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