For the reaction :

2A + B A2B


the Rate = k[A][B]2 with k = 2.0 × 10-6 mol-2 L2 sec-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1 and [B] = 0.2 mol L-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1.

a) Rate = k[A][B]2, Rate = 2.0 × 10-6[0.1][0.2]2


Rate = 8 × 10-9 mol L-1sec-1


b)


Situation when A is remained 0.06 mol L-1


Now, According to rate law, Rate = k[A][B]2


Rate = 2 × 10-6[0.06][0.18]2


i.e. Rate = 3.888 × 10-9 mol L-1sec-1


Initial rate of reaction is 8 × 10-9 mol L-1sec-1. and rate when concentration of A is 0.06 mol L-1 is 3.888 × 10-9 mol L-1sec-1.


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