A) \[{{\log }_{e}}A={{\log }_{e}}K+\frac{{{E}_{a}}}{RT}\]
B) \[\log \,K=A\,\frac{{{E}_{a}}}{RT}\]
C) \[{{\log }_{e}}K={{\log }_{e}}A-\frac{{{E}_{a}}}{R{{T}^{2}}}\]
D) \[\log A=RT\,\ln \,{{E}_{a}}-\ln \,K\]
Correct Answer: A
Solution :
Arrhenius equation is \[\log k=\log A-\frac{{{E}_{a}}}{RT}\]
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