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

`2 A+B to A_2B `
(i) Rate = `k [A]^2`
`= 0.5 xx 0.36 `
`= 0.18 mol l^(-1) sec^(-1)`
(ii) Rate = `k [A]^2`
`[A] = 0.60` then
`1/4 [A] = 0.15 `
Rate = `0.05 xx 0.225 `
`= 1.12 xx 10^(-12) mol l^(-1)`