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

To determine the ratio \(\dfrac{k_2}{k_1}\) for two chemical reactions A and B with a difference in activation energy of 20 kJ at 300 K, we use the Arrhenius equation: 
Eqs.₁: \(k = A e^{-\dfrac{E_a}{RT}}\)
For each reaction, the rate constants are \(k_1\) and \(k_2\) with activation energies \(E_{a1}\) and \(E_{a2}\) respectively. Given \(E_{a2} - E_{a1} = 20,000 \text{ J mol}^{-1}\).
Rearrange the Arrhenius equations:
\(\dfrac{k_2}{k_1} = \dfrac{A e^{-\dfrac{E_{a2}}{RT}}}{A e^{-\dfrac{E_{a1}}{RT}}} = e^{-\dfrac{E_{a2}}{RT} + \dfrac{E_{a1}}{RT}}\)
This simplifies to:
\(\dfrac{k_2}{k_1} = e^{-\dfrac{E_{a2} - E_{a1}}{RT}} = e^{-\dfrac{20,000}{8.3 \times 300}}\)
Calculate the exponent:
\(\dfrac{20,000}{8.3 \times 300} = \dfrac{20,000}{2490} \approx 8.032\)
\(\dfrac{k_2}{k_1} = e^{-8.032} \approx 0.000325\)
The provided range is 8,8, which seems to be misinterpreted or incorrect, as the calculated ratio should naturally be a value less than 1 due to the nature of the problem.
Therefore, \(\dfrac{k_2}{k_1} \approx 0.000325\) satisfies our calculation.