At 27°C in presence of a catalyst, activation energy of a reaction is lowered by 10 kJ mol⁻¹. The logarithm of ratio of k(catalysed) / k(uncatalysed) is…. (Consider that the frequency factor for both the reactions is same)

At 27°C in presence of a catalyst activation energy of a reaction is lowered by 10 kJ mol⁻¹ The logarithm of ratio of k catalysed by k uncatalysed is

Q. At 27°C in presence of a catalyst, activation energy of a reaction is lowered by 10 kJ mol⁻¹. The logarithm of ratio of \[ \frac{k(\text{catalysed})}{k(\text{uncatalysed})} \] is….

(Consider that the frequency factor for both the reactions is same)

(A) 1.741

(B) 0.1741

(D) 3.482

Correct Answer: (A)

Explanation

According to the Arrhenius equation,

\[ k = A e^{-\frac{E_a}{RT}} \]

Since the frequency factor \(A\) is same for both catalysed and uncatalysed reactions, the ratio of rate constants is given by

\[ \frac{k(\text{catalysed})}{k(\text{uncatalysed})} = e^{\frac{E_{a,\text{uncat}} - E_{a,\text{cat}}}{RT}} \]

Given that activation energy is lowered by \(10 \text{ kJ mol}^{-1}\),

\[ E_{a,\text{uncat}} - E_{a,\text{cat}} = 10\,000 \text{ J mol}^{-1} \]

Temperature,

\[ T = 27 + 273 = 300 \text{ K} \]

\[ \ln \left(\frac{k(\text{catalysed})}{k(\text{uncatalysed})}\right) = \frac{10\,000}{8.314 \times 300} \approx 4.01 \]

Converting natural logarithm to base 10,

\[ \log \left(\frac{k(\text{catalysed})}{k(\text{uncatalysed})}\right) = \frac{4.01}{2.303} \approx 1.741 \]

Hence, the correct value of the logarithm is 1.741.

Explanation

Related JEE Main Chemistry Questions