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Lattice Enthalpy, Entropy and Free Energy

Standard entropy change calculations

  • 3 things to know
  • Unit A2 1

Overview

Recall questions on the products-minus-reactants expression and why coefficients are applied, worked calculations for the decomposition of calcium carbonate, the synthesis of ammonia, boiling water and the reduction of carbon dioxide, converting to kJ K⁻¹ mol⁻¹, why an element's standard entropy is not zero, and why boiling gives a far larger change than melting.

What you need to know

Answer each one before you reveal it.

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Key term

State the expression used to calculate a standard entropy change for a reaction.

Key term

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

Numbers

Calculate ΔSӨ for CaCO₃(s) → CaO(s) + CO₂(g), given standard entropies of 93, 40 and 214 J K⁻¹ mol⁻¹ respectively.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the expression used to calculate a standard entropy change for a reaction.

    Model answer

    The standard entropy change equals the total standard entropy of the products minus the total standard entropy of the reactants.

  2. 2

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

    Model answer

    Standard entropies are quoted per mole, so a species present as two moles contributes twice its molar value.

  3. 3

    Calculate ΔSӨ for CaCO₃(s) → CaO(s) + CO₂(g), given standard entropies of 93, 40 and 214 J K⁻¹ mol⁻¹ respectively.

    Model answer

    The products total 40 + 214 = 254 J K⁻¹ mol⁻¹, so ΔSӨ is 254 − 93 = +161 J K⁻¹ mol⁻¹.

  4. 4

    Calculate ΔSӨ for N₂(g) + 3H₂(g) → 2NH₃(g), given standard entropies of 192, 131 and 192 J K⁻¹ mol⁻¹ respectively.

    Model answer

    The products total 2(192) = 384 J K⁻¹ mol⁻¹ and the reactants 192 + 3(131) = 585 J K⁻¹ mol⁻¹, so ΔSӨ is 384 − 585 = −201 J K⁻¹ mol⁻¹.

  5. 5

    Calculate ΔSӨ for H₂O(l) → H₂O(g), given standard entropies of 70 and 189 J K⁻¹ mol⁻¹.

    Model answer

    The entropy change is 189 − 70 = +119 J K⁻¹ mol⁻¹.

  6. 6

    Calculate ΔSӨ for 2H₂(g) + O₂(g) → 2H₂O(l), given standard entropies of 131, 205 and 70 J K⁻¹ mol⁻¹ respectively.

    Model answer

    The products total 2(70) = 140 J K⁻¹ mol⁻¹ and the reactants 2(131) + 205 = 467 J K⁻¹ mol⁻¹, so ΔSӨ is 140 − 467 = −327 J K⁻¹ mol⁻¹.

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