Spontaneous processes: why ΔH is not enough
A spontaneous process is one that happens on its own once it has started, without continuous outside help: water flowing downhill, gas filling a room, iron rusting. Spontaneous does not mean fast (diamond turning into graphite is spontaneous but extremely slow).
Many spontaneous reactions are exothermic, so one might think 'ΔH < 0 means spontaneous'. But ice melting at room temperature, ammonium nitrate dissolving and water evaporating are all endothermic and still spontaneous. So we need a second factor: entropy.
Entropy and the second and third laws
Entropy (S) measures how spread out (random) the particles and their energy are. It is a state function. S increases from solid → liquid → gas, when a solid dissolves, when the number of gas moles increases, and when temperature rises.
For a reversible change at temperature T: ΔS = q_rev / T (unit J K⁻¹ mol⁻¹). Heat added at a low temperature causes a bigger increase in randomness than the same heat at a high temperature.
Entropy change of reactions
ΔrS° = ΣS°(products) − ΣS°(reactants). Note: elements do not have S° = 0 (unlike ΔfH°).
Second law
The entropy of an isolated system (the universe) increases in every spontaneous change: ΔS_total = ΔS_sys + ΔS_surr > 0. At equilibrium ΔS_total = 0. For the surroundings, ΔS_surr = −ΔH_sys / T.
Third law
The entropy of a perfectly crystalline pure substance is zero at absolute zero (0 K). This lets us find absolute entropies of substances.
Gibbs energy change: ΔG = ΔH − TΔS
Measuring the surroundings is hard, so J. W. Gibbs defined a function of the system alone: G = H − TS. At constant temperature and pressure: ΔG = ΔH − TΔS. It comes from ΔS_total = ΔS_sys − ΔH_sys/T; multiplying by −T gives −TΔS_total = ΔH − TΔS = ΔG. So ΔS_total > 0 is the same as ΔG < 0.
- ΔG < 0: spontaneous.
- ΔG = 0: at equilibrium.
- ΔG > 0: not spontaneous (the reverse is spontaneous).
Effect of temperature (four cases)
- ΔH −, ΔS + → ΔG always negative: spontaneous at all T.
- ΔH +, ΔS − → ΔG always positive: never spontaneous.
- ΔH −, ΔS − → spontaneous at low T (below T = ΔH/ΔS).
- ΔH +, ΔS + → spontaneous at high T (above T = ΔH/ΔS).
ΔG is also the maximum useful (non-expansion) work the system can do, which is why it is called free energy.
Gibbs energy and equilibrium
As a reaction proceeds, G of the mixture falls until it reaches a minimum. At that point the forward and backward reactions balance: equilibrium, with ΔG = 0.
The general relation is ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. At equilibrium ΔG = 0 and Q = K, so ΔG° = −RT ln K = −2.303 RT log K.
- ΔG° negative → K > 1 → products are favoured.
- ΔG° positive → K < 1 → reactants are favoured.
- ΔG° = 0 → K = 1.
This lets us predict how far a reaction goes from thermal data alone. Also ΔG° = ΔH° − TΔS°, so both routes can be combined.
In the exam
Expect questions on the sign table (2 marks), ΔG and the crossover temperature (3 marks), and K from ΔG° (3 marks). Keep units the same: change ΔS from J to kJ, and use R = 8.314 J K⁻¹ mol⁻¹ with ΔG° in J.
Try it: predict, then check
At home: put a drop of food colour into still water and watch it spread by itself (ΔS > 0). Now think: could the colour gather back into one drop on its own? In the 3D free play, set ΔH = +40 kJ and ΔS = +110 J/K (like boiling water). Predict the temperature where ΔG becomes zero (40 000 ÷ 110 ≈ 364 K), then slide T and check where ΔG changes sign.
Key formulas and definitions
- ΔS = q_rev / T
- ΔrS° = ΣS°(products) − ΣS°(reactants)
- ΔS_total = ΔS_sys + ΔS_surr > 0 (spontaneous); ΔS_surr = −ΔH_sys / T
- G = H − TS; ΔG = ΔH − TΔS (constant T, p)
- ΔG < 0 spontaneous; ΔG = 0 equilibrium; ΔG > 0 non-spontaneous
- Crossover temperature: T = ΔH / ΔS
- ΔG = ΔG° + RT ln Q; ΔG° = −RT ln K = −2.303 RT log K
Worked examples
1. Find ΔS when 1 mol of ice melts at 273 K (ΔfusH = 6.01 kJ mol⁻¹).
ΔS = q_rev / T = 6010 ÷ 273 = +22.0 J K⁻¹ mol⁻¹. Positive: the liquid is more disordered.
2. Find the entropy of vaporisation of water at 373 K (ΔvapH = 40.79 kJ mol⁻¹).
ΔS = 40 790 ÷ 373 = +109.4 J K⁻¹ mol⁻¹.
3. A reaction at 300 K has ΔH = −100 kJ and ΔS_sys = −200 J K⁻¹. Is it spontaneous? Use total entropy.
ΔS_surr = −ΔH/T = +100 000 ÷ 300 = +333 J K⁻¹. ΔS_total = −200 + 333 = +133 J K⁻¹ > 0, so it is spontaneous.
4. Check the same reaction using ΔG.
ΔG = ΔH − TΔS = −100 − 300 × (−0.200) = −100 + 60 = −40 kJ. Negative, so spontaneous. (Also −TΔS_total = −300 × 0.133 = −40 kJ, the same.)
5. For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178 kJ and ΔS° = +160.6 J K⁻¹. Is it spontaneous at 298 K?
ΔG° = 178 − 298 × 0.1606 = 178 − 47.9 = +130.1 kJ. Positive, so not spontaneous at room temperature.
6. Above what temperature does the decomposition of CaCO₃ (Example 5) become spontaneous?
At the crossover ΔG = 0: T = ΔH/ΔS = 178 000 ÷ 160.6 = 1108 K (about 835 °C). Above this ΔG < 0.
7. The equilibrium constant of a reaction at 298 K is 10. Find ΔG°.
ΔG° = −2.303 RT log K = −2.303 × 8.314 × 298 × log 10 = −5706 J ≈ −5.71 kJ mol⁻¹.
8. ΔG° of a reaction at 298 K is −11.42 kJ mol⁻¹. Find K.
log K = −ΔG° ÷ (2.303 RT) = 11 420 ÷ (2.303 × 8.314 × 298) = 11 420 ÷ 5706 = 2.00. K = 10² = 100.
9. For N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K: S° (J K⁻¹ mol⁻¹) N₂ 191.6, H₂ 130.7, NH₃ 192.5; ΔH° = −92.2 kJ. Find ΔS°, ΔG° and K.
ΔS° = 2(192.5) − [191.6 + 3(130.7)] = 385.0 − 583.7 = −198.7 J K⁻¹. ΔG° = −92.2 − 298 × (−0.1987) = −92.2 + 59.2 = −33.0 kJ. log K = 33 000 ÷ 5706 = 5.78, so K ≈ 6 × 10⁵. Spontaneous at 298 K; products are favoured.
Common mistakes
- Mixing units in ΔG = ΔH − TΔS: ΔH is in kJ but ΔS is in J K⁻¹. Divide ΔS by 1000 first.
- Thinking 'spontaneous' means 'fast'. Thermodynamics tells whether a change can happen, not how quickly.
- Taking S° of elements as zero. Only ΔfH° of elements is zero; S° of elements is positive.
- Using ΔG° = −2.303 RT log K with ΔG° in kJ and R = 8.314. Keep ΔG° in J with R in J K⁻¹ mol⁻¹.