Two distinct competitive reactions, A and B, share the exact same pre-exponential factor (). However — Chemical Kinetics Chemistry Question
Question
Two distinct competitive reactions, A and B, share the exact same pre-exponential factor ($A_A = A_B$). However, the activation energy of reaction A exceeds that of reaction B by strictly $1.0 \text{ kcal mol}^{-1}$ ($E_{a,A} - E_{a,B} = 1000 \text{ cal mol}^{-1}$). At an operating temperature of $300 \text{ K}$, what is the exact kinetic ratio $k_B / k_A$? (Assume $R \approx 2 \text{ cal K}^{-1} \text{ mol}^{-1}$)
Answer: B
💡 Solution & Explanation
Applying the Arrhenius equation: $k_A = A \cdot e^{-E_{a,A}/RT}$ and $k_B = A \cdot e^{-E_{a,B}/RT}$. Dividing the two yields $\frac{k_B}{k_A} = e^{(E_{a,A} - E_{a,B})/RT}$. Plugging in the values gives $\frac{k_B}{k_A} = \exp\left(\frac{1000}{2 \times 300}\right) = \exp\left(\frac{1000}{600}\right) = e^{1.66}$.
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