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JEE Main & Advanced JEE Main Paper (Held On 12 May 2012)

  • question_answer
    In a chemical reaction A is converted into B. The rates of reaction, starting with initial concentrations of A as \[2\times {{10}^{-3}}M\] and \[1\times {{10}^{-3}}M,\]are equal to \[2.40\times {{10}^{-4}}M{{s}^{-1}}\]and \[0.60\times {{10}^{-4}}M{{s}^{-1}}\] respectively. The order of reaction with respect to reactant A will be     JEE Main Online Paper (Held On 12 May 2012)

    A) 0      

    B)                               1.5

    C)                        1                                             

    D)                        2

    Correct Answer: D

    Solution :

                    \[A\xrightarrow[{}]{{}}B\] Initial concentration        Rate of reaction \[2\times {{10}^{-3}}M\]                              \[2.40\times {{10}^{-4}}M{{s}^{-1}}\] \[1\times {{10}^{-3}}M\]                              \[0.60\times {{10}^{-4}}M{{s}^{-1}}\] rate of reaction\[r=k{{[A]}^{x}}\]where x = order of reaction hence \[2.40\times {{10}^{-4}}=k{{[2\times {{10}^{-3}}]}^{x}}\]                               .....(i) \[0.60\times {{10}^{-4}}=k{{[1\times {{10}^{-3}}]}^{x}}\]                                               ....(ii) on dividing eqn.(i) from eqn. (ii) we get\[4={{(2)}^{x}}\] \[\therefore \]\[x=2\]i.e. order of reaction = 2


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