Calculating entropy changes

Overview

Fifteen recall questions on calculating entropy changes.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   State the relationship used to calculate an entropy change from absolute entropy values.

   Model answer

   The entropy change is ΔS = S(final) - S(initial). For a reaction this means the total entropy of the products minus the total entropy of the reactants, each entropy being multiplied by the number of moles in the balanced equation.

2. 2

   Why must each absolute entropy be multiplied by the coefficient from the balanced equation?

   Model answer

   Standard entropies are quoted per mole, so a species present as three moles contributes three times its molar entropy. Omitting the coefficients gives a value that does not correspond to the equation as written.

3. 3

   The standard entropy of liquid water is 69.9 and that of steam is 188.7 J K⁻¹ mol⁻¹. Calculate the entropy change on vaporising one mole of water.

   Model answer

   The entropy change is 188.7 - 69.9 = +118.8 J K⁻¹ mol⁻¹. It is positive because the freely moving gas molecules are far more disordered than the liquid.

4. 4

   Using standard entropies of N₂ 191.6, H₂ 130.6 and NH₃ 192.3 J K⁻¹ mol⁻¹, calculate the entropy change for N₂(g) + 3H₂(g) → 2NH₃(g).

   Model answer

   The products contribute 2 × 192.3 = 384.6 J K⁻¹ mol⁻¹. The reactants contribute 191.6 + 3 × 130.6 = 191.6 + 391.8 = 583.4 J K⁻¹ mol⁻¹. The entropy change is 384.6 - 583.4 = −198.8 J K⁻¹ mol⁻¹.

5. 5

   Using standard entropies of CaCO₃ 92.9, CaO 39.7 and CO₂ 213.6 J K⁻¹ mol⁻¹, calculate the entropy change for CaCO₃(s) → CaO(s) + CO₂(g).

   Model answer

   The products contribute 39.7 + 213.6 = 253.3 J K⁻¹ mol⁻¹ and the reactant 92.9 J K⁻¹ mol⁻¹. The entropy change is 253.3 - 92.9 = +160.4 J K⁻¹ mol⁻¹.

6. 6

   Using standard entropies of H₂ 130.6, O₂ 205.0 and liquid water 69.9 J K⁻¹ mol⁻¹, calculate the entropy change for 2H₂(g) + O₂(g) → 2H₂O(l).

   Model answer

   The products contribute 2 × 69.9 = 139.8 J K⁻¹ mol⁻¹. The reactants contribute 2 × 130.6 + 205.0 = 261.2 + 205.0 = 466.2 J K⁻¹ mol⁻¹. The entropy change is 139.8 - 466.2 = −326.4 J K⁻¹ mol⁻¹, strongly negative because three moles of gas become two moles of liquid.

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