For the reaction : 2A+B to A_(2)B, the rate =k[A][B]^(2) with k=2.0xx10^(-6)"mol"^(-2)L^(2)s^(-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).

For the reaction : 2A+B to A_(2)B, the rate =k[A][B]^(2) with k=2.0xx1

1.	For the reaction : 2A+B to A_(2)B, the rate =k[A][B]^(2) with k=2.0xx10^(-6)"mol"^(-2)L^(2)s^(-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).
Answer» Solution :SUBSTITUTING the values in the rate equation, we get INITIAL rate `=K[A][B]^(2)` `=(2.0xx10^(-6)"mol"^(-2)L^(2)s^(-1))(0.1 "mol L"^(-1))(0.2" mol L"^(-1))^(2)` When [A] is reduced from 0.10 mol `L^(-1)` to 0.06 mol `L^(-1)`, 0.04 mol `L^(-1)` of A has reacted. Amount of B reacted `=(1)/(2)xx0.04 "mol L"^(-1)=0.02"mol L"^(-1)` Hence, [B] at that TIME `=0.2-0.02=0.18"mol L"^(-1)` Now,rate `=(2.0xx10^(-6)"mol"^(-2)L^(2)s^(-1))(0.06"mol L"^(-1))(0.18" mol L"^(-1))^(2)` `=3.89xx10^(-9)" mol L"^(-1)s^(-1)`.