A) \[2\times {{10}^{18}}{{s}^{-\,1}}\]
B) \[6\times {{10}^{14}}{{s}^{-\,1}}\]
C) \[3.6\times {{10}^{30}}{{s}^{-\,1}}\]
D) infinity
Correct Answer: B
Solution :
| Given, \[{{T}_{1}}=25{}^\circ C=298K\,{{T}_{2}}=T\] |
| \[{{E}_{a}}=104.4kJ\,mo{{l}^{-1}}=104.4\times {{10}^{3}}J\,mo{{l}^{-\,1}}\] |
| \[{{k}_{1}}=3\times {{10}^{-\,4}}{{s}^{-\,1}},{{k}_{2}}=?\] |
| According to Arrhenius equation \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] |
| \[\log \frac{{{k}_{2}}}{3\times {{10}^{-\,4}}}=\frac{104.4\times {{10}^{3}}}{2.303\times 8.314}\times \frac{1}{298}\]\[\left[ asT\to \infty ,\frac{1}{T}\to 0 \right]\] |
| \[\log \frac{{{k}_{2}}}{3\times {{10}^{-4}}}=18.297\] |
| \[\frac{{{k}_{2}}}{3\times {{10}^{-\,4}}}=1.98\times {{10}^{18}}\] |
| \[{{k}_{2}}=1.98\times {{10}^{18}}\times 3\times {{10}^{-\,4}}\]\[=5.94\times {{10}^{4}}\approx 6\times {{10}^{14}}{{s}^{-\,1}}\] |
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