The activation energies of the two reactions are 18 kJ mol^–1 and 4.0 kJ mol^–1 respectively. Assuming the pre-exponential factor to be the same for both reactions, the ratio of their rate constants at 27°C is:

3.656 × 10^–3

The correct answer is option 1. 3.656 × 10^–3

According to Arrhenius equation,

\(lnk_1 = lnA − \frac{18}{RT}\) -----(1)

\(lnk_2 = lnA − \frac{4}{RT}\) -------(2)

Subtracting (2) from (1),

\(lnk_1 − lnk_2= −\frac{18}{RT} + \frac{4}{RT}\)

\(⇒⇒lnk_1 − lnk_2= \frac{1}{RT}[4 − 18]\)

\(⇒lnk_1 − lnk_2= −\frac{14kJ}{RT}\)

\(⇒ log\left(\frac{k_1}{k_2}\right) = − \frac{−14 × 10^3}{2.303 × 8.314 × 300}\)

\(⇒ log\left(\frac{k_1}{k_2}\right) = − \frac{−14 × 10^3}{5744.1426}\)

\(⇒ log\left(\frac{k_1}{k_2}\right) = − 2.437\)

\(⇒ \left(\frac{k_1}{k_2}\right) = antilog(− 2.437)\)

\(⇒ \left(\frac{k_1}{k_2}\right) = 0.0036559\)

\(⇒ \left(\frac{k_1}{k_2}\right) = 3.656 × 10^{−3}\text{ J/mol}\)