Acid hydrolysis of ester is first-order reaction and rate constant is given by
$\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{V}_{\infty}-\mathrm{V}_{0}}{\mathrm{~V}_{\infty}-\mathrm{V}_{\mathrm{t}}}$
where $\mathrm{V}_{0}, \mathrm{~V}_{\mathrm{t}}$ and $\mathrm{V} \infty$ are the volume of standard $\mathrm{NaOH}$ required to neutralize acid present at a given time; if ester is $50 \%$ hydrolysed then

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Aicharya · Accepted Answer

$\mathrm{RCOOR}^{\prime}+\mathrm{H}_{2} \mathrm{O} \stackrel{H^{+}}{\longrightarrow} \mathrm{RCOOH}+\mathrm{R}^{\prime} \mathrm{OH}$
$\begin{array}{lll}\text { at } \mathrm{t}=0 \text { : } \mathrm{a} & 0 & 0 \\ \text { at time t: }(\mathrm{a}-\mathrm{x}) & \mathrm{x} & \mathrm{x} \\ \text { at time } \infty:(\mathrm{a}-\mathrm{a}) & \mathrm{a} & \mathrm{a}\end{array}$
at $\mathrm{t}=0, \mathrm{~V}_{0}=$ volume of $\mathrm{NaOH}$ due to $\mathrm{H}^{+}$ (catalyst)
$\mathrm{V}_{\mathrm{t}}=\mathrm{x}+\mathrm{V}_{0}$
$\mathrm{V} \infty=\mathrm{a}+\mathrm{V}_{0}$
If ester is $50 \%$ hydrolysed then $x=\frac{a}{2}$
$V_{t}=\frac{a}{2}+V_{0}$
$a=2 V_{t}-2 V_{0}$
$\therefore V_{\infty}=2 V_{t}-2 V_{0}+V_{0}=2 \mathrm{~V}_{\mathrm{t}}-\mathrm{V}_{0}$ ,

Acid hydrolysis of ester is first-order reaction and rate constant is given by $\mathrm{k}=\frac{2…

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