A chemical reaction procceds into the following steps, Step I, 2A⇌X(fast) Step II, X+B→Y(slow) Step III, Y+BA⇌X[Product](fast) The rate law for the overall reaction is:
A
Rate =k[A]2
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B
Rate =k[B]2
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C
Rate =k[A][B]
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D
Rate =k[A]2[B]
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Solution
The correct option is D Rate =k[A]2[B]
Step I
2A⇋X
Rate constant for forward reaction r=k[A]2 (EQ 1)
Rate constant for backward reactionr=k[X] (EQ 2]
Equating 1 & 2
[A]2=[X]
For step II
X+Y⇋Y
r=[X][B]
Putting the value of [X]=[A]2
r=[A]2B is the correct answer because it is the slowest step which is rate determining.