JEE Main & Advanced Chemistry Chemical Kinetics Question Bank Rate of a reaction

  • question_answer The velocity constant of a reaction at 290 K was found to be \[3.2\times {{10}^{-3}}\]. At 300 K it will be  [MP PMT 2004]

    A)                 \[1.28\times {{10}^{-2}}\]            

    B)                 \[6.4\times {{10}^{-3}}\]

    C)                 \[9.6\times {{10}^{-3}}\]              

    D)                 \[3.2\times {{10}^{-4}}\]

    Correct Answer: B

    Solution :

                    As we know that the velocity constant become double by increasing the temperature by 10°C so if at 290 K, velocity constant \[=3.2\times {{10}^{-3}}\] then at 300 K, velocity constant \[=2({{K}_{290}})\]\[=2\times 3.2\times {{10}^{-3}}\]\[=6.4\times {{10}^{-3}}\].


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