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Question

For the reaction 2A+BC, the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is:

[A](molL1) [B](molL1) Initial Rate(molL1s1)
0.05 0.05 0.045
0.10 0.05 0.090
0.20 0.10 0.72

A
Rate = k[A]2[B]2
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B
Rate = k[A][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 B Rate = k[A][B]2
Let, r=k[A]x[B]y
0.045=k(0.05)x(0.05)y.... (1)
0.090=k(0.10)x(0.05)y..... (2)
0.72=k(0.20)x(0.10)y..... (3)
Dividing (2) by (1),
2=2xx=1
Dividing (3) by (2),
8=(21)(2)yy=2
So, rate= k[A][B]2

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