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}}=1.632\,{{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%=63% $