The Foundations of Chemistry

1.Reactant:More is consumed than is formed.

2.Product:More is formed than is consumed.

3.Reaction intermediate:Formed in earlier steps, then consumed in an equal amount

in later steps.

The gas-phase reaction of nitrogen oxide and bromine is known to be second order

in NO and first order in Br 2.

2NO(g)
B 2 (g)88n2NOBr(g) ratek[NO]^2 [Br 2 ]

A one-step collision involving two NO molecules and one Br 2 molecule would be consis-

tent with the experimentally determined rate-law expression. However, the likelihood of

all threemolecules colliding simultaneously is far less than the likelihood of two colliding.

Routes involving only bimolecular collisions or unimolecular decompositions are thought to be more

favorable in reaction mechanisms.The mechanism is believed to be

(1) NO
Br 2 34 NOBr 2 (fast, equilibrium)

(2) NOBr 2 
NO88n2NOBr (slow)

2NO
Br 2 88n2NOBr (overall)

The first step involves the collision of one NO molecule (reactant) and one Br 2 molecule

(reactant) to produce the intermediate species NOBr 2. The NOBr 2 can react rapidly,

however, to re-form NO and Br 2. We say that this is an equilibrium step.Eventually another

NO molecule (reactant) can collide with a short-lived NOBr 2 molecule and react to

produce two NOBr molecules (product).

To analyze the rate law that would be consistent with this proposed mechanism, we

again start with the slow (rate-determining) step, step 2. Denoting the rate constant for

this step as k 2 , we could express the rate of this step as

ratek 2 [NOBr 2 ][NO]

However, NOBr 2 is a reaction intermediate, so its concentration at the beginning of the

second step may not be easy to measure directly. Because NOBr 2 is formed in a fast equi-

librium step, we can relate its concentration to the concentrations of the original reactants.

When a reaction or reaction step is at equilibrium,its forward (f ) and reverse (r) rates are

equal.

rate1frate1r

Because this is an elementary step, we can write the rate expression for both directions

from the equation for the elementary step

k1f[NO][Br 2 ]k1r[NOBr 2 ]

and then rearrange for [NOBr 2 ].

[NOBr 2 ]

k

k

1

1

r

f[NO][Br

2 ]

When we substitute the right side of this equation for [NOBr 2 ] in the rate expression for

the rate-determining step, ratek 2 [NOBr 2 ][NO], we arrive at the experimentally deter-

mined rate-law expression.

ratek 2 


k

k

1

1

r

f

[NO][Br 2 ][NO] or ratek[NO]^2 [Br 2 ]

Think how unlikely it is for three
moving billiard balls to collide
simultaneously.

Any fast step that precedes a slow step
reaches equilibrium.

682 CHAPTER 16: Chemical Kinetics

The product and quotient of constants
k 2 , k1f, and k1ris another constant, k.

The rate-law expression of step 2 (the
rate-determining step) determines the
rate law for the overall reaction.