Question

The first order reaction is 20% complete in 20 minutes. Calculate the time it will take the reaction to complete 80%.

Answer 1

For first order reaction we have:-

$K=\frac{1}{t}ln\left(\frac{a}{a-x}\right)$ $-\left(i\right)$

where, $K$= rate constant

$t$= time

$a$= initial concentration of the reactant

$x$= amount of reactant reacted in time $t$

Case I: $t=20$ min

$x=\frac{20}{100}\times a$$\Rightarrow x=0.2a$ $(\because 20$ % reaction is complete$)$

$\Rightarrow$ $a-x=0.8a$

Putting the values in $e{q}^{n}1$ :-

$K=\frac{1}{20}ln\left(\frac{a}{0.8x}\right)$ $-\left(ii\right)$

Case II: $t=?$ , $x=0.8a$

$\Rightarrow a-x=0.2a$

Putting the values in (i):-

$\Rightarrow K=\frac{1}{t}ln\left(\frac{x}{0.2x}\right)$ $-\left(iii\right)$

From $\left(ii\right)$ and $\left(iii\right)$:-

$\frac{1}{20}ln\left(\frac{10}{8}\right)=\frac{1}{t}ln\left(\frac{10}{2}\right)$

$\Rightarrow t=20\times \frac{lnS}{ln\left(S/4\right)}=144.25$ min