Question

For a reaction at $300 \text{ K}$, on addition of catalyst, activation energy of reaction lowered by $10 \text{ kJ}$. Then calculate the value of $\log \frac{K_{\text{catalysed}}}{K_{\text{uncatalysed}}}$

• A. $1.74$
• B. $0.174$
• C. $17.4$
• D. $3.48$

Hint:

A catalyst increases the rate constant exponentially with respect to the decrease in activation energy, as defined by the Arrhenius equation derivation: $\log(K_{\text{cat}}/K_{\text{uncat}}) = \Delta E_a/(2.303RT)$.

Answer 1

The Correct Option is A

Change in activation energy $\Delta E_a = 10 \text{ kJ} = 10000 \text{ J}$. $T = 300 \text{ K}$.
The ratio of rate constants is related to $\Delta E_a$ by:
$\log \left( \frac{K_{\text{cat}}}{K_{\text{uncat}}} \right) = \frac{\Delta E_a}{2.303 R T}$.
Substitute $R = 8.314 \text{ J/mol-K}$:
$\log \left( \frac{K_{\text{cat}}}{K_{\text{uncat}}} \right) = \frac{10000}{2.303 \times 8.314 \times 300}$.
$2.303 \times 8.314 \times 300 \approx 5744.17$.
$\log \left( \frac{K_{\text{cat}}}{K_{\text{uncat}}} \right) = \frac{10000}{5744.17} \approx 1.741$.
The value is $1.74$.