice as many spots, so W doubles once per particle.
Ratio of ways Wfinal / Wstart = 2NA
Take the log ΔS = k ln(2NA) = NAk ln 2 = R ln 2
Finish ΔS = (8.314 J/mol·K)(0.6931) = +5.76 J/mol·K
Check The gas spread out, so ΔS came out positive. Note that NAk is just R, the same gas constant from Chapter 9.
Energy spreads too
Matter is one half of the story. Energy is the other half. Particles store energy by moving, spinning and vibrating. Each of those motions has its own ladder of energy levels. The sample's energy can be shared over those levels in many ways, and each way is another microstate.
This explains why entropy climbs with temperature, since added heat lifts particles onto higher rungs. More rungs come into play, so there are more ways to park the same energy.
Four patterns follow from all this, and they answer most questions you will face.
- Gas beats liquid beats solid. Water shows it: 70.0 J/mol·K as a liquid, 188.8 J/mol·K as steam.
- Warmer beats cooler for the same substance in the same phase.
- Larger molecules beat smaller ones, since they hold more ways to vibrate. Ethane sits at 229.2 J/mol·K and ethene at 219.3 J/mol·K.
- A dissolved solid beats the solid sitting in a pile, because its particles mix through the liquid.
- Entropy S measures the spread of matter and of energy.
- A microstate is one full arrangement of where the particles are and which levels hold the energy.
- S = k ln W, with k = 1.381 × 10−23 J/K.
- Doubling the volume of one mole of gas adds R ln 2 = 5.76 J/mol·K.
- Entropy units are J/mol·K, not kJ/mol·K. That detail matters later.
Section 3The second and third laws
The second law
The second law of thermodynamics gives the real test. A change is spontaneous when the total entropy of the universe goes up.
ΔSuniv = ΔSsys + ΔSsurr > 0
Two ledgers are open at once. The system is the part you care about, such as the chemicals in a flask. The surroundings are everything else, such as the glass, the bench and the air. Either ledger may show a loss. Only the sum has to grow.
That rule is how an endothermic change can still be spontaneous. Melting ice raises the entropy of the water by a lot. It lowers the entropy of the room by a little less. The sum comes out positive, so the ice melts.
Heat and the surroundings
The surroundings are enormous, so their temperature barely budges when they take in heat. That makes their entropy change easy to write.
ΔSsurr = −qsys / T = −ΔHsys / T
Read the minus sign with care, because heat that leaves the system lands in the surroundings. An exothermic change has a negative ΔH, so −ΔH/T turns positive and the surroundings gain entropy.
Now look again at the T sitting on the bottom of that fraction. The same joule of heat buys more entropy in a cold room than in a hot one. A dollar means more to a person with almost nothing. This single fact is why temperature decides so many reactions.
Worked example · melting ice at three temperatures
One mole of ice melts. Fusion takes in 6.01 kJ/mol, which is 6010 J/mol. The melt itself happens at 273.15 K, so the system gains the same entropy each time.
System ΔSsys = 6010 J / 273.15 K = +22.0 J/K
Surroundings at 0 °C ΔSsurr = −6010 J / 273.15 K = −22.0 J/K. Total = 22.0 − 22.0 = 0 J/K, so ice and water sit together and neither one wins.
Surroundings at +10 °C ΔSsurr = −6010 J / 283.15 K = −21.2 J/K. Total = 22.0 − 21.2 = +0.8 J/K, so the ice melts.
Surroundings at −10 °C ΔSsurr = −6010 J / 263.15 K = −22.8 J/K. Total = 22.0 − 22.8 = −0.8 J/K, so the ice holds. Freezing is the spontaneous way instead.
What moved The system term never changed. Only the warmth of the room changed, and that was enough to flip the sign.
The third law
The third law of thermodynamics sets the zero point. A perfect crystal of a pure substance at 0 K has an entropy of exactly zero.
The reason is pure counting, because at absolute zero each particle sits in one fixed spot. All the energy rests in the lowest level. There is only one microstate, so W = 1 and S = k ln 1 = 0.
This matters more than it first looks, since enthalpy has no natural zero and tables can only list changes such as ΔH°f. Entropy, by contrast, does have a true zero of its own. So a table can list a real, absolute value for every substance. Those values are standard molar entropies, written S°, measured at 298.15 K and 1 bar. Every one of them is positive, even for elements. The next section spends them.
- Second law: a change is spontaneous when ΔSsys + ΔSsurr is positive.
- ΔSsurr = −ΔHsys/T at steady pressure.
- Cold surroundings gain more entropy per joule than warm ones.
- ΔSuniv = 0 is the fingerprint of equilibrium.
- Third law: a perfect pure crystal at 0 K has S = 0, so S° values are absolute and positive.
Section 4Standard entropy change
Adding up the table
Because entropy has a true zero, a table can list one absolute value per substance. Those S° values are what you spend here. Every number in this chapter is taken from the standard thermodynamic table in Appendix G of OpenStax Chemistry 2e, measured at 298.15 K and 1 bar.
The standard entropy change of a reaction is built the same way as a heat of reaction.
ΔS°rxn = Σ n S°(products) − Σ n S°(reactants)
Multiply each S° by its coefficient from the balanced equation. Then subtract the reactant total from the product total. Two habits will save you here. Elements do not drop out, because their S° is not zero. And the answer arrives in J/K, since S° is tabulated in J/mol·K.
| Substance | S° (J/mol·K) | Substance | S° (J/mol·K) |
|---|---|---|---|
| H2(g) | 130.7 | H2O(l) | 70.0 |
| O2(g) | 205.2 | H2O(g) | 188.8 |
| N2(g) | 191.6 | NH3(g) | 192.8 |
| Cl2(g) | 223.1 | HCl(g) | 186.9 |
| C(graphite) | 5.7 | CO2(g) | 213.8 |
| CaCO3(s) | 91.7 | CaO(s) | 38.1 |
| I2(s) | 116.1 | I2(g) | 260.7 |
| C2H4(g) | 219.3 | C2H6(g) | 229.2 |
Worked example · burning hydrogen to liquid water
Equation 2 H2(g) + O2(g) → 2 H2O(l)
Products 2 mol × 70.0 J/mol·K = 140.0 J/K
Reactants 2 mol × 130.7 + 1 mol × 205.2 = 261.4 + 205.2 = 466.6 J/K
Subtract ΔS° = 140.0 J/K − 466.6 J/K = −326.6 J/K
Does it make sense? Three moles of gas became zero moles of gas. A big drop was the only possible answer.
Worked example · roasting limestone
Equation CaCO3(s) → CaO(s) + CO2(g)
Products 38.1 + 213.8 = 251.9 J/K
Reactants 91.7 J/K
Subtract ΔS° = 251.9 J/K − 91.7 J/K = +160.2 J/K
Does it make sense? A solid released a mole of gas. One free gas molecule is worth far more entropy than the solid it came from. Hold on to this value, because Section 5 uses it again.
Reading the sign without a table
Often the sign is all you need, so count the moles of gas on each side of the balanced equation. Gas holds so much more entropy than liquid or solid that the gas count almost always decides.
- More moles of gas among the products means ΔS° is positive.
- Fewer moles of gas among the products means ΔS° is negative.
- Equal moles of gas means the value will be small, so look at the other clues.
When the gas count ties, reach for the backup clues. A solid that melts or dissolves gains entropy. Two small molecules joining into one large one lose it. More total particles beats fewer.
Try the tie case on HCl. For H2(g) + Cl2(g) → 2 HCl(g) the table gives 2(186.9) − (130.7 + 223.1) = 373.8 − 353.8 = +20.0 J/K. Two moles of gas went to two moles of gas, and the value is tiny next to the −326.6 J/K above.
- ΔS°rxn = sum of nS° for products minus sum of nS° for reactants.
- S° for an element is positive, so elements never drop out of the sum.
- ΔS° comes out in J/K, while ΔH° comes out in kJ.
- Gas moles rising means a positive ΔS°. Gas moles falling means a negative one.
- When the gas count ties, ΔS° is small and the phases decide.
Section 5Gibbs free energy
One number, system only
Tracking the surroundings every time is a chore, and Josiah Willard Gibbs found a way around it. Start from the second law, then swap in the heat term from Section 3.
ΔSsys − ΔHsys/T > 0
Now multiply every term by −T. Because T is always positive on the Kelvin scale, multiplying by a negative number flips the direction of the sign.
ΔHsys − TΔSsys < 0
Gibbs gave that left side a name. Gibbs free energy carries the symbol G, and its change is the whole answer.
ΔG = ΔH − TΔS
At steady temperature and pressure, one sign settles the case. A negative ΔG means the forward change is spontaneous. A positive ΔG means the reverse one is. A ΔG of zero means the system is at equilibrium. Best of all, every term belongs to the system, so the surroundings never enter the work.
You can also build ΔG° straight from a table, the same way as ΔH°. Appendix G lists a standard free energy of formation, ΔG°f, for each substance. Sum products minus reactants. That shortcut only works at 298.15 K, since T is baked into the listed values.
The unit trap
This equation is where units bite hardest. ΔH° comes out of the table in kJ. ΔS° comes out in J/K. You cannot subtract one from the other until they match.
Pick one fix and stick with it: divide ΔS° by 1000 to reach kJ/K, or multiply ΔH° by 1000 to reach J. Forget this and your answer is wrong by a factor of a thousand. The sign often comes out wrong too.
Worked example · roasting limestone, cold and hot
Equation CaCO3(s) → CaO(s) + CO2(g)
Enthalpy ΔH° = (−634.9) + (−393.5) − (−1207.6) = +179.2 kJ
Entropy ΔS° = +160.2 J/K from Section 4. Convert it: 160.2 J/K ÷ 1000 = 0.1602 kJ/K
At 298.15 K ΔG° = 179.2 kJ − (298.15 K)(0.1602 kJ/K) = 179.2 − 47.8 = +131.4 kJ
Check it another way From the free energies of formation: (−603.3) + (−394.4) − (−1129.1) = +131.4 kJ. The two routes agree.
At 1200 K ΔG° = 179.2 kJ − (1200 K)(0.1602 kJ/K) = 179.2 − 192.2 = −13.0 kJ
Meaning Limestone sits safely on a cold shelf. Push a lime kiln near 1200 K and it breaks apart on its own. Nothing about the chemicals changed. Only the size of the TΔS term changed.
The four sign cases
Since ΔH° and ΔS° each have two possible signs, there are four combinations. Two of them give the same answer at every temperature, while the other two switch over somewhere.
| ΔH | ΔS | Sign of ΔG | Verdict |
|---|---|---|---|
| negative | positive | always negative | Spontaneous at every temperature. Both terms help. |
| positive | negative | always positive | Never spontaneous. Both terms fight you. |
| negative | negative | negative when T is low | Spontaneous when cold. Heat release has to outweigh TΔS. |
| positive | positive | negative when T is high | Spontaneous when hot. TΔS has to outgrow ΔH. |
- ΔG = ΔH − TΔS, and the sign of ΔG decides direction.
- Negative means forward, positive means reverse, zero means equilibrium.
- Always convert ΔS° from J/K to kJ/K before subtracting.
- Two sign cases hold at all temperatures and two of them switch over.
- ΔG°f sums give ΔG° directly, but only at 298.15 K.
Section 6Free energy, K and temperature
Free energy sets the size of K
Chapter 13 gave you the equilibrium constant K. Free energy is what fixes its value.
ΔG° = −RT ln K
Here R = 8.314 J/mol·K and T is in kelvin. Since R is written in joules, put ΔG° in joules as well. Rearranging gives K = e−ΔG°/RT.
The link is easy to read off. A negative ΔG° makes K larger than 1, so products win at equilibrium. A positive ΔG° makes K smaller than 1, so reactants win. A ΔG° of exactly zero gives K = 1.
Size matters as much as sign, because at 298.15 K every 5.71 kJ of ΔG° moves K by a factor of ten. That figure is just RT ln 10. So a modest −30 kJ already buys a huge K.
One caution before the math: ΔG° describes the standard state, with each gas at 1 bar and each solute at 1 mol/L. The push you feel at some other mixture is ΔG = ΔG° + RT ln Q. When Q climbs to K, the log term cancels ΔG° and ΔG drops to zero. That is equilibrium, seen from the free energy side.
Worked example · K for making ammonia at 298.15 K
Equation N2(g) + 3 H2(g) → 2 NH3(g)
Free energy ΔG° = 2(−16.4) − 0 − 0 = −32.8 kJ = −32800 J
Set up ln K = −ΔG°/RT = −(−32800 J) / [(8.314 J/mol·K)(298.15 K)]
Denominator (8.314)(298.15) = 2478.8 J/mol
Divide ln K = 32800 / 2478.8 = 13.23
Finish K = e13.23 = 5.6 × 105
Read it Ammonia is strongly favored at room temperature. The reaction is simply far too slow to use there.
Finding the crossover temperature
In the two switching cases a single temperature marks the turn, and you find it by setting ΔG to zero in ΔG = ΔH − TΔS.
T = ΔH° / ΔS°
That value is the crossover temperature. On one side of it ΔG° is negative, on the other side it is positive, and at the crossover itself K = 1. The method assumes ΔH° and ΔS° hardly drift with temperature, which holds well enough over a few hundred kelvin.
Test the idea on boiling water, where the table gives ΔH° = +44.0 kJ and ΔS° = +118.8 J/K for H2O(l) → H2O(g). So T = 44000 J ÷ 118.8 J/K = 370 K. The real boiling point is 373 K, which is a fine result from two table entries.
Worked example · where ammonia gives up
Data ΔH° = −91.8 kJ and ΔS° = −198.1 J/K, so ΔS° = −0.1981 kJ/K
Crossover T = (−91.8 kJ) ÷ (−0.1981 kJ/K) = 463 K, which is 190 °C
Now go hotter At 700 K, ΔG° = −91.8 − (700)(−0.1981) = −91.8 + 138.7 = +46.9 kJ
Turn that into K ln K = −46900 / [(8.314)(700)] = −46900 / 5819.8 = −8.06
Finish K = e−8.06 = 3.2 × 10−4
The real squeeze K fell from 5.6 × 105 to 3.2 × 10−4. Heat is needed for speed, yet heat ruins the yield. Plants answer by running at high pressure, which Chapter 13 showed pushes a gas reaction toward the smaller side.
- ΔG° = −RT ln K, with R = 8.314 J/mol·K and ΔG° in joules.
- Negative ΔG° gives K above 1. Positive gives K below 1. Zero gives K = 1.
- At 298.15 K, 5.71 kJ of ΔG° is worth one power of ten in K.
- ΔG = ΔG° + RT ln Q, and ΔG = 0 once Q = K.
- The crossover temperature is T = ΔH°/ΔS°, where K = 1.
The chapter in one breath
- Spontaneous means a change goes forward on its own. It says nothing about how fast.
- Entropy measures the spread of matter and of energy, and S = k ln W ties it to counting.
- The second law is the real test. A change happens when ΔSsys + ΔSsurr is positive.
- The surroundings keep score as −ΔH/T, so cold surroundings gain more per joule.
- The third law sets S = 0 for a perfect crystal at 0 K, which is why S° values are absolute.
- ΔS°rxn is products minus reactants, and the moles of gas usually fix the sign.
- ΔG = ΔH − TΔS folds both books into one number for the system alone.
- Four sign cases: always, never, cold only, hot only. The last two have a crossover at T = ΔH°/ΔS°.
- ΔG° = −RT ln K, so the sign of ΔG° and the size of K are the same fact twice.
ReferenceWords worth knowing
These are the terms this chapter leans on. Each one shows up in the sections above, and the list runs in alphabetical order.
- Boltzmann constant (k)
- The gas constant per particle, 1.381 × 10−23 J/K. It equals R/NA.
- Crossover temperature
- The temperature where ΔG° equals zero, found from T = ΔH°/ΔS°. On its two sides the sign of ΔG° is different.
- Endothermic
- Taking heat in, so ΔH is positive. Melting ice is one example.
- Entropy (S)
- A measure of how spread out the matter and energy of a sample are. Units are J/mol·K.
- Exothermic
- Giving heat off, so ΔH is negative. Burning fuel is one example.
- Gibbs free energy (G)
- The quantity ΔH − TΔS. Its change decides direction at steady temperature and pressure.
- Kinetics
- The study of how fast a change goes. It is a separate question from whether the change is favored.
- Microstate
- One complete arrangement of where every particle sits and which level holds each piece of energy.
- Nonspontaneous change
- A change that needs steady work from outside to keep going. Its reverse is spontaneous.
- Reaction quotient (Q)
- The same ratio as K, built from the amounts present right now instead of the equilibrium amounts.
- Second law of thermodynamics
- The total entropy of a system plus its surroundings rises in any spontaneous change.
- Spontaneous change
- A change that keeps going on its own, with no outside drive. It carries no promise of speed.
- Standard entropy change (ΔS°)
- The entropy change for a reaction with every substance in its standard state. Found as products minus reactants.
- Standard free energy of formation (ΔG°f)
- The free energy change for making one mole of a substance from its elements at 298.15 K and 1 bar. It is zero for an element in its standard form.
- Standard molar entropy (S°)
- The absolute entropy of one mole of a substance at 298.15 K and 1 bar. Every listed value is positive.
- Surroundings
- Everything outside the system, such as the flask, the bench and the air.
- System
- The part of the world you are tracking, such as the chemicals inside the flask.
- Third law of thermodynamics
- A perfect crystal of a pure substance at 0 K has an entropy of exactly zero.
- Universe
- In thermodynamics, the system plus its surroundings. The second law applies to that total.