Question

For a given reaction, the energy of activation for the forward reaction $\left(E_{af}\right)$ is $80{\text{kJ mol}}^{-1}$ and $\Delta H=-40{\text{kJ mol}}^{-1}$. A catalyst lowers $E_{af}$ to $20{\text{kJ mol}}^{-1}$. The ratio of energy of activation for the reverse reaction before and after the addition of the catalyst is:

• A. $1.0$
• B. $0.5$
• C. $1.2$
• D. $2.0$

Answer 1

The correct option is D $2.0$

Given,
Change in enthalpy of the reaction $\Delta H=-40{\text{kJ mol}}^{-1}$
Energy of activation for the forward reaction $Ea=80{\text{kJ mol}}^{-1}$,
Energy of activation for the backward reaction $E_b=?$
We know,
$\Delta H=E_f-E_b$
$\Rightarrow -40=80-E_b$
$\Rightarrow E_b=120{\text{kJ mol}}^{-1}$

A catalyst lowers the $E_f$ to $20{\text{kJ mol}}^{-1}$.
We know that a catalyst decreases the activation energy in equal amount in the both the directions.
Thus,
$E_b^{\prime }=(120-60)=60{\text{kJ mol}}^{-1}$

$\therefore \frac{E_b}{E_b^{\prime }}=\frac{120}{60}=2.0$