NCEA Scholarship Chemistry free practice questions with answers

Question 1 (8 marks) [Thermochemistry, entropy and Gibbs energy]

(a) Lime (calcium oxide) is made by heating limestone in a kiln: CaCO₃(s) → CaO(s) + CO₂(g). Use the thermodynamic data in the table below (also Table 2 of the resource booklet); treat ΔH° and S° as constant with temperature. Calculate ΔrH° (kJ mol⁻¹) and ΔrS° (J K⁻¹ mol⁻¹) for the decomposition. Explain why ΔrS° has the sign you found.

(a) Answer: ΔrH° = +178 kJ mol⁻¹; ΔrS° = +162 J K⁻¹ mol⁻¹

• Entry step: worth marks in the 1-4 range of the 8-point question score.

• ΔrH° = [ΔfH°(CaO) + ΔfH°(CO₂)] − ΔfH°(CaCO₃) = (−635) + (−394) − (−1207) = +178 kJ mol⁻¹. [1 mark]

• ΔrS° = 40 + 214 − 92 = +162 J K⁻¹ mol⁻¹. It is positive because the reaction produces a gas from a solid: the particles of CO₂(g) are free to move and can be arranged in many more ways than in a solid lattice, so the number of accessible microstates (the dispersal of energy) increases. [1 mark]

(b) Calculate the minimum temperature, in K and in °C, at which the decomposition is thermodynamically spontaneous under standard conditions. Explain why the reaction is spontaneous above this temperature and not below it.

(b) Answer: T ≈ 1099 K (826 °C)

• Scholarship band (5-6): the part needs your own method and accurate reasoning.

• Spontaneous when ΔG° = ΔH° − TΔS° < 0. At the changeover ΔG° = 0, so T = ΔH° ÷ ΔS° = 178 000

J mol⁻¹ ÷ 162 J K⁻¹ mol⁻¹ = 1099 K. [1 mark]

• In °C: 1099 − 273 = 826 °C. [1 mark]

• ΔH° is positive (unfavourable) and ΔS° is positive. At low T the term TΔS° is too small to outweigh ΔH°, so ΔG° > 0; as T rises TΔS° grows until it exceeds ΔH° and ΔG° becomes negative, so the reaction is spontaneous above 1099 K. [1 mark]

(c) A sample of limestone is heated at 1000 K in a sealed, evacuated vessel until the pressure stops changing. Calculate the equilibrium constant K at 1000 K for the decomposition and hence the equilibrium pressure of CO₂(g) in bar. (K = p(CO₂) ÷ p°, with p° = 1 bar.) In an open kiln the CO₂ is carried away, so its partial pressure stays near 4.0 × 10⁻⁴ bar, the value in air. Calculate the lowest temperature at which limestone can decompose when p(CO₂) is held at this value, and explain why it is far below your answer to part (b).

(c) Answer: K = 0.146; p(CO₂) = 0.146 bar; T ≈ 784 K (511 °C)

• Outstanding band (7-8): the part needs insight and a complete, well-supported argument.

• ΔG° at 1000 K = 178 000 − 1000 × 162 = +16000 J mol⁻¹. K = e^(−ΔG°/RT) = e^(−16000 ÷ (8.314 × 1000)) = 0.1460, so p(CO₂) = 0.146 bar. [1 mark]

• Reaction is just possible when ΔG = 0 with Q = 4.0 × 10⁻⁴: ΔH° − TΔS° + RT ln(4.0 × 10⁻⁴) = 0, so T = ΔH° ÷ [ΔS° − R ln(4.0 × 10⁻⁴)] = 178 000 ÷ (162 + 65.05) = 784 K (511 °C). [1 mark]

• Part (b) assumes standard conditions (1 bar CO₂). In an open kiln the product gas is removed, so Q is tiny and ΔG = ΔG° + RT ln Q is much more negative than ΔG°; by Le Chatelier's principle removing CO₂ shifts the equilibrium to the right. The decomposition therefore starts at about 784 K, far below 1099 K, and is driven further by continuous removal of CO₂. [1 mark]

Question 2 (8 marks) [Aqueous equilibria and titration]