For the reaction :

2A + B → A_{2}B

the Rate = k[A][B]^{2} with k = 2.0 × 10^{-6} mol^{-2} L^{2} 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^{-1}sec^{-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^{-1}sec^{-1}

__Initial rate of reaction is 8 × 10 ^{-9} mol L^{-1}sec^{-1}. and rate when concentration of A is 0.06 mol L^{-1} is 3.888 × 10^{-9} mol L^{-1}sec^{-1}.__

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