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Question: Answered & Verified by Expert
According to Arrhenius equation, the rate constant $(\mathrm{k})$ is related to temperature $(\mathrm{T})$ as
ChemistryChemical KineticsVITEEEVITEEE 2007
Options:
  • A $\operatorname{In} \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$
  • B In $\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$
  • C In $\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}\right]$
  • D $\operatorname{In} \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}\right]$
Solution:
2828 Upvotes Verified Answer
The correct answer is: $\operatorname{In} \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$
$\ln \frac{k_{2}}{k_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$

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