The rate law expression of a reaction `2A+B rarr A_(2)B` is rate =`k[A][B]^(2)` with rate constant k =0.65 `mol^(-1) L s^(-1)`
calculate rate of reaction when concentration of A has been reduced to `1//4` and concentration of B has been reduced by `1//4`

Explanation

`[A]=1/4 xx0.25 mol L^(-1) =0.0625 mol L^(-1)` and `[B] =(1-1/4)xx1.84 mol L^(-1)=1.38 mol l^(-1)`
Rate `=k[A]^(2) =0.65 mol^(-1) L s^(-1) xx(0.0625) mol L^(-1)^(2) =0.002539 mol L^(-1) s^(-1)`
`=0.254 xx10^(-2) mol L^(-1) s^(-1)`