Entropy change, ΔS — practice questions

1. Identify products and reactants: Products: 2NH₃(g) Reactants: N₂(g), 3H₂(g)
2. Write the formula: $ΔS^\circ_{reaction} = ΣS^\circ(products) - ΣS^\circ(reactants)$
3. Substitute values, including stoichiometry: $ΔS^\circ = [2 \times S^\circ(\text{NH}_3)] - [ (1 \times S^\circ(\text{N}_2)) + (3 \times S^\circ(\text{H}_2)) ]$ $ΔS^\circ = [2 \times 192.8] - [191.6 + (3 \times 130.7)]$
4. Calculate the sums for products and reactants: $ΣS^\circ(products) = 385.6 \text{ J K}^{-1}$ $ΣS^\circ(reactants) = 191.6 + 392.1 = 583.7 \text{ J K}^{-1}$
5. Calculate the final ΔS°: $ΔS^\circ = 385.6 - 583.7 = -198.1 \text{ J K}^{-1} \text{mol}^{-1}$

Comment: The entropy change is negative, which is expected as 4 moles of gas react to form only 2 moles of gas, representing a significant increase in order.

1. Identify products and reactants: Products: CaO(s), CO₂(g) Reactants: CaCO₃(s)
2. Write the formula: $ΔS^\circ_{reaction} = ΣS^\circ(products) - ΣS^\circ(reactants)$
3. Substitute values: $ΔS^\circ = [ (1 \times S^\circ(\text{CaO})) + (1 \times S^\circ(\text{CO}_2)) ] - [1 \times S^\circ(\text{CaCO}_3)]$ $ΔS^\circ = [38.1 + 213.8] - [92.9]$
4. Calculate the sums: $ΣS^\circ(products) = 251.9 \text{ J K}^{-1}$ $ΣS^\circ(reactants) = 92.9 \text{ J K}^{-1}$
5. Calculate the final ΔS°: $ΔS^\circ = 251.9 - 92.9 = +159.0 \text{ J K}^{-1} \text{mol}^{-1}$

Comment: The entropy change is large and positive. This is dominated by the production of one mole of gas from solid reactants, leading to a massive increase in disorder.