Chapter 12
Chemical Kinetics
Honors Chemistry 2
How fast or slow is a reaction?
Let’s look at the reaction
 CO(g) + NO2 (g) → CO2 (g) + NO (g)
 At the start, there are only reactants and
no products. As times goes on, there are
less reactants and more products. In the
end, only product remains.
 OK, but how fast did the reaction occur?

Factors that affect reaction rate.
Spontaneity isn’t one of them, don’t
confuse spontaneous and instantaneous.
 The nature of reactants.
 Concentration
 Surface Area
 Temperature
 Presence of a catalyst.

Reaction Rate
The average rate of a reaction is
 quantity = mol/L =  concentration
 time
s
 time

We use brackets [ ] to denote
concentration in mol/L or M.
 Average reaction rate = [NO]t2 - [NO]t1
t2 - t1

Reaction Rate Laws
The equation that represents the
relationship between the rate and the
concentration of reactants is called the
rate law.
 For the reaction A → B, the rate law is:
Rate = k [A], where k is the specific rate
constant for that reaction. k is unique for
each reaction.

This is the differential rate law (rate law) – rate
of a reaction depends on concentration.
Reaction order.
Rate = k [A] is a first order reaction
because [A] = [A]1. This means that the
concentration and the rate are a direct
relationship.
 For the reaction a A + b B → products, the
general rate law is: Rate = k [A]m [B]n,
where m and n are the reaction orders of A
and B. Rarely does m=a or n=b. The rate
order is m + n.

Determining the rate order.
 Rate
laws are determined
experimentally, one method uses
initial rates. This is done by
comparing the initial rates of a
reaction with varying reactant
concentration. See page 535.
 You can graph them.
Determining the Reaction Order
For Experiment 1:
Rate = 5.4 x 10-7 mol/L*s = k(0.0100 mol/L)n (0.200 mol/L)m
For Experiment 2:
Rate = 10.8 x 10-7 mol/L*s = k (0.0200 mol/L)n (0.200 mol/L)m
Rate 2 = 10.8 x 10-7 mol/L*s = k(0.0200 mol/L)n (0.200 mol/L)m
Rate 1
5.4 x 10-7 mol/L*s
= (0.0200 mol/L) = (2.0)n
(0.0100 mol/L)
Rate 2 = 2.00 = (2.0)
Rate 1
n=1
k(0.0100 mol/L)n (0.200 mol/L)m
Determining the Reaction Order,
cont'd
Similarly,
Rate 5 = 10.8 x 10-7 mol/L*s = k(0.200mol/L)n (0.0202mol/L)m
Rate 6
21.6 x 10-7 mol/L*s = k(0.200mol/L)n (0.0404mol/L)m
=0.0202 mol/L = (0.50)m
0.0404 mol/L
Rate 5 = 0.500 = (0.50)m
Rate 6
m=1
Rate = k[NH4+ ] [NO2- ]
5.4 x 10-7 mol/L*s = k(0.0100 mol/L) (0.200 mol/L)
k = 2.7 x 10-4 L/mol*s
The Integrated Rate Law
The integrated rate law shows the
concentration of species in the reaction
depend on time.
 For a chemical reaction
aA products
where the kinetics are first order in [A]
 In[A] = -kt + ln[A]º or ln([A]º) = kt
y = mx + b
([A])
 The reaction is first order in A if a plot of
ln[A] is a straight line.

Half-Life
Half-life of a reactant is the time it takes for
a reactant to reach half its original
concentration.
 For a first-order reaction,
t1/2 = 0.693
k

Second-Order Rate Laws
For a reaction
aA → products
 That is second order in A, the integrated
second-order rate law is
1 = kt + 1
[A]
[A]º
y = mx + b
 t1/2 = 1
k[A]º

Zero-Order Rate Laws
A reaction is said to be zero-order if the rate
is constant.
 The integrated rate law for a zero-order
reaction is
[A] = -kt + [A]º
 t1/2 = [A]º
2k

Instantaneous Reaction Rates.
It sometimes is necessary to know the
rate at any given moment, this is the
instantaneous rate.
 One way is to determine the slope of the
tangent to the curve of the reaction at the
given time.
 Another way is to use the rate law and
determine the rate at the given
concentration of interest.

Graph ln[ ]
1st order
Graph 1/[ ]
2nd order
Temperature
 In
general, increasing temperature
increases the rate of reaction. The
molecules are moving faster and the
number of collisions increases.
 In general, increasing the
temperature by 10 K approximately
doubles the rate of reaction.
Collision Theory
In order for a reactants to come together to
form products, collisions under the right
conditions have to occur.
 If the orientation of the collision is correct
an activated complex or intermediate will
form.
 If the collision has sufficient energy
(activation energy) then the product is
formed.

Arrhenius Equation

For reactants to collide successfully:
The collision must involve enough energy to
produce the reaction; the collision energy must
equal or exceed the activation energy.
 The relative orientation of the reactants must
allow formation of any new bonds necessary to
produce products.



ln(k) = -Ea (1/T) + ln(A)
R
m x + b
y =
k = rate constant; E = activation energy; R = gas law constant, 8.3145
J/K*mol; T = temperature; A = frequency factor
Arrhenius Equation

The graph of Arrhenius equation yields a
straight line of slope= -Ea /R and intercept =
ln(A)
To find Ea , plot ln(k) vs 1/T and calculate the
slope.
 See example page 555


Ea can also be calculated from the values of
k at only two temperatures by
ln (k2/k1 ) = Ea (1/T1 – 1/T2)
R
Reaction energy diagram.
activated complex
Energy
activation Energy
reactants
energy released
by reaction
products
Reaction Progress
Nature of the Reactants
 Some
substances, like halogens
and group 1 and 2 elements, are
more reactive than others. The
more reactive a substance, the
faster the reaction occurs.
Concentration
 In
order for a reaction to occur
collisions must happen. Obviously,
more particles mean more collisions
and this means that the reaction
occurs faster.
 However, too many particles of one
kind tend to get in the way of each
other and the reaction slows.
Surface Area
 Increasing
the surface area allows the
number of collisions between
reactants to increase. As with
dissolving rates, by grinding,
pulverizing, or vaporizing the
reactants increases the rate of
reaction.
Reaction Mechanisms



Most reactions consist of a sequence of steps.
Each of these is called an elementary step.
Several elementary steps make a complex
reaction.
The complete sequence of steps in a complex
reaction is a reaction mechanism.
The product of the first elementary step of a
complex reaction is an intermediate, like
catalysts, these don’t appear in the balanced
equation.
Rate determining step.
 In
a complex reaction the
elementary step that is the
slowest determines the rate of the
overall reaction. This is the rate
determining step. The slowest
step always determines the rate.
Catalyst or Inhibitor
A catalyst is a substance that increases
the rate of reaction without actually taking
part in the reaction (it remains
unchanged). It does this by lowering the
activation energy and thus speeds the
formation of intermediate products.
 An inhibitor raises the activation energy
and slows the reaction down.

The effect of a catalyst.
 Catalysts
don’t change
equilibrium since they only lower
the activation energy. This
speeds up the reaction rate
equally in both directions and
doesn’t provide a stress to either
reactants or products.