Reaction rate

Topic 6:
chemical
Kinetics
6.1 COLLISION THEORY AND
RATES OF REACTIONS
Overview
Major Understandings

Species react as a result of collisions of sufficient
energy and proper orientation.

The rate of reaction is expressed as the chance in
concentration of a particular reactant/product per
unit time.

Concentration changes in a reaction can be followed
indirectly by monitoring the changes in a mass,
volume, and color.

Activation Energy is the minimum energy that colliding
molecules need in order to have successful collisions
leading to a reaction.

By decreasing Ea, a catalyst increases the rate fo a
chemical reaction, without itself being permanently
chemically changed.
We Are Here
Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
13.1
Collision Theory

Particles have to collide to react.

Have to hit hard enough

Things that increase this increase rate

High temp – faster reaction

High concentration – faster reaction

Small particles = greater surface area means
faster reaction
Kinetics

The study of reaction rates.

Spontaneous reactions are reactions that will
happen - but we can’t tell how fast.

Diamond will spontaneously turn to graphite –
eventually.

Reaction mechanism- the steps by which a
reaction takes place.
Factors that Affect
Reaction Rates
1.
The Physical State of the reactants

Reactants must come together to react

When reactants are in different phase the
reaction is limited to their area of contact
Factors that Affect
Reaction Rates
2.
The concentration of the reactants

Most chemical reactions proceed faster if the
concentration of one ore more of the reactants
is increased

As concentration increases the frequency with
which the reactant molecules collide increases,
leading to increase rates
Factors that Affect
Reaction Rates
3.
The temperature at which the reaction occurs

The rates of reaction increase as the
temperature increases

As molecules move more rapidly they collide
more frequently and also with higher energy,
leading to increased reaction rates
Factors that Affect
Reaction Rates
4.
The presence of a catalyst

Catalysts are agents that increase reaction rates
without being used up

Enzymes act as catalysts for several biochemical
reactions
Reaction Rates
The Rate of Reactions

In simple terms, the rate of a reaction can be
thought of as its speed

Some reactions are very fast

Some reactions are very slow
Defining the Rate of
Reaction

The rate of reaction is:

Where:

 [ R] [ P]
Rate 

t
t

∆[R] is change in concentration of
reactants

∆[P] is change in concentration of
products

∆t is change in time
Units: mol dm-3 s-1 (moles per
decimetre cubed per second)
A
[A]
rate = t
[B]
rate =
t
B
Reaction Rate

Change in concentration (conc) of a reactant or
product per unit time.
conc of A at time t 2 - conc of A at time t1
Rate =
t 2 - t1
[A ]

t
Chemical Kinetics
A
B
[A]
rate = t
[A] = change in concentration of A over
time period t
[B]
rate =
t
[B] = change in concentration of B over
time period t
Because [A] decreases with time, [A] is
negative.
Calculating Rates

Average rates are taken over long intervals

Instantaneous rates are determined by finding the
slope of a line tangent to the curve at any given
point because the rate can change over time

C
o
n
c
e
n
t
r
a
t
i
o
n
As the reaction progresses the concentration H2 goes
down
N2 + 3H2 → 2NH3
[H2]
Time

C
o
n
c
e
n
t
r
a
t
i
o
n
As the reaction progresses the concentration N2 goes
down 1/3 as fast
N2 + 3H2 → 2NH3
[N2]
[H2]
Time

C
o
n
c
e
n
t
r
a
t
i
o
n
As the reaction progresses the concentration NH3 goes up
2/3 times
N2 + 3H2 → 2NH3
[N2]
[H2]
[NH3]
Time
12–22
12–23
Br2 (aq) + HCOOH (aq)
2Br– (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
[Br2]
average rate = –
=–
t
tfinal - tinitial
instantaneous rate = rate for specific instance in time
13.1
Rate Laws: An Introduction
Rate Laws
•
Rate laws are always determined experimentally.
•
Reaction order is always defined in terms of reactant
(not product) concentrations.
•
The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the balanced
chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
13.2
Rate Laws

Reactions are reversible.

As products accumulate they can begin
to turn back into reactants.

Early on the rate will depend on only the
amount of reactants present.

We want to measure the reactants as
soon as they are mixed.

This is called the Initial rate method.
The method of Initial Rates

This method requires that a reaction be run
several times.

The initial concentrations of the reactants are
varied.

The reaction rate is measured just after the
reactants are mixed.

Eliminates the effect of the reverse reaction.
Types of Rate Laws

Differential Rate law - describes how rate
depends on concentration.

Integrated Rate Law - Describes how
concentration depends on time.

For each type of differential rate law there is an
integrated rate law and vice versa.
Rate laws can help us better understand reaction
mechanisms.

Rate Laws
Rate
k
=
n
k[NO2]
= rate constant
n = rate order
Rate Laws
12–32
Defining Rate

We can define rate in terms of the disappearance of the
reactant or in terms of the rate of appearance of the
product.

In our example

[H2]
t

[NH3]
t
= 3[N2]
t
= -2[N2]
t
N2 + 3H2
2NH3
RATE LAWS: A SUMMARY
12–34
Summary of the Rate Laws
The Integrated Rate Law
Integrated Rate Law

Expresses the reaction concentration as a
function of time.

Form of the equation depends on the order of the
rate law (differential).

Changes

We will only work with n=0, 1, and 2
n
Rate = [A]
t
Determining Rate Laws

The first step is to determine the form of the rate law
(especially its order).

Must be determined from experimental data.

For this reaction 2 N2O5 (aq)

4NO2 (aq) + O2(g)
The reverse reaction won’t play a role because the gas
leaves
2 NO2
2 NO + O2

You will find that the rate will only depend on the
concentration of the reactants. (Initially)

Rate = k[NO2]

This is called a rate law expression.

k is called the rate constant.

n is the order of the reactant -usually a positive
integer.
n
2 NO2
2 NO + O2

The rate of appearance of O2 can be said to be.

Rate' = [O2] = k'[NO2]

Because there are 2 NO2 for each O2

Rate = 2 x Rate'

So k[NO2] = 2 x k'[NO2]

So k = 2 x k'
n
t
n

Since the rate at twice as fast when the concentration is
twice as big the rate law must be..

First power

Rate = -[N2O5] = k[N2O5]1 = k[N2O5]

t

We say this reaction is first order in N2O5

The only way to determine order is to run the experiment.
An example

-
-
For the reaction BrO3 + 5 Br + 6H
+
3Br2 + 3 H2O

The general form of the Rate Law is Rate =
- n - m[H+]p
k[BrO3 ] [Br ]

We use experimental data to determine the
values of n,m,and p
Reaction order
What is a rate equation?

A rate equation allows us to calculate the rate we would expect for any
concentration of reactants.

For example in the reaction:
A+B+CD+E

The rate equation is:
Rate = k[A]x[B]y[C]z

Where:

[A], [B] an [C] are the concentrations of each reactant

x, y, and z are the order of reaction with respect to each reactant

k is the ‘rate constant’
The rate constant, k



The rate constant k is essentially a measure of how
readily a reaction will take place:

Higher k  faster reaction

Lower k  slower reaction
k is dependent on temperature

As temperature increases, so does k

As temperature decreases, so does k
k is independent of concentration:

Provided the temperature is fixed, k is always the
same
What is reaction order?

Reaction order describes how changes to the concentration
of reactants affect the rate of a reaction

Assuming temperature and pressure are fixed
0th Order (0o)
1st Order (1o)
2nd Order
(2o)
Doubling the concentration
of a reactant would double
the rate, tripling it would
triple the rate
Doubling the concentration
of a reactant would
quadruple the rate, tripling
it would increase rate ninefold
Changing the
concentration does not
affect the rate
Doubling the concentration
of a reactant would have
no effect on rate
For example:
Run #

1.00 M
1.00 M
1.25 x 10-2 M/s
2
1.00 M
2.00 M
2.5 x 10-2 M/s
3
2.00 M
2.00 M
2.5 x 10-2 M/s
Comparing Runs 2 and 3:

[A] doubles but [B] remains fixed

Rate unchanged
The reaction is 1st order w.r.t reactant B


1
The reaction is 0th order w.r.t reactant A


Initial [A] ([A]0) Initial [B] ([B]0) Initial Rate (v0)
Comparing Runs 1 and 2:

[B] doubles but [A] remains fixed

Rate doubles
Overall the reaction is 1st order
Another example:
Experiment

Initial rate /
mol (N2) dm–3 s–1
1
0.100
0.100
2.53×10–6
2
0.100
0.200
3
0.200
0.100
4
0.300
0.100
5.05×10–6
1.01×10–5
2.28×10–5
Comparing Runs 1 and 2:


[H2] doubles but [NO] remains fixed
Rate doubles
The reaction is 2nd order w.r.t reactant NO

Comparing Runs 1 and 3:



Initial [H2] /
mol dm–3
The reaction is 1st order w.r.t reactant H2


Initial [NO] /
mol dm–3
[NO] doubles but [H2] remains fixed
Rate quadruples
Overall the reaction is 3rd order (1st order + 2nd order = 3rd order)
Rate-Concentration
Graphs
0th Order
No
effect
Gradient 0
1st Order
Direct
proportion
Gradient positive
and constant
2nd Order
Squared
relationship
Gradient positive
and increasing
First Order

General form Rate = [A] / t = k[A]

ln[A] = - kt + ln[A]0

In the form y = mx + b

y = ln[A]

x=t

A graph of ln[A] vs time is a straight
line.
m = -k
b = ln[A]0
Zero Order Rate Law

Rate = k[A]0 = k

Rate does not change with concentration.

Integrated [A] = -kt + [A]0

When [A] = [A]0 /2 t = t1/2

t1/2 = [A]0 /2k
Zero Order Rate Law

Most often when reaction happens on a surface
because the surface area stays constant.

Also applies to enzyme chemistry.
First Order

By getting the straight line you can prove it is first
order

Often expressed in a ratio
 [ A] 0 
ln
 = kt
 [ A] 
Second Order

Rate = -[A]/t = k[A]2

integrated rate law

1/[A] = kt + 1/[A]0

y= 1/[A]
m=k

x= t

A straight line if 1/[A] vs t is graphed

Knowing k and [A]0 you can calculate [A] at any time t
b = 1/[A]0
Concentration-Time
Graphs
0th Order
t1/2
t1/
1st Order
t1/2
t1/2
t1/
t1/2
2
2
Half-life
decreases
2nd Order
Half-life
constant
Half-life
increases
Half Life

The time required to reach half the original
concentration.

If the reaction is first order

[A] = [A]0/2 when t = t1/2
Half-Life of Reactions
12–64

The highly radioactive plutonium in nuclear waste undergoes
first-order decay with a half-life of approximately 24,000
years. How many years must pass before the level of
radioactivity due to the plutonium falls to 1/128th (about 1%)
of its original potency?
Reaction Mechanisms
Terms

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.
P
o
t
e
n
t
i
a
l
Reactants
E
n
e
r
g
y
Products
Reaction Coordinate
P
o
t
e
n
t
i
a
l
Activation
Energy Ea
Reactants
E
n
e
r
g
y
Products
Reaction Coordinate
P
o
t
e
n
t
i
a
l
Activated
complex
Reactants
E
n
e
r
g
y
Products
Reaction Coordinate
P
o
t
e
n
t
i
a
l
Reactants
E
n
e
r
g
y
}
Products
Reaction Coordinate
E
Br---NO
P
o
t
e
n
t
i
a
l
Br---NO
Transition
State
2BrNO
E
n
e
r
g
y
2NO + Br2
Reaction Coordinate
Activation
Energy and
Rates
Mechanisms and rates

There is an activation energy for each elementary
step.

Activation energy determines k.
k determines rate

Slowest step (rate determining) must have the
highest activation energy.
This reaction takes place in three steps

Ea
First step is fast
Low activation energy

Ea
Second step is slow
High activation energy

Ea
Third step is fast
Low activation energy
Second step is rate determining
Intermediates are present
Activated Complexes or
Transition States
Reaction Mechanisms

Each of the two reactions is called an elementary
step .

The rate for a reaction can be written from its
molecularity .

Molecularity is the number of pieces that must
come together.

Elementary steps add up to the balanced
equation
Reaction Mechanisms

The series of steps that actually occur in a
chemical reaction.

Kinetics can tell us something about the
mechanism

A balanced equation does not tell us how the
reactants become products.
Unimolecular step involves one molecule Rate is first order.
 Bimolecular step - requires two molecules
- Rate is second order
 Termolecular step- requires three
molecules - Rate is third order
 Termolecular steps are almost never
heard of because the chances of three
molecules coming into contact at the
same time are miniscule.

Catalysis
Catalysts

Speed up a reaction without being used up in the
reaction.

Enzymes are biological catalysts.

Homogenous Catalysts are in the same phase as
the reactants.

Heterogeneous Catalysts are in a different phase
as the reactants.
How Catalysts Work

Catalysts allow reactions to proceed by a different
mechanism - a new pathway.

New pathway has a lower activation energy.

More molecules will have this activation energy.

Does not change E

Show up as a reactant in one step and a product in a later
step
Heterogenous Catalysts
H H

Hydrogen bonds to surface of
metal.

Break H-H bonds
H H
H H
H
H
Pt surface
Heterogenous Catalysts
H
H
H
C
C
H
H H
H
H
Pt surface
Heterogenous Catalysts

The double bond breaks and bonds to the catalyst.
H
H
H
C
H
C
H
H
Pt surface
H H
Heterogenous Catalysts

The hydrogen atoms bond with the carbon
H
H
H
C
H
C
H
H
Pt surface
H H
Heterogenous Catalysts
H
H
H
H
C
C
H
H
H
Pt surface
H
Homogenous Catalysts

Chlorofluorocarbons (CFCs) catalyze the
decomposition of ozone.

Enzymes regulating the body processes. (Protein
catalysts)
Catalysts and rate

Catalysts will speed up a reaction but only to a
certain point.

Past a certain point adding more reactants won’t
change the rate.

Zero Order
Catalysts and rate.
R
a
t
e

Rate increases until the active sites of catalyst are
filled.

Then rate is independent of concentration
Concentration of reactants
Heterogeneous Catalysis
12–97
Homogeneous Catalysis
12–98
Energy Plots for a Catalyzed
and an Uncatalyzed Pathway
for a Given Reaction
12–100
Effect of a Catalyst on the
Number of ReactionProducing Collisions
More Complicated
Reactions
-
-
+

BrO3 + 5 Br + 6H

For this reaction we found the rate law to be

Rate = k[BrO3 ][Br ][H ]

To investigate this reaction rate we need to
control the conditions
-
-
3Br2 + 3 H2O
+2
Rate = k[BrO3-][Br-][H+]2






We set up the experiment so that two of the reactants are
in large excess.
[BrO3 ]0= 1.0 x 10-3 M
+
[H ]
[Br ]0 = 1.0 M
0=
1.0 M
-
As the reaction proceeds [BrO3 ] changes noticeably
-
+
[Br ] and [H ] don’t
Rate = k[BrO3-][Br-][H+]2

This rate law can be rewritten

Rate = k[BrO3 ][Br ]0[H ]0



-
- + 2
- + 2
Rate = k[Br ]0[H ]0 [BrO3 ]
Rate = k’[BrO3 ]
This is called a pseudo first order rate law.
k =
k’
-
+ 2
[Br ]0[H ]0

A reaction is found to have a rate constant of 8.60x10-1sec-1 at
523 K and an activation energy of 120.8 kJ/mol. What is the
value of the rate constant at 270 K?
Which statement is true concerning the plot of
rate constants at various temperatures for a
particular reaction?
A) A steep slope of the ln k versus 1/T plot is
indicative of small changes in the rate
constant for a given increase in
temperature.
B) Different sections of the ln k versus 1/T plot
show different Ea values.
C) The plot of k versus T shows a linear
increase in k as the temperature increases.
D) A steep slope of the ln k versus 1/T plot is
indicative of a large Ea.
E) The y-intercept of the ln k versus 1/T plot is
the Ea value for that reaction