Question

At $27^\circ\text{C}$, in presence of a catalyst, activation energy of a reaction is lowered by $10\,\text{kJ mol}^{-1}$. The logarithm of the ratio $\dfrac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is ___.
(Consider that the frequency factor for both the reactions is same)

• A. $0.1741$
• B. $1.741$
• C. $3.482$
• D. $17.41$

Hint:

A small decrease in activation energy leads to a very large increase in reaction rate.

Answer 1

The Correct Option is D

To solve this problem, we need to understand the concept of activation energy and its effect on the rate constant of a reaction, typically expressed through the Arrhenius equation:

\(k = A \exp\left(\frac{-E_a}{RT}\right)\)

Where:

• \(k\) is the rate constant.
• \(A\) is the frequency factor, assumed to be the same for both reactions here.
• \(E_a\) is the activation energy.
• \(R\) is the universal gas constant (\(8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\)).
• \(T\) is the temperature in Kelvin.

The logarithm of the ratio of the rate constants for catalyzed and uncatalyzed reactions can be found using:

\(\log\left(\frac{k_{\text{catalysed}}}{k_{\text{uncatalysed}}}\right) = \frac{E_a(\text{uncatalysed}) - E_a(\text{catalysed})}{2.303RT}\)

Given:

• Reduction in activation energy: \(\Delta E_a = 10 \, \text{kJ mol}^{-1} = 10000 \, \text{J mol}^{-1}\)
• Temperature: \(T = 27^\circ \text{C} = 27 + 273.15 = 300.15 \, \text{K}\)

Substituting the values into the equation:

\(\log\left(\frac{k_{\text{catalysed}}}{k_{\text{uncatalysed}}}\right) = \frac{10000}{2.303 \times 8.314 \times 300.15}\)

Calculating the value:

\(\log\left(\frac{k_{\text{catalysed}}}{k_{\text{uncatalysed}}}\right) \approx \frac{10000}{5740.2096} \approx 1.741\)

Thus, the logarithm of the ratio of the catalyzed to uncatalyzed rate constant is approximately \(1.741\).

The correct answer is therefore the option: \(1.741\).