A) \[\ln \frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)\]
B) \[\ln \frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right)\]
C) \[\ln \frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{2}}}+\frac{1}{{{T}_{1}}} \right)\]
D) \[\ln \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)\]
Correct Answer: C
Solution :
| [c] According to Arrhenius equation, activation energy \[\text{(}{{\text{E}}_{\text{a}}}\text{)}\] and rate constants (\[{{\text{k}}_{\text{1}}}\] and \[{{\text{k}}_{2}}\]) of a chemical reaction at two different temperatures (\[{{\text{T}}_{\text{1}}}\] and \[{{\text{T}}_{2}}\]) are related as |
| \[\ln \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right)\] |
| \[=-\frac{{{E}_{a}}}{R}\left( \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right)\] |
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