Kinetics.pptx

Chemical kinetics is a branch of chemistry that studies the rate of reaction and
the mechanism of the reaction. The main contents of chemical kinetics include
the following:
1) Determine the rate of the chemical reaction and the
influence of external factors such as temperature, pressure,
catalyst, solvent and light on the reaction rate;
2) Study the chemical reaction mechanism and reveal the
nature of the chemical reaction rate;
3) Explore the relationship and laws between material
structure and reaction ability.

 The Nature of the Reactants
◦ Chemical compounds vary considerably in their
chemical reactivities.
 Concentration of Reactants
◦ As the concentration of reactants increases, so does
the likelihood that reactant molecules will collide.
 Temperature
◦ At higher temperatures, reactant molecules have
more kinetic energy, move faster, and collide more
often and with greater energy.
 Catalysts
◦ Change the rate of a reaction
by changing the mechanism.
3.

The rate of a chemical reaction can be determined by
monitoring the change in concentration of either
reactants or the products as a function of time.
[A] vs t
4.

In this reaction, the
concentration of butyl
chloride, C4H9Cl, was
measured at various
times, t.
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
Rate =
[C4H9Cl]
t
5.

 A plot of concentration vs.
time for this reaction yields
a curve like this.
 The slope of a line tangent
to the curve at any point is
the instantaneous rate at
that time.
6.

 The reaction slows
down with time
because the
concentration of the
reactants decreases.
7.

 In this reaction, the ratio
of C4H9Cl to C4H9OH is
1:1.
 Thus, the rate of
disappearance of C4H9Cl
is the same as the rate
of appearance of
C4H9OH.
Rate =
-[C4H9Cl]
t
=
[C4H9OH]
t
8.

Suppose that the mole ratio is not 1:1?
9.
 H2 (g) + I2 (g)  2 HI (g)
 Disappearance of reactant
 Appearance of product
 2 moles of HI are produced for each mole of H2 used.
 The concentration of HI increases at 2x the rate of that I2 or H2
decreases
 For the overall rate to have the same value when defined with respect
to any reactant the change in HI must be multiplied by 1/2

Rate = 
[H2]
t
or 
[I2]
t

Rate = 1/2
[HI]
t

 For 2N2O (g)  2 N2 (g) + O2 (g)
1.Express the rate in terms of change in concentration of reactants
and products.
2.In the first 15 sec., 0.015 mol of O2 is produced in a 0.500 L
reaction vessel. Determine the average rate of reaction.
3.What is [N2O] during this same time interval?
10
The Rate of a Chemical Reaction

 For 2N2O (g)  2 N2 (g) + O2 (g)
1.
2.
3.
11

Rate = 
1
2
[N2O]
t
= 
1
2
[N2]
t
= 
[O2]
t

0.015 mol/0.500 L = 0.030 M Rate =
0.030 M
15 sec
 0.0020 M/s
Rate = 
1
2
[N2O]
t
 
[O2 ]
t
so
[N2O]
t
=  2
[O2 ]
t
=  0.0040 M/s
Δ [N2O] = - 0.004 ×15 sec. = - 0.06 M

* Each reaction has its own equation that gives its rate as a
function of reactant concentrations.
- This is called its Rate Law
The general form of the rate law is
Rate = k[A]x[B]y
* K: is the rate constant
* [A] and [B] are the concentrations of the reactants.
* X and y are exponents known as rate orders that must be
determined experimentally
 To determine the rate law we measure the rate at different
starting concentrations.
12.

Compare Experiments 1 and 2:
when [NH4
+] doubles, the initial rate doubles.
NH4
+ (aq) + NO2
- (aq)  N2 (g) + 2H2O (l)
13.

Likewise, compare Experiments 5 and 6:
when [NO2
-] doubles, the initial rate doubles.
NH4
+ (aq) + NO2
- (aq)  N2 (g) + 2H2O (l)
14.

This equation is called the rate law, and k is the
rate constant.
NH4
+ (aq) + NO2
- (aq)  N2 (g) + 2H2O (l)
15.

 Spectroscopy
◦ Light is passed thru a sample
◦ Absorption of light is proportional to concentration.
◦ More light absorbed = higher concentration.
16

 Pressure Measurement
◦ Concentrations of gaseous reactants can be
monitored by measuring changes in pressure
17
For gas-phase reactants use PA instead of [A].

 Rates of reaction often depend on the
concentration of one or more reactants
 Consider a simple reaction A  products
 We can express this relationship in a Rate Law
 Rate = k [A]n
 Where k is a proportionality constant called a rate
constant and n is the reaction order
 The value of n determines how the rate depends on
the concentration of the reactant
18
The Rate Law
The Effect of Concentration on Reaction Rate

 Rate = k [A]n
 The value of n determines how the rate
depends on the concentration of the reactant
19
The Rate Law
The Effect of Concentration on Reaction Rate
Value of n Reaction order rate
0 zero Independent of the [A]
1 first Directly proportional to [A]
2 second Proportional to [A]2

 Exponents tell the order of the reaction with
respect to each reactant.
 This reaction is
First-order in [NH4
+]
First-order in [NO2
−]
 The overall reaction order can be found by
adding the exponents on the reactants in the
rate law.
 This reaction is second-order overall.
Rate = K [NH4
+ ]1[NO2
- ]1
20.

 We can examine reaction order by looking at:
 [reactant] vs time
◦ The slope of this line = rate  [R] / t
21
The Rate Law: The Effect of Concentration on Reaction Rate

22
In this figure:Rate vs [reactant]. How does [R] affect
rate?

 Rate = k[A]0 = k
◦ Independent of reactant concentration
 [Reactant] decreases linearly with time
◦ Constant reaction rate
◦ Reaction does not slow down as reaction progresses
 Rate is the same at any [reactant]
23

 Example of a zero-order reaction is decomposition
of NH3 in presence of molybdenum or tungsten
catalyst
[Mo]
 2NH3 (g) N2 (g) + 3H2 (g)
 The surface of the catalyst is almost completely
covered by NH3 molecules. The adsorption of gas
on the surface cannot change by increasing the
pressure or concentration of NH3. Thus, the
concentration of gas phase remains constant
although the product is formed.
25

 Rate = k[A]1 = k
◦ Directly proportional to [reactant]
 [Reactant] decreases with time
◦ Not linear – slope becomes less steep
◦ Reaction slows down as reaction progresses
◦ Rate decreases
 Rate is directly proportional to [reactant]
◦ linear
26

 Rate = k[A]2
◦ Proportional to square of [reactant]
 [Reactant] decreases with time
◦ Not linear – slope becomes less
steep
◦ Reaction slows down as reaction
progresses
◦ Rate decreases
 Rate is proportional to square
[reactant]
28

 Reaction order can only be determined by
experiment.
 It cannot be deduced from the balanced
equation.
 Remember – reaction order is how the rate
depends on reactant concentration.
 So how do we determine this?
29

 Measure the (initial) rate of reaction at several
different [reactant].
 Look for the effect of [reactant] on rate.
 In this set of data when [A] doubles the rate
doubles
 Rate is directly
proportional to [A]
 First order
 Rate = k [A]
30

 This set of data shows no effect on rate with
increasing [A].
 Rate is independent of [A].
 Zero order
 Rate = k
31

 In this set of data when [A] doubles the rate
quadruples.
 Rate is directly
proportional to [A]2
 Second order
 Rate = k [A]2
32

 The math on that last one was obvious but
this might not always be the case.
 May need to actually solve for the value of n
to determine the rate law.
33

34

rate 2
rate 1

k [A]2
n
k[A]1
n =
0.240 M/s
0.060 M/s
=
k(0.40)n
k(0.20)n
4.0 =
0.40
0.20




n
= 2n
log 4.0 = log 2n
= n log 2
n =
log 4
log 2
= 2

The following data was collected for the reaction of
substances A and B to produce products C and D.
Deduce the order of this reaction with respect to A and to B.
Write an expression for the rate law in this reaction and
calculate the value of the rate constant.
[NO] mol dm-3 [O2] mol dm-3 Rate mol dm-3 s-1
0.40 0.50 1.6 x 10-3
0.40 0.25 8.0 x 10-4
0.20 0.25 2.0 x 10-4
35.

 Half-life is defined
as the time
required for one-
half of a reactant
to react.
 Because [A] at t1/2
is one-half of the
original [A],
[A]t = 0.5 [A]0.

For a first-order process, set [A]t=0.5 [A]0
in integrated rate equation:
NOTE: For a first-order
process, the half-life does
not depend on [A]0.

For a second-order process, set
[A]t=0.5 [A]0 in 2nd order equation.

First order Second order Second order
Rate
Laws
Integrate
d Rate
Laws
complicated
Half-life complicated

 The half-life (t1/2)of a reaction is the time
required for the concentration of a reactant to
fall to one-half of its initial value.
 If initial concentration = 1.0 M and the half-
life is 100 seconds then the concentration will
fall to 0.50 M in 100s.
 The half-life expression
◦ defines the dependence of half life on rate constant
and initial concentration
◦ Is different for different reaction orders
41

42
Zero-order half-life varies and depends (directly) on initial
concentration
k
2
[A]
t
[A]
kt
–
1/2[A]
[A]
1/2
[A]t
t
at t
[A]
kt
–
[A]
0
1/2
0
1/2
0
0
1/2
0
t








43
k
0.693
t
kt
0.693
kt
1/2
ln
[A]
1/2[A]
ln
[A]
1/2
[A]
t
at t
kt
[A]
[A]
ln
1/2
1/2
1/2
0
0
0
t
1/2
0
t











First-order half-life is constant and independent of concentration

44
Second -order half-life varies and
depends (inversely) on initial
concentration
0
1/2
0
0
1/2
0
0
1/2
0
1/2
0
0
t
1/2
0
t
k[A]
1
t
[A]
1
[A]
2
kt
[A]
1
1/2[A]
1
kt
[A]
1
kt
1/2[A]
1
[A]
1/2
[A]
t
at t
[A]
1
kt
[A]
1












45
The Integrated Rate Law: The Dependence of Concentration on Time

 Reaction order and rate law determined by
experiment
 Rate law (differential) relates rate of reaction
to concentration
 Integrated rate law relates concentration to
time
 Half-life of first order reaction is independent
of initial concentration
46

 There are several factors affect reaction rates
◦ Concentration
◦ Surface area (Particle size)
◦ Temperature
◦ Catalysts
◦ Inhibitors

 A catalyst is a substance that increases the rate of a
chemical reaction without itself being used up in the
reaction.
 A catalysts provide an alternative energy pathway for the
reaction.
 The different pathway lowers the activation energy
allowing more molecules to overcome the activation energy
and produce products at a faster rate

H
A lower activation energy allows more molecules to
overcome the activation energy, speeding up the
reaction
Potential
Energy

 An inhibitor is a substance that slows down, or inhibits
reaction rates.
 Uses
◦ A preservative
◦ A weed killer

 When a reaction results in complete conversion of reactants to
products chemists say it goes to completion
 Not all reactions go to completion. They appear to stop
because they are reversible
 Reversible reactions can occur in both the forward and reverse
directions

 Forward N2 + 3H2  2NH3
◦ The reactants are N2 and H2
 Reverse N2 + 3H2  2NH3
◦ The reactant is NH3
N2 + 3H2 ⇆ 2NH3
The forward and reverse reactions are happening at the same time.

 Chemical equilibrium is a state in which the
forward and reverse reactions balance each
other because they take place at equal rates.
◦ Rateforward reaction = Ratereverse reaction
 This does not mean the concentrations of the
products and reactants are the same
 Equilibrium is a state of action, not inaction.
This process is dynamic; dynamic
equilibrium.

 Law of chemical equilibrium states that at a
given temperature, a chemical system may
reach a state in which a particular ratio of
reactant and product concentrations has a
constant value known as Keq or equilibrium
constant.

 Keq is the ratio of product concentrations to reactant
concentrations.
 A large Keq, Keq > 1 means the products are favored over the
reactants
 A small Keq, Keq < 1 means the reactants are favored over
the products

 Homogenous equilibrium means all reactants and products
are in the same physical state
 Ex: H2(g) + O2(g)  H20(g)
 Heterogeneous equilibrium is when the reactants and
products are in more than one physical state.
 Ex: H2O(g) + C(s)  H2(g) + CO(g)

aA + bB  cC +dD
products [C]c [D]d
reactants [A]a [B]b
= =
Keq
**only use Keq for gases and aqueous compounds

1. Given N2(g) + 3H2(g)  2NH3(g) write the
equilibrium constant expression for the
reaction.

2. Given SO2(g) + O2(g)  2SO3(g) write the
equilibrium constant expression for the
reaction.

1. Given 2NaHCO3(s)  Na2CO3(s) + CO2(g) + 3H2O(g)
write the equilibrium constant expression for
the reaction.

At equilibrium and 100°C a flask contains:
[PCl5]=0.0325M [H2O]=0.025M
[HCl]=0.375M [POCl3]=0.250M
Calculate the Keq for the reaction
PCl5(g) + H20(g)  2HCl(g) + POCl3(g)

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