Question

The activation energies of two reactions are $ {{E}_{1}} $ and $ {{E}_{2}}({{E}_{1}}>{{E}_{2}}) $ If the temperature of the system is increased from $ {{T}_{1}} $ to $ {{T}_{2}}, $ he rate constant of the reactions changes from $ {{k}_{1}} $ to $ k'{{}_{1}} $ in the first reaction and $ {{k}_{2}} $ to $ k'{{}_{2}} $ in the second reaction. Predict which of the following expression is correct?

• A. $ \frac{k'_{1}}{k_{1}}=\frac{k'_{2}}{k_{2}} $
• B. $ \frac{k'_{1}}{k_{1}} >\frac{k'_{2}}{k_{2}} $
• C. $ \frac{k'_{1}}{k_{1}} < \frac{k'_{2}}{k_{2}} $
• D. $ \frac{k'_{1}}{k_{1}}=\frac{k'_{2}}{k_{2}}=1 $
• E. $ \frac{k'_{1}}{k_{1}}=\frac{k'_{2}}{k_{2}}=0 $

Answer 1

Answer: (B)

For first reaction,
$ {{E}_{1}}=\frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{1}}-{{T}_{2}})}\log \frac{k_{1}^{}}{{{k}_{1}}} $ ..(i)
For second reaction, $ {{E}_{2}}=\frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{1}}-{{T}_{2}})}\log \frac{k_{2}^{}}{{{k}_{2}}} $ ..(ii)
Given, $ {{E}_{1}}>{{E}_{2}} $
$ \Rightarrow $ $ \frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{1}}-{{T}_{2}})}\log \frac{k_{1}^{}}{{{k}_{1}}}>\frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{1}}-{{T}_{2}})}\log \frac{k_{2}^{}}{{{k}_{2}}} $
$ \therefore $ $ \frac{k_{1}^{}}{{{k}_{1}}}>\frac{k_{2}^{}}{{{k}_{2}}} $