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 `=k[A][B]^(2)=(2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.1molL^(-1))(0.2molL^(-1))^(2)=8xx10^(-9)molL^(-1)s^(-1)` When [A] is reduced from `0.10molL^(-1)" to "0.06molL^(-1)`, i.e., `0.04molL^(-1)` of A has reacted, B reacted
`=(1)/(2)xx0.04molL^(-1)=0.02molL^(-1).`
Hence, new `[B]=0.2-0.02=0.18molL^(-1).`
Now, rate `=(2.0xx10^(-6)mol^(-2)L^(2)s^(-1))(0.06molL^(-1))(0.18molL^(-1))^(2)=3.89xx10^(-9)molL^(-1)s^(-1).`