For acid catalysed hydrolysis of ester, rate law obtained is rate= k [ester] $[H^{+}]$
where $k = 0.01 M^{- 1} {hr}^{- 1}$ . What is the half-life of this reaction, if the initial concentrations are 0.02 M for the ester and 0,05 M for the catalyzing acid?

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Detailed Solution

The value of rate constant,

$\begin{array}{l} k^{'} = 0.01 \times 0.05 \end{array}$

The half-life of this reaction

$t_{1 / 2} = \frac{\ln 2}{k^{'}} t_{1 / 2} = \frac{0.693}{0.01 \times 0.05} t_{1 / 2} = 1386 hrs$

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