The activation energy of a reaction is 9 kcal/mole. The increase in the rate constant when its temperature is raised from 295 to 300 is:

28.9%

The correct answer is option 3. 28.9%

Given,

Activation Energy, \(E_a = 9 \text{kcal/mol}\)

\(T_1 = 295 K\)

& \(T_2 = 300K\)

We know,

\(2.303 log \frac{k_2}{k_1} = \frac{E_a}{R}\left[\frac{T_2 − T_1}{T_2T_1}\right]\)

\(⇒ log \frac{k_2}{k_1} = \frac{9 × 10^3}{2.303 × 2}\left[\frac{300 − 295}{295 × 300}\right]\)

\(⇒ log \frac{k_2}{k_1} = 0.11039\)

\(⇒ \frac{k_2}{k_1} = antilog(0.11039)\)

\(⇒ \frac{k_2}{k_1} = 1.289\)

\(⇒ k_2 = 1.289k_1\)

Now,

\(\frac{1.289k_1 − k_1}{k_1} × 100\)%

\(= \frac{0.289k_1}{k_1} × 100\)%

\(= \frac{0.289k_1}{k_1} × 100\)%

\(= 0.289 × 100\)%

\(= 28.9 \) %