Question

At 25° C and at a constant pH of 5, the inversion of sucrose proceeds with a constant half-life of 500 min. At this same temperature, but at a pH of 4, the half-life is constant at 50 min. The rate law is . What are the values of a and b?

Answer 1

Answer:     rate = k [sugar]^-1[H⁺]²---------------> Yes you read is correct, not possible, but question test our knowledge.

The given rate law is

d sucrose-ksucrose dt

I know this question is from

http://faculty.virginia.edu/richardson/thermo/problems/problems/question/9/quest9.htm

:The inversion of cane sugar proceeds with half-life of 500 minute at pH 5 for any concentration of sugar. However ifpH-4, the half-life changes to 50 minute. The rate law expression for the sugar inversion can be written as Solution:Since t2 does not depends upon the sugar concentration means it is first order respect to sugar concentration. 20 sugar] 12 xak I. n 2) H 500 (105 50 104 Hence n =2

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