Chemical Thermodynamics, Chemical Equilibria and Statistical Thermodynamics
2. Entropy, the third law and the criteria of spontaneity
The second law defines entropy through dS = dq_rev/T and states that the entropy of an isolated system never decreases: ΔS_total = ΔS_sys + ΔS_surr ≥ 0, with equality at equilibrium. Standard results: isothermal ideal-gas expansion ΔS = nR ln(V₂/V₁); heating at constant pressure ΔS = nC_P ln(T₂/T₁); a phase transition at its equilibrium temperature ΔS = ΔH_trs/T_trs. Trouton’s rule puts ΔvapS near 85 J K⁻¹ mol⁻¹ for many unassociated liquids at their normal boiling point; hydrogen-bonded liquids such as water (about 109) lie above it.
The third law: the entropy of a perfect crystal is zero at 0 K. This makes absolute entropies measurable, S(T) = ∫₀ᵀ (C_P/T) dT plus ΔH/T for each transition, with the Debye T³ law covering the lowest temperatures. Crystals that freeze in disorder keep a residual entropy S = k ln W: solid CO, whose molecules can point either way, has R ln 2 = 5.76 J K⁻¹ mol⁻¹, and ice, by Pauling’s count of proton arrangements, R ln(3/2) = 3.37 J K⁻¹ mol⁻¹.
| Constraint | Function | Spontaneous change | Equilibrium |
|---|---|---|---|
| isolated (U, V constant) | S | dS > 0 | S maximum |
| T, V constant | Helmholtz A = U − TS | dA < 0 | A minimum |
| T, P constant | Gibbs G = H − TS | dG < 0 | G minimum |
| S, V constant | U | dU < 0 | U minimum |
3. Fundamental equations, Maxwell and Gibbs–Helmholtz relations
For a closed system doing only PV work the four fundamental equations are dU = TdS − PdV, dH = TdS + VdP, dA = −SdT − PdV and dG = −SdT + VdP. Reading off coefficients gives (∂G/∂T)_P = −S and (∂G/∂P)_T = V, so for an ideal gas G(P₂) − G(P₁) = nRT ln(P₂/P₁). Because each is an exact differential, the mixed second derivatives are equal, which gives the Maxwell relations: (∂T/∂V)_S = −(∂P/∂S)_V; (∂T/∂P)_S = (∂V/∂S)_P; (∂S/∂V)_T = (∂P/∂T)_V; (∂S/∂P)_T = −(∂V/∂T)_P. The last two turn unmeasurable entropy derivatives into equation-of-state data, and give the thermodynamic equation of state (∂U/∂V)_T = T(∂P/∂T)_V − P, which is zero for an ideal gas.
The Gibbs–Helmholtz equation, [∂(G/T)/∂T]_P = −H/T², applied to a reaction gives [∂(ΔG/T)/∂T]_P = −ΔH/T², and with ΔG° = −RT ln K it becomes the van ’t Hoff equation d ln K/dT = ΔᵣH°/(RT²). It is also how a free-energy measurement at one temperature, with the enthalpy, predicts the free energy at another.
4. Partial molar quantities, chemical potential, fugacity and activity
In a mixture a partial molar quantity is the change in an extensive property per mole of component added at constant T, P and the other amounts: V̄ᵢ = (∂V/∂nᵢ)_(T,P,nⱼ), and V = Σnᵢ V̄ᵢ. The partial molar Gibbs energy is the chemical potential μᵢ = (∂G/∂nᵢ)_(T,P,nⱼ), and matter flows from high to low μ until μ is equal in every phase — the condition of phase and chemical equilibrium. The Gibbs–Duhem equation Σnᵢ dμᵢ = 0 (at constant T and P) says the chemical potentials of a mixture cannot all change independently: in a binary mixture x_A dμ_A = −x_B dμ_B, so one component’s activity coefficient fixes the other’s.
For an ideal gas μ = μ° + RT ln(P/P°). A real gas keeps this form by replacing pressure with the fugacity f = φP, where the fugacity coefficient φ → 1 as P → 0 and ln φ = ∫₀ᴾ (Z − 1)dP/P. For a component of a condensed mixture the same role is played by the activity: μᵢ = μᵢ* + RT ln aᵢ with aᵢ = γᵢxᵢ, where the activity coefficient γᵢ measures departure from ideality (γ = 1 ideal, γ > 1 positive deviation). For ions in solution the Debye–Hückel law estimates γ (see the electrochemistry chapter).
Thermodynamics of mixing. For ideal gases or an ideal solution mixed at constant T and P: ΔmixG = nRT Σxᵢ ln xᵢ (always negative), ΔmixS = −nR Σxᵢ ln xᵢ (always positive), and ΔmixH = 0, ΔmixV = 0. Mixing is driven entirely by entropy. Equal amounts of two components give the largest ΔmixS, 2R ln 2 = 11.5 J/K for one mole of each. A real solution adds excess functions G^E = RT Σnᵢ ln γᵢ.
5. Chemical equilibria and their dependence on temperature and pressure
For a reaction at constant T and P, ΔᵣG = Σνᵢμᵢ = ΔᵣG° + RT ln Q, where Q is the reaction quotient in activities (partial pressures over 1 bar for ideal gases). At equilibrium ΔᵣG = 0 and Q = K, so ΔᵣG° = −RT ln K: a negative ΔᵣG° means K > 1, and each 5.7 kJ/mol at 298 K is a factor of 10 in K. For gases, Kp = Kc(RT/P°)^Δn in consistent units, and in terms of mole fractions K_x = K_p(P/P°)^(−Δn).
Temperature: by the van ’t Hoff equation, ln(K₂/K₁) = −(ΔᵣH°/R)(1/T₂ − 1/T₁) when ΔᵣH° is constant. K rises with T for an endothermic reaction and falls for an exothermic one, and a plot of ln K against 1/T has slope −ΔᵣH°/R. Pressure: the thermodynamic K of an ideal-gas reaction does not depend on the total pressure, because the standard state is fixed at 1 bar; what changes is the equilibrium composition, through K_x = K_p(P/P°)^(−Δn). Raising P favours the side with fewer moles of gas (Le Chatelier). For condensed phases (∂ ln K/∂P)_T = −ΔᵣV°/(RT), a small effect. Adding an inert gas at constant volume changes nothing; at constant total pressure it acts like lowering P.
6. Statistical thermodynamics: ensembles, the Boltzmann distribution and partition functions
An ensemble is a large imagined collection of copies of the system. The microcanonical ensemble fixes N, V and E (isolated systems; S = k ln W); the canonical ensemble fixes N, V and T (closed systems in a heat bath); the grand canonical ensemble fixes μ, V and T (open systems exchanging both energy and particles, as in adsorption). For independent molecules the most probable distribution over energy levels is the Boltzmann distribution, nᵢ/N = gᵢ e^(−εᵢ/kT)/q, with the molecular partition function q = Σgᵢ e^(−εᵢ/kT) — in effect the number of states thermally accessible at temperature T. The population ratio of two levels is n₂/n₁ = (g₂/g₁)e^(−Δε/kT).
For N indistinguishable independent molecules the canonical partition function is Q = q^N/N!, and every thermodynamic function follows: U − U(0) = kT²(∂ ln Q/∂T)_V, A − A(0) = −kT ln Q, S = (U − A)/T, P = kT(∂ ln Q/∂V)_T, and for an ideal gas G − G(0) = −nRT ln(q/N). Because the energy separates, q = q_trans·q_rot·q_vib·q_elec.
| Motion | Partition function | Typical size at room temperature | Contribution to C_V,m |
|---|---|---|---|
| Translation | q_t = V/Λ³, Λ = h/√(2πmkT) | ~10³⁰ per m³ | 3R/2 |
| Rotation (linear) | q_r = kT/(σhcB), σ = 1 or 2 | 10–1000 | R |
| Vibration (per mode) | q_v = 1/(1 − e^(−hcν̃/kT)) | ≈ 1 | 0 → R as T rises |
| Electronic | q_e ≈ g₀ (ground-state degeneracy) | 1 for most molecules, 3 for O₂ | 0 |
The symmetry number σ is 1 for a heteronuclear diatomic and 2 for a homonuclear one, because rotating a homonuclear molecule by 180° gives an indistinguishable arrangement and half the states are not new. For monatomic gases only translation (and a constant electronic factor) contributes, so C_V,m = 3R/2; for diatomics at room temperature translation and rotation give 5R/2, with vibration switching on only when kT approaches hcν̃ — which is why the equipartition value 7R/2 is reached only at high temperature.
Key takeaways
- ΔU = q + w; reversible isothermal ideal-gas work −nRT ln(V₂/V₁); Hess and Kirchhoff laws; standard state 1 bar.
- ΔS = nR ln(V₂/V₁), ΔH/T at a transition; third law S(0) = 0 with residual entropy k ln W; at constant T and P a change is spontaneous if dG < 0.
- dG = −SdT + VdP; Maxwell (∂S/∂V)_T = (∂P/∂T)_V; Gibbs–Helmholtz ∂(G/T)/∂T = −H/T²; Gibbs–Duhem Σnᵢdμᵢ = 0; ΔmixG = nRTΣx ln x, ΔmixS = −nRΣx ln x.
- ΔG° = −RT ln K; van ’t Hoff ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁); pressure shifts composition, not the thermodynamic K of an ideal-gas reaction.
- Boltzmann nᵢ/N = gᵢe^(−εᵢ/kT)/q; Q = q^N/N!; U = kT²∂lnQ/∂T, A = −kT ln Q; q_r = kT/σhcB, q_v = 1/(1 − e^(−hcν̃/kT)).
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A reaction has ΔᵣG° = −10.0 kJ/mol at 298 K. What is its equilibrium constant, to one decimal place? (R = 8.314 J mol⁻¹ K⁻¹)
Numerical answer — type the value.
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Answer: 56.6
ln K = −ΔᵣG°/RT = 10 000/(8.314 × 298) = 4.036, so K = e^4.036 = 56.6. Leaving ΔᵣG° in kJ while R is in J gives ln K = 0.004 and K ≈ 1.00, the commonest slip; a negative ΔᵣG° must give K > 1.A reaction has K = 1.0 × 10⁻⁵ at 298 K. What is ΔᵣG°, in kJ/mol, to one decimal place? (R = 8.314 J mol⁻¹ K⁻¹)
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Answer: 28.5
ΔᵣG° = −RT ln K = −8.314 × 298 × ln(10⁻⁵) = −8.314 × 298 × (−11.513) = 28 524 J/mol = 28.5 kJ/mol, positive because K < 1. A negative answer signals a lost sign in ln(10⁻⁵).The equilibrium constant of a reaction doubles when the temperature is raised from 298 K to 308 K. Assuming ΔᵣH° constant, what is ΔᵣH°, in kJ/mol, to one decimal place? (R = 8.314 J mol⁻¹ K⁻¹)
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Answer: 52.9
ln(K₂/K₁) = −(ΔᵣH°/R)(1/T₂ − 1/T₁): ln 2 = (ΔᵣH°/8.314)(1/298 − 1/308) = (ΔᵣH°/8.314)(1.0895 × 10⁻⁴), so ΔᵣH° = 8.314 × 0.6931/1.0895 × 10⁻⁴ = 52 900 J/mol = 52.9 kJ/mol. K rising with T means the reaction is endothermic, so the sign is positive.One mole of an ideal gas expands reversibly and isothermally at 300 K from 10 L to 20 L. What is the work done on the gas, in kJ, to two decimal places? (R = 8.314 J mol⁻¹ K⁻¹)
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Answer: -1.73
w = −nRT ln(V₂/V₁) = −1 × 8.314 × 300 × ln 2 = −1729 J = -1.73 kJ. The sign is negative because the gas does work on the surroundings; q = +1.73 kJ since ΔU = 0 for an isothermal ideal gas.What is the entropy change, in J/K, when one mole of an ideal gas doubles its volume isothermally? Give the answer to two decimal places. (R = 8.314 J mol⁻¹ K⁻¹)
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Answer: 5.76
ΔS = nR ln(V₂/V₁) = 8.314 × ln 2 = 5.76 J/K, whether the expansion is reversible or free, because S is a state function. Only ΔS of the surroundings depends on the path.According to Pauling, the residual entropy of ice is R ln(3/2) per mole. What is its value, in J K⁻¹ mol⁻¹, to two decimal places? (R = 8.314 J mol⁻¹ K⁻¹)
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Answer: 3.37
S = R ln(3/2) = 8.314 × 0.4055 = 3.37 J K⁻¹ mol⁻¹, close to the calorimetric 3.4. It exists because the proton positions stay disordered at 0 K; solid CO has R ln 2 = 5.76 for the same kind of reason.One mole of gas A and one mole of gas B, both ideal, are mixed at constant temperature and pressure. What is ΔmixS, in J/K, to one decimal place? (R = 8.314 J mol⁻¹ K⁻¹)
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Answer: 11.5
ΔmixS = −nR Σxᵢ ln xᵢ = −2 × 8.314 × (0.5 ln 0.5 + 0.5 ln 0.5) = 2R ln 2 = 11.5 J/K. Using n = 1 for the total gives half the value; ΔmixH is zero, so the mixing is purely entropy-driven.Benzene boils at 353 K. Using Trouton’s rule with ΔvapS = 85 J K⁻¹ mol⁻¹, estimate its enthalpy of vaporisation, in kJ/mol, to one decimal place.
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Answer: 30.0
ΔvapH = T_b × ΔvapS = 353 × 85 = 30 005 J/mol = 30.0 kJ/mol, close to the measured 30.7. The rule works for unassociated liquids; for water it underestimates, because hydrogen bonding makes ΔvapS larger.Which of the following is a correct Maxwell relation?
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Answer: A — (∂S/∂V)_T = (∂P/∂T)_V
From dA = −SdT − PdV, equality of mixed derivatives gives (∂S/∂V)_T = (∂P/∂T)_V. From dG = −SdT + VdP the partner is (∂S/∂P)_T = −(∂V/∂T)_P, with a minus sign that the second option drops; the fourth also needs a minus sign.For a process at constant temperature and volume, the condition for spontaneity is
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Answer: A — dA < 0
At constant T and V the Helmholtz energy A = U − TS must decrease. dG < 0 is the condition at constant T and P; ΔS_sys > 0 alone is the criterion only for an isolated system, not for one exchanging heat.In a binary liquid mixture at constant T and P, the activity coefficient of A increases as x_A is varied. By the Gibbs–Duhem equation, the chemical potential of B must
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Answer: A — change so that x_A dμ_A + x_B dμ_B = 0
Gibbs–Duhem, Σnᵢdμᵢ = 0 at constant T and P, ties the two: x_A dμ_A = −x_B dμ_B, so μ_B must move opposite to μ_A in proportion x_A/x_B. That is why activity coefficients of a binary mixture cannot be chosen independently.For the ideal-gas equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), which statements are correct?
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Answer: A — Raising the total pressure increases the equilibrium mole fraction of NH₃; C — Since the reaction is exothermic, K_p decreases as the temperature rises
Δn = −2, so K_x = K_p(P/P°)² grows with P and more NH₃ forms, while K_p itself is fixed at a given T. An exothermic reaction has K falling with T (van ’t Hoff). Argon at constant volume leaves every partial pressure unchanged, so nothing shifts.Two non-degenerate energy levels are separated by 400 cm⁻¹. What is the ratio of the population of the upper level to that of the lower at 300 K, to two decimal places? (kT/hc = 208.6 cm⁻¹)
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Answer: 0.15
n₂/n₁ = exp(−Δε/kT) = exp(−400/208.6) = exp(−1.918) = 0.147, i.e. 0.15. If the upper level were doubly degenerate the ratio would double, which is why degeneracies must be checked first.Estimate the rotational partition function of N₂ at 300 K, given B = 1.998 cm⁻¹ and kT/hc = 208.6 cm⁻¹, to the nearest whole number.
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Answer: 52
For a linear rotor at high temperature q_r = kT/(σhcB) with σ = 2 for homonuclear N₂: q_r = 208.6/(2 × 1.998) = 52.2, i.e. 52. Forgetting σ gives 104, the value a heteronuclear molecule with the same B would have.For a vibrational mode with hcν̃/kT = 1, what is the vibrational partition function (energies measured from the zero-point level), to two decimal places?
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Answer: 1.58
q_v = 1/(1 − e^(−hcν̃/kT)) = 1/(1 − e^(−1)) = 1/(1 − 0.368) = 1.58. For typical bond stretches hcν̃/kT is 5–10 at room temperature and q_v is barely above 1, which is why vibrations add little to C_V there.Which statements about statistical ensembles are correct?
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Answer: A — The canonical ensemble has fixed N, V and T; B — The microcanonical ensemble has fixed N, V and E; D — In the canonical ensemble A = −kT ln Q
Canonical: N, V, T fixed, with A = −kT ln Q. Microcanonical: N, V, E fixed, S = k ln W. The grand canonical ensemble fixes μ, V and T and lets both energy and particle number fluctuate, so the third statement is wrong.At room temperature, the molar constant-volume heat capacity of an ideal diatomic gas such as N₂ is close to
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Answer: A — 5R/2
Translation gives 3R/2 and rotation of a linear molecule R, a total of 5R/2 = 20.8 J K⁻¹ mol⁻¹; the N₂ vibration (2359 cm⁻¹) is barely excited at 300 K. 7R/2 is the high-temperature limit with vibration fully active, and 3R/2 belongs to a monatomic gas.