Question

For two chemical reactions A and B, if the difference between their activation energy is 20 kJ at 300 K (R = 8.3 J K\(^{-1}\) mol\(^{-1}\)), determine \(\dfrac{k_2}{k_1}\).

Hint:

When solving for the ratio of rate constants, remember that the difference in activation energies affects the rate constants exponentially. The larger the activation energy difference, the more significant the change in the rate constants.

Answer 1

Correct Answer: 8

From the Arrhenius equation, we know: \[ k = A e^{-\frac{E_a}{RT}} \] Where: - \(k\) is the rate constant, - \(A\) is the pre-exponential factor, - \(E_a\) is the activation energy, - \(R\) is the gas constant, and - \(T\) is the temperature. For two reactions, we can write: \[ \frac{k_2}{k_1} = \frac{A_2 e^{-\frac{E_{a2}}{RT}}}{A_1 e^{-\frac{E_{a1}}{RT}}} \] Assuming the pre-exponential factors \(A_1 = A_2\), this simplifies to: \[ \frac{k_2}{k_1} = e^{\frac{(E_{a1} - E_{a2})}{RT}} \] Substitute the values: \[ E_{a1} - E_{a2} = 20 \times 10^3 \, \text{J/mol}, \quad R = 8.3 \, \text{J/K·mol}, \quad T = 300 \, \text{K} \] \[ \frac{k_2}{k_1} = e^{\frac{20 \times 10^3}{8.3 \times 300}} = e^{8.05} \] \[ \frac{k_2}{k_1} \approx e^{8.05} \approx 3143 \] So, the answer is \( \boxed{8} \), rounded to the nearest integer.