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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    For the two gaseous reactions, following data are given \[A\xrightarrow{{}}B;{{k}_{1}}={{10}^{10}}{{e}^{-20,000/T}}\] \[C\xrightarrow{{}}D;{{k}_{2}}={{10}^{12}}{{e}^{-24,606/T}}\] the temperature at which\[{{k}_{1}}\]becomes equal to \[{{k}_{2}}\]is

    A) 400 K                                    

    B) 1000 K

    C) 800 K                    

    D)        1500 K

    E)  500 K

    Correct Answer: B

    Solution :

    Given    \[{{k}_{1}}={{10}^{10}}{{e}^{-20,000/T}}\] \[{{k}_{2}}={{10}^{12}}{{e}^{-24,606/T}}\] \[{{k}_{1}}={{k}_{2}}\] \[{{10}^{10}}{{e}^{-20,000/T}}={{10}^{12}}{{e}^{-24,606/T}}\] \[{{e}^{\frac{-20,000}{T}+\frac{24,606}{T}}}={{10}^{2}}\]                 \[{{e}^{\frac{4,606}{T}}}={{10}^{2}}\] On taking log both sides                 \[\frac{4606}{2.303T}=\log \,{{10}^{2}}\]                 \[2\,\log 10\times T=\frac{4606}{2.303}\]                 \[T=\frac{4606}{2.303\times 2}\]                 \[=\frac{4606}{4.606}=1000\,K\]


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