Chemical Kinetics and Nuclear ChemistryHard
Question
The inversion of cane sugar proceeds with half life of 500 min at pH 5 for any concentration of sugar. However, if pH = 6, the half life change to 50 min. The rate law expression for the sugar inversion can be written as
Options
A.r = k[sugar]2[H]6
B.r = k[sugar]1[H]0
C.r = k[sugar]0[H+]6
D.r = k[sugar]0[H+]1
Solution
Sincet1/2 does not depends upon the sugar concentration means it is first order w.r.t.[sugar]
∴ t1/2 ∝ [sugar]1
t1/2 × an-1 = k
10 = (10)1-n ⇒ n = 0.
∴ t1/2 ∝ [sugar]1
t1/2 × an-1 = k
10 = (10)1-n ⇒ n = 0.
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