The rate of a reaction `2A+B rarr A_2B`.
has rate law : rate = k `[A]^2` with the rate constant equal to 0.50 `mol^(-1)" L sec"^(-1)`. Calculate the rate of reaction when
(i) [A] = 0.60 mol `L^(-1)`, [B] =-0.05 mol `L^(-1)` and
(ii) When concentration of A and B have been reduced to 1/4 th

Explanation

Rate law is ,
Rate=`k[A]^2`
Where k=0.50 `mol^(-1)Ls^(-1)`
`therefore` Rate=0.50 `mol^(-1)Ls^(-1)[A]^2`
(i) [A]=0.60 mol `L^(-1)`
[B]=0.50 mol `L^(-1)`
`therefore` Rate=`(0.50 mol^(-1)Ls^(-1)) xx (0.60 mol^(-1))^2`
`=0.18" mol L"^(-1)s^(-1)`
(ii) [A]=`(0.60)/(4)" mol L^(-1)=0.15 " mol L"^(-1)`
Rate=k`[A]^2`
`therefore" Rate"=(0.50 mol^(-1)Ls^(-1)`
`(0.15" mol L"^(-1))^2`
`=0.01125" mol L"^(-1)s^(-1)`
=1.125 `xx10^(-2)" mol L"^(-1)s^(-1)`