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.