2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    A reactant  forms two products : \[A\xrightarrow[{}]{{{k}_{1}}}B,\]Activation Energy \[E{{a}_{1}}\] \[A\xrightarrow[{}]{{{k}_{2}}}C,\] Activation Energy \[E{{a}_{2}}\] If \[E{{a}_{2}}=2E{{a}_{1}},\]then \[{{k}_{1}}\]and \[{{k}_{2}}\] are related as :     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) \[{{k}_{2}}={{k}_{1}}{{e}^{E{{a}_{1}}/RT}}\]

    B) \[{{k}_{2}}={{k}_{1}}{{e}^{E{{a}_{2}}/RT}}\]

    C) \[{{k}_{1}}-A{{k}_{2}}{{e}^{E{{a}_{1}}/RT}}\]

    D) \[{{k}_{1}}-2{{k}_{2}}{{e}^{E{{a}_{2}}/RT}}\]

    Correct Answer: C

    Solution :

                                    \[{{K}_{1}}={{A}_{1}}{{e}^{-E{{a}_{1}}/RT}}\]                 \[{{K}_{2}}={{A}_{2}}{{e}^{-E{{a}_{2}}/RT}}\]                 \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{{{A}_{1}}}{{{A}_{2}}}{{e}^{({{E}_{a}}_{2}\,\,\,{{E}_{a}}_{1})RT}}\]                 \[{{K}_{1}}={{K}_{2}}A\times {{e}^{{{E}_{{{a}_{1}}}}/RT}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner