For the hydrolysis of methyl acetate in aqueous solution the above tabulated results were obtained:
(a) Show that it follow pseudo first order reaction, as the concentration of water remains constant.
(b) Calculate the average rate of reaction between the time interval 10 to 20 seconds.
(Given : log 2=0.3010, log 4=0.6021)

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Solution

(a) The reaction is $1^{s t}$ order w.r.t ester since $[H_{2} O]$ remain constant. $T_{\circ} a n d T_{t}$ are concentrations at respective times.
Thus, rate constant $(k) = \frac{1}{t} I n \frac{T_{\circ} - T_{t}}{T_{\circ}}$
$∴ k = \frac{1}{10} I n \frac{0.10 - 0.05}{0.10} = 0.0693$
$k = \frac{1}{10} i n \frac{0.05 - 0.025}{0.05} = 0.0693 s^{- 1}$
It appears from the calculation that value of $\frac{1}{t} I n \frac{T_{0} - T_{t}}{T_{0}}$ is constant. Thus, the reaction is $1^{s t}$ order (as assumed ).
(b) The average rate constant is
$= \frac{Δ [C H_{3} C O O C H_{3}]}{Δ t}$
$= \frac{[0.02 m o l L^{- 1} - 0.05 m o l L^{- 1}]}{20 s_{\circ} - 10 s} = \frac{0.025}{10 s} m o l^{L - 1}$
$= 0.0025 m o l L^{- 1} s$

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log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)

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