Question

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

• A. 10%
• B. 50%
• C. 100%
• D. 28.8%

Answer 1

Answer: (D)

28.8%

Answer 2

$\log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303R}\left\lbrack \frac{T_{2} - T_{1}}{T_{1}T_{2}} \right\rbrack$

= $\frac{9000}{2.303 \times 2}\left\lbrack \frac{300 - 295}{295 \times 300} \right\rbrack$

= $0.1103$

Hence $\frac{k_{2}}{k_{1}} = 1.288$ or $k_{2} = 1.288k_{1}$ i.e., increase = 28.8%.