A) In\[\frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)\]
B) In\[\frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right)\]
C) In\[\frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{2}}}+\frac{1}{{{T}_{1}}} \right)\]
D) In \[\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)\]
Correct Answer: B
Solution :
(b, d) On integrating within limits \[{{k}_{1}}\] to \[{{k}_{2}}\]and \[{{T}_{1}}\]to \[{{T}_{2}}\] \[\int\limits_{{{k}_{1}}}^{{{k}_{2}}}{\ln \,K=-\frac{{{E}_{a}}}{R}}\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{\frac{1}{T}}\] \[\ln \,\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right]=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] or \[\ln \,\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] Options and both are one and the same and are correct.
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