In the acid hydrolyiss reaction `A + H_(2)O + H^(o+) rarr` Product, where `[H^(o+)] = 0.1 mol dm^(-3)` and `H_(2)O` is present in large excess, the apparent rate constant is `1.5 xx 10^(-5) s^(-1)`. Calculate the true rate constant.

Explanation

`r = (-d[A])/(dt) = k[A][H_(2)O][H^(o+)]`
Since `[H^(o+)]` is essentially constant (because the catalyst is regenerated and `[H_(2)O]gt gt [A]`)
`:. r = k_(app)[A]`, where `k_(app) = k[H_(2)O][H^(o+)]`
`k = (k_(app))/([H_(2)O][H^(o+)])`
`[H^(o+)] = 0.1 "mol dm"^(-3)`
`[H_(2)O] = (1000)/(18) = 55.6 mol dm^(-3)`
Hence, `k = (1.5 xx 10^(-5) s^(-1))/((55.6 "mol dm"^(-3))(0.1 "mol dm"^(3)))`
`= 2.7 xx 10^(-6) dm^(6) mol^(-2)s^(-1)`
`= 2.7 xx 10^(-6) L^(2) mol^(-2) s^(-1)`