Chapter 14 Chemical Kinetics
AP CHEMISTRY
14.1 Chemical Kinetics
 Study of how rapidly a reaction will occur. In
addition to speed of reaction, kinetics also deals with
the reaction mechanism (exactly how the reaction
occurs)
1. Factors that affect reaction rate
1.
2.
3.
4.
5.
Concentration of reactants – more particles means more
frequent “effective” collisions occur
Temperature – higher temps, the molecules have more KE,
are moving faster and are more likely to collide
Catalyst – chemical that speeds up a reaction by changing
the reaction mechanism, not consumed in reaction
Physical state of reactants – (phase)
Surface area (for solids) – greater SA = faster reaction
14.2 Reaction Rate
 The change in concentration of reactants or products
over time.

The rate can be expressed as a rate of decomposition of
reactants or rate of production of products.
 What is the average rate of decomposition of “A” in the 1st
20 seconds of the reaction? In the 2nd 20 seconds?
 Rate = Rate =-
∆𝑚𝑜𝑙𝑒 𝐴
−0.46 𝑚𝑜𝑙
== 0.023
𝑡
20 𝑠𝑒𝑐
−0.24 𝑚𝑜𝑙
= 0.012 mol/sec
20 𝑠𝑒𝑐
mol/sec
 What is the average rate of appearance of “B” in the
1st 20 seconds of the reaction? In the 2nd 2o seconds?
∆𝑚𝑜𝑙𝑒 𝐵 0.46 𝑚𝑜𝑙
 Rate=
=
= 0.023 mol/sec
𝑡
20 𝑠𝑒𝑐
0.24 𝑚𝑜𝑙
 Rate =
= -0.012 mol/sec
20 𝑠𝑒𝑐
 Chemical Equation showing mol ratio of A changing
into B: 1A → 1B
Mathematical Expressions for Reactions
 Reaction: N2(g) + 3 H2(g) → 2NH3(g)
***If you want to compare the rates using different
substances, you use the mole ratios
 The rate of reaction is always positive, and is usually
measured in M/s

Under certain conditions, the rate of formation of NH3 is 0.28
M/s. What is the rate of change for N2?

0.28 M NH3
s
1 mol N2
2 mol NH3
= 0.14 M/s
 Indicate how the rate of disappearance of each reactant is
related to the rate of appearance of each product:
1.
H2O2(g) → H2(g) + O2(g)
∆[𝐻2𝑂2]
∆[H2] ∆[O2]
=
=
∆𝑡
∆t
∆t
2.
2SO2(g) + O2(g) → 2SO3(g)
1 ∆[SO2]
∆[O2]
1 ∆[SO3]
(= =
)2
2
∆t
∆t
2
∆t
- ∆[SO2]/∆t = -2∆[O2]/ ∆t = ∆[SO3]/∆t
*SO2 will disappear and SO3 will appear at twice the rate of O2
Average Rate of Change
Avg rate =
- ∆[C4H9Cl]
∆t
Avg rate =
- 0.0905 – 0.1000
50 – 0
= 1.9 x 10-4 M/s
Instantaneous Rate of Change
 The rate of reaction at a given moment in time
 The instantaneous rate can be determined by finding
the slope of a tangent line at that point in time.
All reactions slow down over
time, therefore, the best indicator
of the rate of a reaction is the
instantaneous rate near the
beginning of the reaction.
Slope (rate at t = 0)
𝑦 −𝑦
0.060𝑀 −0.100𝑀
= 2 1 =
𝑥2−𝑥1
210𝑠 −0𝑠
= 1.9 x 10-4 M/s
14.3 Concentration and Rate Laws
 One can gain information about the rate of a rxn by
seeing how the rate changes with changes in
concentration
 The rate of rxn generally decreases as the
concentration of reactant decreases
Determining Rate Laws
 Rate Law- shows how the rate depends on the
concentration of reactants


Must be determined from experimental data
Rate depends only on the concentration of the reactants
2N2O5 → 4 NO2 + O2
 Rate = k[N2O5]m



This is called the rate law
k is called the rate constant (depends on temperature and
solvent used)
m is the order of the reactant – usually a positive integer (not
necessarily the same as the coefficient)
General Terms
 For a general reaction,
aA + bB → cC + dD
 The rate law has the form,
Rate = k [A]m[B]n
 Example: NH4+ + NO2- → N2 + 2H2O
 Rate = k[NH4+]m[NO2-]n
 m and n must be determined experimentally.
Reaction Orders
 The exponents m and n are called reaction orders.
 If the exponent is one, the rate is first order with
respect to that reactant
 If the exponent is two, the rate is second order with
respect to that reactant
 The overall reaction order is the sum of the orders
with respect to each reactant
 Example: Rate = k[A]2[B]1

The rate is 2nd order with respect to [A], and first order with
respect to [B]. The overall reaction order is 3.
Order versus Concentration
 The rate of a zero order is independent of [A].
 Doubling the concentration multiplies the rate of the rxn by 1.
 Tripling the concentration multiplies the rate of the rxn by 1.
 In first order the rate is directly proportional to
the [A].


Doubling the concentration multiplies the rate of the rxn by 2.
Tripling the concentration multiplies the rate of the rxn by 3.
 In second order the rate is proportional to the
square of the [A].


Doubling the concentration multiplies the rate of the rxn by 4.
Tripling the concentration multiplies the rate of the rxn by 9.
Units of Rate Constants
 Units of the rate constant depend on the overall
order of the rate law.
 First order overall: Rate = k[A] or Rate = k[A][B]0

Units of rate = (unit of rate constant)(unit of concentration)1

Units of rate constant =
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑟𝑎𝑡𝑒
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
1 =
𝑀/𝑠
𝑀
= s-1
 Second order overall: Rate = k[A]2 or Rate = k[A][B]
 Units of rate = (unit of rate constant)(units of concentration)2

Unit of rate constant =
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑟𝑎𝑡𝑒
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
2
=
𝑀/𝑠
𝑀2
= M-1s-1
 Consider the hypothetical equation and data:
A + B→C + D
#
[A], M
[B], M
Rate of formation of C
(M/s)
1
0.200
0.200
0.0025
2
0.400
0.200
0.010
3
0.400
0.400
0.010
 Calculate the order of the reaction with respect to each reactant

Second order with respect to [A] and Zero order with respect to [B]
 What is the overall order of the reaction

2nd order overall
 Write the rate law for the reaction.

Rate = k [A]2[B]0
 Calculate the value of k

K = 0.0625 M-1s-1
Example
 The following data were measured for the reaction of nitric
oxide with hydrogen:
2NO(g) + 2H2(g)→ N2(g) + 2H2O(g)
Exp #
[NO] (M)
[H2] (M)
Initial Rate
1
0.10
0.10
1.23 x 10-3
2
0.10
0.20
2.46 x 10-3
3
0.20
0.10
4.92 x 10-3
 Determine the rate law for this reaction.
 rate = k[NO]2[H2]
 What is the overall order of the reaction?
 2 + 1 = 3rd order overall
 Calculate the rate constant.
 k = 1.2 M-2s-1
 Calculate the rate when [NO] = 0.050M and [H2] = 0.150M
 rate = 4.5 x 10-4 M/s
Example 2
 Given the following reaction and data: 2N2O5 → 4NO2 + O2
Time (min) 0
1
2
3
[N2O5], M
0.160
0.112
0.080
0.056
rate
(M/min)
0.056
0.039
0.028
0.020
 Determine the order of the reaction, and write the rate law for the
reaction

Rate = k[N2O5]; 1st order overall
 Calculate the rate constant, k, for the reaction
 K = 0.35 s-1
 Calculate the rate if the [N2O5] is 0.098 M.
 Rate = 0.034 M/s
 Calculate the rate of appearance of O2 at time = 3

0.020 𝑀𝑁2𝑂5/ min
1 𝑚𝑜𝑙 𝑂2
2 𝑚𝑜𝑙 𝑁2𝑂5
= 0.010 M/min
14.4 Change of Concentration with Time
 A first-order reaction is one whose rate depends on
the concentration of a single reactant raised to the
first power.

Rate = k[A]
 Relating concentration of A at the start of the
reaction, [A]0, to its concentration at any other time,
[A]t can be achieved using the following equation
ln[A]t = -kt + ln[A]0
Example
 The decomposition of dimethyl ether, (CH3)2O, at
510⁰C is a first-order process with a rate constant of
6.8 x 10-4 s-1:
(CH3)2O(g) → CH4(g) + H2(g) + CO(g)
 If the initial pressure of (CH3)2O is 135 torr, what is
its partial pressure after 1420 s





lnPt = -kt + lnPo
lnPt = -(6.8 x 10-4 s-1)(1420 s) + ln(135)
lnPt = -0.9656 + 4.91
lnPt = 3.94 = e(3.94)
Pt = 51 sec
Second-Order Reactions
 A second-order reaction is one whose rate depends
on the reactant concentration raised to the second
power or on the concentrations of two different
reactants:

Rate = k[A]2 or Rate = [A][B]
 Relating concentrations at two different times can be
derived using the following equation:
1
𝐴𝑡
= kt +
1
𝐴0
Example
 Consider the decomposition of NO2 which is second
order in NO2 with k = 0.543 M-1s-1. If the initial
[NO2] in a closed vessel is 0.0500 M, what is the
concentration after 0.500 hr? (.5 hr = 1800 s)
1
𝐴𝑡
= kt +
1
𝐴0
1
𝐴𝑡
= (0.543
1
𝐴𝑡
= 997.4
M-1s-1)(1800s)
[A]t = 1.00 x 10-3 M
+
1
0.0500𝑀
1st Order Reaction Graphs
 Ln[A]t = -k t + ln[A]0
y
= mx + b
 For a first order reaction, a graph of ln[A]t vs time
will give a straight line with a slope of –k.

2nd Order Reaction Graphs
1

𝐴𝑡
1
𝐴0
= k t +
y
= mx + b
 For a second order reaction, a graph of
1
𝐴𝑡
will give a straight line with a slope of k.
vs time
Example
 Cyclopentadiene (C5H6) reacts with itself to form
dicyclopentadiene (C10H12). A 0.0400 M solution of C5H6
was monitored as a function of time as the reaction 2C5H6
C10H12 proceeded. The following data was collected:
 What is the order of the reaction?
 What is the value of the rate constant?
Time (s)
[C5H6]
(M)
0.0
0.0400
50.0
0.0300
100.0
0.0240
150.0
0.0200
200.0
0.0174
• This reaction is second order
𝑦 −𝑦
50 −25
• K = slope = 𝑥2 −𝑥 1 = 150−0
2
1
• K = 0.167 M-1s-1
Time
(s)
[C5H6]
(M)
ln[C5H6]
1/[C5H6]
0.0
0.0400
-3.219
25.0
50
0.0300
-3.507
33.3
100
0.0240
-3.730
41.7
150
0.0200
-3.912
50
200
0.0174
-4.051
57.5
-3
0
-3.2
ln[C5H6]
-3.4
-3.6
-3.8
50
100
150
1/[C5H6]
200
70
60
50
40
1/[C5H6]
30
20
-4
-4.2
10
0
0
50
100
150
200
Half-Life
 The half-life of a reaction, t1/2, is the time required
for the concentration of a reactant to drop to one half
its initial value
 Half-Life of First-Order reactions:
 T1/2 =
0.693
𝑘
(half-life not affected by initial concentration)
 Half-Life of Second-Order reactions
 T1/2 =
1
𝑘𝐴
(half-life is dependent on initial concentration)
0
Example
 The isomerization of CH3NC to CH3CN is first order. What
is the half-life if the rate constant is 2.08 x 10-4 s-1?
 T1/2 =
0.693
𝑘
=
0.693
−
2.08 𝑥 10−4 𝑠 1
= 3.33 x 103 s
14.5 Temperature and Rate
 The rates of most chemical reactions increase as
temperature increases.
 Why?
 Collision Model



Molecules must collide to react
The greater the number of collisions occurring per second, the
greater the reaction rate.
Greater concentration and higher temperature both lead to an
increased reaction rate according to the Collision Model.
Orientation Factor
 Molecules must be oriented in a certain way during
collisions in order for a reaction to occur

Atoms must be suitably positioned for bonds to break and
form new bonds
Activation Energy
 In order to react, colliding molecules must have a
total kinetic energy equal to or greater than some
minimum value
 Activation energy – the minimum energy needed to
make a reaction happen.
 Activated Complex or Transition State – The
arrangement of atoms at the top of the energy
barrier.
Activation Energy
Fraction of molecules with Ea
 Said that reaction rate should increase with
temperature.
 At high temperature more molecules have the energy
required to get over the barrier
 The fraction of molecules that has an energy equal to
or greater than the Ea is given by the expression:
f = e-Ea/Rt




E is Euler’s number (opposite of ln)
Ea is activation energy
R is gas constant, 8.314 J/mol
T is temperature is Kelvin
Magnitude of f
 Suppose that Ea is 100 kJ/mol, a typical value of
many reactions, and that T is 300K, around room
temp.
Calculate the value of the fraction of molecules that has
energy equal to or greater than Ea
 f = e-Ea/Rt
 -Ea/Rt = -100 x 103 J/mol/(8.314 J/mol)(300K)

= -40.1
 e-40.1 = 3.87 x 10-18

Arrhenius Equation
 Arrhenius found that most reaction rate data obeyed
an equation based on three factors:



The fraction of molecules possessing an energy of Ea or greater
The number of collisions occurring per second
The fraction of collisions that have the appropriate orientation
Arrhenius Equation
 He incorporated the 3 factors into his equation:
 K = Ae-Ea/RT
 K is the rate constant
 Ea is activation energy
 R is the gas constant, 8.314 kJ/mol
 T is absolute temperature
 A = frequency factor, (is constant as temperature is varied)

As the magnitude of Ea increases, k decreases because the
fraction of molecules that possess the required energy is
smaller.
Determining Activation Energy
 Reaction rate decreases as activation energy
increases
 Calculating Activation Energy:
𝐸𝑎
𝑅𝑡
lnk = -
+ ln A
If we know the rate constant at two or more
temperatures, we can determine the activation energy
as well:
𝑘1
ln
𝑘2
=
𝐸𝑎
𝑅
1
(
𝑇2
−
1
𝑇1
)
Example
 The activation energy of a certain reaction is 34.7
kJ/mol. How many times faster will the reaction
occur at 30⁰C than at 0⁰C?
 T1 = 30 + 273 = 323 K; T2 = 273 K




𝑘1
ln
𝑘2
𝑘1
ln
𝑘2
𝑘1
ln
𝑘2
𝑘1
=
𝑘2
=
=
𝐸𝑎 1
1
( - )
𝑅 𝑇2 𝑇1
34.7 𝑥 103 𝐽/𝑚𝑜𝑙 1
(
8.314 𝐽/𝑚𝑜𝑙
273
-
1
)
323
= 2.37
10.7 =11
 The reaction will occur 11 times faster at 30⁰C than at
0⁰C.
Activation Energy from Graphs
 Ea can be determined from a graph.
 If the ln k vs 1/T is plotted, it will be a line with the
slope equal to –Ea/R and a y-intercept equal to lnA
𝐸𝑎
(ln k = - + ln A)
𝑅𝑇
 Slope =
𝐸𝑎
𝑅
 Ea =-(1.07 x 104)(8.314)
= 88960 J/mol
= 89.0 kJ/mol
14.6 Reaction Mechanisms
 The process by which a reaction occurs is called a
reaction mechanism
 There is activation energy for each elementary step.
 Slowest step (rate determining) must have the
highest activation energy
 The reaction to the right
occurs in 2 steps. The
first step is the rate
determining step.
Reaction Mechanisms
 2NO2+ F2 → 2NO2F
 Rate = k[NO2][F2]
 The proposed mechanism is
NO2 + F2 → NO2F + F (slow)
F + NO2 → NO2F
(fast)
 •F is called an intermediate. It is formed then
consumed in the reaction
Molecularity
 Each of the 2 reactions is called an elementary step.
 The number of molecules that participate in an
elementary step defines the molecularity.



If a single molecule, the reaction is unimolecular
If 2 reactant molecules, the reaction is bimolecular
If 3, the reaction is termolecular (very rare)
Rate Law for Elementary Reactions
 If we know that a reaction is an elementary step,
then we know its rate law
 The rate of a unimolecular process will be first order.
 The rate of a bimolecular process will be second
order
Multi-Step Mechanisms are most common
 Most chemical reactions occur by mechanisms that
involve two or more elementary steps.
 Each step has its own rate constant and activation
energy
 The overall rate of the reaction cannot exceed the
rate of the slowest elementary step = rate
determining step.
14.7 Catalysts
 Speed up a reaction without being used up in the
reaction
 Catalysts allow reactions to follow an alternate
pathway that has a lower activation energy



Enzymes are biological catalysts
Homogeneous catalysts are in the same phase as the reactants
Heterogeneous catalysts are in a different phase as the
reactants.
 Catalysts can speed up and slow down reactions
How Catalysts work
 Catalysts allow reactions to proceed by a different
mechanisms – a new pathway
 New pathway has a lower activation energy
Examples of Catalysts
 Catalase is an enzyme in the livers of mammals that
catalyzes the decomposition of hydrogen peroxide
into water and oxygen.
 Nitrogenase is an enzyme in bacteria that live on the
roots of certain plants. It converts N2 into NH3,
which can then be used by the plants.
 Manganese dioxide will also act as a catalyst in the
decomposition of hydrogen peroxide to water and
oxygen gas.