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2.3.7 (HL)—K and ΔG

Syllabus
First assessment 2025
Objective
2.3.7
Level
HL

K and Gibbs Energy

HL only

ΔG°=−RTlnKΔG° = −RT ln K

K and ΔG both describe equilibrium position. K > 1 corresponds to product-favoured equilibrium and ΔG° < 0; K = 1 corresponds to ΔG° = 0 and approximately equal equilibrium tendencies.

Use a dimensionless K, T in kelvin and R = 8.31 J mol⁻¹ K⁻¹. The sign link applies to ΔG°: the actual ΔG away from standard conditions also depends on the reaction quotient.

Worked Gibbs example: for 2 NO(g)⇌NX2OX2(g)\ce{2NO(g) <=> N2O2(g)} at 298 K, local spectroscopic data give K=1.39×10−5K=1.39\times10^{-5}. ΔG∘=−(8.31 J K−1 mol−1)(298 K)ln⁡(1.39×10−5)=+2.77×104 J mol−1=+27.7 kJ mol−1\Delta G^\circ=-(8.31\,\mathrm{J\,K^{-1}\,mol^{-1}})(298\,\mathrm{K})\ln(1.39\times10^{-5})=+2.77\times10^4\,\mathrm{J\,mol^{-1}}=+27.7\,\mathrm{kJ\,mol^{-1}}. The positive value agrees with K<1K<1: reactants are favoured under standard conditions at this temperature.

Calculating ΔG° from K

HL only

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Determine the equilibrium constant, K , for this reaction at 25∘C25^{\circ} \mathrm{C}, referring to section 1 of the data booklet.

If you did not obtain an answer in (c)(iii), use ΔG=−43.5 kJ mol−1\Delta G=-43.5 \mathrm{~kJ} \mathrm{~mol}^{-1}, but this is not the correct answer.

Extent of Chemical Change Summary

Retrieve the route: define dynamic equilibrium, write K, interpret its magnitude, predict Le Châtelier shifts, compare Q with K, solve a RICE table, and connect K with ΔG.

Check closed-system and equal-rate language, exponents and direction, whether a change affects K, current versus equilibrium concentrations, stoichiometric x changes, and kelvin/unit consistency in ΔG calculations.