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

Answer

`8xx10^(-9)mol.L^(-1)s^(-1)3.89xx10^(-9)mol.L^(-1)s^(-1)`

Explanation

`a. 2A+B rarr A_(2)B`
Rate `=k[A][B]^(2)`
Rate `=2.0 xx 10^(-6)(0.1)(0.2)^(2)`
`=8xx10^(-9) mol . L^(-1)s^(-1)`
`b. (-d[A])/(2dt)=(-d[B])/(DT)=(d[A_(2)B])/(dt)`
`[A]_(Initial)=0.1M,[A]_(F i nal)=0.06M`
`[A]_(Reacted)=0.1-0.06=0.04M`
`-(d[B])/(dt)=(-d[A])/(2dt)=(1)/(2)xx0.04=0.02Ms^(-1)`
`[B]_(Initial)=0.2M`
`[B]_(Reacted)=0.02M`
`[B]_(Final)=0.2-0.02=0.18M`
`c.` Rate `=k[A][B]^(2)`
`=2.0xx10^(-6)(0.06)(0.18)^(2)`
`3.89xx10^(-9)mkol. L^(-1)s^(-1)`