Question

Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given :

It follows a pseudo first order reaction, as the concentration of water remains nearly constant ($55mol{L}^{-1}$), during the course of the reaction. What is the value of k' in this equation? $Rate=k^{\prime }[C{H}_{3}COOC{H}_3][H_{2}O]$

• A. $3.64\times 10^{-5}mo{l}^{-1}Lmi{n}^{-1}$
• B. $2.5\times 10^{-5}mo{l}^{-1}Lmi{n}^{-1}$
• C. $4\times 10^{-2}mo{l}^{-1}Lmi{n}^{-1}$
• D. $5\times 10^{-5}mo{l}^{-1}Lmi{n}^{-1}$

Answer 1

Answer: (A)

For a pseudo first order reaction, the reaction should be first order with respect to ester when $[H_{2}O]$ is constant. The rate constant k for pseudo first order reaction is
$k=\frac{2.303}{t}log\frac{C_0}{C}wherek=\left[H_{2}O\right]$
From the above data we note

t/min	$C/mol{L}^{-1}$	$k/mi{n}^{-1}$
0	0.8500	-
30	0.8004	2.004 $\times 10^{-3}$
60	0.7538	2.002 $\times 10^{-3}$
90	0.7096	2.005 $\times 10^{-3}$

Average value of $k=2.004\times 10^{-3}mi{n}^{-1}$

$\begin{array}{ll}\therefore & k^{\prime }\left[H_{2}O\right]=2.004\times 10^{-3}mi{n}^{-1}\\ & k^{\prime }\left[55mol{L}^{-1}\right]=2.004\times 10^{-3}mi{n}^{-1}\\ & k^{\prime }=3.64\times 10^{-5}mo{l}^{-1}L{\min }^{-1}\end{array}$