Chemical Thermodynamics

Thermodynamics governs energy changes in chemical processes. The first law states ΔU = q + w (IUPAC: w positive when work done ON the system, q positive when heat absorbed BY the system). For different processes:

• Free expansion (against vacuum): w = 0 (no opposing pressure)
• Irreversible against constant external pressure: w = −$P_{ext}$ × ΔV
• Reversible isothermal: w = −nRT ln(V₂/V₁) = −2.303 nRT log(V₂/V₁)

Enthalpy H = U + PV. For reactions involving gases, the key relationship is: Derivation: H = U + PV. For ideal gases, PV = nRT, so ΔH = ΔU + Δ(PV) = ΔU + Δ$n_{g}$ RT where Δ$n_{g}$ = moles of gaseous products − moles of gaseous reactants (only gaseous species count). Dimensional analysis: mol × J/(mol·K) × K = J ✓

Heat capacities: $C_{p}$ − $C_{v}$ = R (for 1 mol ideal gas).
Monoatomic: $C_{v}$ = 3R/2, $C_{p}$ = 5R/2, γ = $\frac{5}{3}$.
Diatomic: $C_{v}$ = 5R/2, $C_{p}$ = 7R/2, γ = $\frac{7}{5}$.

Solved Example 1: Work done during reversible isothermal expansion of 2 mol ideal gas from 10 L to 20 L at 300 K. w = −nRT ln(V₂/V₁) = −2 × 8.314 × 300 × ln($\frac{20}{10}$) Dimensional analysis: mol × J/(mol·K) × K = J ✓ w = −2 × 8.314 × 300 × 0.693 = −3458 J = −3.458 kJ (negative = work done BY the system)

Solved Example 2: ΔH for C₂H₆(g) → C₂H₄(g) + H₂(g) using bond enthalpies. BE values: C−H = 414, C−C = 347, C=C = 611, H−H = 436 kJ/mol. Bonds broken: 1(C−C) + 6(C−H) = 347 + 2484 = 2831 kJ (energy absorbed) Bonds formed: 1(C=C) + 4(C−H) + 1(H−H) = 611 + 1656 + 436 = 2703 kJ (energy released) ΔH = 2831 − 2703 = +128 kJ/mol (endothermic) ✓

Hess's law states that enthalpy change is path-independent (state function): Δ$H_{rxn}$ = ΣΔ$H_{f}$(products) − ΣΔ$H_{f}$(reactants).
Using bond enthalpies: ΔH = Σ($\text{BE}_{broken}$) − Σ($\text{BE}_{formed}$).

Entropy S measures disorder. ΔS = $q_{rev}$/T. Entropy increases: solid → liquid → gas, or when moles of gas increase. The second law: Δ$S_{universe}$ > 0 for spontaneous processes.

Gibbs free energy G = H − TS determines spontaneity: ΔG = ΔH − TΔS. Spontaneous when ΔG < 0; equilibrium when ΔG = 0; non-spontaneous when ΔG > 0.

Four spontaneity cases: (1) ΔH < 0, ΔS > 0 → ΔG always < 0 (always spontaneous) (2) ΔH > 0, ΔS < 0 → ΔG always > 0 (never spontaneous) (3) ΔH < 0, ΔS < 0 → spontaneous at low T (below T = ΔH/ΔS) (4) ΔH > 0, ΔS > 0 → spontaneous at high T (above T = ΔH/ΔS)

Solved Example 3: For ΔH° = −40 kJ/mol, ΔS° = −150 J/(mol·K). At what temperature does the reaction become non-spontaneous? At crossover: ΔG = 0 → T = ΔH/ΔS = −40000 J/mol ÷ (−150 J/(mol·K)) = 266.7 K Above 266.7 K: ΔG > 0 (non-spontaneous). Below 266.7 K: ΔG < 0 (spontaneous).

The standard Gibbs energy relates to the equilibrium constant: ΔG° = −RT ln K = −2.303 RT log K. At equilibrium, ΔG = 0 (not ΔG° = 0).

The key testable concept is applying the Gibbs equation ΔG = ΔH − TΔS to predict spontaneity and calculate the crossover temperature.