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.
Create a free account to view solution
View Solution FreeTopic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions
More Chemical Kinetics and Nuclear Chemistry Questions
A reaction takes place in three steps having rate constants K1, K2, K3 respectively. The overall rate constant K = . If ...A zero-order reaction is one...What is the final product in the disintegrations series of starting from 92U235 :...The decomposition of hydrogen peroxide in aqueous solution is a first-order reactionTime in minutes 0 10Volume (V, in ml...In a zero order reaction for every 10o rise of temperature, the rate is doubled. If the temperature is increased from 10...