# How to derive rate law expression for inversion of sucrose with changing pH?

I came across a question worded as follows- "The inversion of cane sugar proceeds with half life of 50 minutes at pH=5 at any concentration of sugar. However, if the pH=6, the half life changes to 500 minutes at any concentration of sugar, the rate law expression for inversion of sugar can be written as?"

"at any concentration of sugar" suggests time period is independent of concentration of sugar and therefore of order 1 with respect to sugar. Reducing H+ concentration by 10 increases half life by 10, so half life is inversely proportional to [H+], also,as a rule- half life is inversely proportional to [a]^n-1 for any order n and concentration of reactant [a], comparing exponents of a with [H+]- we get n=2, but this is clearly not the correct way to go about it since the rate law is actually given by k(sugar)(H+) while we get k(sugar)(h+)^2; that is order 1 and 2 w.r.t to reactants- So my question is what is the correct method to find rate law expression in such conditions?

• Please edit punctuation to make your question more intelligible. As it reads now there seems to be a line ending missing. Also please provide definitions of [a] and n. Overall you have the right rate law it seems though. – Buck Thorn Nov 19 '18 at 13:55
• I cleaned it up as much as I could. The rate law is not correct as per my textbook though. – hiten pandya Nov 19 '18 at 14:05
• d[a]/dt = -k[a][h+], where a=sugar. You seem to confuse [h+] and [a]. – Buck Thorn Nov 19 '18 at 15:40
• can you explain the method behind how you obtained that expression from the information given? – hiten pandya Nov 19 '18 at 18:00
• You actually answer this yourself: half life is independent of [a], a =sugar, thus n=1. As for [h+], as you explain half life is inversely proportional to [H+]. Solve the diff eq: separate terms and integrate. Then apply boundary condition [sugar](t=0) = [sugar]0, then solve the resulting equation for the half-life. – Buck Thorn Nov 19 '18 at 20:26

## 1 Answer

It is clear that order of reaction w.r.t. sugar is 1.

Let rate of reaction r=k(sugar)^1(H+)^z Then you have to define a new half life -d(A)÷dt=k(A)(H+)^z let (sugar)=(A) -d(A)÷d(A)=k(H+)^z dt Integrate both side .
on left put (initial concentration)to ((initial concentration)÷2) and left side from 0 to (half life )

Then ln2=k(H+)^z *(half life) From here half life =ln2÷k(H+)^z

Then put the values given 500=ln2÷k10^-5z

50=ln2÷k10^-6z

Solve and you will get z=1