AIIMS AIIMS Solved Paper-2010

  • question_answer
    At a certain temperature, the first order rate constant, \[{{k}_{1}}\] is found to be smaller than the second order rate constant, \[{{k}_{2}}\]. If the energy of activation, \[{{E}_{1}}\] of the first order reaction is greater than energy of activation, \[{{E}_{2}}\] of the second order reaction then with increase in temperature

    A)  \[{{k}_{1}}\] will increase faster than\[{{k}_{2}}\] , but always will remain less than \[{{k}_{2}}\]

    B) \[{{k}_{2}}\] will increase faster than \[{{k}_{1}}\]

    C) \[{{k}_{1}}\] will increase faster than \[{{k}_{2}}\] and becomes equal to \[{{k}_{2}}\]

    D)  \[{{k}_{1}}\]will increase faster than \[{{k}_{2}}\] and becomes greater than \[{{k}_{2}}\]

    Correct Answer: A

    Solution :

    \[\frac{d(In\,k)}{dt}=\frac{{{E}_{a}}}{R{{T}^{2}}}\] As \[{{E}_{a}}\] increases, rate of increase in k also increases so \[{{k}_{1}}\] will increase faster than \[{{k}_{2}}\] but always will remain less than \[{{k}_{2}}\].


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