A) \[4.17\,KJ\,mo{{l}^{-1}}\]
B) \[41.7\,KJ\,mo{{l}^{-1}}\]
C) \[417.0\,KJ\,mo{{l}^{-1}}\]
D) \[4170\,KJ\,mo{{l}^{-1}}\]
Correct Answer: B
Solution :
Arrhenius equation, \[\frac{\log {{k}_{2}}}{\log {{k}_{1}}}=\frac{{{E}_{a}}[{{T}_{2}}-{{T}_{1}}]}{2.303R[{{T}_{2}}.{{T}_{1}}]}\] \[\log \frac{6}{1}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}.{{T}_{1}}} \right]\] \[\log 2+\log 3=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{50}{400\times 350} \right]\] \[{{E}_{a}}=\frac{\left[ \begin{align} & (0.3010+0.4771)\times 2.303 \\ & \times 8.314\times 400\times 350 \\ \end{align} \right]}{50}\] \[=41715.49Jmo{{l}^{-1}}\] \[=41.7kJmo{{l}^{-1}}\]
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