| Given is the information: |
| Reaction 1 : \[k=A{{e}^{-{{E}_{1}}/RT}}\] |
| Reaction 2 : \[k'=A{{e}^{-{{E}_{2}}/RT}}\] |
| If \[{{E}_{1}}=2{{E}_{2}},\] then the ratio \[\ell n\left[ \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right]\] for the reaction will be |
A) Equal to that of reaction 2
B) greater than that of reaction 2
C) lesser than that of reaction 2
D) may be greater or lesser than reaction 2 depending upon the temperature T.
Correct Answer: B
Solution :
| \[\ell n{{k}_{T}}=\ell nA-\frac{{{E}_{1}}}{RT}\] |
| or \[\ell n{{k}_{T}}+10k=\ell nA-\frac{{{E}_{1}}}{R(T+10k)}\] |
| or \[\ell n{{k}_{T}}+10k-\ell n\,{{k}_{T}}=\frac{{{E}_{1}}}{R}\left[ \frac{1}{T}-\frac{1}{T+10k} \right]\] |
| or \[\ell n\left[ \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right]=\frac{{{E}_{1}}(10k)}{RT(T+10k)}\] |
| similarly, \[\ell n\left( \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right)=\frac{{{E}_{2}}(10k)}{RT(T+10k)}\] |
| Dividing both, we get, |
| \[\frac{\ell n\left( \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right)}{\ell n\left( \frac{{{k}_{T}}+10k}{{{k}_{T}}} \right)}=\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{2{{E}_{2}}}{{{E}_{2}}}=2\] |
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