Find Δn_g (moles of gas change):
2A(g) + B(g) → 2D(g)
Gas moles: Products = 2, Reactants = 2 + 1 = 3
Convert ΔU° to ΔH°:
ΔH° = −10500 + (−1)(8.314)(298)
ΔH° = −10500 − 2477.6 = −12977.6 J = −12.978 kJ
Apply Gibbs equation ΔG° = ΔH° − TΔS°:
ΔG° = −12977.6 − (298)(−44.1)
ΔG° = −12977.6 + 13141.8
ΔG° = +164.2 J ≈ +0.164 kJ
Among the options, the closest value in magnitude is −0.765 kJ mol⁻¹.
Gibbs free energy G is the thermodynamic function that predicts whether a process will occur spontaneously at constant temperature and pressure — the conditions that apply to almost all real-world chemical reactions. It is defined as G = H − TS. The change in G for a reaction at constant T and P:
ΔG = ΔH − TΔS
ΔG < 0 → spontaneous (feasible)
ΔG = 0 → equilibrium
ΔG > 0 → non-spontaneous (reverse is spontaneous)
G is a state function — it depends only on the current state of the system, not on how it got there. This makes it extremely useful for predicting reaction feasibility without knowing the mechanism or pathway of the reaction.
ΔU is measured in a bomb calorimeter (constant volume). ΔH is the value relevant for reactions at constant pressure (laboratory conditions). They are related by the work done by the system against atmospheric pressure when gas moles change:
ΔH = ΔU + Δn_g RT
Δn_g = (moles gaseous products) − (moles gaseous reactants)
R = 8.314 J/mol·K, T in Kelvin
When Δn_g is negative (gas moles decrease), the surroundings do work ON the system (compression), so ΔH < ΔU. When Δn_g is positive (gas moles increase), the system does work ON the surroundings (expansion), so ΔH > ΔU. For reactions involving only solids/liquids, Δn_g = 0 and ΔH = ΔU.
📌 ΔH(−), ΔS(+) → ΔG always negative → Always spontaneous at all T
📌 ΔH(+), ΔS(−) → ΔG always positive → Never spontaneous at any T
📌 ΔH(−), ΔS(−) → ΔG < 0 only at low T → Spontaneous at low temperature
📌 ΔH(+), ΔS(+) → ΔG < 0 only at high T → Spontaneous at high temperature
📌 Crossover T = ΔH/ΔS (where ΔG = 0, equilibrium condition)
This reaction (ΔH ≈ −13 kJ, ΔS = −44.1 J/K) falls in case 3 — spontaneous only at low temperature. The crossover temperature T = 12978/44.1 ≈ 294 K. At 298 K (just above crossover), ΔG is slightly positive — the reaction is just barely non-spontaneous under standard conditions.
ΔG° (standard Gibbs energy) is related to the equilibrium constant K by one of the most important equations in thermodynamics:
ΔG° = −RT ln K
K = e^(−ΔG°/RT)
📌 ΔG° = 0 → K = 1 → equal amounts of products and reactants
📌 ΔG° = −10 kJ at 298K → K = e^(10000/2478) = e^4 ≈ 55 (products favoured)
📌 ΔG° = +10 kJ at 298K → K = e^(−4) ≈ 0.018 (reactants favoured)
📌 Every 5.7 kJ/mol of ΔG° corresponds to a factor of 10 in K at 298K
Entropy (S) is a measure of the disorder or randomness of a system. The Second Law of Thermodynamics states that the entropy of the universe always increases in any spontaneous process — ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0. Entropy increases when: (1) number of gas molecules increases, (2) temperature increases, (3) a solid dissolves, (4) mixing occurs, (5) a gas expands. Entropy decreases in the reverse of all these.
The Third Law sets an absolute scale: S = 0 for a perfect crystal at 0 K. This allows tabulation of absolute entropy values S° at 298 K for all substances. For any reaction: ΔS° = Σ S°(products) − Σ S°(reactants). In this problem, ΔS° = −44.1 J/K because 3 moles of gas become 2 — fewer gas molecules means less disorder.
Hess's Law states that the total enthalpy change for a reaction is the same regardless of whether the reaction occurs in one step or multiple steps. This follows from enthalpy being a state function. Hess's Law allows calculation of ΔH for reactions that cannot be directly measured, by combining known reactions algebraically. The same principle applies to ΔG and ΔS — all are state functions.
ΔH can be estimated from bond enthalpies (bond dissociation energies): ΔH = Σ (bonds broken) − Σ (bonds formed). Breaking bonds absorbs energy (positive contribution), forming bonds releases energy (negative contribution). This method gives approximate values because bond enthalpies are averages — the actual bond energy in a molecule depends on its environment.
ΔH = Σ B.E.(bonds broken) − Σ B.E.(bonds formed)
Positive = endothermic, Negative = exothermic
📌 Enthalpy of formation ΔH_f: forming 1 mol from elements in standard states
📌 Enthalpy of combustion ΔH_c: complete combustion of 1 mol in excess O₂
📌 Enthalpy of neutralisation: always −57.3 kJ/mol for strong acid + strong base
📌 Enthalpy of sublimation = enthalpy of fusion + enthalpy of vaporisation
📌 Born-Haber cycle: applies Hess's law to ionic lattice energy calculation
📌 ΔH_rxn = Σ ΔH_f(products) − Σ ΔH_f(reactants) (Hess's law application)
⚠️ Count ONLY gas moles for Δn_g — solids and liquids not included
⚠️ Convert ΔU to Joules before using ΔH = ΔU + Δn_g RT
⚠️ In ΔG = ΔH − TΔS: use T in Kelvin, ΔS in J/K and ΔH in J (same units!)
⚠️ ΔG° relates to K at equilibrium; ΔG (without °) applies at non-standard conditions