- Calculate ΔH° from enthalpies of formation and with Hess's law
- Estimate and calculate the entropy change of a reaction
- Use ΔG = ΔH − TΔS for spontaneity and ΔG° = −RT ln K for the equilibrium constant
Iron rusts by itself, yet water does not split into hydrogen and oxygen on its own. Most exothermic reactions are spontaneous, but ice melts at room temperature by absorbing heat — an endothermic process can be spontaneous too. So heat alone does not decide. Thermodynamics combines two quantities, enthalpy and entropy, into the Gibbs energy, which tells us exactly whether a reaction is “allowed”.
Enthalpy and Hess's law
The first law of thermodynamics: ΔU = q + w. At constant pressure the expansion work is w = −pΔV, so qp = ΔU + pΔV. This combination is called enthalpy, H = U + pV, so the heat absorbed at constant pressure equals ΔH. ΔH < 0 means exothermic, ΔH > 0 endothermic. H is a state function: its change depends only on the initial and final states, not on the path. This is Hess's law — chemical equations can be added like algebraic ones.
- ΔfH°standard enthalpy of formation, kJ/mol; zero for elements in their standard states
- νstoichiometric coefficient in the equation
A consequence of Hess's law: products minus reactants
| Substance | ΔfH°, kJ/mol | S°, J/(mol·K) |
|---|---|---|
| CO₂(g) | −393.5 | 213.7 |
| H₂O(l) | −285.8 | 69.9 |
| CH₄(g) | −74.8 | 186.3 |
| NH₃(g) | −46.1 | 192.5 |
| CaCO₃(s) | −1206.9 | 92.9 |
| CaO(s) | −635.1 | 39.8 |
| N₂(g) / H₂(g) / O₂(g) | 0 | 191.6 / 130.7 / 205.1 |
Calculate the enthalpy of combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).
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= −965.1 + 74.8 = −890.3 kJ/mol.
That is the heat released by burning 1 mol (16 g) of methane on a gas stove.
C + O₂ → CO₂, ΔH₁ = −393.5 kJ; CO + ½O₂ → CO₂, ΔH₂ = −283.0 kJ. Find the enthalpy of formation of CO (hard to measure directly, because some CO₂ always forms as well).
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Add: C + O₂ + CO₂ → CO₂ + CO + ½O₂ ⇒ C + ½O₂ → CO.
ΔfH°(CO) = −393.5 + 283.0 = −110.5 kJ/mol.
Average bond energies: E(H–H) = 436, E(Cl–Cl) = 243, E(H–Cl) = 432 kJ/mol. Estimate ΔH for H₂ + Cl₂ → 2HCl.
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Formed: 2 H–Cl = 2 · 432 = 864 kJ (energy released).
ΔH ≈ 679 − 864 = −185 kJ — almost exactly the tabulated value (2 · (−92.3) = −184.6 kJ).
Entropy
Entropy measures the number of microstates W of a system: S = k · ln W (Boltzmann's formula). The more disorder, the larger S: solid < liquid < gas. Reactions that increase the number of moles of gas have ΔS > 0. The second law of thermodynamics: in a spontaneous process the entropy of the universe (system + surroundings) increases. The surroundings' entropy change depends on the heat the system gives them: ΔS(surr) = −ΔH / T.
- kBoltzmann constant, 1.381·10⁻²³ J/K (k = R / Nₐ)
- Wnumber of microstates consistent with the macrostate
- S°standard molar entropy, J/(mol·K)
For example, N₂ + 3H₂ → 2NH₃: ΔS° = 2 · 192.5 − (191.6 + 3 · 130.7) = −198.7 J/K (4 mol of gas → 2 mol)
Gibbs energy and spontaneity
Derivation. For a spontaneous process ΔS(universe) = ΔS − ΔH / T > 0. Multiply the inequality by −T (which flips it): ΔH − TΔS < 0. Defining G = H − TS gives ΔG = ΔH − TΔS at constant T and p. So the condition for spontaneity is simply ΔG < 0, expressed entirely through properties of the system.
- ΔGchange in Gibbs energy, kJ/mol; ΔG < 0 means spontaneous
- ΔSentropy change, J/(mol·K)
- Tabsolute temperature, K
- R8.314 J/(mol·K)
- Kequilibrium constant
| ΔH | ΔS | Result |
|---|---|---|
| − | + | spontaneous at every temperature |
| + | − | never spontaneous |
| − | − | at low temperature: T < ΔH / ΔS |
| + | + | at high temperature: T > ΔH / ΔS |
CaCO₃(s) → CaO(s) + CO₂(g). Using the table, calculate ΔH°, ΔS° and ΔG° at 298 K. Above what temperature does limestone decompose spontaneously?
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ΔS° = (39.8 + 213.7) − 92.9 = +160.6 J/K = 0.1606 kJ/K.
ΔG°(298) = 178.3 − 298 · 0.1606 ≈ +130.4 kJ > 0 — no decomposition at room temperature.
Temperature where ΔG = 0: T = ΔH / ΔS = 178 300 / 160.6 ≈ 1110 K (≈ 840 °C). The estimate assumes ΔH and ΔS do not depend on temperature.
For N₂ + 3H₂ ⇌ 2NH₃ at 298 K, ΔG° = −32.8 kJ. Find the equilibrium constant.
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K = exp(13.24) ≈ 5.6·10⁵ — the equilibrium lies far to the right. Yet at 298 K the reaction hardly proceeds: thermodynamics says “allowed”, but kinetics decides how fast.
Key points
- At constant pressure q = ΔH; ΔᵣH° = Σ ΔfH°(products) − Σ ΔfH°(reactants).
- Hess's law: ΔH is path-independent; equations can be added, reversed and scaled.
- S = k ln W; more moles of gas give ΔS > 0; ΔS(surr) = −ΔH / T.
- ΔG = ΔH − TΔS < 0 means spontaneous; crossover temperature T = ΔH / ΔS.
- ΔG° = −RT ln K; at 298 K every −5.7 kJ raises K tenfold.
Check yourself
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