A) \[\text{342}\,\text{kJ}\,\text{mo}{{\text{l}}^{\text{-1}}}\]
B) \[269\,\text{kJ}\,\text{mo}{{\text{l}}^{\text{-1}}}\]
C) \[34.7\,\text{kJ}\,\text{mo}{{\text{l}}^{\text{-1}}}\]
D) \[15.1\,\text{kJ}\,\text{mo}{{\text{l}}^{\text{-1}}}\]
Correct Answer: C
Solution :
Given, initial temperature, \[{{T}_{1}}=20+273=293K\] Final temperature \[{{T}_{2}}=35+273\] \[=308K\] \[R=8.314J\,mo{{l}^{-1}}\,{{K}^{-1}}\] Since, are 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 Arrhenius equation, we know that \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=-\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}{{T}_{2}}} \right]\] \[\log 2=-\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{293-308}{293\times 308} \right]\] \[0.3010=-\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{-15}{293\times 308} \right]\] \[\therefore \]\[{{E}_{a}}=\frac{0.3010\times 2.303\times 8.314\times 293\times 308}{15}\] \[=34673.48\,J\,mo{{l}^{-1}}\,=34.7\,kJ\,mo{{l}^{-1}}\]
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