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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The activation energies of two reactions are \[{{E}_{1}}\]and\[{{E}_{2}}({{E}_{1}}>{{E}_{2}})\]If the temperature of the system is increased from\[{{T}_{1}}\]to\[{{T}_{2}},\]he rate constant of the reactions changes from\[{{k}_{1}}\]to \[k{{}_{1}}\] in the first reaction and \[{{k}_{2}}\] to \[k{{}_{2}}\] in the second reaction. Predict which of the following expression is correct?

    A) \[\frac{k_{1}^{}}{{{k}_{1}}}=\frac{k_{2}^{}}{{{k}_{2}}}\]                

    B) \[\frac{k_{1}^{}}{{{k}_{1}}}>\frac{k_{2}^{}}{{{k}_{2}}}\]

    C) \[\frac{k_{1}^{}}{{{k}_{1}}}<\frac{k_{2}^{}}{{{k}_{2}}}\]

    D)        \[\frac{k_{1}^{}}{{{k}_{1}}}=\frac{k_{2}^{}}{{{k}_{2}}}=1\]

    E) \[\frac{k_{1}^{}}{{{k}_{1}}}=\frac{k_{2}^{}}{{{k}_{2}}}=0\]

    Correct Answer: B

    Solution :

    For first reaction, \[{{E}_{1}}=\frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{1}}-{{T}_{2}})}\log \frac{k_{1}^{}}{{{k}_{1}}}\] ?..(i) For second reaction,                 \[{{E}_{2}}=\frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{1}}-{{T}_{2}})}\log \frac{k_{2}^{}}{{{k}_{2}}}\] ??(ii) Given, \[{{E}_{1}}>{{E}_{2}}\] \[\Rightarrow \]\[\frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{1}}-{{T}_{2}})}\log \frac{k_{1}^{}}{{{k}_{1}}}>\frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{1}}-{{T}_{2}})}\log \frac{k_{2}^{}}{{{k}_{2}}}\] \[\therefore \]  \[\frac{k_{1}^{}}{{{k}_{1}}}>\frac{k_{2}^{}}{{{k}_{2}}}\]


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