When inversion of sucrose is studied at pH = 5, the half-life period is always found to be 500 minutes irrespective of any initial concentration but when it is studied at pH = 6, the half-life period is found to be 50 minutes. Derive the rate law expression for the inversion of sucrose.

Explanation

At `pH=5,` as half-life period is found to be independent of initial concentration of sucrose, this means with respect to sucrose, it is a reaction of first order, i.e., Rate = k [Sucrose].
If n is the order with respect to `H^(+)` ion, `t_(1//2)prop[H^(+)]^(1-n),`
`{:(i.e.",",,,500prop(10^(-5))^(1-n),,,[pH=5" means "[H^(+)]=10^(-5)M]" "...(i)),("and",,,50prop(10^(-6))^(1-n),,,[pH=6" means "[H^(+)]=10^(-6)M]" "...(ii)):}`
Dividing (i) by (ii), `10=(10)^(1-n)"i.e. "1-n=1" or "n=0," i.e., order with respect to "H^(+)" ion = 0. Hence, overall rate law is Rate = k [Sucrose] "[H^(+)]^(0).`