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.

When inversion of sucrose is studied at pH=5, the half-life period is

1.	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.
Answer» Solution :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).`