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NEET Sample Paper NEET Sample Test Paper-70

  • question_answer
    On introducing a catalyst at 500 K the rate of a first-order reaction increases by 1.718 times. The activation energy in the presence of catalyst is\[4.15\text{ }kJ\text{ }mo{{l}^{-}}^{1}\]. The slope of the plot of ln k\[(in\,\,se{{c}^{-1}})vs.\,\frac{1}{T}(T\,\,in\,\,Kelvin)\] in the absence of catalyst is \[(R=8.3\,\,J\,mo{{l}^{-1}}\,{{K}^{-1}})\]

    A) \[+1\]                         

    B) \[-1\]

    C) \[+1000\]                   

    D) \[-1000\]

    Correct Answer: D

    Solution :

    \[k=A{{e}^{-{{E}_{a}}/RT}}\] \[k'=A{{e}^{-E{{'}_{a}}/RT}}\] where k = rate constant for non-catalysed reaction, and k? = rate constant for catalysed reaction. \[{{E}_{a}}\] = activation energy for non-catalysed reaction and \[E{{'}_{a}}\] = activation energy for catalysed reaction) \[\frac{k'}{k}={{e}^{{{E}_{a}}-E{{'}_{a}}}}\] Also given \[\operatorname{k}'=k+1.718\,\,k=2.718\,\,k\] \[\therefore \,\,\,\,2.718={{e}^{\frac{{{E}_{a}}-E{{'}_{a}}}{RT}}}\] \[{{\log }_{e}}\,2.718=\frac{{{E}_{a}}-E{{'}_{a}}}{8.314\times {{10}^{-3}}\times 500}\] \[{{E}_{a}}-E_{a}^{'}=4.11\] \[E_{a}^{'}=\,\,4.15\] \[\therefore \,\,\,\,\,{{E}_{a}}=8.3\,\,kJ/mo{{l}^{-1}}\] \[k=A{{e}^{-{{E}_{a}}/RT}}\] \[\therefore \,\,\,\,\,\,\,\,{{\log }_{e}}k={{\log }_{e}}A-\frac{Ea}{RT}\] This is an equation for straight line with slope \[=-\frac{Ea}{R}=-\frac{8.3}{8.3\times {{10}^{-3}}}\] \[=\,\,-1000\]


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