Transcript K - PHA Science

22 Nov. 2010
1
Objective: SWBAT write rate expressions
and calculate reaction rates for chemical
reactions.
 Do now: Describe one very slow reaction
that you’ve seen, and one very fast
reaction.

Agenda
Do now
II. Kinetics notes
III. Reaction Rates Demonstrations
IV. Rate constant and reaction rates
problems.
Homework: p. 602 #2, 3, 5, 7, 12, 13, 15, 17,
19
I.
3
How can we predict whether or not a
reaction will take place?
 Thermodynamics
 Once started, how fast does the reaction
proceed?
 Chemical kinetics: this unit!
 How far will the reaction go before it
stops?
 Equilibrium: next unit

Chemical Kinetics


The area of chemistry concerned with the
speeds, or rates, at which a chemical reaction
occurs.
reaction rate: the change in the concentration of
a reactant or product with time (M/s)
 Why do reactions have such very different
rates?
 Steps in vision: 10-12 to 10-6 seconds!
 Graphite to diamonds: millions of years!
 In chemical industry, often more important to
maximize the speed of a reaction, not
necessarily yield.
Chemical Kinetics
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A
B
D[A]
rate = Dt
D[A] = change in concentration of A over
time period Dt
D[B]
rate =
Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
A
D[A]
rate = Dt
D[B]
rate =
Dt
6
B
red-brown
2Br- (aq) + 2H+ (aq) + CO2 (g)
Br2 (aq) + HCOOH (aq)
time
t 1< t 2 < t 3
393 nm
light
7
Detector
D[Br2] a D Absorption
Br2 (aq) + HCOOH (aq)
2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
D[Br2]
average rate = =Dt
tfinal - tinitial
8
instantaneous rate = rate for specific instance in time
rate a [Br2]
rate = k [Br2]
rate
= rate constant
k=
[Br2]
= 3.50 x 10-3 s-1
2H2O2 (aq)
2H2O (l) + O2 (g)
PV = nRT
n
P=
RT = [O2]RT
V
1
[O2] =
P
RT
D[O2]
1 DP
rate =
=
RT Dt
Dt
measure DP over time
11
Reaction Rates and Stoichiometry
2A
B
Two moles of A disappear for each mole of B that is formed.
1 D[A]
rate = 2 Dt
aA + bB
D[B]
rate =
Dt
cC + dD
1 D[A]
1 D[B]
1 D[C]
1 D[D]
rate = ==
=
a Dt
b Dt
c Dt
d Dt
12
Example
Write the rate expression for the following
reaction:
 CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (g)

Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (g)
D[CH4]
D[CO2]
1 D[O2]
1 D[H2O]
rate = =
==
Dt
Dt
Dt
2 Dt
2
14
Practice Problems

Write the rate expressions for the following
reactions in terms of the disappearance of the
reactants and appearance of products.
a. I-(aq) + OCl-(aq)  Cl-(aq) + OI-(aq)
b. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Using Rate Expressions
Consider the reaction:
 4NO2(g) + O2(g)  2N2O5(g)
Suppose that, at a particular moment during
the reaction, molecular oxygen is reacting
at the rate of 0.024 M/s.
a. At what rate is N2O5 being formed?
b. At what rate is NO2 reacting?
Consider the reaction:
4PH3(g)  P4(g) + 6H2(g)
Suppose that, at a particular moment during
the reaction, molecular hydrogen is being
formed at the rate of 0.078 M/s.
a. At what rate is P4 being formed?
b. At what rate is PH3 reacting?
The Rate Law
The rate law expresses the relationship of the rate of a reaction
to the rate constant and the concentrations of the reactants
raised to some powers.
aA + bB
cC + dD
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x +y)th order overall
x and y are determined experimentally!
F2 (g) + 2ClO2 (g)







2FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
rate = k [F2][ClO2]
x=1
Quadruple [ClO2] with [F2] constant
Rate quadruples
y=1
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
20
Determine the rate law and calculate the rate constant for the
following reaction from the following data:
S2O82- (aq) + 3I- (aq)
2SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
0.08
0.017
1.1 x 10-4
3
0.16
0.017
2.2 x 10-4
Determine the rate law and calculate the rate constant for the
following reaction from the following data:
S2O82- (aq) + 3I- (aq)
2SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
0.08
0.017
1.1 x 10-4
3
0.16
0.017
2.2 x 10-4
rate = k [S2O82-]x[I-]y
y=1
x=1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
22
2.2 x 10-4 M/s
rate
k=
=
= 0.08/M•s
2[S2O8 ][I ] (0.08 M)(0.034 M)
Practice Problems
The reaction of nitric oxide with hydrogen at
1280oC:
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
From the following data collected at this
temperature, determine (a) the rate law, (b)
the rate constant and (c) the rate of the
reaction when [NO] = 12.0x10-3 M and [H2]
= 6.0x10-3 M

23
Experiment
[NO] M
[H2] M
Initial Rate (M/s)
1
5.0x10-3
2.0x10-3
1.3x10-5
2
10.0x10-3
2.0x10-3
5.0x10-5
3
10.0x10-3
4.0x10-3
10.0x10-5
23 Nov. 2010
24
Objective: SWBAT write rate expressions
and calculate reaction rates for chemical
reactions.
 Do now: Calculate the rate constant (k)
from the data below:

Agenda
25
Do now
II. Practice Problems
Homework: Revisit last night’s
assignment
 #19 hint: Graph lnP vs. time. If linear, it
is 1st order. slope = -k
 Excel works great.
I.
Write the reaction rate expressions for the
following in terms of the disappearance of
the reactants and the appearance of
products:
a) 2H2(g) + O2(g)  2H2O(g)
b) 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Consider the reaction
N2(g) + 3H2(g)  2NH3(g)
Suppose that at a particular moment during
the reaction molecular hydrogen is
reacting at a rate of 0.074 M/s.
a) At what rate is ammonia being formed?
b) At what rate is molecular nitrogen
reacting?
Calculate the rate of the reaction at the
time when [F2] = 0.010 M and [ClO2] =
0.020 M.
 F2(g) + 2ClO2(g)  2FClO2(g)

[F2] (M)
[ClO2] (M)
Initial Rate (M/s)
0.10
0.010
1.2x10-3
0.10
0.040
4.8x10-3
0.20
0.010
2.4x10-3
Consider the reaction X + Y  Z
From the following data, obtained at 360 K,
a) determine the order of the reaction
b) determine the initial rate of disappearance
of X when the concentration of X is 0.30 M
and that of Y is 0.40 M
29
Initial Rate of Disappearance
of X (M/s)
[X] (M)
[Y] (M)
0.053
0.10
0.50
0.127
0.20
0.30
1.02
0.40
0.60
0.254
0.20
0.60
0.509
0.40
0.30
30
Consider the reaction A B.
The rate of the reaction is 1.6x10-2 M/s
when the concentration of A is 0.35 M.
Calculate the rate constant if the reaction
is
a. first order in A
b. second order in A
24 Nov. 2010
Objective: SWBAT determine reaction
order graphically and relate concentration
of 1st order reactions to time.
 Do now: What is the overall order of
reaction for the data shown below?
 What is the rate constant?

Run # Initial [A] ([A]0)
Initial [B] ([B]0)
Initial Rate (v0)
1
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
Agenda
Do now
II. Homework solutions
III. First order reactions
IV. Time calculations for first-order
reactions.
Homework: p. 603 #19, 21, 22, 23, 25-29
I.

The rate laws can be used to determine the
concentrations of any reactants at any
time during the course of a reaction.
29 Nov. 2010

34


Take Out Homework p. 603 #19, 21, 22, 23, 2529
Objective: SWBAT compare 1st order, 2nd order,
and zero order reactions, and describe how
temperature and activation energy effect the rate
constant.
Do now: Calculate the half life of the reaction
F2(g) + 2ClO2(g)  2FClO2(g), with rate data shown
[ClO2] (M)
Initial Rate (M/s)
below: [F2] (M)
0.10
0.010
1.2x10-3
0.10
0.040
4.8x10-3
0.20
0.010
2.4x10-3
Agenda
35
Homework solutions
st
II. Review 1 order reactions
III. Second and zero order reactions
IV. Activation Energy
V. Problem Set
Quiz Weds.
p. 31, 32, 35, 37, 39, 42
+ Problem set
I.
First-Order Reactions
A
k=
product
D[A]
rate = Dt
rate
M/s
=
= 1/s or s-1
M
[A]
[A] = [A]0e−kt
rate = k [A]
D[A]
= k [A]
Dt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
ln[A] = ln[A]0 - kt
Graphical Determination of k
2N2O5
37
4NO2 (g) + O2 (g)
The reaction 2A
B is first order in A with a rate
constant of 2.8 x 10-2 s-1 at 800C. How long will it
take for A to decrease from 0.88 M to 0.14 M ?
The reaction 2A
B is first order in A with a rate constant
of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease
from 0.88 M to 0.14 M ?
[A]0 = 0.88 M
ln[A] = ln[A]0 - kt
[A] = 0.14 M
kt = ln[A]0 – ln[A]
ln[A]0 – ln[A]
=
t=
k
39
ln
[A]0
[A]
k
ln
=
0.88 M
0.14 M
2.8 x
10-2
s-1
= 66 s
The conversion of cyclopropane to propene in the
gas phase is a first order reaction with a rate
constant of 6.7x10-4 s-1 at 500oC.
a)
b)
c)
If the initial concentration of cyclopropane was
0.25 M, what is the concentration after 8.8
minutes?
How long, in minutes, will it take for the
concentration of cyclopropane to decrease
from 0.25 M to 0.15 M?
How long, in minutes, will it take to convert
74% of the starting material?
Practice Problem
The reaction 2A → B is first order in A with a
rate constant of 2.8×10-2s-1 at 80oC. How
long, in seconds, will it take for A to decrease
from 0.88 M to 0.14 M?
The rate of decomposition of azomethane (C2H6N2)
is studied by monitoring partial pressure of the
reactant as a function of time:
CH3-N=N-CH3(g) → N2(g) + C2H6(g)
The data obtained at 300oC are shown here:
Time (s)
Partial Pressure of Azomethane (mmHg)
0
284
100
220
150
193
200
170
250
150
300
132
Are these values consistent with first-order kinetics? If
so, determine the rate constant.

43
The following gas-phase reaction was studied at
290oC by observing the change in pressure as a
function of time in a constant-volume vessel:
 ClCO2CCl3(g)  2COCl2(g)
 Determine the order of the reaction and the
rate constant based on the following data,
where P is the total pressure
Time (s)
P (mmHg)
0
400
2,000
316
4,000
248
6,000
196
8,000
155
10,000
122
Ethyl iodide (C2H5I) decomposes at a certain
temperature in the gas phase as follows:
C2H5I(g) → C2H4(g) + HI(g)
From the following data, determine the order of the
reaction and the rate constant:
Time (min)
[C2H5I] (M)
0
0.36
15
0.30
30
0.25
48
0.19
75
0.13
First-Order Reactions
The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
ln 2
0.693
=
=
k
k
What is the half-life of N2O5 if it decomposes with a rate constant
of 5.7 x 10-4 s-1?
How do you know decomposition is first order?
First-Order Reactions
The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
ln 2
0.693
=
=
k
k
What is the half-life of N2O5 if it decomposes with a rate constant
of 5.7 x 10-4 s-1?
0.693
t½ = ln 2 =
= 1200 s = 20 minutes
-4
-1
k
5.7 x 10 s
How do you know decomposition is first order?
46
units of k (s-1)
First-order reaction
A
product
# of
half-lives [A] = [A]0/n
1
2
47
2
4
3
8
4
16
48
The decomposition of ethane (C2H6) to
methyl radicals is a first-order reaction
with a rate constant of 5.36x10-4 s-1 at
700oC:
C2H6(g)  2CH3(g)
Calculate the half-life of the reaction in
minutes.

49

Calculate the half-life of the decomposition
of N2O5:
2N2O5  4NO2(g) + O2(g)
t (s)
[N2O5] (M)
ln [N2O5]
0
0.91
-0.094
300
0.75
-0.29
600
0.64
-0.45
1200
0.44
-0.82
3000
0.16
-1.83
Second-Order Reactions
A
product
D[A]
rate = Dt
rate
M/s
=
= 1/M•s
k=
2
2
M
[A]
1
1
=
+ kt
[A]
[A]0
50
D[A]
= k [A]2
Dt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
1
t½ =
k[A]0
rate = k [A]2
Iodine atoms combine to form molecular
iodine in the gas phase:
I(g) + I(g)  I2(g)
This reaction follows second-order kinetics
and has the high rate constant 7.0x109/M·s
at 23oC.
a. If the initial concentration of I was 0.086
M, calculate the concentration after 2.0
minutes.
b. Calculate the half-life of the reaction if the
initial concentration of I is 0.60 M and if it
is 0.42 M.
The reaction 2A → B is second order with a rate
constant of 51/M·min at 24oC.
a. Starting with [A]o = 0.0092 M, how long
will it take for [A]t = 3.7x10-3 M?
b. Calculate the half-life of the reaction.
Zero-Order Reactions
A
product
D[A]
rate = Dt
D[A]
=k
Dt
rate
= M/s
k=
0
[A]
[A] = [A]0 - kt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
[A]0
t½ =
2k
53
rate = k [A]0 = k
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
0
54
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] = [A]0 - kt
ln[A] = ln[A]0 - kt
1
1
=
+ kt
[A]
[A]0
Half-Life
t½ =
[A]0
2k
t½ = ln 2
k
1
t½ =
k[A]0
55
Activation Energy and Temperature
Dependence of Rate Constants
Reaction rates increase with increasing
temperature
 Ex: Hard boiling an egg
 Ex: Storing food
 How do reactions get started in the first
place?

Collision Theory
56

Chemical reactions occur as a result of
collisions between reacting molecules.
reaction rate depends on concentration
 But, the relationship is more complicated
than you might expect!
 Not all collisions result in reaction (p.
582-583)

57
In order to react, colliding molecules must
have a total KE ≥ activation energy (Ea)
 Ea: minimum amount of energy required
to initiate a chemical reaction
 activated complex (transition state):
a temporary species formed by the
reactant molecules as a result of the
collision before they form the product.

A+B
Exothermic Reaction
+
AB+
C+D
Endothermic Reaction
The activation energy (Ea ) is the minimum amount of energy
required to initiate a chemical reaction.
=a barrier that prevents less energetic molecules from reacting
Temperature Dependence of the Rate Constant
k  A  e( Ea / RT )
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor (constant
across wide range of temps.)
Alternate format (y=mx+b):
Ea 1
ln k = + lnA
R T
59
Alternate Form of the Arrhenius Equation
At two temperatures, T1 and T2
or
60
The rate constants for the decomposition of
acetaldehyde:
CH3CHO(g) → CH4(g) + CO(g)
were measured at five different temperatures. The
data are shown below. Plot lnk versus 1/T, and
determine the activation energy (in kJ/mol) for the
3
reaction. (Note: the reaction
is 2 order in CH3CHO,
1
so k has the units of 1 / M 2  s )
1
2
k (1 / M  s)
0.011
0.035
T (K)
700
730
0.105
0.343
0.789
760
790
810
0.00
0.0012
0.00125
0.0013
0.00135
0.0014
0.00145
-0.50
-1.00
-1.50
slope = -2.19x104 K
 slope =  E a

ln K
-2.00
-2.50
-3.00
R
-3.50
y = -21881x + 26.662
-4.00
-4.50
-5.00
1/T (K-1)
The second order rate constant for the
decomposition of nitrous oxide (N2O) into
nitrogen molecule and oxygen atom has been
measured at different temperatures.
Determine graphically the activation energy
for the reaction.
63
k (1 / M  s)
1.87x10-3
0.0113
T (oC)
600
650
0.0569
0.244
700
750
30 Nov. 2010
Take Out Homework p. 605# 31, 32, 35,
37, 39
 Objective: SWBAT use the Arrhenius
equation to solve for rate constants and
temperatures, and solve practice problems
on kinetics.
 Do now: Match

64
Order Rate Law
2
rate = k[A]
Conc-Time Eq. Half Life Eq.
[A]=[A]0-kt
t1/2=1/k[A]o
1
rate = k[A]2 1/[A]=1/[A]0 + kt
0
rate = k
ln[A]=ln[A]0 –kt
t1/2=ln2/k
t1/2=[A]0/2k
Agenda
65
Homework solutions
II. Using the Arrhenius equation part 2
III. Molecular orientation
IV. Problem Set work time
Homework: Complete problem set and
p. 605 #40, 42
Quiz tomorrow
I.
66

The rate constant of a first order reaction
is 3.46x10-2 /s at 298 K. What is the rate
constant at 350 K if the activation energy
for the reaction is 50.2 kJ/mol?
67

The first order rate constant for the
reaction of methyl chloride (CH3Cl) with
water to produce methanol (CH3OH) and
hydrochloric acid (HCl) is 3.32x10-10/s at
25oC. Calculate the rate constant at 40oC
if the activation energy is 116 kJ/mol.
Homework
68
Quiz Weds.
p. 604 #40, 42
+ Problem set #2, 18, 20, 25, 29, 32, 37, 42,
55, 57: Weds.
Problem Set
69
Handout #2, 18, 20, 25, 29, 32, 37, 42, 55,
57: Weds.
 Quiz Weds.

2 Dec. 2010
70
Objective: SWBAT describe a reaction in
terms of elementary steps and determine the
rate law.
 Do now: Which graph represents an
exothermic reaction? Which is
endothermic? Why?

Agenda
71
Do now
II. Elementary steps in reactions
III. Catalysts
Homework: p. 605 #45, 47, 48, 52, 54, 55,
56, 58, 62: Mon.
I.
72
For simple reactions (between atoms, for
example) the frequency factor (A) is
proportional to the frequency of collision
between the reacting species.
 “Orientation factor” is also important.

Importance of Molecular Orientation
effective collision
ineffective collision
73
74
A balanced chemical equation doesn’t tell
us much about how the reaction actually
takes place.
 It may represent the sum of elementary
steps
 Reaction mechanism: the sequence of
elementary steps that leads to product
formation.

Reaction Mechanisms
The overall progress of a chemical reaction can be represented
at the molecular level by a series of simple elementary steps
or elementary reactions.
The sequence of elementary steps that leads to product
formation is the reaction mechanism.
2NO (g) + O2 (g)
2NO2 (g)
N2O2 is detected during the reaction!
75
Elementary step:
NO + NO
N 2O 2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
2NO (g) + O2 (g)
Mechanism:
76
2NO2 (g)
Intermediates are species that appear in a reaction
mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step
and consumed in a later elementary step.
Elementary step:
NO + NO
N 2O 2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
The molecularity of a reaction is the number of molecules
reacting in an elementary step.
•
Unimolecular reaction – elementary step with 1 molecule
•
Bimolecular reaction – elementary step with 2 molecules
•
Termolecular reaction – elementary step with 3 molecules
77
Rate Laws and Elementary Steps
Unimolecular reaction
A
products
rate = k [A]
Bimolecular reaction
A+B
products
rate = k [A][B]
Bimolecular reaction
A+A
products
rate = k [A]2
Writing plausible reaction mechanisms:
•
The sum of the elementary steps must give the overall
balanced equation for the reaction.
•
The rate-determining step should predict the same rate
law that is determined experimentally.
The rate-determining step is the slowest step in the
sequence of steps leading to product formation.
78
Sequence of Steps in Studying a Reaction Mechanism
79
The experimental rate law for the reaction between NO2 and CO
to produce NO and CO2 is rate = k[NO2]2. The reaction is
believed to occur via two steps:
Step 1:
NO2 + NO2
NO + NO3
Step 2:
NO3 + CO
NO2 + CO2
What is the equation for the overall reaction?
What is the intermediate?
What can you say about the relative rates of steps 1 and 2?
The experimental rate law for the reaction between NO2 and CO
to produce NO and CO2 is rate = k[NO2]2. The reaction is
believed to occur via two steps:
Step 1:
NO2 + NO2
NO + NO3
Step 2:
NO3 + CO
NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO
NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
81
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
82

rate determining step: the slowest step in
the sequence of steps leading to product
formation.
Decomposition of Hydrogen Peroxide
83
2H2O2(aq)  2H2O(l) + O2(g)
Can be catalyzed using iodide ions (I-)
rate = k[H2O2][I-] Why?!
Determined experimentally.
Step 1: H2O2 + I-  H2O + IOStep 2: H2O2 + IO-  H2O + O2 + I-
7 Dec. 2010
84
Objective: SWBAT describe catalysis and
how ammonia is synthesized through the
Haber process.
Do now: The rate for the reaction
2H2O2(aq)  2H2O(l) + O2(g) is
rate = k[H2O2][I-]
Step 1: H2O2 + I-  H2O + IOStep 2: H2O2 + IO-  H2O + O2 + IWhich is the rate determining step? Why?
Agenda
85
Do now
II. Homework solutions
III. Catalysts and catalyzing mechanisms
IV. AP Kinetics problem set
Homework: p. 605 #55, 56, 58
Bring your book on Thursday!
I.
Example
The gas-phase decomposition of nitrous oxide
(N2O) is believed to occur via two elementary
steps:
Step 1:
N2O  N2 + O
Step 2 N2O + O  N2 + O2
Experimentally the rate law is found to be
rate = k[N2O].
a) Write the equation for the overall reaction.
b) Identify the intermediates.
c) What can you say about the relative rates of
steps 1 and 2?

86
87
For the decomposition for H2O2, the
reaction rate depends on the
concentration of I- ions, even though Idoesn’t appear in the overall equation.
 I- is a catalyst for the reaction.

A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
k  A  e( Ea / RT )
Ea
Catalyzed
k
Uncatalyzed
ratecatalyzed > rateuncatalyzed
88
Ea′ < Ea
Catalysts
89
forms an alternative reaction pathway
 for example, it might form an
intermediate with the reactant.
 Ex: 2KClO3(s)  2KCl(s) + 3O2(g)
Very slow, until you add MnO2, a catalyst.
The MnO2 can be recovered at the end of
the reaction!

In heterogeneous catalysis, the reactants and the catalysts
are in different phases (usually, catalyst is a solid, reactants
are gases or liquids).
•
Haber synthesis of ammonia
•
Ostwald process for the production of nitric acid
•
Catalytic converters
In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
90
•
Acid catalysis
•
Base catalysis
Haber Process
Synthesis of Ammonia
Extremely slow at room temperature. Must be fast
and high yield!
Process occurs on the surface of the Fe/Al2O3/K2O
catalyst, which weakens the covalent N-N and H-H
bonds.
Fe/Al2O3/K2O
N2 (g) + 3H2 (g)
catalyst
91
2NH3 (g)
Ostwald Process
4NH3 (g) + 5O2 (g)
Pt catalyst
2NO (g) + O2 (g)
2NO2 (g) + H2O (l)
92
4NO (g) + 6H2O (g)
2NO2 (g)
HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process
Catalytic Converters
catalytic
CO + Unburned Hydrocarbons + O2 converter
CO2 + H2O
catalytic
2NO + 2NO2 converter
2N2 + 3O2
93
Enzyme Catalysis
biological catalysts
94
Binding of Glucose to Hexokinase
95
96

p. 605 #45, 47, 48, 52, 54, 55, 56, 58, 62