Transcript Rates of Reaction - Derry Area School District
Rates of Reaction
1
Reaction Rates
Chemical reactions require varying lengths of time for completion.
– This
reaction rate
depends on the
characteristics of the reactants and products
and the
conditions under which the reaction is run.
– By understanding how the rate of a reaction is affected by changing conditions, one can learn the details of what is happening at the molecular level.
2
Reaction Rates
The questions posed in this chapter will be: – – –
How is the rate of a reaction measured?
What conditions will affect the rate of a reaction?
–
How do you express the relationship of rate to the variables affecting the rate?
What happens on a molecular level during a chemical reaction?
3
Reaction Rates
Chemical kinetics
is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction.
–
What variables affect reaction rate?
• • • • Concentration of reactants. Concentration of a catalyst Temperature at which the reaction occurs.
Surface area of a solid reactant or catalyst.
4
Reaction Rates
Chemical kinetics
is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction.
–
What variables affect reaction rate?
Let’s look at each in more detail.
5
Factors Affecting Reaction Rates
Concentration of reactants.
– More often than not, the rate of a reaction
increases
when the concentration of a reactant is increased.
– Increasing the population of reactants increases the likelihood of a successful collision.
– In some reactions, however, the rate is unaffected by the concentration of a particular reactant, as long as it is present at some concentration.
6
Factors Affecting Reaction Rates
Concentration of a catalyst.
– A
catalyst
is a
substance that increases the rate of a reaction without being consumed in the overall reaction.
– The catalyst generally does not appear in the overall balanced chemical equation (although its presence may be indicated by writing its formula over the arrow).
2
H
2
O
2 (
aq
) 2
H
2
O
(
l
)
O
2 (
g
) 7
Factors Affecting Reaction Rates
Concentration of a catalyst.
– Figure 14.2 shows the HBr catalyzed decomposition of H 2 O 2 to H 2 O and O 2 .
2 H 2 O 2 ( aq ) ( 2 H 2 O ( l )
O 2 ( g )
– A catalyst speeds up reactions by reducing the “activation energy” needed for successful reaction.
– A catalyst may also provide an alternative mechanism, or pathway, that results in a faster rate.
8
Factors Affecting Reaction Rates
Temperature at which a reaction occurs.
– Usually reactions speed up when the temperature increases.
– A good “rule of thumb” is that reactions approximately double in rate with a 10 o C rise in temperature.
9
Factors Affecting Reaction Rates
Surface area of a solid reactant or catalyst.
– Because the reaction occurs at the surface of the solid, the rate increases with increasing surface area.
– Figure 14.3 shows the effect of surface area on reaction rate.
10
Definition of Reaction Rate
The
reaction rate
is
the increase in molar concentration of a product of a reaction per unit time.
– Always a positive value.
– It can also be expressed as the
decrease
molar concentration of a
reactant
per unit in time.
11
Definition of Reaction Rates
Consider the gas-phase decomposition of dintrogen pentoxide.
2 N 2 O 5 ( g )
4 NO 2 ( g )
O 2 ( g )
– If we denote molar concentrations using brackets, then the change in the molarity of O represented as 2 would be symbol, D
[ O 2 ]
where the D (capital Greek delta), means “the change in.” 12
Definition of Reaction Rates
Then, in a given time interval, D t , the molar – – concentration of O 2 D [O 2 ].
– would increase by The rate of the reaction is given by:
Rate of formation of oxygen
D
[O
2
]
D
t
This equation gives the average rate over the time interval, D t .
If D t is short, you obtain an
instantaneous rate
, that is, the rate at a particular instant. (Figure 14.4) 13
Figure 14.4 The instantaneous rate of reaction In the reaction
2 N 2 O 5 ( g )
4 NO 2 ( g )
O 2 ( g )
The concentration of O 2 increases over time. You obtain the instantaneous rate from the slope of the tangent at the point of the curve corresponding to that time.
Definition of Reaction Rates
Figure 14.5 shows the increase in concentration of O 2 during the decomposition of N 2 O 5 .
• Note that the
rate decreases
as the reaction proceeds.
15
Figure 14.5 Calculation of the average rate.
When the time changes from 600 s to 1200 s, the average rate is 2.5 x 10 -6 mol/(L
.
s). Later when the time changes from 4200 s to 4800 s, the average rate has slowed to 5 x 10 -7 mol/(L
.
s). Thus, the rate of a reaction decreases as the reaction proceeds.
Definition of Reaction Rates
Because the amounts of products and reactants are related by stoichiometry, any substance in the reaction can be used to
Rate
• express the rate.
of decomposit ion of N
2
O
5
-
D
[N
D
2
t O
Note the negative sign. This results in a
5
]
positive rate as reactant concentrations decrease. 17
Definition of Reaction Rates
The rate of decomposition of N 2 O 5 are easily related.
2 N 2 O 5 ( g )
4 NO 2
and the formation of O 2
( g )
O 2 ( g )
D
[O 2
D
t ]
1 2 ( -
D
[N
D
2 O t 5 ] )
– Since two moles of N 2 O 5 decompose for each mole of O 2 formed, the rate of the decomposition of N 2 O 5 is twice the rate of the formation of O 2 .
18
Experimental Determination of Reaction Rates
To obtain the rate of a reaction you must determine the concentration of a reactant or product during the course of the reaction.
– One method for slow reactions is to withdraw samples from the reaction vessel at various times and analyze them.
– More convenient are techniques that continuously monitor the progress of a reaction based on some physical property of the system.
19
Experimental Determination of Reaction Rates
Gas-phase partial pressures.
– When dinitrogen pentoxide crystals are sealed in a vessel equipped with a manometer (see Figure 14.6) and heated to 45 o C, the crystals vaporize and the N 2 O 5 (g) decomposes.
2 N 2 O 5 ( g )
4 NO 2 ( g )
O 2 ( g )
– Manometer readings provide the concentration of N 2 O 5 during the course of the reaction based on partial pressures.
20
This device indicates the difference between two pressures (differential pressure), or between a single pressure and atmosphere (gage pressure), when one side is open to atmosphere. If a U-tube is filled to the half way point with water and air pressure is exerted on one of the columns, the fluid will be displaced. Thus one leg of water column will rise and the other falls. The difference in height "h" which is the sum of the readings above and below the half way point, indicates the pressure .
21
Experimental Determination of Reaction Rates
Colorimetry
– Consider the reaction of the hypochlorite ion with iodide.
ClO
( aq )
I
( aq )
IO
( aq )
Cl
( aq )
– The hypoiodate ion, IO , absorbs near 400 nm. The intensity of the absorbtion is proportional to [IO ], and you can use the absorbtion rate to determine the reaction rate.
22
Dependence of Rate on Concentration
Experimentally, it has been found that the rate of a reaction depends on the concentration of certain reactants as well as catalysts. – Let’s look at the reaction of nitrogen dioxide with fluorine to give nitryl fluoride.
2 NO 2 ( g )
F 2 ( g )
2 NO 2 F ( g )
– The rate of this reaction has been observed to be proportional to the concentration of nitrogen dioxide.
23
Dependence of Rate on Concentration
• When the concentration of nitrogen dioxide is doubled, the reaction rate doubles.
– The rate is also proportional to the concentration of fluorine; doubling the concentration of fluorine also doubles the rate.
–
We need a mathematical expression
to relate the rate of the reaction to the concentrations of the reactants.
24
Dependence of Rate on Concentration
A
rate law
is an
equation that relates the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers.
Rate
k [NO 2 ][F 2 ]
– The
rate constant
,
k
, is a proportionality constant in the relationship between rate and concentrations.
25
Dependence of Rate on Concentration
As a more general example, consider the reaction of substances A and B to give D and E.
aA
bB
–
dD
eE C
You could write the rate law in the form
Rate
k [A] m [B] n [C] p catalyst
– The exponents m, n, and p are frequently, but not always, integers. They must be determined experimentally and cannot be obtained by simply looking at the balanced equation.
26
Find the rate law expression and evaluate k for 2H 2 + 2NO 2H 2 O + N 2 @800K; all gases 1 2 3 4 5 6 H 2 .001
.002
.003
.009
.009
.009
NO .006
.006
.006
.001
.002
.003
Rate (atm/min .025
Rate = k [A] x [B] y [A] x [B] y Trial 2 = .002
x x .006
y = .050
.050
.075
Trial 1 .001 .006 .025
2 x x 1 y = 2 .0063
2 x = 2 .025
x = 1 .056
27
1 2 3 4 5 6 Find the rate law expression and evaluate k for 2H2 + 2NO 2H2O + N2 @800K; all gases H 2 .001
.002
.003
.009
.009
.009
NO .006
.006
.006
.001
.002
.003
Rate (atm/min .025
.050
.075
.0063
.025
.056
[A] x [B] y Trial 6 = .009
1 x .003
y = .056
Trial 4 .009 .001 .0063
3 y = 8.89
ylog3 =log 8.89
y = 2 Rxn is first order for x second order for y , and (2 + 1 =) third order overall.
28
Rate = k[H 2 ][NO] 2 From Trial 1: .025atm = k(.001M)(.006M) 2 min .025atm = k(.001M)(.006M) 2 min 6.94 x 10 5 atm = k min M 3 29
1 2 3 Find the rate law expression and evaluate k for H 2 + 2NO 2H 2 O + N 2 O @1100K; all gases P NO [atm] P H2 [atm] Rate [atm/min] .150
.400
.020
.075
.400
.005
.150
.200
.101
30
Find the rate law expression and evaluate k for H
2
+ 2NO
2H
2
O + N
2
O
@1100K; all gases ANSWERS: x = 2 y = 1 k = 2.22(min atm) -1 31
Dependence of Rate on Concentration
Reaction Order – The
reaction order
with respect to a given reactant species
equals the exponent of the concentration of that species in the rate law, as determined experimentally.
– The
overall order
of the reaction
equals the sum of the orders of the reacting species in the rate law.
32
Dependence of Rate on Concentration
Reaction Order – Consider the reaction of nitric oxide with hydrogen according to the following equation.
2 NO ( g )
2 H 2 ( g )
N 2 ( g )
2 H 2 O ( g )
– The experimentally determined rate law is
Rate
k [NO] 2 [H 2 ]
– Thus, the reaction is second order in NO, first order in H 2 , and third order overall.
33
Dependence of Rate on Concentration
Reaction Order – Although reaction orders frequently have whole number values (particularly 1 and 2), they can be
fractional.
–
Zero
and
negative
orders are also possible.
– The concentration of a reactant with a zero order dependence has
no effect
on the rate of the reaction.
34
Dependence of Rate on Concentration
Determining the Rate Law.
– One method for determining the order of a reaction with respect to each reactant is the
method of initial rates
.
– It involves running the experiment multiple times, each time varying the concentration of only one reactant and measuring its initial rate.
– The resulting change in rate indicates the order with respect to that reactant.
35
Dependence of Rate on Concentration
Determining the Rate Law.
– If
doubling the concentration
of a reactant has a
doubling
effect on the rate, then one would deduce it was a
first-order dependence
. 2:2 correspondence – If
doubling the concentration
had a
quadrupling
effect on the rate, one would deduce it was a
second-order dependence
. 2:4 correspondence – A
doubling of concentration
that results in an
eight-fold
increase in the rate would be a
third order dependence
. 2:8 correspondence 36
A Problem to Consider
Iodide ion is oxidized in acidic solution to
H
2
O
triiodide ion, I3- , by hydrogen peroxide.
2 (
aq
) 3
I
(
aq
) 2
H
(
aq
)
I
3 (
aq
) 2
H
2
O
(
l
) – – – A series of four experiments was run at different concentrations, and the initial rates of I 3 were determined. formation From the following data, obtain the reaction orders with respect to H 2 O 2 , I , and H + .
Calculate the numerical value of the rate constant.
37
A Problem to Consider
Exp. 1 Exp. 2 Exp. 3 Exp. 4
Initial Concentrations (mol/L)
H 2 O 2 0.010
0.020
0.010
0.010
I 0.010
0.010
0.020
0.010
H + 0.00050
0.00050
0.00050
0.00100
Initial Rate [mol/(L .
s)]
1.15 x 10 -6 2.30 x 10 -6 2.30 x 10 -6 1.15 x 10 -6 – – Comparing Experiment 1 and Experiment 2, you see that when the H 2 O 2 concentration doubles (with other concentrations constant), the rate doubles.
This implies a
first-order dependence with respect to H 2 O 2 .
38
A Problem to Consider
Exp. 1 Exp. 2 Exp. 3 Exp. 4
Initial Concentrations (mol/L)
H 2 O 2 0.010
0.020
0.010
0.010
I 0.010
0.010
0.020
0.010
H + 0.00050
0.00050
0.00050
0.00100
Initial Rate [mol/(L .
s)]
1.15 x 10 -6 2.30 x 10 -6 2.30 x 10 -6 1.15 x 10 -6 – – Comparing Experiment 1 and Experiment 3, you see that when the I concentration doubles (with other concentrations constant), the rate doubles.
This implies a
first-order dependence with respect to I .
39
A Problem to Consider
Exp. 1 Exp. 2 Exp. 3 Exp. 4
Initial Concentrations (mol/L)
H 2 O 2 0.010
0.020
0.010
0.010
I 0.010
0.010
0.020
0.010
H + 0.00050
0.00050
0.00050
0.00100
Initial Rate [mol/(L .
s)]
1.15 x 10 -6 2.30 x 10 -6 2.30 x 10 -6 1.15 x 10 -6 – – Comparing Experiment 1 and Experiment 4, you see that when the H + concentration doubles (with other concentrations constant), the rate is unchanged.
This implies a
zero-order dependence with respect to H + .
40
A Problem to Consider
Exp. 1 Exp. 2 Exp. 3 Exp. 4
Initial Concentrations (mol/L)
H 2 O 2 0.010
0.020
0.010
0.010
I 0.010
0.010
0.020
0.010
H + 0.00050
0.00050
0.00050
0.00100
Initial Rate [mol/(L .
s)]
1.15 x 10 -6 2.30 x 10 -6 2.30 x 10 -6 1.15 x 10 -6 – Because [H + ] 0 = 1, the rate law is:
Rate
k [ H 2 O 2 ][ I
]
– The reaction orders with respect to H 2 O 2 , I , and H + , are
1
,
1
, and
0
, respectively.
41
A Problem to Consider
Exp. 1 Exp. 2 Exp. 3 Exp. 4 –
Initial Concentrations (mol/L)
H 2 O 2 0.010
0.020
0.010
0.010
I 0.010
0.010
0.020
0.010
H + 0.00050
0.00050
0.00050
0.00100
Initial Rate [mol/(L .
s)]
1.15 x 10 2.30 x 10 2.30 x 10 1.15 x 10 -6 -6 -6 -6 You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:
1 .
15
10
6 mol L
s
k
0 .
010 mol L
0 .
010 mol L
42
A Problem to Consider
Exp. 1 Exp. 2 Exp. 3 Exp. 4 –
Initial Concentrations (mol/L)
H 2 O 2 0.010
0.020
0.010
0.010
I 0.010
0.010
0.020
0.010
H + 0.00050
0.00050
0.00050
0.00100
Initial Rate [mol/(L .
s)]
1.15 x 10 -6 2.30 x 10 -6 2.30 x 10 -6 1.15 x 10 -6 You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:
k
1 .
15 10 6
s
1 0 .
010 0 .
010
mol
/
L
1 .
2 10 2
L
/(
mol
s
) 43
Change of Concentration with Time
A rate law simply tells you how the rate of reaction changes as reactant concentrations change.
– A more useful mathematical relationship would show how a reactant concentration changes over a period of time.
44
Change of Concentration with Time
A rate law simply tells you how the rate of reaction changes as reactant concentrations change.
– Using calculus we can transform a rate law into a mathematical relationship between concentration and time.
– This provides a
graphical method
determining rate laws.
for 45
Concentration-Time Equations
First-Order Integrated Rate Law – You could write the rate law in the form
Rate
D
[ A
D
t ]
k [ A ]
46
Concentration-Time Equations
First-Order Integrated Rate Law – Using calculus, you get the following equation.
[ A ] ln [ A ] o t
kt
– – Here
[A] t
and
[A] o
is the concentration of reactant A at time t, is the initial concentration.
The ratio
[A] t /[A] o
time
t
.
is the fraction of A remaining at 47
A Problem to Consider
The decomposition of N 2 O 5 to NO 2 and O 2 order with a rate constant of 4.8 x 10 -4 s -1 is first . If the initial concentration of N 2 O 5 is 1.65 x 10
what is the concentration of N 2 O 5
-2 mol/L,
after 825 seconds?
– The first-order time-concentration equation for this reaction would be:
[ N ln [ N 2 2 O 5 O 5 ] ] o t
kt
48
A Problem to Consider
The decomposition of N 2 O 5 to NO 2 and O 2 order with a rate constant of 4.8 x 10 -4 s -1 is first . If the initial concentration of N 2 O 5 is 1.65 x 10
what is the concentration of N 2 O 5
-2 mol/L,
after 825 seconds?
– Substituting the given information we obtain:
ln 1 .
65 [
N 2 O 10
2 5 ] t mol / L
(4.80
10 4 s 1 )
( 825 s)
49
A Problem to Consider
The decomposition of N 2 O 5 to NO 2 and O 2 order with a rate constant of 4.8 x 10 -4 s -1 is first . If the initial concentration of N 2 O 5 is 1.65 x 10
what is the concentration of N 2 O 5
-2 mol/L,
after 825 seconds?
– Substituting the given information we obtain:
ln 1 .
65 [
N 2 O
10 2 5 ] t mol / L
0.396
50
A Problem to Consider
The decomposition of N 2 O 5 to NO 2 and O 2 order with a rate constant of 4.8 x 10 -4 s -1 is first . If the initial concentration of N 2 O 5 is 1.65 x 10
what is the concentration of N 2 O 5
-2 mol/L,
after 825 seconds?
– Taking the inverse natural log of both sides obtain: we
1 .
65 [
N 2 O
10 2 5 ] t mol / L
e 0.396
0 .
673
51
A Problem to Consider
The decomposition of N 2 O 5 to NO 2 and O 2 order with a rate constant of 4.8 x 10 -4 s -1 is first . If the initial concentration of N 2 O 5 is 1.65 x 10
what is the concentration of N 2 O 5
-2 mol/L,
after 825 seconds?
– Solving for [N 2 O 5 ] at 825 s we obtain:
[N 2 O 5 ]
( 1.65
10 2 mol / L )
( 0 .
673 )
0 .
0111 mol / L
52
Concentration-Time Equations
Second-Order Integrated Rate Law – You could write the rate law in the form
Rate
D
[ A
D
t ]
k [ A ] 2
53
Concentration-Time Equations
Second-Order Integrated Rate Law – Using calculus, you get the following equation.
1 [
A
]
t
1 [A] o kt – Here
[A] t
is the concentration of reactant A at time
t
, and
[A] o
is the initial concentration.
54
Concentration-Time Equations
Zero-Order Integrated Rate Law – The Zero-Order Integrated Rate Law equation is:.
[
A
] o [
A
]
kt
55
Concentration-Time Equations
First order integrated rate law [
A
] ln [
A
]
o t
kt Second order integrated rate law 1 1 kt [
A
]
t
[A] o Zero order integrated rate law [
A
] o [
A
]
kt
56
Concentration-Time Equations
k values General k formula: Units k = (L/mol) order-1 unit of time First order: s -1 Second order: L/mol•sec, or L•mol -1 •sec -1 Zero order: L 2 /mol 2 •sec, or L 2 •mol -2 •sec -1 57
Half-life
The
half-life
of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value.
– For a first-order reaction, the half-life is independent of the initial concentration of reactant.
– In one half-life the amount of reactant decreases
ln( 1 2 ) kt 1 2
by one-half. Substituting into the first-order concentration-time equation, we get: ln( 1 / 2 ) .
693
kt
1 2 58
Half-life
The
half-life
of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value.
– Solving for
t 1/2
we obtain:
t 1 2
0 .
693 k
– Figure 14.8 illustrates the half-life of a first-order reaction.
59
Half-life
Sulfuryl chloride, SO 2 Cl 2 , decomposes in a first-order reaction to SO
SO 2 Cl 2 ( g )
SO 2 (
2 and Cl
g )
Cl
2 .
2 ( g )
– At 320 o C, the rate constant is 2.2 x 10 -5 s -1 .
What is the half-life of SO 2 Cl 2 vapor at this temperature?
– Substitute the value of k into the relationship between k and t 1/2.
t 1 2
0 .
693 k
60
Half-life
Sulfuryl chloride, SO 2 Cl 2 , decomposes in a first-order reaction to SO
SO 2 Cl 2 ( g )
SO 2 (
2 and Cl
g )
Cl
2 .
2 ( g )
– At 320 o C, the rate constant is 2.20 x 10 -5 s -1 . What is the half-life of SO 2 Cl 2 vapor at this temperature?
– Substitute the value of k into the relationship between k and t 1/2.
t 1 2
0 .
693 2 .
20
10
5 s 1
61
Half-life
Sulfuryl chloride, SO 2 Cl 2 , decomposes in a first-order reaction to SO
SO 2 Cl 2 ( g )
SO 2 (
2 and Cl
g )
Cl
2 .
2 ( g )
– At 320 o C, the rate constant is 2.20 x 10 -5 s -1 . What is the half-life of SO 2 Cl 2 vapor at this temperature?
– Substitute the value of k into the relationship between k and t 1/2.
t 1 2
3 .
15
10 4 s
62
Half-life
For a second-order reaction, half-life depends on the initial concentration and becomes larger as time goes on.
– Again, assuming that [A] t = ½[A] o after one half-life, it can be shown that: –
t 1 2
1 k [ A ] o
Each succeeding half-life is twice the length of its predecessor.
63
Half-life
For Zero-Order reactions, the half-lite is dependent upon the initial concentration of the reactant and becomes shorter as the reaction proceeds.
t 1 2
[ A ] o 2 k
64
Half-life
First order reactions ln( 1 / 2 ) .
693 Second order reactions
kt
1 2
t 1 2
1 k [ A ] o
Zero order reactions
t 1 2
[ A ] o 2 k ln( 1 2 )
kt 1
65
Graphing Kinetic Data
In addition to the method of initial rates, rate laws can be deduced by
graphical methods.
– If we rewrite the first-order concentration time equation in a slightly different form, it can be identified as the equation of a straight line.
ln[ A ] t
kt
ln[ A ] o y = mx + b
66
Graphing Kinetic Data
In addition to the method of initial rates, rate laws can be deduced by
graphical methods.
– If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.
– This means if you
plot ln[A] versus time,
you will
straight line for a first-order reaction
67
Graphing Kinetic Data
In addition to the method of initial rates, rate laws can be deduced by
graphical methods.
– If we rewrite the
second-order
concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.
[ 1 A ] t
kt
1 [ A ] o y = mx + b
68
Graphing Kinetic Data
In addition to the method of initial rates, rate laws can be deduced by
graphical methods.
– If we rewrite the
second-order
concentration-time equation in a slightly different form, it can be identified – as the equation of a straight line.
This means if you
plot 1/[A] versus time ,
you will get a
straight line for a second-order reaction
.
– Figure 14.10 illustrates the graphical method of deducing the order of a reaction.
69
70
71
Collision Theory
Rate constants vary with temperature. Consequently, the actual rate of a reaction is very temperature dependent.
• Why the rate depends on temperature can by explained by
collision theory
.
72
Collision Theory
Collision theory
assumes that for a reaction to occur, reactant molecules must collide with sufficient energy and the proper orientation.
• The minimum energy of collision required for two molecules to react is called the
activation energy , E a
.
73
Transition-State Theory
Transition-state theory
explains the reaction resulting from the collision of two molecules in terms of an
activated complex.
– An
activated complex
(transition state) is an unstable grouping of atoms that can break up to form products .
– A simple analogy would be the collision of three billiard balls on a billiard table.
74
Transition-State Theory
Transition-state theory
explains the reaction resulting from the collision of two molecules in terms of an
activated complex.
– Suppose two balls are coated with a – slightly stick adhesive.
We’ll take a third ball covered with an extremely sticky adhesive and collide it with our joined pair.
75
Transition-State Theory
Transition-state theory
explains the reaction resulting from the collision of two molecules in terms of an
activated complex.
– At the instant of impact, when all three spheres – are joined, we have an unstable
transition-state complex.
The “incoming” billiard ball would likely stick to one of the joined spheres and provide sufficient energy to dislodge the other,
resulting in a new “pairing.”
76
Transition-State Theory
Transition-state theory
explains the reaction resulting from the collision of two molecules in terms of an
activated complex.
– If we repeated this scenario several times, some collisions would be successful and others (because of either insufficient energy or improper orientation) would not be successful.
– We could compare the energy we provided to the billiard balls to the
activation energy, E a
.
77
Potential-Energy Diagrams for Reactions
To illustrate graphically the formation of a transition state, we can plot the
potential energy
of a reaction versus
time.
– – – Figure 14.13
illustrates the endothermic reaction of nitric oxide and chlorine gas.
Note that the forward activation energy is the energy necessary to form the activated complex.
The D
H
of the reaction is the net change in energy between reactants and products.
78
Figure 14.13 Potential-energy curve for the endothermic reaction of nitric oxide and chlorine.
Potential-Energy Diagrams for Reactions
The
potential-energy diagram
for an exothermic reaction shows that the products are more stable than the reactants.
– – – Figure 14.14 illustrates the potential-energy diagram for an exothermic reaction.
We see again that the forward activation energy is required to form the transition-state complex.
In both of these graphs, the reverse reaction must still supply enough activation energy to form the activated complex.
80
Figure 14.14 Potential-energy curve for an exothermic reaction.
Collision Theory and the Arrhenius Equation
Collision theory maintains that the rate constant for a reaction is the product of three factors.
1.
Z
, the collision frequency
2.
f
, the fraction of collisions with sufficient energy to react
3.
p
, the fraction of collisions with the proper orientation to react
k
Zpf
82
Collision Theory and the Arrhenius Equation
Z
is only slightly temperature dependent.
– This is illustrated using the kinetic theory of gases, which shows the relationship between the velocity of gas molecules and their absolute temperature
.
velocity
3 RT abs M m or velocity
T abs
83
Collision Theory and the Arrhenius Equation
Z
is only slightly temperature dependent.
– –
This alone does not account for the observed increases in rates with only small increases in temperature.
From kinetic theory, it can be shown that a 10 o C rise in temperature will produce only a 2% rise in collision frequency.
84
Collision Theory and the Arrhenius Equation
On the other hand,
f
, the fraction of molecules with sufficient activation energy, turns out to be very temperature dependent.
– – It can be shown that following expression.
f
is related to
E a
by the
f
e E a RT
Here e = 2.718… , and R is the ideal gas constant, 8.31 J/(mol
.
K).
85
Collision Theory and the Arrhenius Equation
On the other hand,
f
, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent.
– From this relationship,
as temperature increases, f increases.
f
e E a RT
Also, a
decrease in the activation energy, E a
,
increases the value of f .
86
Collision Theory and the Arrhenius Equation
On the other hand,
f
, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent.
– This is the primary factor relating temperature increases to observed rate increases.
f
e E a RT
87
Collision Theory and the Arrhenius Equation
The reaction rate also depends on
p
, the fraction of collisions with the proper orientation.
– This factor is
independent of temperature changes
.
– So, with changes in temperature, Z and p remain fairly constant.
– We can use that fact to derive a mathematical relationship between the rate constant,
k
, and the
absolute temperature
.
88
The Arrhenius Equation
If we were to combine the relatively constant terms,
Z
let’s call it
A.
and
p
, into one constant, We obtain the
Arrhenius equation: k
A e E a RT
– The Arrhenius equation expresses the dependence of the rate constant on absolute temperature and activation energy.
89
The Arrhenius Equation
It is useful to recast the Arrhenius equation in logarithmic form.
– Taking the natural logarithm of both sides of the equation, we get:
ln k
ln A E a RT
90
The Arrhenius Equation
It is useful to recast the Arrhenius equation in logarithmic form.
We can relate this equation to the (somewhat rearranged) general formula for a straight line.
ln k
ln A E a ( 1 R T
y = b + m x
)
A plot of ln k versus (1/T) should yield a straight line with a
slope of (-E
a /R)
and an
intercept of ln A
. 91
The Arrhenius Equation
A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k 1 , (1/T 1 )) and (k 2 , (1/T 2 )).
– The two equations describing the relationship at each coordinate would be
ln k 1
ln A and ln k 2
ln A E a R E a R ( 1 T 1 ) ( 1 T 2 )
92
The Arrhenius Equation
A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k 1 , (1/T 1 )) and (k 2 , (1/T 2 )).
– We can eliminate ln A by subtracting the two equations to obtain
ln k k 1 2
E a R ( 1 T 1
1 T 2 )
93
The Arrhenius Equation
A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k 1 , (1/T 1 )) and (k 2 , (1/T 2 )).
– With this form of the equation, given the activation energy and the rate constant k 1 given temperature T 1 , we can find the rate at a constant k 2 at any other temperature, T 2 .
ln k k 1 2
E a R ( 1 T 1
1 T 2 )
94
A Problem to Consider
The rate constant for the formation of hydrogen iodide from its elements
H 2 ( g )
I 2 ( g )
2 HI ( g )
is 2.7 x 10 -4 L/(mol
.
s) at 600 K and 3.5 x 10 -3 L/(mol
.
s) at 650 K. Find the activation energy, E a .
– Substitute the given data into the Arrhenius equation.
ln 3.5
2 .
7
10 10 3
4
8.31
E a J/(mol
K) ( 1 600 K
1 650 K )
95
A Problem to Consider
The rate constant for the formation of hydrogen iodide from its elements
H 2 ( g )
I 2 ( g )
2 HI ( g )
is 2.7 x 10 -4 L/(mol
.
s) at 600 K and 3.5 x 10 -3 L/(mol
.
s) at 650 K. Find the activation energy, E a .
– Simplifying, we get:
ln ( 1.30
10 1 )
1 .
11
8.31
E a J/(mol)
( 1 .
28
10
4 )
96
A Problem to Consider
The rate constant for the formation of hydrogen iodide from its elements
H 2 ( g )
I 2 ( g )
2 HI ( g )
is 2.7 x 10 -4 L/(mol
.
s) at 600 K and 3.5 x 10 -3 L/(mol
.
s) at 650 K. Find the activation energy, E a .
– Solving for E a :
E a
1 .
11 1 .
8 .
31 J 28
/ 10
4 mol
1 .
66
10 5 J
97
Reaction Mechanisms
Even though a balanced chemical equation may give the ultimate result of a reaction, what actually happens in the reaction
may take place in several steps.
– This “pathway” the reaction takes is referred to as the
reaction mechanism
.
– The individual steps in the larger overall reaction
elementary reactions. (See animation: Decomposition of N 2 O 5 Step 1)
98
Elementary Reactions
Consider the reaction of nitrogen dioxide with carbon monoxide.
NO 2 ( g )
CO ( g )
NO ( g )
CO 2 ( g )
– This reaction is believed to take place in two steps.
NO 2 ( g )
NO 2 ( g )
NO 3 ( g )
NO ( g ) (elementary reaction) NO 3 ( g )
CO ( g )
NO 2 ( g )
CO 2 ( g ) (elementary reaction)
99
Elementary Reactions
Each step is a singular molecular event resulting in the formation of products.
– Note that NO 3 does not appear in the overall equation, but is formed as a temporary
reaction intermediate
.
100
Elementary Reactions
Each step is a singular molecular event resulting in the formation of products.
– The overall chemical equation is obtained by adding the two steps together and canceling any species common to both sides.
NO 2 ( g )
NO 2 ( g )
NO 3 ( g )
NO ( g ) NO 3 ( g )
CO ( g )
NO 2 ( g )
CO 2 ( g ) NO 2 ( g )
NO 2 ( g )
NO 3 ( g )
CO ( g )
NO 3 ( g )
NO ( g )
NO 2 ( g )
CO 2 ( g )
101
Molecularity
We can classify reactions according to their
molecularity ,
that is, the number of molecules that must collide for the elementary reaction to occur.
– A
unimolecular
reaction involves only one reactant molecule.
– A
bimolecular
reaction involves the collision of two reactant molecules.
– A
termolecular
reaction requires the collision of three reactant molecules.
102
Molecularity
We can classify reactions according to their
molecularity,
that is, the number of molecules that must collide for the elementary reaction to occur.
– Higher molecularities are rare because of the small statistical probability that four or more molecules would all collide at the same instant.
103
Rate Equations for Elementary Reactions
Since a chemical reaction may occur in several steps, there is no easily stated relationship between its overall reaction and its rate law.
• For elementary reactions, the rate is proportional to the concentrations of all reactant molecules involved.
104
Rate Equations for Elementary Reactions
For example, consider the generic equation below.
A
products
The rate is dependent only on the concentration of A; that is,
Rate
k[A]
105
Rate Equations for Elementary Reactions
However, for the reaction
A
B
products
the rate is dependent on the concentrations of both A and B.
Rate
k[A][B]
106
Rate Equations for Elementary Reactions
For a termolecular reaction
A
B
C
products
the rate is dependent on the populations of all three participants.
Rate
k[A][B][C]
107
Rate Equations for Elementary Reactions
Note that if two molecules of a given reactant are required, it appears twice in the rate law. For example, the reaction
2A
B
products
would have the rate law:
Rate
k[A][A][B] or Rate
k[A] 2 [B]
108
Rate Equations for Elementary Reactions
So, in essence, for an
elementary reaction, the coefficient of each reactant becomes the power to which it is raised
in the rate law for that reaction.
– Note that many chemical reactions occur in multiple steps and it is, therefore,
impossible
to predict the rate law
based solely on the overall reaction
.
109
Rate Laws and Mechanisms
Consider the reaction below.
2 NO 2 (g)
F 2 (g)
2 NO 2 F(g)
– Experiments performed with this reaction show that the rate law is
Rate
k[NO 2 ][F 2 ]
– The reaction is first order with respect to each reactant, even though the coefficient for NO 2 in the overall reaction is 2.
110
Rate Laws and Mechanisms
Consider the reaction below.
2 NO 2 (g)
F 2 (g)
2 NO 2 F(g)
– Experiments performed with this reaction show that the rate law is
Rate
k[NO 2 ][F 2 ]
– This implies that the reaction above is not an elementary reaction but rather the result of multiple steps.
111
Rate-Determining Step
In multiple-step reactions, one of the elementary reactions in the sequence is often slower than the rest.
– The overall reaction cannot proceed any faster than this slowest
rate-determining step
.
112
Rate-Determining Step
In multiple-step reactions, one of the elementary reactions in the sequence is often slower than the rest.
– Our previous example occurs in two elementary steps where the first step is much slower.
NO 2 (g)
F 2 (g)
NO 2 F(g)
F(g) NO 2 (g)
F(g)
NO 2 F(g) 2 NO 2 (g)
F 2 (g)
2 NO 2 F(g) (slow) (fast)
113
Rate-Determining Step
In multiple-step reactions, one of the elementary reactions in the sequence is often slower than the rest.
– Since the overall rate of this reaction is determined by the slow step, it seems logical that the observed rate law is
Rate = k 1 [NO 2 ][F 2 ].
NO 2 (g)
F 2 (g)
NO 2 F(g)
F(g) (slow)
114
Rate-Determining Step
In a mechanism where the first elementary step is the rate-determining step,
the overall rate law is simply expressed as the elementary rate law for that slow step.
– A more complicated scenario occurs when the rate-determining step contains a reaction intermediate, as you’ll see in the next section.
115
Rate-Determining Step
Mechanisms with an Initial Fast Step – There are cases where the rate determining step of a mechanism contains a
reaction intermediate
that does not appear in the overall reaction.
– The experimental rate law, however,
can be expressed only in terms of substances that appear in the overall reaction
.
116
Rate-Determining Step
Consider the reduction of nitric oxide with H 2 .
2 NO ( g )
2 H 2 ( g ) N 2 ( g )
2 H 2 O ( g )
– A proposed mechanism is: –
2 k 1 NO N k -1 2 O 2 (fast, equilibrium) N 2 O 2
H 2
N 2 O
H 2 O (slow) N 2 O
H 2 k 3 N 2
H 2 O (fast)
It has been experimentally determined that the rate law is
Rate = k [NO] 2 [H 2 ]
117
Rate-Determining Step
The rate-determining step (step 2 in this case) generally outlines the rate law for the overall reaction.
Rate
k 2 [ N 2 O 2 ][ H 2 ] (Rate law for the rate-determining step)
– As mentioned earlier, the overall rate law can be expressed only in terms of substances represented in the overall reaction and cannot contain reaction intermediates.
118
Rate-Determining Step
The rate-determining step (step 2 in this case) generally outlines the rate law for the overall reaction.
Rate
k 2 [ N 2 O 2 ][ H 2 ] (Rate law for the rate-determining step)
– It is necessary to reexpress this proposed rate law after eliminating [N 2 O 2 ].
119
Rate-Determining Step
The rate-determining step (step 2 in this case) generally outlines the rate law for the overall reaction.
Rate
k 2 [ N 2 O 2 ][ H 2 ] (Rate law for the rate-determining step)
– We can do this by looking at the first step, which is fast and establishes equilibrium.
120
Rate-Determining Step
The rate-determining step (step 2 in this case) generally outlines the rate law for the overall reaction.
Rate
k 2 [ N 2 O 2 ][ H 2 ] (Rate law for the rate-determining step)
– At equilibrium, the forward rate and the reverse rate are equal.
k 1 [ NO ] 2
k
1 [ N 2 O 2 ]
121
Rate-Determining Step
The rate-determining step (step 2 in this case) generally outlines the rate law for the overall reaction.
Rate
k 2 [ N 2 O 2 ][ H 2 ] (Rate law for the rate-determining step)
– Therefore,
[ N 2 O 2 ]
( k 1 / k
1 )[ NO ] 2
– If we substitute this into our proposed rate law we obtain: 122
Rate-Determining Step
The rate-determining step (step 2 in this case) generally outlines the rate law for the overall reaction.
Rate
k 2 [ N 2 O 2 ][ H 2 ] (Rate law for the rate-determining step)
–
Rate
k k 2 k
1 1 [ NO ] 2 [ H 2 ]
If we replace the constants (k 2 k 1 /k -1 ) with k, we obtain the observed rate law:
Rate = k[NO] 2 [H 2 ].
123
Rate-Determining Step
Determine the overall reaction and the rate expression that corresponds to the following reaction mechanism.
2A + B ↔ D D + B → E + F F → G Whitten 6 th Ed. Ch 16 # 66a 124
Rate-Determining Step
Determine the overall reaction and the rate expression that corresponds to the following reaction mechanism.
2A + B ↔ D D + B → E + F F → G Answer: 2A + 2B → E + G Rate = k[A] 2 [B] 2 125
Catalysts
A
catalyst
is a substance that provides a good “environment” for a reaction to occur, thereby increasing the reaction rate without being consumed by the reaction.
– To avoid being consumed, the catalyst must participate in at least one step of the reaction and then be regenerated in a later step.
126
Catalysts
A
catalyst
is a substance that provides a good “environment” for a reaction to occur, thereby increasing the reaction rate without being consumed by the reaction.
– Its presence increases the rate of reaction by either increasing the frequency factor,
A
(from the Arrhenius equation) or lowering the activation energy,
E a
.
127
Catalysts
Homogeneous catalysis
is the use of a catalyst in the same phase as the reacting species.
– The oxidation of sulfur dioxide using nitric oxide as a catalyst is an example where all species are in the gas phase.
2 SO 2 ( g )
O 2 ( g )
2 SO 3 ( g )
128
Catalysts
Heterogeneous catalysis
is the use of a catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a liquid or gaseous solution of reactants.
– Such surface catalysis is thought to occur by chemical adsorbtion of the reactants onto the surface of the catalyst.
– Adsorbtion is the attraction of molecules to a surface.
129
Enzyme Catalysis
Enzymes have enormous catalytic activity.
– The substance whose reaction the enzyme catalyzes is called the
illustrates the reduction in acivation energy resulting from the formation of an enzyme-substrate complex. 130
Operational Skills
Relating the different ways of expressing reaction rates Calculating the average reaction rate Determining the order of reaction from the rate law Determining the rate law from initial rates Using the concentration-time equation for first order reactions Relating the half-life of a reaction to the rate constant 131
Operational Skills
• • • • Using the Arrhenius equation Writing the overall chemical equation from a mechanism Determining the molecularity of an elementary reaction Writing the rate equation for an elementary reaction Determining the rate law from a mechanism 132
Figure 14.1: Combining product of formaldehyde with hydrogen sulfite ion.
Photo courtesy of James Scherer.
133
Figure 14.9: A plot of ln [N
2
O
5
] versus time.
134
Figure 14.15: Plot of ln
k
versus 1/
T
.
135
Animation: Decomposition of N
2
O
5
(Click here to open QuickTime animation)
136
Figure 14.20: Enzyme action (lock-and-key model).
137
Figure 14.21 Potential-energy curves for the reaction of substrate, S, to products, P.
138