2. Solved Papers
  3. Manipal Medical

Manipal Medical Manipal Medical Solved Paper-2015

  • question_answer
    In the following reaction,\[0.02=1.61\exp .\left( \frac{-18260.8}{8.314\times T} \right)\] The temperature at which the reaction occurs is

    A) 400 K                                        

    B) 200 K

    C) 500 K                                        

    D) 800 K

    Correct Answer: C

    Solution :

    \[0.02=1.61\,\exp .\left( -\frac{18230.8}{8.314\times T} \right)\]                 In \[0.02=In\,\,1.61\,+\left( \frac{-18230.8}{8.314\times T} \right)\]                  \[\log \frac{0.02}{1.61}=\frac{-18230.8}{8.314\times T}\]                   \[T=-\frac{18230.8}{2.303\times 8.314\times (-1.90)}\]                  \[T=500\,K\]


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