The questions are on the entropy Gibbs energy test. Take it under real conditions before reading the mark scheme.
Quick answer: Entropy measures how many ways particles and energy can be arranged; it rises from solid to liquid to gas. This mark scheme gives the model answers and the marks for the 8 questions of the Chemistry 11-12 test on entropy Gibbs energy, from meaning of entropy to entropy of the surroundings, with the equivalent answers to accept.
Use these model answers and the marking notes to score yourself out of 20. Marks are given for consistent units and for the logic of setting ΔG = 0 at the changeover, as well as for the numerical result.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R1.4
Question 1 Meaning of entropy (2 marks)
In the liquid the particles are close together; in the gas they are far apart and move freely in all directions, so there are many more ways to arrange the particles and distribute energy. Entropy increases.Marking: 1 mark for particles more free or further apart; 1 mark for more ways of arranging the energy.
Question 2 Predicting the sign of ΔS (3 marks)
- Positive: gas formed from solid.
- Negative: 2 mol of gas become 1 mol.
- Negative: ions in solution form a solid.
Marking: 1 mark each for the correct sign with a reason.
Question 3 Steam reforming: entropy (3 marks)
Products: 197.7 + 3 × 130.7 = 589.8. Reactants: 186.3 + 188.8 = 375.1. ΔS° = 589.8 − 375.1 = +214.7 J K⁻¹ mol⁻¹.Marking: 1 mark for the sum of products; 1 mark for the sum of reactants; 1 mark for the answer with sign and unit.
Question 4 Steam reforming: feasibility (2 marks)
ΔG = 206 − 298 × 0.2147 = 206 − 64.0 = +142 kJ/mol. ΔG is positive, so the reaction is not spontaneous at 298 K.Marking: 1 mark for the calculation with ΔS in kJ; 1 mark for the conclusion.
Question 5 Steam reforming: temperature (3 marks)
At the changeover ΔG = 0, so T = ΔH ÷ ΔS = 206 000 ÷ 214.7 = 959 K (about 960 K).Marking: 1 mark for ΔG = 0; 1 mark for T = ΔH/ΔS with consistent units; 1 mark for the answer.
Question 6 Reading a ΔG–T graph (3 marks)
- ΔH is negative (ΔG is negative at T = 0) and ΔS is negative (ΔG rises as T increases).
- ΔG = 0 at 400 K.
- Spontaneous below 400 K.
Marking: 1 mark for each part.
Question 7 From ΔG to K (2 marks)
ln K = 17 000 ÷ (8.31 × 298) = 6.865. K = e6.865 = 9.6 × 10².Marking: 1 mark for ln K; 1 mark for K.
Question 8 Entropy of the surroundings (2 marks)
An exothermic reaction transfers heat to the surroundings, giving their particles more energy and more ways to arrange it. ΔS(surroundings) = −ΔH/T is positive when ΔH is negative.Marking: 1 mark for heat transferred to the surroundings; 1 mark for the link to −ΔH/T and a positive entropy change.
