Q.
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 ($55\, mol\, L^{-1}$), during the course of the reaction. What is the value of k' in this equation?
$Rate = k' [CH_3COOCH_3] [H_2O] $
Chemical Kinetics
Solution:
For a pseudo first order reaction, the reaction should be first order with respect to ester when $[H_2O] $ is constant. The rate constant k for pseudo first order reaction is
$k=\frac{2.303}{t } log\frac{ C_{0}}{C}where k =\left[H_{2}O\right]$
From the above data we note
t/min
$ C/molL^{-1} $
$ k/ min^{-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 $k=2.004 \times 10^{-3} min ^{-1}$
$\begin{array}{ll}\therefore \quad & k^{\prime}\left[ H _{2} O \right]=2.004 \times 10^{-3} min ^{-1} \\ & k^{\prime}\left[55 mol L ^{-1}\right]=2.004 \times 10^{-3} min ^{-1} \\ & k^{\prime}=3.64 \times 10^{-5} mol ^{-1} L \min ^{-1}\end{array}$
| t/min | $ C/molL^{-1} $ | $ k/ min^{-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} $ |