At constant temperature and constant `pH` of `4`, the inverison of sucrose proceeds with a constant half life of `300 min`. At the same temperature but at `pH` of `3` the half life is constant at `30 min`. Calculate the order of reaction w.r.t. sucrose and w.r.t. `[H^(o+)]`.

Explanation

Sucrose `[A] +H_(2)Ooverset(H^(o+))rarr "Glucose + Fructose"`
`(-d[A])/(dt) = Rate = k[A]^(a)[H^(o+)]^(b)`
(Does not depend on `H_(2)O`, Since it is in large excess)
During any experiment, `pH` is constant, hence
`(-d[A])/(dt) = k'[A]^(a)`, where `k' = k[H^(o+)]^(b)`
Since half life is independent of the initial concentration of `a`, hence `a = 1`, consequently `k'`, is a first order rate constant and is given by
`k' = (0.0693)/(t_(1//2))`
`:. ((t_(50))_(1))/((t_(50))_(2)) = (k'_(2))/(k'_(1)) = (k[H^(o+)]_(2)^(b))/(k[H^(o+)]_(1)^(b)) = ([H^(o+)]_(2)^(b))/([H^(o+)]_(1)^(b))`
`((300)/(30)) = ((10^(-3))/(10^(-4)))^(b) or b = 1`
Therefore, the rate of hydrolyiss of sucrose has a first order dependence on `[H^(o+)`, so that
`(-d[A])/(d t) = k[A][H^(o+)]`
`:.` Order w.r.t. sucrose `= 1` , order w.r.t. `[H^(o+)]=1`