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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2007

  • question_answer
    \[3A\xrightarrow{{}}2B,\] rate of reaction \[\frac{+d(B)}{dt}\]is equals to

    A)  \[-\frac{3}{2}\frac{d(A)}{dt}\]                  

    B)  \[-\frac{2}{3}\frac{d(A)}{dt}\]

    C)  \[-\frac{1}{3}\frac{d(A)}{dt}\]                  

    D)  \[+2\frac{d(A)}{dt}\]

    Correct Answer: B

    Solution :

    Given reaction is \[3A\xrightarrow{{}}2B\] \[-\frac{1}{3}\frac{d(A)}{dt}=\frac{1}{2}\frac{d(B)}{dt}\]                 \[\therefore \]  \[\frac{d(B)}{dt}=-\frac{2}{3}\frac{d(A)}{dt}\]


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