A) 15kJ
B) 30kJ
C) 68 kJ
D) 120 kJ
Correct Answer: D
Solution :
From Arrhenius equation, \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] \[\log \frac{4.5\times {{10}^{-10}}}{1.3\times {{10}^{-11}}}=\frac{{{E}_{a}}}{2.303\times 8.314}\] \[\left[ \frac{1}{(270+273)}-\frac{1}{(350+273)} \right]\] \[\log \,\,34.62=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{1}{543}-\frac{1}{623} \right]\] \[1.539=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{80}{543\times 623} \right]\] \[{{E}_{a}}=\frac{1.539\times 2.303\times 8.314\times 543\times 623}{80}\] \[=\frac{9.968\times {{10}^{6}}}{80}\] \[=1.246\times {{10}^{5}}J=124\,kJ\approx 120\,kJ\]
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