A) \[2.0\times {{10}^{18}}{{\operatorname{s}}^{-1}}\]
B) \[6.0\times {{10}^{14}}{{\operatorname{s}}^{-1}}\]
C) Infinity
D) \[3.6\times {{10}^{30}}{{\operatorname{s}}^{-1}}\]
Correct Answer: B
Solution :
| \[{{T}_{2}}=T(say),{{T}_{1}}=25{}^\circ C=298K,\] |
| \[{{E}_{a}}=104.4kJmo{{l}^{-1}}=104.4\times {{10}^{3}}Jmo{{l}^{-1}}\] |
| \[{{k}_{1}}=3\times {{10}^{-4}},{{k}_{2}}=?,\] |
| \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] |
| \[\log \frac{{{k}_{2}}}{3\times {{10}^{-4}}}=\frac{104.4\times {{10}^{3}}J\,mo{{l}^{-1}}}{2.303\times (8.314J{{K}^{-1}}mo{{l}^{-1}})}\] |
| \[\left[ \frac{1}{298}-\frac{1}{\operatorname{T}} \right]\] |
| As \[T\to \infty ,\frac{1}{T}\to 0\] |
| \[\therefore \]\[\log \frac{{{k}_{2}}}{3\times {{10}^{-4}}}=\frac{104.4\times {{10}^{3}}J\,mo{{l}^{-1}}}{2.303\times 8.314\times 298}\] |
| \[\log \frac{{{k}_{2}}}{3\times {{10}^{-4}}}=18.297,\frac{{{k}_{2}}}{3\times {{10}^{-4}}}=1.98\times {{10}^{18}}\] |
| \[{{k}_{2}}=(1.98\times {{10}^{18}})\times (3\times {{10}^{-4}})=6\times {{10}^{14}}{{s}^{-1}}\] |
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