The acid-catalysed hydrolysis of ethyl acetate, $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$, is carried out in a large excess of water so that $[{\mathrm{H}}_2\mathrm{O}]$ is essentially constant at $55.5\mathrm{mol}{\mathrm{L}}^{-1}$. Under these conditions the reaction follows pseudo-first-order kinetics with a measured rate constant $k^{\prime }=1.10\times 10^{-3}{\mathrm{s}}^{-1}$ at $298\mathrm{K}$.

Step-by-step solution

1. Step 1: Identify the relationship. The true rate law is $\text{rate}=k[\text{ester}][{\text{H}}_2\text{O}]$. Because $[{\text{H}}_2\text{O}]\approx \text{const}$, we absorb it into $k^{\prime }$: $\text{rate}=k^{\prime }[\text{ester}]$ where $k^{\prime }=k[{\text{H}}_2\text{O}]$. Hence $k=\frac{k^{\prime }}{[{\text{H}}_2\text{O}]}$.
2. Step 2: Calculate $k$ for part (a): $k=\frac{1.10\times 10^{-3}{\mathrm{s}}^{-1}}{55.5\mathrm{mol}{\mathrm{L}}^{-1}}=1.98\times 10^{-5}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{L}{\mathrm{s}}^{-1}$. Round to two significant figures: $k=2.0\times 10^{-5}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{L}{\mathrm{s}}^{-1}$.
3. Step 3: For part (b), the temperature is unchanged so the true constant $k$ is the same. The new pseudo-first-order constant is $k_{\text{new}}^{\prime }=k[{\text{H}}_2\text{O}{]}_{\text{new}}=(2.0\times 10^{-5})(54.0)=1.08\times 10^{-3}{\mathrm{s}}^{-1}$.
4. Step 4: For part (c), compare $55.5$ and $54.0\mathrm{mol}{\mathrm{L}}^{-1}$: the ratio is $54.0/55.5=0.973$, a change of only $2.7\%$. Since the ester is consumed but water's concentration changes negligibly, $[{\text{H}}_2\text{O}]$ is effectively constant throughout the reaction, validating the pseudo-first-order treatment.

Concept tested: Pseudo-first-order kinetics and conversion to true rate constant

Published 16 September 2026 · How this solution was prepared and checked