Question

The activation energy for a reaction is 9.0 kcal/mol. The increase in the rate constant when its temperature is increased from 298 K to 308 K is:

• A. 63%
• B. 50%
• C. 100%
• D. 10%

Answer 1

Answer: (A)

From Arrhenius equation, $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$ $=\frac{9\times {10}^3}{2.303\times 2}\left(\frac{308-298}{308\times 298}\right)=0.2129$ $\therefore$ $\frac{k_2}{k_1}=\text{antilog}\text{0}\text{.2129}\text{=}\text{1}\text{.632}$ $\therefore$ $k_2=1.632k_1$ Hence, increase in rate constant $=k_2-k_1$ $=1.632k_2-k_1=0.632k_1$ $=\frac{0.632k_1}{k_1}\times 100=63.2$