Question

For following reactions, find the activation energy for catalyzed reaction

$\mathrm{A}\stackrel{700\mathrm{K}}{\to }$Product

A $\underset{catalyzed}{\overset{500\mathrm{K}}{\to }}$Product

It was found that the ${\mathrm{E}}_{\mathrm{a}}$ is decreased by $30\mathrm{KJ}/\mathrm{mol}$ in the presence of catalyst. If the rate remains unchanged, the activation energy for catalyzed reaction is (Assume pre exponential factor is same).

• A. $75\mathrm{kJ}/\mathrm{mol}$
• B. $135\mathrm{kJ}/\mathrm{mol}$
• C. $105\mathrm{kJ}/\mathrm{mol}$
• D. $198\mathrm{kJ}/\mathrm{mol}$

Answer 1

The correct option is A

$75\mathrm{kJ}/\mathrm{mol}$

The above question can be solved as:

Step 1:

$T_1=700K$ ,$T_2=500K$

Step 2: From the formulae

$$\begin{gathered} {\mathrm{K}}_1={\mathrm{Ae}}^{{\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1} \\ {\mathrm{K}}_1={\mathrm{Ae}}^{{\mathrm{E}}_{\mathrm{a}}/\mathrm{R}700} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_1={\mathrm{Ae}}^{{\mathrm{E}}_{\mathrm{a}}/\mathrm{R}700} \\ {\mathrm{K}}_2={\mathrm{Ae}}^{{\mathrm{E}}_{\mathrm{a}}-30/\mathrm{R}500} \end{gathered}$$

Step 3:

The rate of the reaction is same, thus ${\mathrm{K}}_1={\mathrm{K}}_2$

${\mathrm{e}}^{{\mathrm{E}}_{\mathrm{a}}/\mathrm{R}700}={\mathrm{e}}^{{\mathrm{E}}_{\mathrm{a}}-30/\mathrm{R}500}$

$E_a/700\mathrm{R}=E_a-30/500\mathrm{R}$

$5E_a=7E_a-210$

$210=2E_a$

$E_a=105\mathrm{kJ}/\mathrm{mol}$

Step 3:

The activation energy of catalyzed reaction $$\begin{gathered} E_a-30=105-30\mathrm{kJ}/\mathrm{mol} \\ {E}_a=75\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Therefore, the activation energy of catalyzed reaction is $75\mathrm{kJ}/\mathrm{mol}$.