JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Sample Paper Topic Test - Chemical Kinetics

  • question_answer
    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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