Question

A certain reaction is of first order. After $540$ seconds, $32.5$% of the reactant remains.
(a) Calculate the rate constant.
(b) How long would it require for $25$% of the reactant to be decomposed?

Answer 1

For first order reaction we have:-

$K=\frac{1}{t}ln\left(\frac{[A{]}_o}{[A{]}_t}\right)$ $-\left(i\right)$

where, $K$= rate constant

$t$= time

$[A{]}_o$= initial concentration of the reactant

$[A{]}_t$=concentration of reaction at time $t$

Now, as per question,

After, $t=540seconds$, $[A{]}_t=32.5$% of ${[A{]}_o}^{-}$

Putting these values in equation $\left(i\right)$:-

$K=\frac{1}{540}ln\left(\frac{[A{]}_o}{0.325[A{]}_o}\right)$

$\Rightarrow K=\frac{1}{540}ln\left(\frac{1}{0.325}\right)=0.002s^{-1}$

Now, in second case as $25$% of the reactant is to be decomposed so $75$% of it will be remaining after time ${}^{\prime }{t}^{\prime }$

$\therefore$ $t=\frac{1}{k}ln\left(\frac{\left[A_o\right]}{0.75[A{]}_o}\right)$

$=\frac{1}{0.002}ln\left(\frac{1}{0.75}\right)=143.8s$