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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.  [JEE Online 08-04-2017]

    A)  4.92 K                  

    B)  9.84 K

    C)  19.67 K                                

    D)  2.45 K

    Correct Answer: A

    Solution :

    \[2=\frac{Eq}{R}\left\{ \frac{1}{300}-\frac{1}{310} \right\}\]                                         ?(i) \[2={{e}^{2}}\frac{Ea}{R}\left\{ \frac{1}{300}-\frac{1}{T} \right\}\]                                            ?(ii) \[\frac{2Ea}{R}\left\{ \frac{1}{300}-\frac{1}{T} \right\}=\frac{{{E}_{a}}}{R}\left\{ \frac{1}{300}-\frac{1}{310} \right\}\] \[\frac{1}{300}+\frac{1}{310}=\frac{2}{T}\Rightarrow T=\frac{300\times 310}{610}\times 2\] \[=304.92\]


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