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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    A certain reaction rate increases 1000 folds in the presence of a catalyst at\[{{27}^{o}}C\]. The activation energy of the original pathway is\[98\,\,kJ/mol\]. What is the activation energy of the new pathway?

    A) \[80.77\,\,kJ\]

    B)                        \[56.38\,\,kJ\]

    C) \[24.67\,\,kJ\]  

    D)        \[90.43\,\,kJ\]

    Correct Answer: A

    Solution :

    \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\log 1000\]             \[=\frac{{{E}_{1}}-{{E}_{2}}}{RT}\]             \[=\frac{98000-{{E}_{2}}}{8.314\times 300}\] \[{{E}_{2}}\](catalyst)\[=80.77\,\,kJ\]


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