A) \[342\,kJ\,mo{{l}^{-1}}\]
B) \[269\,kJ\,mo{{l}^{-1}}\]
C) \[34.7\,kJ\,mo{{l}^{-1}}\]
D) \[15.1\,kJ\,mo{{l}^{-1}}\]
Correct Answer: C
Solution :
Given, initial temperature, \[{{T}_{1}}=20+273=293K\] Final temperature \[{{T}_{2}}=35+273\] \[=308\,K\] \[R=8.314\,J\,mo{{l}^{-1}}\,{{K}^{-1}}\] Since, rate becomes double on raising temperature, \[\therefore \] \[{{r}_{2}}=2{{r}_{1}}\]or \[\frac{{{r}_{2}}}{{{r}_{1}}}=2\] As rate constant, \[k\propto r\] \[\therefore \] \[\frac{{{k}_{2}}}{{{k}_{1}}}=2\] From Arrnhenius equations we know that \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}{{T}_{2}}} \right]\] \[\log 2=-\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{293-308}{293\times 308} \right]\] \[0.3010=-\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{-15}{293\times 308} \right]\] \[\therefore \] \[{{E}_{a}}=\frac{0.3010\times 2.303\times 8.314\times 293\times 308}{15}\]
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