For the reaction 2A+B rarr A_(2)B. The rate = k[A] [B]^(2) with 2.0xx10^(-6)mol^(-2)L^(2)s^(-1). Calculate the initial rate of the reaction, when [A] = 0.1 "mol L"^(-1), [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 rarr A_(2)B. The rate = k[A] [B]^(2) with 2.0xx1

1.	For the reaction 2A+B rarr A_(2)B. The rate = k[A] [B]^(2) with 2.0xx10^(-6)mol^(-2)L^(2)s^(-1). Calculate the initial rate of the reaction, when [A] = 0.1 "mol L"^(-1), [B] = 0.2 "mol L"^(-1). Calculate the rate of reaction after [A] is reduced to 0.06 "mol L"^(-1).
Answer» Solution :The initial rate of the REACTION is 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)` `= 8.0xx10^(-9)mol^(-2)L^(2)s^(-1)` When [A] is reduced from `0.1 "mol L"^(-1)` to `0.06 mol^(-1)`, the concentration of A reacted `=(0.1-0.06)"mol L"^(-1)=0.04 "mol L"^(-1)`. `therefore` The concentration of B reaction `=(1)/(2)xx0.04 "mol L"^(-1)`. `=0.02 " mol L"^(-1)`. Then, concentration of B available `[B]=(0.2-0.02)"mol L"^(-1)`. `= 0.18 " mol L"^(-1)`. Aften [A] is reduced to `0.06 "mol L"^(-1)`, the rate of the reaction is given by, Rate `= k[A][B]2` `=(2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.06"mol L"^(-1)) (0.18 "mol L"^(-1))^(2)` `= 3.89"mol L"^(-1)s^(-1)`.