Question

A reaction takes place at 300 K. When the catalyst is added, the rate of reaction increases. The change in temperature $\left(in{}^{\circ }C\right)$ which can create the same effect as produced by catalyst (experimentally it is known that catalyst change the activation energy by $20\%$) is $x$, then find the value of $\frac{x+6}{9}$:

Answer 1

$ln\frac{K_2}{K_1}=\frac{E_a}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$ ............(i)

$ln{K}_C=\frac{\left(E_{ac}\right)}{R}\times T_1$

For effect of catalyst new $K_C=\frac{K_2}{K_1}$

$ln\frac{K_2}{K_1}=\frac{\left(E_{ac}\right)}{R}\times T_1$ ...........(ii)

from (i) and (ii) $\frac{\left(E_{ac}\right)}{R}\times T_1=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$[\frac{1}{T_1}-\frac{1}{T_2}]=\frac{\frac{E_a}{E_{ac}}}{E_a\times T_1}$

$\frac{1}{300}-\frac{1}{T_2}=\frac{(100-80)}{100}\times 300=\frac{20}{100}\times 300=\frac{1}{1500}$

$\frac{1}{T_2}=\frac{1}{300}-\frac{1}{1500}=\frac{5-1}{1500}=\frac{4}{1500}$

$T_2=375$

Change in temp $=375-300=75$