The inversion of cane sugar proceeds with half life of `500 min` at `pH 5` for any concentration of sugar. However, if `pH = 6`, if the half life changes to `50 min`. The rate law expression for the sugar inversion can be written as

Options

A. `r =k["sugar"]^2[H^+]^0`
B. `r =k["sugar"]^1[H^+]^0`
C. `r =k["sugar"]^1[H^+]^1`
D. `r =k["sugar"]^0[H^+]^1`

Explanation

At pH = 5, the half- life is 500 for all concentrations of sugar, i.e., `t_(1//2) prop["sugar"]^0`.
Thus, the reaction is 1 order with respect to sugar.
Now, rate `=k["sugar"]^1[H^+]^(1-m)`
Also for `[H^(+)]` `t_(1//2)prop[H^+]^(1-m)`
Given , pH = 5, `:. [H^+]= 10^(-5)`
`:.500 prop [10^(-6)]^(1-m) ...(i)`
pH = 6 `:. [H^+]= 10^6`
`:. 50 prop[10^(-6)]^(1-m) ...(ii)`
`:.` Dividing (i) by (ii)
`:. 10 =(10)^((1-m))`
or (1 - m) = 1 `:. m = 0`
Therefore,
rate `=k ["sugar"]^1[H^+]^0`