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NEET Sample Paper NEET Sample Test Paper-43

  • question_answer
    The mechanism of the reaction: \[A+2B+C\to D\]is
    Step 1     \[A+BX\](Fast equilibrium)
    Step 2    \[X+C\to \]Y(slow)
    Step 3    \[Y+B\to \] D(fast)
    Which rate law is correct

    A) \[r=k\left[ C \right]\]

    B) \[r=k\left[ A \right]{{\left[ B \right]}^{2}}\left[ C \right]\]

    C) \[r=k\left[ A \right]\left[ B \right]\left[ C \right]\]

    D) \[r=k\left[ D \right]\]

    Correct Answer: C

    Solution :

    Step-2 is the rate determining step (slow) \[\therefore \] rate law is- v=k?[X][C] X can be evaluated from step -1 \[K=\frac{[X]}{[A][B]}\] or K[A][B] = [X] here K- equilibrium constant on putting the value of [X] the rate law becomes v=k?K [A] [B] [C] or v=k[A] [B] [C]


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