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Passage / Comprehension

Hydrolysis of ester follows first-order kinetics in the presence of an acid catalyst.

$RCOOR^{'} + H_{2} O \to RCOOH + R^{'} OH$

To study its kinetics, a fixed amount of reaction mixture is taken and titrated with a strong base like NaOH, which reacts with the acid formed and the catalyst present. By measuring the volume of NaOH required at different intervals of time, the following observations are made for one such reaction's progress.

If same reaction is assumed to be performed without catalyst $(H^{+})$ and titrated with same reagent, then relationship which holds true for rate constant ' K ' (Given : $V_{0} = 0$ )

Options:

Answer:

Solution:

Ester hydrolysis follows pseudo-first-order kinetics:

$k = \frac{1}{t} ln \frac{[ A ] _{0}}{[ A ] _{t}}$

In titration terms, the concentration of ester is proportional to $V_{\infty} - V_{t}$ .

So, $[A]_{t} \propto (V_{\infty} - V_{t})$ and $[A]_{0} \propto (V_{\infty} - V_{0})$ .

Hence,

$k = \frac{1}{t} ln \frac{V _{\infty} - V _{0}}{V _{\infty} - V _{t}}$

If $V_{0} = 0$ , then:

$k = \frac{1}{t} ln \frac{V _{\infty}}{V _{\infty} - V _{t}}$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Rate of Reactions/Rate Law

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