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:
Answer & explanation
Correct answer: option 1
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),
\(lnk_1 − lnk_2= −\frac{18}{RT} + \frac{4}{RT}\)
\(⇒⇒lnk_1 − lnk_2= \frac{1}{RT}[4 − 18]\)
\(⇒lnk_1 − lnk_2= −\frac{14kJ}{RT}\)
\(⇒ log\left(\frac{k_1}{k_2}\right) = − \frac{−14 × 10^3}{2.303 × 8.314 × 300}\)
\(⇒ log\left(\frac{k_1}{k_2}\right) = − \frac{−14 × 10^3}{5744.1426}\)
\(⇒ log\left(\frac{k_1}{k_2}\right) = − 2.437\)
\(⇒ \left(\frac{k_1}{k_2}\right) = antilog(− 2.437)\)
\(⇒ \left(\frac{k_1}{k_2}\right) = 0.0036559\)
\(⇒ \left(\frac{k_1}{k_2}\right) = 3.656 × 10^{−3}\text{ J/mol}\)