Question

consider a second order aqueous phase reaction 2A-B which is maintained at 300k isothermally. calculate the ratio of time required for 90% conversion to 65% CONVERSION

Answer 1

$[A_o] = 1$Given:

$2A \to B$ is a second order reaction

To find:

Ratio of time from 90% to 65% conversion

Explanation:

The rate of such a reaction can be written either as $r = k [A]^2$, or as $r = k [A][B]$.

Integrated rate expression of second order reaction is given by:

$\frac{1}{A}-\frac{1}{A_o}= kT$   $\to$  $1$

Considering 90% conversion in the first case:

$[A] = 0.1$

$[A_o ]$ $=1$

Substituting these values in equation 1;

we get , $kT_1 = 9$    $\to$  $2$

Similarly when 65% conversion occurs :

$[A] = 0.35$

$[A_o] = 1$

we get, $kT_2 = 1.85$   $\to$   $3$

dividing equations $2$ and $3$ to determine the required ratio :

$\frac{T_1}{T_2} = 4.86$

Therefore ratio of time required for 90% conversion to 60% conversion is 4.86.