| What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? |
| (\[\text{R=}\,\text{8}\text{.314}\,\text{J}\,\text{mo}{{\text{l}}^{\text{-1}}}\,{{\text{K}}^{\text{-1}}}\])[NEET 2013] |
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 :
| [c] 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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