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JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    Reaction rate between two substance A and B is expressed as following:\[\text{rate}\,\text{=k }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{\text{n}}}{{\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }}^{\text{m}}}\] If the concentration of A is doubled and concentration of B is made half of initialcon centration, the ratio of the new rate to the earlier rate will be:     JEE Main Online Paper (Held On 07 May 2012)

    A) \[m+n\]                              

    B)                        \[n-m\]

    C)                        \[\frac{1}{{{2}^{\left(m+n \right)}}}\]                    

    D)                        \[{{2}^{\left( n-m \right)}}\]

    Correct Answer: D

    Solution :

                    Rate\[_{1}=k{{[A]}^{n}}{{[B]}^{m}}\] Rate\[_{2}=k{{[2A]}^{n}}{{\left[ \frac{1}{2}B \right]}^{m}}\] \[\therefore \]\[\frac{\text{Rat}{{\text{e}}_{2}}}{\text{Rat}{{\text{e}}_{1}}}=\frac{k{{[2A]}^{n}}{{\left[ \frac{1}{2}B \right]}^{m}}}{k{{[A]}^{n}}{{[B]}^{m}}}={{(2)}^{n}}{{\left( \frac{1}{2} \right)}^{m}}\] \[={{2}^{n}}.{{(2)}^{-m}}={{2}^{n-m}}\]


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