For the reaction `:`
`2A+B rarr 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^(c-),[B]=0.2 mol L^(-1)`. Calculate the rate of reaction after `[A]` is reduced to `0.06 mol L^(-1)`.

Explanation

Initial rate is obtained by substituting the values of rate constant (k) and the concentration of both the reactants (A and B) into the rate law expression:
Rate `= k[A][B]^(2)`
Initial rate `= (2.0xx10^(-6)mol^(2)L^(2)s^(-1)(0.1mol L^(-1))(0.2mol L^(-1))(2)`
`= 0.008xx10^(-6)mol L(-1)s^(-1)`
`= 8.0xx10^(-9)mol L^(-1)s^(-1)`.
In the next calculation, we are given the concntration of the reactant, A. It is `0.06 mol L^(-1)` . But to calculate the reaction rate we also need the concentration of the other reactant B. According to the balanced equation, when 2 mol of A are consumed, only 1 mol of B disappears. Thus, when [A] is reduced from `0.1mol L^(-1) to 0.06 mol L^(-1), i.e., 0.04 mol L^(-1)` of A has reacted, concentration of B consumed will be half of the concentration of A reacted:
`= (1)/(2) (0.04 mol L^(-1))`
`= 0.02 mol L^(-1)`
Thus, final `[B] = "Intial" [B]- "Reacted" [B]`
`= (0.2mol L^(-1)) - (0.02 mol L^(-1))`
`= 0.81mol L^(-1)`
and
`"Rate" = (2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.06mol L^(-1))(0.18mol L^(-1))^(2)`
`= 0.00389xx10^(-6)mol^(-2)L^(2)s^(-1)`
`= 3.89xx10^(-9)mol L^(-1)s^(-1)`