Hydrolysis of methyl acetate in aqueous solution has been studied by titrating acetic acid against NaOH. The concentration of ester at different times is given below.
Time in minutes 0 23.03 46.06 69.09
Ester mole/lit 40 20 10 5
The concentration of water remain constant at 55 M. Find Rate constant (K)?

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3 $\times$ 10^-2 lit/mole/min

3 $\times$ 10^-2 min^-1

5.4 $\times$ 10^-4min^-1

5.4 $\times$ 10^-4 lit/mole/min

answer is D.

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Detailed Solution

K $[H_{2} O] = \frac{2.303}{t} \log \frac{a}{a - x}$

K $[H_{2} O] = \frac{2.303}{23.03} \log \frac{40}{20} = 0.03$

Average value for different time intervals K[H₂O] = 0.03 min^-1

It is pseudo first order reaction

K $[\frac{1000}{18}]$ = 0.03

K = 5.4 $\times$ 10^-4 lit/mole/min

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