A) \[40\]
B) \[70\]
C) \[83.8\]
D) \[200\]
Correct Answer: C
Solution :
Given, \[{{T}_{1}}={{25}^{o}}C=25+273=298K\] \[{{T}_{2}}=(25+{{10}^{o}}C)={{35}^{o}}C=308K\] Rate \[\propto \]k \[\Rightarrow \] \[\frac{{{(Rate)}_{{{25}^{o}}}}}{{{(Rate)}_{{{35}^{o}}}}}=\frac{{{k}_{{{25}^{o}}}}}{{{k}_{{{35}^{o}}}}}\] \[\frac{{{k}_{{{25}^{o}}}}}{{{k}_{{{35}^{o}}}}}=\frac{1}{3}\] From Arrhenius equation, \[\log \frac{{{k}_{{{35}^{o}}}}}{{{k}_{{{25}^{o}}}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] \[\log 3=\frac{{{E}_{a}}}{2.303\times 8.314\times {{10}^{-3}}}\left[ \frac{1}{298}-\frac{1}{308} \right]\] \[0.477=\frac{{{E}_{a}}}{2.303\times 8.314\times {{10}^{-3}}}\left[ \frac{10}{298\times 308} \right]\] \[\therefore \] \[{{E}_{a}}=83.8kJ\,mo{{l}^{-1}}\]
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