2021 H2 Chem P1 Q12 — Chemical equilibrium Solution

The answer is A.

CaO and \(\text{CaCO}_3\) are solids, so the rate of the forward reaction depends only on the partial pressure of \(\text{CO}_2\): rate of forward reaction \(\propto p_{\text{CO}_2}\).

• At time \(t\) the plunger is pulled out quickly, so the pressure, and hence \(p_{\text{CO}_2}\), falls suddenly. The rate of the forward reaction drops sharply.
• The position of equilibrium shifts left, and \(\text{CaCO}_3\) decomposes to produce more \(\text{CO}_2\) to counter this decrease, so \(p_{\text{CO}_2}\) rises again and the forward rate increases.
• The pressure is allowed to return slowly to atmospheric pressure at the same temperature. \(K_p = \frac{1}{p_{\text{CO}_2}}\) is unchanged, so \(p_{\text{CO}_2}\) returns to its original value and the forward rate slowly returns to its original value.
Graph A shows exactly this: a sharp fall at \(t\) followed by a gradual rise back to the original rate.

• B is wrong: it shows the rate falling at \(t\) and staying at the lower value. The pressure returns to atmospheric and \(p_{\text{CO}_2}\) returns to its equilibrium value, so the rate recovers to its original value.
• C is wrong: it shows the rate rising at \(t\). Lowering the pressure lowers \(p_{\text{CO}_2}\), so the forward rate falls, it does not rise.
• D is wrong: it shows no change at all, but the sudden fall in \(p_{\text{CO}_2}\) lowers the forward rate at time \(t\).