A) \[{{\log }_{10}}\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
B) \[{{\log }_{10}}\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right]\]
C) \[{{\log }_{10}}\frac{1}{2}=\frac{{{E}_{a}}}{2.303}\left[ \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right]\]
D) \[{{\log }_{10}}2=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
Correct Answer: D
Solution :
According to question, the rate of reaction is doubled on increasing temperature from\[{{T}_{1}}K\]to\[{{T}_{2}}K\]. At \[{{T}_{1}},\,\,{{k}_{1}}=k\](say) and \[{{T}_{2}},\,\,{{k}_{2}}=2k\] On applying, \[{{\log }_{10}}\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] \[{{\log }_{10}}\frac{2k}{k}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] \[{{\log }_{10}}2=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
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