For the non-equilibrium process, `A + B rarr` Product, the rate is first-order w.r.t. `A` and second order w.r.t. `B`. If `1.0 mol` each of `A` and `B` were inrofuced into `1.0 L` vessel and the initial rate was `1.0 xx 10^(-2) mol L^(-1) s^(-1)`, calculate the rate when half the reactants have been turned into Products.

Explanation

`r = k[A][B]^(2)`
`10^(-2) = k[1.0] [1.0]^(2)`
`k = 10^(-2) L mol^(-1) s^(-1)`
When half of the reactants have been turned into Products, the concentration of `A` and `B` left are `0.5 mol`.
`:. r = k[0.5] [0.5]^(2)`
`= 10^(-2) xx 0.5 xx 0.5 xx 0.5`
`= 1.25 xx 10^(-3) mol L^(-1) s^(-1)`