Unit 3: Kinetics
ANTHONY GATES
Chemical Kinetics
 The
rate of reaction is defined as the amount of
reactant converted into product over time
 The
change in the concentration of a reactant or
product over a unit of time.
Rates
[ A] at time t 2  [ A] at time t1
Rate 
t 2  t1
[ A]
Rate 
t
Reaction Rates…

Can measure disappearance of reactants

Can measure appearance of products

Are proportional stoichiometrically

Are equal to the slope of the tangent of
any given point

Change as the reaction proceeds, if the
rate is dependent upon concentration
[ NO2 ]
 constant
t
Rate Laws
 Differential
rate law expresses how
the rate depends on concentration.
Typically
called “the rate law.”
 Integrated
rate law expresses how
the concentrations depend on time.
The Rate Law
∆[𝑁𝑂2 ]
𝑅𝑎𝑡𝑒 =
= 𝑘[𝑁𝑂2 ]𝑛
∆𝑡
 k is a proportionality constant called the rate constant.


n is the order of the reactant
 Has
to be a whole number
 The
sum of all of the powers of the reactants is the rate order.
Both k and n must be determined by experiment.
Rate Constant

Important measurable quantity that characterizes the
reaction.

Vary over many order of magnitudes because
reaction rates vary widely
Method of Initial Rates

The initial rate of a reaction is the instantaneous rate determined
just after the reaction begins.

Since the concentration of a product will influence the rate of a
reaction, calculating the initial rate of reaction before much of
the reactant has changed into product, allows us to circumvent
the impact the concentration of the product will have on the rate
of reaction, when calculating various rates in experiments.

Investigation of data for initial rates enables prediction of how
concentration will vary as the reaction progresses.
Writing a (differential) Rate Law

Problem - Write the rate law, determine the value of the
rate constant, k, and the overall order for the following
reaction:
2 NO(g) + Cl2(g)  2 NOCl(g)
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
11.4 x 10-6
Writing a Rate Law

Part 1- Determine the values for the exponents in the rate law:
R = k[NO]x [Cl2]y

Looking at Experiment 1 and 2, [Cl2] is constant and as [NO]
doubles, the rate quadruples. So the rate is second order with
respect to [NO].
Quick math
0.250𝑚𝑜𝑙/𝐿 𝑛 0.250𝑚𝑜𝑙 𝑚
𝑘(
) (
)
1.43 𝑥 10 𝑚𝑜𝑙/𝐿𝑠
𝐿
𝐿
=
−6
0.500𝑚𝑜𝑙 𝑛 0.250𝑚𝑜𝑙 𝑚
5.72 𝑥 10 𝑚𝑜𝑙/𝐿𝑠
𝑘(
) (
)
𝐿
𝐿
−6
0.25 = (0.500)𝑛
𝑛=2
Writing the Rate Law

Part 1: Determine the values for the exponents in the rate law:
R = k[NO]x [Cl2]y

Looking at Experiment 2 and 4, [NO] is constant and as [Cl2]
doubles, the rate doubles. So the rate is first order with respect to
[Cl2].
Finishing it off…
R = k[NO]2 [Cl2]1

The rate was determined to be second order according
to [NO] and first order according to [Cl2].
1.43 x 10-6mol/Ls = k (0.250mol/L)2 (0.250mol/L)
k = 9.15 x 10-5 L2/mol2s

Since the overall rate order is equal to the sum of the
exponents, this reaction’s rate order is 3rd order.
Graphing with your calculator

L1:


L2:


0; 120; 300; 600; 1200; 1800; 2400; 3000; 3600;
1.00; 0.91; 0.78; 0.59; 0.37; 0.22; 0.13; 0.082; 0.050
L3:

1/(L2)

Graph L1 vs L2

Graph L1 vs L3
Integrated Rate Law
 Relating
concentration to time
 The
rate order (sum of the exponents) can be
determined by graphing experimental data
 Zero
 First
Order: Concentration vs. time is linear
Order: ln[Concentration] vs. time is linear
 Second
Order: 1/Concentration vs. time is linear
Relationship of the Integrated Rate Law

Concentration of A depends on time. As long
as the initial concentration of A and the rate
constant k are known, the concentration at any
time t can be found.
Zero Order … very rare circumstances
 𝑅𝑎𝑡𝑒
=
∆[𝐴]
∆𝑡
= 𝑘[𝐴]0 = 𝑘
 Integrated
 [A]=
Rate Law:
-kt + [A]0
Where
[A0]
[A] is the concentration at time t
is the initial concentration at t=0
Zero Order cont.

[A]= -kt + [A]0 is a form of y=mx+b, where…

y=[A]
m= -k
x=t
b= [A0]

A reaction is zero order if and only if [A] vs. t
produces a straight line.
First Order
 𝑅𝑎𝑡𝑒

=
= 𝑘[𝐴]
Integrated rate law:
 ln[A]

∆[𝐴]
∆𝑡
= -kt + ln[A0]
ln[A] = -kt + ln[A0] is a form of y=mx+b, where…

y= ln[A]
m=-k
x=t
b= ln[A0]

A reaction is first order if and only if ln[A] vs. t produces a
straight line.
First Order cont.
 This
law can also be expressed as
ln
𝐴0
𝐴
= 𝑘𝑡
Second Order
∆[𝐴]
∆𝑡
= 𝑘[𝐴]2

𝑅𝑎𝑡𝑒 =

Integrated Rate Law:


𝟏
𝑨
= 𝒌𝒕 +
1
𝐴0
𝟏
𝑨𝟎
is a form of y=mx+b, where…
1
𝐴
= 𝑘𝑡 +

y= 1/[A]

A reaction is second order if and only if 1/[A] vs. t produces a straight
line.
m=k
x=t
b= 1/[A0]
Solving an Integrated Rate Law
Time (s)
[H2O2] (mol/L)
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Problem: Find the
integrated rate law
and the value for the
rate constant, k
Time vs. [H2O2]
Time (s)
[H2O2]
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Time vs. ln[H2O2]
Time (s)
ln[H2O2]
0
0
120
-0.0943
300
-0.2485
600
-0.5276
1200
-0.9943
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Time vs. 1/[H2O2]
Time (s)
1/[H2O2]
0
1.00
120
1.0989
300
1.2821
600
1.6949
1200
2.7027
1800
4.5455
2400
7.6923
3000
12.195
3600
20.000
And the winner is… Time vs. ln[H2O2]
1. As a result, the reaction is 1st order
2. The (differential) rate law is:
R  k[ H 2O2 ]
3. The integrated rate law is:
ln[ H 2O2 ]   kt  ln[ H 2 02 ]0
4. But…what is the rate constant, k ?
Finding the Rate Constant, k
Method #1: Calculate the slope from the
Time vs. ln[H2O2] table.
 ln[ H 2O2 ]  2.996
slope 

t
3600 s
4 1
slope   8.32 x10 s
Now remember:
ln[ H 2O2 ]   kt  ln[ H 2 02 ]0
k = -slope
k = 8.32 x
10-4s-1
Time (s)
ln[H2O2]
0
0
120
-0.0943
300
-0.2485
600
-0.5276
1200
-0.9943
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Finding the Rate Constant, k
Method #2: Obtain k from the linear regression analysis.
4
slope   8.32 x10 s
1
Now remember:
ln[ H 2O2 ]   kt  ln[ H 2 02 ]0
k = -slope
k = 8.32 x 10-4s-1
Half Life

The time required for a
reactant to reach half of
its original concentration.

It does not matter what
the original amount is…

Half life is specific to each
reactant.
Half Life Cont.



Zero Order:
First Order:
𝑡1/2 =
𝐴0
2𝑘
𝑡1/2 =
0.693
𝑘
Second Order:
𝑡1/2 =
1
𝑘𝐴0
Pseudo-first order rate law

If there are multiple reactants, you can write the first
order rate law with respect to the concentration of one
of the reactants and assume the others to be constants
with values equal to their initial concentrations.

The rate constant (k’) for the pseudo-first order rated
law is not the rate constant (k) for the entire reaction,
but solely with respect to the chosen reactant.
Pseudo cont.
Rate = k [A]n [B]m [C]p

Can be reduced to:
Rate = k’ [A]n

To find the rate constant for the entire reaction:
 k’
= k [B]0m[C]0p
 When
 [B]0
k’ is the rate constant for the pseudo-first order rate law
and [C]0 are any reactants not chosen.
Rate Laws Summary
Zero Order
First Order
Second Order
Rate = k
Rate = k[A]
Rate = k[A]2
Integrated
Rate Law
[A] = -kt + [A]0
ln[A] = -kt + ln[A]0
1
1
 kt 
[ A]
[ A]0
Plot the
produces a
straight line
[A] versus t
ln[A] versus t
1
versus t
[ A]
Slope = -k
Slope = -k
Slope = k
0.693
t1/ 2 
k
1
t1/ 2 
k [ A]0
Rate Law
Relationship of
rate constant
to slope of
straight line
Half-Life
[ A]0
t1/ 2 
2k
Bell Work

Sketch a straight line graph for zero order,
first order and second order.

Label the axis appropriately

For each graph, relate the slope to “k”
Pre-lecture task complete while waiting for the bell!

Write down a description of the
path you take to get from your class
to lunch and what parts of your
path cause you to go faster or
slower.

List any issues you can think of that
prevents you from getting to eat
your lunch sooner (ex. lines, hallway
traffic, maze-like path, etc.)
Learning Objective

I can describe the relationship between a
mechanism and its rate law.
Reaction Mechanisms

A + 3B  AB3
Rate=k[A][B]2

A “reaction mechanism” is a series of steps by which the reaction occurs, and the steps can
occur at different rates:

Step 1
A + 2B AB2
(Slow)

Step 2
AB2 + B  AB3
(Fast)

Overall
A + 3B  AB3
(must add up to overall reaction)

Each step is called an “elementary step”

AB2 is an “intermediate” – a substance that is part of the mechanism but doesn’t show up in the
balanced equation
Reaction Mechanism cont.

A + 3B  AB3
Rate=k[A][B]2

Step 1
A + 2B AB2
(Slow)

Step 2
AB2 + B  AB3
(Fast)

Overall
A + 3B  AB3

The rate law for an elementary step can be written from the coefficients
in the elementary step

For Step 1 the rate law is:
Rate=k[A][B]2

For Step 2 the rate law is:
Rate=k[AB2][B]
A little something to consider…

Key Concept – the slow step is the rate
determining step
 The
rate of the slow step is the rate of the overall
reaction
Let’s try one out!

Step 1
Step 2
Step 3

What would the overall reaction be (balanced equation)?

Overall

What would the rate law be??

Rate = k[A]2

What are A2, X and A2X?

Intermediates!!

Note – intermediates do not appear in the rate law


2AA2
A2 + XA2X
A2X + B  A2B + X
(SLOW!)
(Fast!)
(Fast!)
2A + B  A2B
Your turn!!

For the reaction 3X+2YX3Y2 the rate law is Rate=k[X]2[Y]

The following mechanism is proposed:

Step 1:
X+YXY

Step 2:
2X+YX2Y

Step 3:
XY+X2YX3Y2

Which is the rate limiting step?

Step 2

What is (are) the intermediates?

XY and X2Y
Try this on for size!!

For the reaction 2A + 2B A2B2 the rate law is Rate=k[A]2

Two proposed mechanisms are:

Mechanism 1: Step 1
2A + B  A2B
Slow
Step 2
A2B + BA2B2
Fast
2A A2
Slow
Step 2
A2+BA2B
Fast
Step 3
A2B+BA2B2
Fast


Mechanism 2: Step 1
Which mechanism is consistent with the rate law?
For a hypothetical chemical reaction that has the stoichiometry 2 X + Y  Z, the following initial rate data were
obtained. All measurements were made at the same temperature.
Initial Rate of
Formation of Z, Initial [X]o, Initial [Y]o,
(mol.L-1)
(mol.L-1)
(mol.L-1.sec-1)
7.010-4
0.20
0.10
-3
1.410
0.40
0.20
2.810-3
0.40
0.40
-3
4.210
0.60
0.60
(a) Give the rate law for this reaction from the data above.
(b) Calculate the specific rate constant for this reaction and specify its units.
(c) How long must the reaction proceed to produce a concentration of Z equal to 0.20 molar, if the initial reaction
concentrations are [X]o = 0.80 molar, [Y]o = 0.60 molar and [Z]0 = 0 molar?
(d) Select from the mechanisms below the one most consistent with the observed data, and explain your choice.
In these mechanisms M and N are reaction intermediates.
(1) X + Y  M
(slow)
X+MZ
(fast)
(2) X + X  M
(fast)
Y+MZ
(slow)
(3) Y  M
(slow)
M+XN
(fast)
N+XZ
(fast)
Factors that Affect Reaction Rates

Temperature

Concentration (except in zero order rxns)

Surface Area

Other environmental factors


Presence of other chemicals
https://www.youtube.com/watch?v=OttRV5ykP7A
Temperature

Generally as the
temperature
increases, so does
the reaction rate.

The reaction
constant (k) is
temperature
dependent.
Concentration & Surface Area
Other than a zero order… as
you increase the
concentration, the rate
should increase.
 Increasing the amount of
reactant present increases
the likelihood of two
reactants colliding and
reacting.



Increasing the surface area
should increase the reaction
rate.
Increasing the amount of
surface area present will
allow for more of a reactant
to make collisions without
being shielded by more of
that reactant.
Collision Theory/Model

In chemical reactions bonds are either broken or formed.

Molecules must collide in order to react.


More Collisions More Reactions
Qualitatively explains the order to elementary reactions and the temperature
dependence of the rate constant.
Collision Theory/Model

Molecules must collide with the correct orientation in space
and with enough energy in order to cause the original bonds
to break and for new bonds to form.
Activation Energy

The minimum amount of energy required for a reaction to occur is
called the activation energy, Ea

Just as a ball cannot go over a hill if it does not have sufficient
energy, a reaction cannot occur unless the molecules possess
sufficient energy to get over the activation energy barrier.
Correct Orientation and Energy
2NOCl2NO + Cl2
Reaction Coordinate Diagrams

We can see how
the energy changes
throughout a
process on a
reaction coordinate
diagram.
Reaction Coordinate Diagrams

Shows the change in energy
throughout the reaction (ΔE).

The species present at the local
maximum is called a transition
state.

The energy gap between the
initial energy and that of the
transition state is the activation
energy.
Unimolecular Reactions

Unimolecular reactions occur because collisions with solvent
or background molecules activate the molecule in a way that
can be understood in terms of a Maxwell-Boltzmann thermal
distribution of particle energies.
Maxwell Boltzmann Diagrams



Another way to “see” how
energy (temperature) affects a
reaction rate.
Temperature is defined as a
measure of the average kinetic
energy of the molecules in a
sample.
At any temperature there is a
wide distribution of kinetic
energies.
Maxwell Boltzmann Diagrams

As the temperature increases,
the curve flattens and
broadens.

Thus at higher temperatures,
a larger population of
molecules has higher energy.
Maxwell Boltzmann Diagrams

If the dotted line represents the
activation energy, as the
temperature increases, so does
the fraction of molecules that
can overcome the activation
energy barrier.

As a result, the reaction rate
increases.
Maxwell Boltzmann Diagrams

The fraction of the molecules present in a
gas which have energies equal to or in
excess of the activation energy at a
particular temperature can be found
through the expression:

where R is the gas constant and T is the
temperature in Kelvin .
Arrhenius Equation

Svante Arrhenius developed a mathematical relationship
between k and Ea:

where A is the frequency factor, a number that represents the
likelihood that collisions would occur with the proper orientation
for reaction.

reaction
= the probability that any given collision will result in a
Arrhenius Equation

Taking the natural logarithm of both sides, the equation becomes…
y = mx + b
Arrhenius Equation

k = number of collisions that result in a reaction per
second

A = the total number of collisions (leading to a reaction
or not) per second
Arrhenius Equation

It can be seen that either increasing the temperature or
decreasing the activation energy (for example through
the use of catalysts) will result in an increase in rate of
reaction.
Thought provoking ideas…
One generalization derived from
the Arrhenius' equation is that, for
many common chemical
reactions at room temperature, a
10 degree Celsius increase in
temperature doubles the reaction
rate.
Catalyst

Catalysts can stabilize the transition
state, lowering the activation energy
and thus increasing the reaction
rate.

Catalysts can increase the reaction
rate by participating in the formation
of a new reaction intermediate,
thereby providing a new reaction
pathway or mechanism.

Catalysts are present throughout the
entire reaction and do not change
concentration.
Catalysis: types of catalysts

In acid-base catalysis, a reactant either gains or loses a proton;
this changes the rate of the reaction.

In surface catalysis, either a new reaction intermediate is
formed, or the probability of successful collisions is modified.

Some enzymes accelerate reactions by binding to the reactants
in a way that lowers the activation energy.

Other enzymes react with the reactant species to form a new
reaction intermediate.
Enzymes

Enzymes are catalysts in
biological systems.

The substrate fits into the
active site of the enzyme
much like a key fits into a
lock.