JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    For the reaction, \[3A+2B\to C+D,\] the differential rate law can be written as:     JEE Main Online Paper (Held On 19 April 2016)

    A) \[\frac{1}{3}\frac{d\left[ A \right]}{dt}=\frac{d\left[ C \right]}{dt}=k{{\left[ A \right]}^{n}}{{\left[ B \right]}^{m}}\]

    B) \[-\frac{d\left[ A \right]}{dt}=\frac{d\left[ C \right]}{dt}=k{{\left[ A \right]}^{n}}{{\left[ B \right]}^{m}}\]

    C) \[+\frac{1}{3}\frac{d\left[ A \right]}{dt}=-\frac{d\left[ C \right]}{dt}=k{{\left[ A \right]}^{n}}{{\left[ B \right]}^{m}}\]

    D) \[-\frac{1}{3}\frac{d\left[ A \right]}{dt}=\frac{d\left[ C \right]}{dt}=k{{\left[ A \right]}^{n}}{{\left[ B \right]}^{m}}\]

    Correct Answer: D

    Solution :

    For the reaction\[3A+2B\xrightarrow[{}]{{}}C+D\] Rate of disappearance of A = Rate of appearance of C reaction \[=-\frac{1}{3}\frac{d[A]}{dt}=\frac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}\]


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