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:
Answer & explanation
Correct answer: option 3
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,
\(\frac{1.289k_1 − k_1}{k_1} × 100\)%
\(= \frac{0.289k_1}{k_1} × 100\)%
\(= \frac{0.289k_1}{k_1} × 100\)%
\(= 0.289 × 100\)%
\(= 28.9 \) %