Chemical equilibrium — Question 120

Answer
The volume is halved. Immediately $Q_c={0.400}^2/0.400=0.400>K_c$, so the net reaction is reverse. Initially and finally $K_c=0.200$.

Working
At 4.0 min both concentrations double from 0.200 to 0.400 mol L⁻¹, indicating compression to half the volume. Initially $K_c={0.200}^2/0.200=0.200$. Immediately $Q_c=0.400$, so B is consumed and A is formed. The final values give $K_c={0.300}^2/0.450=0.200$. Only the volume changed, so the equilibrium constant is unchanged.

Marking criteria

• Identifies halving the vessel volume. [1 mark]
• Uses the simultaneous doubling of both concentrations as evidence. [1 mark]
• Calculates the immediate Qc = 0.400. [1 mark]
• Compares Qc with Kc and predicts the reverse reaction. [1 mark]
• Calculates initial Kc = 0.200. [1 mark]
• Calculates final Kc = 0.200 from the new concentrations. [1 mark]