Ap chem unit 12 presentation

 Reaction Rates
 Rate Laws: An Introduction
 Determining the Form of the Rate Law
 The Integrated Rate Law
 Reaction Mechanisms
 A Model for Chemical Kinetics
 Catalysis

 Identities of products and reactants
 Stoichiometric quantities
 Spontaneity
› Refers to the inherent tendency for the
process to occur. Does not imply anything
about speed. Spontaneous does not mean
fast. A reaction can be considered
spontaneous but take years to occur.

 The area of chemistry that concerns
rates is called chemical kinetics.
› One of the main goals of chemical kinetics is
to understand the steps by which a reaction
takes place.
› This series of steps is called the reaction
mechanism.
› Understanding the mechanism allows us to
find ways to change or improve the rate of a
reaction.

We start with a flask of NO2 gas at 300°C.
But NO2 dioxide decomposes to nitric
oxide (a source of air pollution) and O2.
 2NO2(g)  2NO(g) + O2(g)
 If we were to measure the
concentrations of the three gases over
time we would see a change in the
amount of reactants and products over
time.

 The reactant NO2
decreases with time
and the concentrations
of the products (NO
and O2) increase with
time.
 2NO2(g)  2NO(g) + O2(g)

The speed, or rate, of a process is defined
as the change in a given quantity
(concentration in Molarity) over a specific
period of time.
 Reaction rate = [concentration of A at
time (t2) – concentration of A at time
(t1)]/ (t2 – t1)
 (Final – initial)
D[A]
Rate = -
Dt

D[A]
Rate = -
Dt
 The square brackets indicate
concentration in mol/l
 In Kinetics, rate is always defined as
positive. Since the concentration of
reactants decreases over time, a negative
sign is added to the equation.

 Looking at the NO2
table, calculate the
average rate at which
NO2 changes over the
first 50 seconds of the
reaction.

 4.2 x 10-5 mol/ls

Reaction rates are
often not constant
through the course of a
reaction. For example,
average rates for NO2
are not constant but
decreases with time.

The value of the rate at
a particular time can
be obtained by
computing the slope of
a line tangent to the
curve at that point in
time.

Dy
slope =
Dx

 When considering rates of a reaction
you must also take into account the
coefficients in the balanced equation for
the reaction.
 The balanced reaction determines the
relative rates of consumption of
reactants and generation of products.

2NO2(g)  2NO(g) + O2(g)

 In this example, both
the reactant NO2 and
the product NO have a
coefficient of 2, so NO is
produced at the same
rate NO2 is consumed.

2NO2(g)  2NO(g) + O2(g)

 The product O2 has a
coefficient of 1, which
means it is produced
half as fast as NO.

Chemical reactions are reversible. Often
times as products are formed, they
accumulate and react to form what was
the reactant(s).
 The previous example:
› 2NO2(g) <-> 2NO(g) + O2(g)
› As NO and O2 accumulate, they can react
to re-form NO2.

Chemical reactions are reversible. Often
times as products are formed, they
accumulate and react to form what was
the reactant(s).
 Now the Δ[NO2] depends on the
difference in the rates of the forward
and reverse reactions.
 In this unit we will not take into account
the reverse reactions.

If the reverse reaction can be neglected, the
reaction rate will depend only on the
concentrations of the reactants.
 A rate law shows how concentrations
relate to the rate of a reaction.
D[A]
Rate = - = k[A]n

Dt
› A is a reactant. k is a proportionality constant
and n is called the order of the reactant. Both
are usually determined by experiment.

Most simple reactions, the rate orders are
often positive integers, but they can be 0
or a fraction.
 The concentrations of the products do
not appear in the rate law because the
reaction rate is being studied under
conditions where the reverse reaction
does not contribute to the overall rate.

Most simple reactions, the orders are often
positive integers, but they can be 0 or a
fraction.
 The value of the exponent n must be
determined by experiment; it cannot be
written from the balanced equation.
 The rate law constant is dependent
upon species in a reaction.

There are two types of rate laws:
1. The differential rate law (often called
simply the rate law) shows how the rate
of a reaction depends on
concentration.
2. The integrated rate law shows how the
concentrations of species in the
reaction depend on time.

 The differential and integrated rate laws
for a given reaction are related and
knowing the rate law for a reaction is
important because we can usually infer
the individual steps involved in a
reaction from the specific form of the
rate law.

The first step in understanding how a given
chemical reaction occurs is to determine
the form of the rate law.
 Reaction rate form is described as orders.
› Example: first order, second order, zero
order
 A first order reaction: concentration of
the reactants are reduced by half, the
overall rate of the reaction will also be
half.
 First Order: A direct relationship exists
between concentration and rate.

 One common method for experimentally
determining the form of the rate law for
a reaction is the method of initial rates.
› The initial rate of a reaction is the
instantaneous rate determined just after the
reaction begins. Before the initial
concentrations of reactants have changed
significantly.

NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l)
+
D[NH 4 ]
Rate = = k[NH + ]n [NO2 ]m
-

Dt
4

 Rate 1 = 1.35 x10-7 mol/ls = k(.100M)n(.0050M)m

NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l)

 n and m can be determined by dividing known
rates.

NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l)
 Example: Find the form of the rate law for each
reactant and the overall reaction order:

 n and m are both 1 (unrelated), overall reaction
order is 2 (n+m=2)

NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l)
 Example: Find the rate constant for this reaction

 k = 2.7 x 10-4 L/mols

The reaction between bromate ions and
bromide ions in acidic aqueous solution is
given by the equation:
 BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l)
 Using the experimental data, determine
the orders for all three reactants, the overall
reaction order, and the value of the rate
constant.

 BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l)

 n=1, m=1, p=2, overall = 4, k=8.0 L3/mol3s

The rate laws we have considered so far
express the rate as a function of the
reactant concentrations. The integrated
rate law expresses the reactant
concentrations as a function of time.

D[A]
First order rate law: Rate = - = k[A]1
Dt

 The above rate law can be put into a
different form using calculus (integration)
 Integrated rate law for first order:

ln[A] = -kt + ln[Ao ]
› this equation is of the form y = mx + b
› first order slope = -k, ln[A] vs t is a straight line.

D[A]
First order rate law: Rate = - = k[A]1
Dt

 Integrated rate law for first order can
also be written:

æ [Ao ] ö
ln ç
è [A] ø ÷ = kt

The decomposition of N2O5 in the gas phase
was studied at constant temperature:
2N2O5(g)  4NO2(g) + O2(g)
The following results were collected:
[N2O5] (mol/l) Time (s)

Verify that this is first order 0.1000 0

for N2O5. Calculate k. 0.0707 50
0.0500 100
0.0250 200
0.0125 300
0.00625 400

2N2O5(g)  4NO2(g) + O2(g) [N2O5] ln[N2O5] Time (s)
(mol/l)
0.1000 0
0.0707 50
0.0500 100
0.0250 200
0.0125 300
0.00625 400

 k= -slope= 6.93 x 10-3s-1

[N2O5] Time (s)
(mol/l)
0.1000 0
0.0707 50
2N2O5(g)  4NO2(g) + O2(g) 0.0500 100
0.0250 200
Using the data from practice
0.0125 300
problem 2, calculate [N2O5] at 0.00625 400

150 s after the start of the reaction.
k= 6.93 x 10-3s-1

 .0353 mol/L

The time required for a reactant to reach
half its original concentration is called the
half life of a reactant.
 t1/2

Using the data from the [N2O5] (mol/l) Time (s)

previous example we
0.1000 0
can observe half-life.
0.0707 50
0.0500 100
0.0250 200
0.0125 300
0.00625 400

The general equation for the half-life of a
first-order reaction is

0.693
t1/2 =
k

 The half-life does not depend on
concentration

A certain first-order reaction has a half-life of
20.0 minutes.
Calculate the rate constant for this reaction
and determine how much time is required for
this reaction to be 75% complete?

 k=3.47 x 10-2min-1 or 5.78 x 10-4s-1
 t= 40 min.

D[A]
 second order rate law: Rate = - = k[A]2
Dt
 integrated second order rate law:
1 1
= kt +
[A] [Ao ]
› A plot of 1/[A] vs t is a straight line, slope =k
 half-life of a second order reaction:
1
t1/2 =
k[Ao ]

Butadiene reacts to form its dimer
according to the equation: [C H ]
4 6 Time (+/-
(mol/l) 1s)
2C4H6(g)C8H12(g) 0.01000 0
0.00625 1000
The following data was
0.00476 1800
collected: 0.00370 2800
0.00313 3600
Is this reaction first or second
0.00270 4400
order? What is the rate 0.00241 5200
0.00208 6200
constant? What is the half-life?

[C4H6] 1/[C4 ln Time
(mol/l) H6] [C4H6] (+/-1s)
2C4H6(g)C8H12(g) 0.01000 0
0.00625 1000
0.00476 1800
0.00370 2800
0.00313 3600
0.00270 4400
0.00241 5200
0.00208 6200

[C4H6] 1/[C4 ln Time
(mol/l) H6] [C4H6] (+/-1s)
2C4H6(g)C8H12(g) 0.01000 0
0.00625 1000
0.00476 1800
0.00370 2800
0.00313 3600
0.00270 4400
0.00241 5200
0.00208 6200

 second order, k =6.14x10-2L/mols
 half life= 1630s

It is important to recognize the difference
between the half-life for a first-order
reaction and the half-life for a second-
order reaction.
 First order half life depends only on k
› half life remains constant throughout the rxn.
 Second order half life depends on k and
the initial concentration
› each successive half-life doubles the
preceding one.

 First order  Second order
[N2O5] Time (s) [C4H6] Time (+/-
(mol/l) (mol/l) 1s)

0.1000 0 0.01000 0
0.00625 1000
0.0707 50
0.00476 1800
0.0500 100
0.00370 2800
0.0250 200 0.00313 3600
0.0125 300 0.00270 4400
0.00625 400 0.00241 5200
0.00208 6200

D[A]
zero order rate law: Rate = - = k[A]0 = k
Dt
 integrated zero order rate law:

[A] = -kt +[A0 ]
› A plot of [A] vs t is a straight line, slope = -k
 half-life of a zero order reaction:
[A0 ]
t1/2 =
2k

 Zero-order reactions are most often
encountered when a substance such as
a metal surface or enzyme (catalyst) is
required for the reaction to occur.

Most reactions with more than one
reactant are simplified by varying the
concentrations of the reactants.
 the reactant in which the form of the
rate law is being determined will have a
low concentration.
 the other reactants are given a much
higher concentration so that the use of
the other reactants are not limiting the
first.

The reaction from the first example problem:
BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l)
 If the rate form for BrO3- is being
determined, the concentration of BrO3- is
set at 0.001M
 The other two reactants have
concentrations of 1.0M so that their initial
and final concentrations are very similar
and can be disregarded.

Most chemical reactions occur by a series
of steps called the reaction mechanism.
 to really understand a reaction, the
mechanism is studied.
 Kinetics includes the study of possible
steps in a reaction

NO2(g) + CO(g)  NO(g) + CO2(g)

The rate law for this reaction= k[NO2]2
 The reaction is actually more
complicated than it appears
› The balanced equation only gives us
products, reactants and the stoichiometry,
but does not give direct information about
the mechanism.

NO2(g) + CO(g)  NO(g) + CO2(g)

 For this reaction, the mechanism involves
the following steps:
› NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
› NO3(g) + CO(g)  NO2(g) + CO2(g) , k2

 k1 and k2 are the rate constants for the
individual reactions

NO2(g) + CO(g)  NO(g) + CO2(g)

› NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
› NO3(g) + CO(g)  NO2(g) + CO2(g) , k2

 NO3(g) is an intermediate, a species that
is neither a reactant nor a product but is
formed and consumed during the
reaction sequence.

NO2(g) + CO(g)  NO(g) + CO2(g)

› NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
› NO3(g) + CO(g)  NO2(g) + CO2(g) , k2

 Each of these two reactions is called an
elementary step.
 Elementary steps have a rate law that is
written from its molecularity.

Molecularity is defined as the number of
species that must collide to produce the
reaction.
 A unimolecular step involves one
molecule
 A bimolecular reaction involves the
collision of two species
 A termolecular reaction involves the
collision of three species
› These steps are rare due to the probability of
three molecules colliding simultaneously.

A reaction mechanism is a series of
elementary steps that must satisfy two
requirements:
1. The sum of the elementary steps must
give the overall balanced equation for
the reaction.
2. The mechanism must agree with the
experimentally determined rate law

Example: NO2(g) + CO(g)  NO(g) + CO2(g)

For this reaction, the mechanism involves
› NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
› NO3(g) + CO(g)  NO2(g) + CO2(g) , k2

In order for a mechanism to meet rule 2,
the rate-determining step is considered.
 Multistep reactions often have one step
that is much slower than all the others.
 The overall reaction cannot be faster
than the slowest, or rate-determining
step.


If we assume the first step is rate-
determining:
› NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
› NO3(g) + CO(g)  NO2(g) + CO2(g) , k2
 The overall rate is equal to the rate of
production of NO3.


If we assume the first step is rate-
determining:
› NO2(g) + NO2(g)  NO3(g) + NO(g) , k1
› NO3(g) + CO(g)  NO2(g) + CO2(g) , k2
 The rate law for the first step is written
from its molecularity:
› Overall Rate = k1[NO2]2

The balanced equation for the reaction of
the gases nitrogen dioxide and fluorine is:
2NO2(g) + F2(g)  2NO2F(g)
The experimentally determined rate law is:
Rate = k[NO2][F2]
A suggested mechanism for this reaction is:
NO2 + F2 NO2F + F (slow, k1)
F + NO2 NO2F (fast, k2)
Is this an acceptable mechanism?

2NO2(g) + F2(g)  2NO2F(g)
The experimentally determined rate law is:
Rate = k[NO2][F2]
A suggested mechanism for this reaction is:
NO2 + F2 NO2F + F (slow, k1)
F + NO2 NO2F (fast, k2)

 The mechanisms satisfy both
requirements.

Kinetics depend on concentration, time
and reaction mechanisms. There are other
factors that affect reaction rates:
 Temperature: chemical reactions speed
up when the temperature is increased

 Rate constants
show an
exponential
increase with
absolute
temperature.

This observed increase in reaction rate with
temperature is explained with the Collision
model.
 Molecules must collide to react.
› higher concentration means more molecules
to collide with each other.
› KMT states that an increase in temperature
raises molecular velocities and increases
collision frequency.

Only a small amount of collisions produce
a reaction
 Activation energy is the minimum amount of
energy required for a collision to produce a
reaction.
 The kinetic energy of a molecule is changed
into potential energy as the molecules are
distorted during a collision
 The arrangement of this distortion is called the
activated complex or transition state.

In addition to the activation energy
required for a reaction to take place,
another variable must be considered.
 Molecular orientations must be correct
for collisions to lead to reactions.
 A correction factor is included in our
activation energy formula to account for
nonproductive molecular collisions

Two requirements must be satisfied for
reactants to collide successfully and
rearrange to form products:
1. The collision must involve enough
energy to produce the reaction; the
collision energy must equal or exceed
activation energy.
2. The relative orientation of the reactants
must allow formation of new bonds.

- Ea /RT
k = Ae
Ea æ R ö
ln(k) = ç ÷ + ln(A)
R èTø

 The first equation can be rewritten in a
linear equation (y=mx+b). Slope = -Ea/R,
x=1/T, y=ln(k)
 R = 8.3145 J/Kmol

The most common procedure for finding Ea
is to plot ln(k) vs 1/T and find the slope
 It can also be found using only two
temperatures:

æ k2 ö E a æ 1 1 ö
ln ç ÷ = ç - ÷
è k1 ø R è T1 T2 ø

The reaction 2N2O5(g)  4NO2(g) + O2(g) was
studied at several temperatures, and the
following values of k were obtained:
Calculate the value of Ea for this reaction.

k(s-1) T (°C)
2.0x10-5 20
7.3x10-5 30
2.7x10-4 40
 1.0x105 J/mol 9.1x10-4 50
2.9x10-3 60

A catalyst is a substance that speeds up a
reaction without being consumed in the
reaction.
There are two different types of catalysts:
1. homogeneous – a catalyst that is
present in the same phase as the
reacting molecules
2. heterogeneous – exists in a different
phase as the reacting molecules
(usually a solid).

Heterogeneous catalysis most often
involves gaseous reactants being
adsorbed on the surface of a solid
catalyst. This process is called
adsorption.
 Adsorption is the collection of one
substance on the surface of
another substance.

Biologically important reactions in our body
our usually assisted by a catalyst/enzyme.
 specific proteins are needed by the
human body, the proteins in food must
be broken into their constituent amino
acids that are then used to construct
new proteins in the body’s cells.
 Without enzymes in human cells these
reactions would be much to slow to be
useful.

Ap chem unit 12 presentation

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Ap chem unit 12 presentation