A) In \[\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
B) In \[\frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
C) In \[\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}+\frac{1}{{{T}_{2}}} \right]\]
D) In \[\frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}+\frac{1}{{{T}_{2}}} \right]\]
Correct Answer: A
Solution :
Arrhenius equation is written as: \[K=A{{e}^{-{{E}_{a}}/RT}}\] Taking logarithm, above equation may be written as: In K = In \[A-\frac{{{E}_{a}}}{R}\times \frac{1}{T}\] \[\therefore \] In \[{{K}_{1}}=\] In \[A-\frac{{{E}_{a}}}{R}\times \frac{1}{{{T}_{1}}}\] ??.(i) In \[{{K}_{2}}=\]In \[A-\frac{{{E}_{a}}}{R}\times \frac{1}{{{T}_{2}}}\] ??.(ii) Subtracting the Eq. (i) from Eq. (ii) In \[{{K}_{2}}=\] In \[{{K}_{1}}=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] In \[\frac{{{K}_{2}}}{{{K}_{1}}}=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
