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Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    \[\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] The activation energy for the forward reaction is 50 kcal. What is the activation energy for the back word reaction?

    A)  -72 kcal                               

    B) -28 kcal

    C)  +28kcal                               

    D) +72kcal

    Correct Answer: D

    Solution :

    \[x-y+1=0\] \[(p\to \tilde{\ }p\wedge )(\tilde{\ }p\to p)\]Therefore, weight of solute = 480 g \[\frac{6!}{3!}\]The activation energy for the forward reaction = 50 kcal \[\underset{x\to -1}{\mathop{\lim }}\,\left( \frac{{{x}^{4}}+{{x}^{2}}=x+1}{{{x}^{2}}-x+1} \right)\frac{1-\cos (x+1)}{{{(x+1)}^{2}}}\]The activation energy for the backward reaction = 50 + 72 = + 72 kcal


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