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Question:

For the reaction 2A+B→C, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reactions is:
[A](mol L^-1) [B](mol L^-1) Initial Rate(mol L^-1s^-1)
0.05 0.05 0.045
0.10 0.05 0.090
0.20 0.10 0.71

Rate=k[A][B]

Rate=k[A][B]2

Rate=k[A]2[B]

Rate=k[A]2[B

Solution:

Correct option is C. Rate=k[A][B]2
Solution:- (C)Rate=k[A][B]2
r=K[A]x[B]y
0.045=K(0.05)x(0.05)y(1)
0.090=K(0.010)x(0.05)y(2)
0.72=K(0.20)x(0.10)y(3)
From(1)÷(2), 0.045/0.090=(0.05/0.10)x ⇒x=1
From(2)÷(3), 0.090/0.720=(0.10/0.20)1×(0.05/0.10)y ⇒y=2
Hence, r=K[A][B]2