A) \[1\,\,h\]
B) \[1.68\,\,h\]
C) \[1.28\,\,h\]
D) \[1.11\,\,h\]
Correct Answer: B
Solution :
Given,\[A=1.11\times {{10}^{11}}{{s}^{-1}}\,\,T=573\,\,K\] \[{{E}_{a}}=39.3\times {{10}^{3}}cal\,\,mo{{l}^{-1}};\,\,R=1.987\,\,cal;\] \[\because \] \[k=A{{e}^{-{{E}_{a}}/RT}}\] \[\therefore \] \[\ln k=\ln A-\frac{{{E}_{a}}}{RT}\] or \[{{\log }_{10}}k={{\log }_{10}}A-\frac{{{E}_{a}}}{2.303RT}\] or \[{{\log }_{10}}k={{\log }_{10}}1.11\times {{10}^{11}}\] \[-\left\{ \frac{39.3\times {{10}^{3}}}{2.303\times 1987\times 573} \right\}\] or \[k=1.11\times {{10}^{11}}\] \[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{1.14\times {{10}^{-4}}}\] \[=6078s=1.68\,\,h\]
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