AQA A-Level Chemistry: Entropy and Gibbs Free Energy in Chemical Reactions — mark scheme explained

The question

Calculate the entropy change (∆S) for the reaction: H 2 (g) + 1/2 O 2 (g) → H 2 O(g), given the absolute entropy values: S ° (H 2 ) = 130.68 J K -1 mol -1 , S ° (O 2 ) = 205.0 J K -1 mol -1 , and S ° (H 2 O) = 188.84 J K -1 mol -1 . [Paraphrased for study — not reproduced from any exam paper.]

Mark scheme, decoded

What each mark is really for — in plain English — and the wording trap that loses it.

• S1

  Calculate the total entropy of the reactants: S reactants ° = 130.68 J K -1 mol -1 + 0.5 × 205.0 J K -1 mol -1 = 233.18 J K -1 mol -1

• S2

  Calculate the total entropy of the products: S products ° = 188.84 J K -1 mol -1

• S3

  Calculate the entropy change: ∆S = S products ° - S reactants ° = 188.84 J K -1 mol -1 - 233.18 J K -1 mol -1 = -44.34 J K -1 mol -1

Model answer

Worked through, with each step tagged to the mark it earns.

1. S1

   Calculate the total entropy of the reactants: S reactants ° = 130.68 J K -1 mol -1 + 0.5 × 205.0 J K -1 mol -1 = 233.18 J K -1 mol -1

2. S2

   Calculate the total entropy of the products: S products ° = 188.84 J K -1 mol -1

3. S3

   Calculate the entropy change: ∆S = S products ° - S reactants ° = 188.84 J K -1 mol -1 - 233.18 J K -1 mol -1 = -44.34 J K -1 mol -1

4. Final answer: -44.34 J K -1 mol -1

Common mistakes

• Using the wrong units for entropy (S) and temperature (T). — Always ensure that entropy is in joules per kelvin per mole (J K -1 mol -1 ) and temperature is in Kelvin when using the Gibbs free energy equation.
• Forgetting to convert enthalpy (∆H) from kJ to J. — Always convert ∆H from kilojoules (kJ) to joules (J) when substituting into the Gibbs free energy equation.
• Confusing the signs of ∆H and ∆S in the Gibbs free energy equation. — Remember that a positive ∆H indicates an endothermic reaction, while a negative ∆H indicates an exothermic reaction. A positive ∆S indicates an increase in disorder, while a negative ∆S indicates a decrease in disorder.
• Misinterpreting the feasibility of a reaction based on ∆G. — Always check if ∆G is less than or equal to zero to determine the feasibility of a reaction. If ∆G is positive, the reaction is non-spontaneous.
• Incorrectly calculating T∆S in the Gibbs free energy equation. — Double-check your calculations for T∆S and ensure that you are multiplying the correct values. Use a calculator if necessary to avoid simple mistakes.
• Failing to explain the significance of entropy in determining reaction feasibility. — Practice explaining the role of entropy in reaction feasibility. A positive ∆S indicates an increase in disorder, which is generally favorable for a reaction to proceed. Conversely, a negative ∆S suggests a decrease in disorder, making the reaction less likely to occur spontaneously.

Where the marks go

• Full worked solution (all marking points)3 marks

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