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]You need to login to perform this action.
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