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