Question

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

t in s	0	30	60	90
[Ester] in mol $L^{-1}$	0.55	0.31	0.17	0.085

Calculate the average rate of reaction between the time interval 30 to 60 seconds.

Answer 1

The given reaction pseudo first order reaction.

Now, for first order reaction we have:-

$K=\frac{1}{t}ln\frac{[A{]}_0}{[A{]}_t}$

& Rate=$K\left[Ester\right]$

Now, at $t=0,\left[Ester\right]=0.55mol{L}^{-1}$

$t=30s,\left[Ester\right]=0.31mol{L}^{-1}$

$\Rightarrow K=\frac{1}{30}ln\left(\frac{0.55}{0.31}\right)=\frac{0.53}{30}=0.019s^{-1}$

Now, st $t=30s$, $\left[Ester\right]=0.31mol{L}^{-1}$

so, Rate=$0.019\times 0.31=0.02s$ $mol{L}^{-1}{s}^{-1}$

At $t=60s$, $\left[Ester\right]=0.17$ $mol{L}^{-1}$

$\therefore Rate=0.019\times 0.17=0.015$ $mol{L}^{-1}{s}^{-1}$

Average rate of reaction from $30$ to $60s$

$=\frac{0.028+0.015}{2}$ $mol{L}^{-1}{s}^{-1}$

$=\frac{0.043}{2}=0.022$ $mol{L}^{-1}{s}^{-1}$