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. 0.63
• B. 0.5
• C. 1
• D. 0.1

Answer 1

Answer: (A)

0.63

Answer 2

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