A) 40
B) 70
C) 83.8
D) 200
Correct Answer: C
Solution :
Given, \[{{\text{T}}_{1}}=25{}^\circ C=25+273=298\text{ }K\] \[{{T}_{2}}=(25+10{}^\circ C)=35{}^\circ C=308\text{ }K\] Rate\[\propto k\] \[\Rightarrow \] \[\frac{{{(Rate)}_{25{}^\circ }}}{{{(Rate)}_{35{}^\circ }}}=\frac{{{k}_{25{}^\circ }}}{{{k}_{35{}^\circ }}}\] \[\frac{{{k}_{25{}^\circ }}}{{{k}_{35{}^\circ }}}=\frac{1}{3}\] From Arrhenius equation, \[\log \frac{{{k}_{25{}^\circ }}}{{{k}_{35{}^\circ }}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] \[\log 3=\frac{{{E}_{a}}}{2.30\times 8.314\times {{10}^{-3}}}\left[ \frac{1}{298}-\frac{1}{308} \right]\] \[0.477=\frac{{{E}_{a}}}{2.30\times 8.314\times {{10}^{-3}}}\left[ \frac{10}{298\times 308} \right]\] \[\therefore \] \[{{E}_{a}}=83.8\,kJ\,mo{{l}^{-1}}\]
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