Question

t/s	0	30	60
$\left[C{H}_{3}COOC{H}_3\right]/mol{L}^{-1}$	0.60	0.30	0.15

For the hydrolysis of methyl acetate in an aqueous solution, the above-tabulated results were obtained:

(i) Show that it follows pseudo-first-order reaction as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2=0.3010,log 4=0.6021)

Answer 1

(i) $k=\frac{2.303}{t}log\frac{[A{]}_0}{\left[A\right]}$
Where, $[A{]}_n=$ Initial concentration of reactant
$\left[A\right]=$ Final concentration of reactant
At $t_1=30sec$,

$k=\frac{2.303}{30}log\frac{0.60}{0.30}$

$k=0.07677log2$

$k=0.0231s^{-1}$

for t=60 sec,
$k=\frac{2.303}{30}log\frac{0.30}{0.15}$

$k=0.07677log2$

$k=0.0231s^{-1}$

$\therefore$ K is same for both the case hence it is pseudo first order reaction.

(ii) Average rate of reaction between the time interval of 30-60 seconds is given by
$k=\frac{-\Delta \left[C{H}_{3}COOC{H}_3\right]}{\Delta t}$

$k=-\left(\frac{0.15-0.30}{60-30}\right)$

$=\frac{0.15}{30}=0.005mol{L}^{-1}{s}^{-1}$