 Zumdahl Chapter 12 Section 12.6
 What is a reaction mechanism?
 Define reaction mechanism, elementary reaction, intermediate,
 molecularity, and rate-determining step.
Evaluate simple reaction mechanisms.
In chemistry, we sometimes find that the overall reaction does not provide us with accurate information about the the kinetics of a reaction. Thus, we may need of way to study a reaction in greater detail than just its reactants and products. Reaction mechanisms act as tools for this by allowing us to split an overall reaction into a series of intermediate reactions. These intermediate steps can then be examined individually and can collectively tell us much about the properties of the overall reactions that we see. The study of reaction mechanisms is an important field in chemistry that can tell us much about the kinetics of a reaction.
A reaction mechanism is an exact step-by-step description of what occurs molecularly in a given chemical reaction. Each step is called an elementary reaction . These elementary steps are the basic building blocks of a complex reaction and cannot be broken down any further. The set of elementary reactions whose overall effect is given by the net chemical equation is called the reaction mechanism .

Consider the reaction, CO(g) + NO
2
(g) → CO
2
(g) + NO(g). A look at this overall reaction would suggest that the carbon monoxide reacts directly with nitrogen dioxide to form the products. However, its reaction mechanism shows this is not the case. This reaction is known to have the following mechanism.
1) 2NO
2
(g) → NO
3
(g) + NO(g)
2) CO(g) + NO
3
(g) → CO
2
(g) + NO
2
(g)
Overall Net) CO(g) + NO
2
(g) → CO
2
(g) + NO(g)
According to the reaction mechanism above, 2NO
2 collide and react to give the product molecule NO and the reaction intermediate NO
3
. A reaction intermediate is a species produced during a reaction that does not appear in the net equation. Often reaction intermediates are short-lived and cannot be isolated from the reaction mixtures. The overall reaction, is obtained by adding the steps together and canceling intermediates.
Example: Writing the overall chemical equation from a mechanism
Obtain the net, or overall, chemical equation from the following reaction mechanism.
CH
4
(g) + Cl
2
(g) → CH
3
(g) + HCl (g)
CH
3
(g) + Cl
2
(g) → CH
3
Cl (g) + Cl (g)
By looking at the elementary reactions, we can see that CH
3
appears on both sides. Therefore, it must be a reaction intermediate. Canceling this species will give us the net reaction.
CH
4
(g) + 2Cl
2
(g) → CH
3
Cl (g) + HCl (g) + Cl -
Elementary reactions are classified according to their molecularity. The molecularity is the number of molecules on the reactant side of an elementary reaction.
Elementary Step Molecularity
A → product unimolecular
2A → product
A + B → product
2A + B → product
bimolecular
bimolecular
termolecular

3A → product
A + B + C → product
termolecular
termolecular
Higher molecularities are not encountered. As an example, lets consider again the following reaction mechanism,
1) 2NO
2
(g) → NO
3
(g) + NO(g)
2) CO(g) + NO
3
(g) → CO
2
(g) + NO
2
(g)
Overall Net) CO(g) + NO
2
(g) → CO
2
(g) + NO(g)
The 1st elementary reaction is unimolecular while the second is bimolecular.
There is no simple relationship between the overall order of a reaction and the rate law. As we stressed before, the rate law must be obtained experimentally.
However, when you are dealing with an elementary reaction, the rate does have a simple predicable form. Unlike the overall equation, the rate law for each elementary step can be deduced from the chemical equation through inspection. A unimolecular elementary step has a first order rate law, whereas a bimolecular elementary step has a second order rate law. The table below summarizes the types of elementary steps and the rate laws that they follow.
A, B, and C here represent the reactants or reaction intermediates.
Elementary Steps and Rate Laws
Molecularity Elementary Step Rate Law for Elementary Step
Unimolecular rate = k [A]
A products
Bimolecular rate = k [A][B]
A + B products rate = k [A] 2
A + A products
Termolecular rate = k [A] 2 [B]
A + A + B products rate = k [A] 3
A + A + A products rate = k [A][B][C]
A + B + C products
Remember, this only applies to the elementary reaction. It doesn't work for the

overall equation, because it is the sum of the elementary reactions!
Example: Writing the rate equation for an elementary reaction
Write the rate equations for each of the following elementary reactions. a) Ozone is converted to O
2
by NO in a single step, O
3
+ NO → O
2
+ NO
2
. b) The recombination of iodine atoms occurs as follows, I + I + M → I
2
+ M*
(*M is some atom or molecule that absorbs energy from the reaction. c) An H
2
O molecule absorbs energy; enough of this energy flows into one
O-H atom to break it, H
2
O → H + O-H
Solution: Unlike the overall equation, for elementary reactions the rate equation can be written directly from the elementary reaction (but only for an elementary reaction). Obviously, I can't stress this enough! Therefor, a) Rate = k [O
3
][NO] b) Rate = k [I] 2 [M] c) Rate = k [H
2
O]
The mechanism for a reaction cannot be observed directly. A mechanism is devised to explain the experimental observations. It is like the explanation provided by a detective to explain a crime in terms of the clues found. Other explanations may be possible, and further clues may make one of the other explanations seem more plausible than the currently accepted one. So it is with reaction mechanisms. They are accepted provisionally, with the understanding that further experiments may lead you to accept another mechanism as the more probable explanation.
A plausible mechanism must satisfy the following two requirements:
1.
2.
The elementary steps must add up to give the overall balanced equation for the reaction.
The rate law for the rate-determining step must agree with the experimentally determined rate law.
An important clue in finding the reaction mechanism is the rate law. The reason for its importance is that once you assume a mechanism, you can predict the rate law. If this prediction does not agree with the experimental rate law, the assumed mechanism must be wrong. For example, consider the

overall equation
2NO
2
(g) + F
2
(g) → 2NO
2
F(g) (overall reaction)
If you follow the rate of disappearance of F
2
, you observe that it is directly proportionally to the concentration of both NO
2 and F
2
. Thus, the experimentally determined rate law is Rate = k [NO
2
][F
2
]. This rate law is a summary of the experimental data. Assume that the reaction occurs in a single elementary reaction.
NO
2
+ NO
2
+ F
2
→ NO
2
F + NO
2
F (elementary reaction)
This, then, is your assumed mechanism. Because we can write the rate law from an elementary reaction, you would predict that the rate law is Rate
= k [NO
2
] 2 [F
2
]. However, this does not agree with the experiment rate law, and your assumed mechanism must be discarded. You must therefore conclude that the reaction occurs in more than one step.
Rate-Determining Step
The reaction above is actually believed to occur in the following steps
1) NO
2
+ F
2
→ NO
2
F + F ( k
1
, slow step)
2) F + NO
2
→ NO
2
F ( k
2
, fast step)
One of the requirements of for a reaction mechanism is that the net results must be equivalent to the overall reaction. By adding the two equations above, you can see this is the case.
1) NO
2
+ F
2
→ NO
2
F + F ( k
1
, slow step)
2) F + NO
2
→ NO
2
F ( k
2
, fast step)
2NO
2
+ F
2
→ 2NO
2
F
The F atom is an intermediate.
The second requirement for a mechanism is that it must agree with the experimental rate law. Lets take a look at that. Note that the second step is assumed to be much faster than the first, so that as soon as the NO
2
and

the F
2
react, the F atom that is formed reacts with another NO
2
molecule to give another NO
2
F molecule. Therefore, the rate of disappearance of F
2
(and therefore the rate law) is determined completely by the slow step, or the rate-determining step. The rate-determining step is the slowest step in the reaction mechanism. The rate-determining step in the reaction above is the fist step in the mechanism. The rate equation for this rate-determining step of the mechanism is Rate = k [NO
2
][F
2
]. This should equal the experimental rate law, which it does.
Example: Determining the rate law from a mechanism with an initial slow step
Consider the reaction mechanism below. What is the rate law predicted by this mechanism?
1) slow reaction: H
2
+ ICl
→
HI + HCl
Rate(1): k
1
[H
2
] [ICl]
2) fast reaction: HI + ICl
→
I
2
+ HCl
Rate(2): k
2
[HI] [ICl]
Overall reaction: H
2
+ 2 ICl
→
I
2
+ 2 HCl
Solution: The first step in the mechanism is rate determining, because its rate is much slower than the second step. We can write the rate law directly from this step.
Rate = k [ H
2
][ICl]
HOMEWORK: Book questions page 567 questions
11,21,23,25,27,29,31,33,39,49,51,52
Practice exercises 14.11 – 14.15