Question

In the presence of a catalyst at a given temperature of 27$^\circ$C, the Activation energy of a specific reaction is reduced by 100 J/mol. What is the ratio between the rate constants for the catalysed ($k_2$) and uncatalysed ($k_1$) reactions?

• A. $1.04 \times 10^{-2}$
• B. $2.32 \times 10^{-2}$
• C. $1.97$
• D. $1.04$

Hint:

The Arrhenius equation is helpful for understanding the relationship between the activation energy and the rate constant of a reaction. A lower activation energy leads to a higher rate constant.

Answer 1

The Correct Option is D

The ratio of the rate constants for the catalysed and uncatalysed reactions can be determined using the Arrhenius equation: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Where: - \(k_2\) and \(k_1\) are the rate constants for the catalysed and uncatalysed reactions, respectively. - \(E_a\) is the activation energy difference (100 J/mol here). - \(R\) is the universal gas constant (8.314 J/mol·K). - \(T_1\) and \(T_2\) are the temperatures in Kelvin for the uncatalysed and catalysed reactions, respectively. Since the temperature change is negligible, we can approximate the ratio of rate constants as: \[ \frac{k_2}{k_1} = e^{-\frac{E_a}{RT}} \] Substituting the values: - \(E_a = 100\) J/mol, - \(R = 8.314\) J/mol·K, - \(T = 27^\circ C = 300\) K, \[ \frac{k_2}{k_1} = e^{-\frac{100}{8.314 \times 300}} = e^{-0.0401} \approx 1.04 \]
Thus, the ratio of the rate constants for the catalysed to uncatalysed reactions is approximately 1.0
4.