Structured Questions
Complete each question on paper, then check your answers against the mark scheme.
Question 1: Le Chatelier's Principle
8 marksMethanol is manufactured industrially by the reaction of carbon monoxide with hydrogen: CO(g) + 2H₂(g) ⇌ CH₃OH(g) (ΔH = −91 kJ mol⁻¹). Predict and explain the effect of the following changes on the position of equilibrium and the rate of the reaction:
(a) The temperature of the system is increased. [4]
(b) The pressure of the system is increased. [4]
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(a) Temperature increase:
- Rate: Increases because particles have more kinetic energy and move faster, leading to a higher collision frequency. More importantly, a much larger fraction of colliding molecules have energy greater than or equal to the activation energy. [2]
- Equilibrium position: Shifts to the left. [1] Since the forward reaction is exothermic, the reverse reaction is endothermic. The system shifts in the endothermic direction to absorb heat and oppose the temperature rise. [1]
(b) Pressure increase:
- Rate: Increases because gaseous reactant molecules are compressed into a smaller volume, increasing the number of molecules per unit volume, which increases the collision frequency. [2]
- Equilibrium position: Shifts to the right. [1] There are 3 moles of gas on the left-hand side (1 CO + 2 H₂) and only 1 mole of gas on the right-hand side (1 CH₃OH). The system shifts to the side with fewer gas moles to decrease the pressure and oppose the pressure rise. [1]
Question 2: Equilibrium Constant Kc calculation
5 marksA student mixes 2.00 mol of ethanol (CH₃CH₂OH) and 1.50 mol of ethanoic acid (CH₃COOH) in a closed flask. The mixture is left to reach equilibrium at a constant temperature: CH₃COOH(l) + CH₃CH₂OH(l) ⇌ CH₃COOCH₂CH₃(l) + H₂O(l). At equilibrium, the mixture contains 0.85 mol of ethyl ethanoate (ester). Calculate the value of the equilibrium constant \(K_c\) for this reaction at this temperature. [5]
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Set up ICE table (moles):
| Species | CH₃COOH | CH₃CH₂OH | CH₃COOCH₂CH₃ | H₂O |
|---|---|---|---|---|
| Initial | 1.50 | 2.00 | 0 | 0 [1] |
| Change | -0.85 | -0.85 | +0.85 | +0.85 |
| Equilibrium | 0.65 [1] | 1.15 [1] | 0.85 | 0.85 [1] |
Calculate Kc:
- \(K_c\) expression: \(K_c = \frac{[\text{CH}_3\text{COOCH}_2\text{CH}_3][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{CH}_3\text{CH}_2\text{OH}]}\)
- Since the total volume \(V\) cancels out (same number of moles on both sides):
\(K_c = \frac{(0.85/V) \times (0.85/V)}{(0.65/V) \times (1.15/V)} = \frac{0.85 \times 0.85}{0.65 \times 1.15}\) - \(K_c = \frac{0.7225}{0.7475} = 0.967\) (or 0.97) [1]
- Units: None (all units cancel).
Question 3: Factors Affecting Kc
4 marksExplain why the value of the equilibrium constant \(K_c\) for the following reaction: N₂(g) + O₂(g) ⇌ 2NO(g) (ΔH = +180 kJ mol⁻¹) increases as the temperature increases, but remains unchanged when the total pressure is increased. [4]
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- The forward reaction is endothermic (ΔH is positive). [1]
- When temperature increases, the position of equilibrium shifts to the right (in the endothermic direction) to absorb the added heat and oppose the change. [1]
- This increases the equilibrium concentrations of the products (NO) and decreases those of the reactants (N₂ and O₂), resulting in a larger value for the ratio in the \(K_c\) expression, so \(K_c\) increases. [1]
- Increasing pressure increases the concentrations of all species, but they adjust so that the ratio in the \(K_c\) expression remains constant, meaning \(K_c\) is unaffected by pressure changes. [1]
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