University:
Georgia Southern University
Course:
CHEM 1310 | Comprehensive General Chemistry
Academic year:

2024
Views:

249

Pages:

2
Author:

Jalen W.

Determination of a Kinetic Mechanism for a Reaction Given the reaction: Cl2 (g) + CHCl3 (g)  HCl (g) + CCl4 (g) Experimentally determined Rate Law: Rate = k [Cl2]1/2 [CHCl3] Proposed Mechanism: k1 Cl2 (g) <-------> 2 Cl (g) FAST k-1 k2 Cl (g) + CHCl3 (g) --------> HCl (g) + CCl3 (g) k3 CCl3 (g) + Cl (g) -------> CCl4 (g) Is this mechanism acceptable? FAST SLOW Mechanism is acceptable. Reasoning for answer: Given the reaction: Cl2 (g) + CHCl3 (g)  HCl (g) + CCl4 (g) Experimentally determined Rate Law: Rate = k [Cl2]1/2 [CHCl3] FIRST CRITERIA: When the individual elementary steps are “summed” does the actual stoichiometric reaction result? Answer: Yes Reasoning: Cl2 (g)   2 Cl (g) (fast) Cl (g) + CHCl3 (g)  HCl (g) + CCl3 (g) (slow) CCl3 (g) + Cl (g)  CCl4 (g) (fast) _______________________________________________________ Cl2 (g) + Cl (g) + CHCl3 (g) + CCl3 (g) + Cl (g)  2 Cl (g) + HCl (g) + CCl3 (g) + CCl4 (g) Cl2 (g) + CHCl3 (g)  HCl(g) + CCl4 (g) which is the given reaction *Note Cl (g) and CCl3 are intermediates in this mechanism. SECOND CRITERIA: Does the slowest elementary step in the mechanism (e.g. rate determining step) give the experimentally determined rate law? Answer Yes. Reasoning: The slowest step in this reaction is the second one listed (k2). k2 Cl (g) + CHCl3 (g)  HCl (g) + CCl3 (g) SLOW If you write a rate law for this elementary step you get Rate = k2 [Cl] [CHCl3] But experimentally you have is this Rate = k [Cl2]1/2 [CHCl3] *NOTE Cl (g) is an INTERMEDIATE; intermediate cannot appear in the rate law - so you must eliminate [Cl] in the rate law - via substitution. Looking at the first step of the reaction: k1 Cl2 (g)   2 Cl (g) k-1 Solving for [Cl]2 [Cl]2 = or k1 [Cl2] = k-1 [Cl]2 k1 [Cl2] k-1 Take the square root of both sides: [Cl] = (k1 /k-1)1/2 [Cl2]1/2 Then substitute [Cl] = (k1 /k-1)1/2 [Cl2]1/ into the rate law derived in the proposed mechanism Rate law from mechanism: Rate = k2 [Cl] [CHCl3] Substitute [Cl] = (k1 /k-1)1/2 [Cl2]1/2 Substitute this into the rate law: Rate = k2 [Cl] [CHCl3] Rate = {(k2)(k1 /k-1)1/2 [Cl2]1/2}[CHCl3] All of this {(k2)(k1 /k-1)1/2} becomes “k’ So you get: Rate = k [Cl2]1/2 [CHCl3] which is the experimentally determined rate law.