A) Double
B) 28 times
C) 7.38 times
D) \[7.38\times {{10}^{3}}\]times
Correct Answer: C
Solution :
[c] \[\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{A{{e}^{-{{E}_{{{a}_{2}}}}/RT}}}{A{{e}^{-{{E}_{{{a}_{1}}}}/RT}}}={{e}^{({{E}_{{{a}_{1}}}}-{{E}_{{{a}_{2}}}})/RT}}\] \[2.303\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{{{a}_{1}}}}-{{E}_{a}}}{RT}\] \[=\frac{\left( 83.314-75 \right)\times {{10}^{3}}}{8.314\times 500}=2\] \[\log {{k}_{2}}=\frac{2}{2.303}=0.868\] Taking Antilog \[{{k}_{2}}=7.38\]You need to login to perform this action.
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