The experimental rate law for the decomposition of X2Y to X2 and Y2 is Rate = kexp(X2Y]². Two mechanisms are proposed. Mechanism A: X2Y <--?-> X2 + Y X2Y + Y- X2 + Y2 Mechanism B: 2X2Y <-?-> X4Y2 X4Y2 2X2 + Y2

Physical Chemistry
2nd Edition
ISBN:9781133958437
Author:Ball, David W. (david Warren), BAER, Tomas
Publisher:Ball, David W. (david Warren), BAER, Tomas
Chapter20: Kinetics
Section: Chapter Questions
Problem 20.85E
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The experimental rate law for the decomposition of X2Y to X2 and Y2 is Rate = kexp[X2Y]?. Two mechanisms are proposed.
Mechanism A:
X2Y <--?-> X2 +Y
X2Y + Y- X2 + Y2
Mechanism B:
2X2Y <-?--> X4Y2
X4Y2 - 2X2 + Y2
Where <--?--> can be - (a simple forward reaction) or = (an equilibrium) depending on how the choices below are described. Which of the following could be a correct mechanism?
I. Mechanism A with the first step being a fast equilibrium and the second step being slow.
II. Mechanism A with the first step being a slow forward reaction and the second step being fast.
II. Mechanism B with the first step being a fast equilibrium and the second step being slow.
IV. Mechanism B with the first step being a slow forward reaction and the second step being fast.
Transcribed Image Text:The experimental rate law for the decomposition of X2Y to X2 and Y2 is Rate = kexp[X2Y]?. Two mechanisms are proposed. Mechanism A: X2Y <--?-> X2 +Y X2Y + Y- X2 + Y2 Mechanism B: 2X2Y <-?--> X4Y2 X4Y2 - 2X2 + Y2 Where <--?--> can be - (a simple forward reaction) or = (an equilibrium) depending on how the choices below are described. Which of the following could be a correct mechanism? I. Mechanism A with the first step being a fast equilibrium and the second step being slow. II. Mechanism A with the first step being a slow forward reaction and the second step being fast. II. Mechanism B with the first step being a fast equilibrium and the second step being slow. IV. Mechanism B with the first step being a slow forward reaction and the second step being fast.
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