Question

In a pseudo first order hydrolysis of ester in water, the following results were obtained:
$\begin{array}{ccccc}t/s& 0& 30& 60& 90\\ \left[Ester\right]/mol{L}^{-1}& 0.55& 0.31& 0.17& 0.085\end{array}$
Calculate the average rate of reaction between the time interval 30 to 60 s.

$\begin{array}{ccccc}t/s& 0& 30& 60& 90\\ \left[Ester\right]/mol{L}^{-1}& 0.55& 0.31& 0.17& 0.085\end{array}$
Calculate the psuedo first order rate constant for the hydrolysis of ester.

Answer 1

Average rate during the interval 30 - 60s
$=\frac{C_2-C_1}{t_2-t_1}$
$=\frac{(0.17-0.31)mol{L}^{-1}}{(60-30)s}$
$=\frac{0.14}{30}mol{L}^{-1}{s}^{-1}$
Average rate $=4.67\times {10}^{-3}mol{L}^{-1}{s}^{-1}$

$k^{{}^{\prime }}=\frac{2.303}{t}log\frac{\left[A_0\right]}{\left[A\right]}$ in which $\left[A_0\right]=0.55M$
At $t=30s;k^{{}^{\prime }}=\frac{2.303}{\left(30s\right)}log\frac{0.55}{0.31}=1.91\times {10}^{-2}{s}^{-1}$
At $t=60s;k^{{}^{\prime }}=\frac{2.303}{\left(60s\right)}log\frac{0.55}{0.17}=1.96\times {10}^{-2}{s}^{-1}=1.96\times {10}^{-2}{s}^{-1}$
At $t=90s;k^{{}^{\prime }}=\frac{2.303}{\left(90s\right)}log\frac{0.55}{0.085}=2.07\times {10}^{-2}{s}^{-1}$
Average $k^{{}^{\prime }}=\frac{(1.91+1.96+2.07)\times {10}^{-2}{s}^{-1}}{3}$
Pseudo first order rate constant $=1.98\times {10}^{-2}{s}^{-1}$