Question

In a pseudo first order hydrolysis of ester in water the following results were obtained:

$t/s$	$0$	$30$	$60$	$90$
$\left[Ester\right]/M.$	$0.55$	$0.31$	$0.17$	$0.085$

(i) Calculate the average rate of reaction between the time interval $30$ to $60$ seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Answer 1

(i) The average rate of reaction between the time interval 30 to 60 is $\frac{0.31-0.17}{60-30}=4.67\times {10}^{-3}M/s$.

(ii) $k=\frac{2.303}{t}log\frac{[Ester{]}_0}{\left[Ester\right]}$

When t = 30 s
$k=\frac{2.303}{30}\times log\left(\frac{0.55}{0.31}\right)=1.91\times {10}^{-2}/s$

When t = 60 s
$k=\frac{2.303}{60}\times log\left(\frac{0.55}{0.17}\right)=1.96\times {10}^{-2}/s$

When t = 90 s
$k=\frac{2.303}{90}\times log\left(\frac{0.55}{0.085}\right)=2.07\times {10}^{-2}/s$

Average rate constant $k=\frac{k1+k2+k3}{3}=1.98\times {10}^{-2}/s$