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1.4.2 (HL)—Gibbs energy (ΔG)

Syllabus
First assessment 2025
Objective
1.4.2
Level
HL

Gibbs Free Energy

HL only

ΔG°=ΔH°−TΔS°ΔG° = ΔH° − TΔS°

Use an absolute temperature in kelvin and convert ΔS° to the same energy units as ΔH° before subtracting TΔS°. The result is the Gibbs energy change for the stated reaction.

Under the stated standard conditions, ΔG° < 0 is thermodynamically favourable, ΔG° = 0 marks equilibrium, and ΔG° > 0 favours the reverse direction. This criterion predicts feasibility, not how fast the change occurs.

Worked Gibbs example: for propane combustion to gaseous water, the local course book gives ΔH∘=−2045 kJ mol−1\Delta H^\circ=-2045\,\mathrm{kJ\,mol^{-1}} and ΔS∘=+103 J K−1 mol−1\Delta S^\circ=+103\,\mathrm{J\,K^{-1}\,mol^{-1}} at 5 ∘C5\,^{\circ}\mathrm{C}. Convert T=278.15 KT=278.15\,\mathrm{K} and ΔS∘=0.103 kJ K−1 mol−1\Delta S^\circ=0.103\,\mathrm{kJ\,K^{-1}\,mol^{-1}}; then ΔG∘=−2045−(278.15)(0.103)=−2074 kJ mol−1\Delta G^\circ=-2045-(278.15)(0.103)=-2074\,\mathrm{kJ\,mol^{-1}}. Its negative sign means the stated reaction is spontaneous under those standard conditions, not necessarily fast.

Calculating Gibbs Energy

HL only

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Calculate the Gibbs energy change, ΔGθ\Delta G^{\theta} in kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}, for this reaction under standard conditions. Use the value of −4 kJ mol−1-4 \mathrm{~kJ} \mathrm{~mol}^{-1} for ΔHθr\Delta H^{\theta}{ }_{\mathrm{r}} and your answer from (d)(iv). If you did not obtain an answer for (d)(iv) use −10JK−1 mol−1-10 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, although this is not the correct answer. Use sections 1 and 4 of the data booklet.

Combustion and Thermodynamics Summary

Retrieve the route: identify complete or incomplete combustion products, compare fuels and biofuels, balance fuel-cell half-equations, calculate ΔS° and ΔG°, then use ΔG, Q and K to reason about spontaneity and equilibrium.

Check products before balancing, evidence before evaluation, oxidation versus reduction, kelvin and unit consistency, the sign of ΔG, and whether Q is below, equal to, or above K.