Question

Two reactions, A → Products and B → Products, have rate constants $K_a$ and $K_b$ at temperature T and activation energies $E_a$ and $E_b$ respectively. If $K_a > K_b$ and $E_a < E_b$ and assuming that A for both the reactions is same then :

• A. At higher temperatures K a will be lesser than K b
• B. At lower temperature K a and K b will differ more and K a > K b
• C. As temperature rises K a and K b will be close to each other in magnitude
• D. None of these

Answer 1

Answer: (B), (C)

Options (B) and (C).

Answer 2

Using the Arrhenius equation:

$k = A \exp(-E/RT)$

If the pre‐exponential factors (A) are the same for both reactions, then:

$k_a/k_b = \exp[(E_b – E_a)/(RT)] > 1$ since $E_a < E_b$

Thus, at any temperature $k_a > k_b$. However, note that:

• At lower temperatures, the exponential term becomes very significant, so the difference between $k_a$ and $k_b$ is large.
• As temperature increases, the factor $(E/RT)$ decreases, making the difference between $k_a$ and $k_b$ smaller (their magnitudes get closer).