Question

The energy of activation for a reaction is $50$ kJ/ mol. The presence of a catalyst lowers the energy of activation by $25$ %. What will be the effect on the rate of reaction at ${30}^{0}C$ ? If the other things remains same.

• A. $142.75$
• B. $242.75$
• C. $342.75$
• D. $442.75$

Answer 1

The correct option is A $142.75$

The catalyst lowers the energy of activation from 50 kJ/mol to 37.5 kJ/mol.
The Arrhenius expression is $k=A{e}^{-Ea/RT}.$
When activation energy is 50 kJ/mol, the expression becomes $k_1=A{e}^{-50\times \left({10}^3/RT\right)}$. ......(1)
When activation energy is 37.5 kJ/mol, the expression becomes $k_2=A{e}^{-37.5\times \left({10}^3/RT\right)}$ ......(2)
Divide equation (1) with equation (2).
$\frac{k_1}{k_2}=e^{-(37.5+50)\times {10}^3/RT}=e^{12.5\times {10}^3/RT}$

$2.303log\frac{k_2}{k_1}=\frac{12.5\times {10}^3}{8.314\times 303}$

$\frac{k_2}{k_1}=142.75$

$\frac{r_2}{r_1}=\frac{k_2}{k_1}=\mathrm{142.75.}$