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AIIMS AIIMS Solved Paper-2005

  • question_answer
    For reaction \[aA\to xP,\], when \[\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }=2.2\text{ }mM\], the rate was found to be \[2.4\text{ }mM\text{ }{{s}^{-1}}\]. On reducing concentration of A to half, the rate changes to \[0.6\text{ }mM\text{ }{{s}^{-1}}\]. The order of reaction with respect to A is:

    A) 1.5                                         

    B) 2.0                         

    C) 2.5                         

    D)        3.0

    Correct Answer: B

    Solution :

    \[aA\xrightarrow{{}}xP\]               - rate \[\propto {{[A]}^{a}}=k{{[A]}^{a}}\] Initially  \[{{[A]}_{1}}=2.2\] millimol, \[{{r}_{1}}=2.4mM/s\], If \[{{[A]}_{2}}=\frac{2.2}{2}=1.1\] millimol, \[{{r}_{2}}=0.6mM/s\] We have \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{[A]_{1}^{a}}{{{[{{A}_{2}}]}^{a}}}\]  (a is the order). \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{(2.2)}^{a}}}{{{(1.1)}^{a}}}\]  or \[\frac{2.4}{0.6}={{(2)}^{a}}\] \[4={{2}^{a}}\] \[\Rightarrow \]  \[{{2}^{2}}={{2}^{a}}\]  or  \[a=2\] The reaction is of second order.


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