| A reactant [a] forms two products |
| \[A\xrightarrow{{{k}_{1}}}B\] Activation energy \[{{E}_{{{a}_{1}}}}\] |
| \[A\xrightarrow{{{k}_{2}}}C\] Activation energy \[E{{\,}_{{{a}_{2}}}}\] |
| If \[{{E}_{{{a}_{2}}}}=2{{E}_{{{a}_{1}}}},\] then \[{{k}_{1}}\] and \[{{k}_{2}}\] will be related as |
A) \[{{k}_{2}}={{k}_{1}}{{e}^{-{{E}_{{{a}_{1}}}}/RT}}\]
B) \[{{k}_{2}}={{k}_{1}}{{e}^{-{{E}_{{{a}_{2}}}}/RT}}\]
C) \[{{k}_{1}}={{k}_{2}}{{e}^{-{{E}_{{{a}_{1}}}}/RT}}\]
D) \[{{k}_{1}}=2{{k}_{2}}{{e}^{{{E}_{{{a}_{2}}}}/RT}}\]
Correct Answer: A
Solution :
| \[{{k}_{2}}=A{{e}^{-{{E}_{a2}}/RT}}\] |
| \[{{k}_{1}}=A{{e}^{-{{E}_{at}}/RT}}\] |
| \[\frac{{{k}_{2}}}{{{k}_{1}}}={{e}^{-{{E}_{at}}/RT}}{{k}_{2}}={{k}_{1}}{{e}^{-Eat/RT}}\] |
| (Since, \[{{E}_{a2}}=2{{E}_{at}}\]) |
| or \[\frac{{{k}_{2}}}{{{k}_{1}}}={{e}^{-{{E}_{at}}/RT}}{{k}_{2}}={{k}_{1}}{{e}^{-{{E}_{at}}/RT}}\] |
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