The inversion of cane sugar proceeds with constant half life of 500 minute at pH = 5 for any concentration of sugar. However, if pH = 6, the half life changes to 50 minute. Derive the rate law for inversion of cane sugar.

This is explained as followsWe are given that the half life at pH = 5 and at any concentration of sugar is 50 minutes while the half life at pH = 6 and any concentration of sugar is 500 minutes. This clearly means that the half life is independent of the concentration of sugar and depends on the concentration of hydrogen ion. The fact that order of reaction with respect to sugar is 1 further justifies this as for first order reaction t1/2 is independent of initial concentration.

As pH = -log[H+], therefore for pH = 5, [H+] = 10-5 and for pH = 6, [H+] = 10-6. The general equation for t1/2 with respect to any reactant is written as follows

t1/2 ∝ 1 / ([A]0 n-1)

where [A]0 is the initial concentration of the reactant A and n is the order of reaction with respect to A.

Thus we have the following relations

t1/2 ∝ 1 / ([H+]y-1)

50 ∝ 1 / ( 10-5) y-1 (1)

and

500 ∝ 1 / ( 10-6) y-1 (2)

Dividing equation (1) by (2) and solving for y gives y = 0.

Thus y = 0

The correct answer is r = K [sugar][H+]^0 with y = 0. This is explained as followsWe are given that the half life at pH = 5 and at any concentration of sugar is 50 minutes while the half life at pH = 6 and any concentration of sugar is 500 minutes. This clearly means that the half life is independent of the concentration of sugar and depends on the concentration of hydrogen ion. The fact that order of reaction with respect to sugar is 1 further justifies this as for first order reaction t1/2 is independent of initial concentration. As pH = -log[H+], therefore for pH = 5, [H+] = 10-5 and for pH = 6, [H+] = 10-6. The general equation for t1/2 with respect to any reactant is written as follows t1/2 ∝ 1 / ([A]0n-1)where [A]0 is the initial concentration of the reactant A and n is the order of reaction with respect to A. Thus we have the following relations t1/2 ∝ 1 / ([H+]y-1) 50 ∝ 1 / ( 10-5) y-1 (1)and 500 ∝ 1 / ( 10-6) y-1 (2)Dividing equation (1) by (2) and solving for y gives y = 0.Thus y = 0