Thermochemistry and Equilibrium Challenge

Introduction

Thermochemistry tells you how much heat a reaction releases or absorbs; equilibrium tells you how far it goes. The bridge between them is the Gibbs energy. A single problem can start with enthalpy and entropy data, produce an equilibrium constant, and end with a prediction about which way a real mixture will move. Getting the units and the distinction between ΔG° and ΔG right is the heart of this challenge.

Core explanation

From heat and entropy to ΔG°. At temperature T, ΔG° = ΔH° − TΔS°. Enthalpies are usually tabulated in kJ/mol and entropies in J K⁻¹ mol⁻¹, so one of them must be converted before subtracting. A sign pattern gives a first qualitative view: an exothermic reaction with an entropy increase is favourable at all temperatures, while an endothermic reaction with an entropy increase becomes favourable only above a certain temperature.

From ΔG° to K. The relation ΔG° = −RT ln K gives K = exp(−ΔG°/RT). A negative ΔG° gives K > 1 and a positive ΔG° gives K < 1, but a positive ΔG° does not mean "no reaction"; it means that at equilibrium reactants are favoured. Because ΔG° is defined with a standard pressure of 1 bar for gases, the K obtained is a dimensionless constant in which partial pressures are expressed in bar.

ΔG° versus ΔG. ΔG° refers to the hypothetical case in which every species is in its standard state. The real driving force for a particular mixture is ΔG = ΔG° + RT ln Q. Combining the two relations gives ΔG = RT ln(Q/K). So ΔG is negative when Q < K, positive when Q > K and zero at equilibrium. This is the thermodynamic justification for the rule that compares Q with K.

Temperature dependence. If ΔH° and ΔS° are treated as roughly constant over a modest range, ΔG° changes linearly with T, and the crossover temperature is T ≈ ΔH°/ΔS°. Above this for an endothermic reaction with positive ΔS°, K exceeds 1. This is consistent with Le Chatelier's principle: raising the temperature favours the endothermic direction.

Formulae

ΔG° = ΔH° − TΔS°. ΔG° = −RT ln K, so K = e^(−ΔG°/RT). ΔG = ΔG° + RT ln Q = RT ln(Q/K). Crossover: T ≈ ΔH°/ΔS°. R = 8.314 J K⁻¹ mol⁻¹.

Step-by-step reasoning

1. Convert ΔS° to kJ K⁻¹ mol⁻¹ (or ΔH° to J/mol). 2. Calculate ΔG° at the stated temperature. 3. Convert ΔG° to J/mol and find K from the exponential. 4. Calculate Q for the actual mixture, with pressures in bar. 5. Compare Q with K, or calculate ΔG, and state the direction.

Visual explanation

Plot ΔG° against T as a straight line with intercept ΔH° and slope −ΔS°. For the dissociation of N₂O₄, the line starts positive and falls, crossing zero near 325 K. Beside it, sketch the Gibbs energy of the mixture against extent of reaction: a curve with a minimum at the equilibrium composition, where ΔG = 0.

Real-world analogy

ΔG° is like the listed price difference between two products under standard conditions, while ΔG is the real deal on the day, adjusted for how many of each are already in your basket. A good standard deal can be cancelled by a bad current situation, and the reverse can also happen.

Real-world example

In industrial ammonia synthesis, the reaction is exothermic with a large entropy decrease, so K falls as the temperature rises. Engineers choose a compromise temperature, using thermodynamic data to estimate K, and then remove ammonia so that Q stays below K and the forward reaction keeps running.

Why?

Why does a reaction with a positive ΔG° still proceed a little? When no products are present, Q is zero and RT ln Q is very large and negative, so ΔG is negative at the start. The reaction runs forward until Q rises to K, which may be at a small but non-zero extent.

Common misconception

"A positive ΔG° means the reaction cannot happen." It means K < 1, so reactants predominate at equilibrium; some product always forms, and a mixture with Q below K will still move forward.

Worked example

Question: For N₂O₄(g) ⇌ 2NO₂(g), ΔH° = +57.2 kJ/mol and ΔS° = +175.8 J K⁻¹ mol⁻¹. Find K at 298 K and decide the direction for a mixture with p(NO₂) = 0.20 bar and p(N₂O₄) = 0.50 bar.

Reasoning: ΔG° = 57.2 − 298 × 0.1758 = 57.2 − 52.4 = +4.8 kJ/mol. K = exp(−4810 ÷ (8.314 × 298)) = exp(−1.94) ≈ 0.14. Q = 0.20² ÷ 0.50 = 0.080. Then ΔG = 4810 + 8.314 × 298 × ln 0.080 ≈ 4810 − 6260 ≈ −1.4 kJ/mol.

Answer: K ≈ 0.14; since Q < K and ΔG is negative, more N₂O₄ dissociates.

Quick check

1. Using the worked-example data, above what temperature does ΔG° for N₂O₄ dissociation become negative? Answer: T ≈ ΔH°/ΔS° = 57 200 ÷ 175.8 ≈ 325 K, so above about 325 K the standard Gibbs energy change is negative.

Exam focus

The most common lost mark is a unit mismatch between kJ and J. State explicitly whether you are using ΔG° or ΔG, express gas pressures in bar for Q, and write a direction conclusion in words. Show the exponential step clearly when finding K.

Advanced insight

Treating ΔH° and ΔS° as constant is only an approximation; heat-capacity differences make both drift with temperature. The van 't Hoff equation, d ln K/dT = ΔH°/RT², expresses the same physics and gives K at a new temperature directly. It also shows why the effect of temperature on K is largest for reactions with large ΔH° .

Summary

Thermodynamic data link heat to equilibrium: ΔG° = ΔH° − TΔS° gives K through ΔG° = −RT ln K, and ΔG = ΔG° + RT ln Q = RT ln(Q/K) gives the direction for a real mixture. Watch the kJ/J conversion, use bar for gas pressures, and use T ≈ ΔH°/ΔS° to locate the crossover temperature.

Practice questions

1. Using the N₂O₄ data, find ΔG° and K at 350 K. Answer: ΔG° = 57.2 − 350 × 0.1758 ≈ −4.3 kJ/mol, so K = exp(4330 ÷ (8.314 × 350)) ≈ 4.4. 2. A reaction has K = 1.0 × 10⁻³ at 298 K. Calculate ΔG°. Answer: ΔG° = −8.314 × 298 × ln(1.0 × 10⁻³) ≈ +17.1 kJ/mol. 3. For a reaction at equilibrium, what are the values of ΔG and Q/K? Answer: ΔG = 0 and Q/K = 1. 4. A reaction has ΔH° < 0 and ΔS° < 0. How does K change as the temperature rises, and why? Answer: K decreases, because ΔG° = ΔH° − TΔS° becomes less negative as −TΔS° grows more positive; equivalently, heating favours the endothermic reverse reaction.