Chemical Thermodynamics, Chemical Equilibria and Statistical Thermodynamics
1. The first law, standard states and thermochemistry
The first law: the internal energy U is a state function and ΔU = q + w, with w = −∫P_ext dV the work done on the system. At constant volume ΔU = q_V; at constant pressure the enthalpy H = U + PV gives ΔH = q_P, and ΔH = ΔU + Δn_gas RT for a reaction of ideal gases. For an ideal gas C_P − C_V = R per mole. A reversible isothermal expansion of an ideal gas has ΔU = 0 and w = −nRT ln(V₂/V₁); a reversible adiabatic one has q = 0 and obeys PV^γ = constant and TV^(γ−1) = constant, with γ = C_P/C_V.
The standard state of a substance is its pure form at 1 bar (for a solute, unit activity at 1 mol/kg or 1 mol/L on the chosen scale); the temperature is not part of the definition, although tables are usually at 298.15 K. Standard enthalpies of formation of elements in their reference forms are zero. Hess’s law follows from H being a state function: ΔᵣH° = ΣνΔfH°(products) − ΣνΔfH°(reactants), and reaction enthalpies can be assembled from combustion data or bond enthalpies. Kirchhoff’s law moves ΔH to another temperature: ΔᵣH°(T₂) = ΔᵣH°(T₁) + ∫ΔC_P dT.
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.
Show answer
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)
Numerical answer — type the value.
Show answer
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)
Numerical answer — type the value.
Show answer
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)
Numerical answer — type the value.
Show answer
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)
Numerical answer — type the value.
Show answer
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)
Numerical answer — type the value.
Show answer
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)
Numerical answer — type the value.
Show answer
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.
Numerical answer — type the value.
Show answer
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?
Show answer
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
Show answer
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
Show answer
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?
Show answer
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)
Numerical answer — type the value.
Show answer
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.
Numerical answer — type the value.
Show answer
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?
Numerical answer — type the value.
Show answer
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?
Show answer
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
Show answer
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.