Chapter 14 Chemical Kinetics Dr. Peter Warburton [email protected] .
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Transcript of Chapter 14 Chemical Kinetics Dr. Peter Warburton [email protected] .
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Rates
Rates can be measured as the concentration change of a chemical (X)
over a period (change) of time
Rate = [X] / t
Often use molL-1 (or M) as units of concentration
so rate often has units molL-1s-1 (or Ms-1 )
01
tt
tt
XXRate 01
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Rates can be expressed in two different ways. We can look at the
formation of a product
(increase in the concentration over time)
or the
disappearance of a reactant
(decrease in the concentration over time).
Rate
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Rate of formation of a product
2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)
t0 = 0.0 s [Fe2+]t0 = 0.0000 M
t1 = 38.5 s [Fe2+]t1 = 0.0010 M
Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0.0000) M
152
2 sM 10 x 2.6s 38.5
M 0.0010
Δt
FeΔFe offormation of rate
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Rate of disappearance of a reactant
2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)
t0 = 0.0 s [Sn2+]t0 = 0.0015 M
t1 = 38.5 s [Sn2+]t1 = 0.0010 M
Δt = 38.5 s Δ[Sn2+] = (0.0010 – 0.0015) M 15
22 sM 10 x 3.1
s 38.5
M) (-0.0005
Δt
SnΔSn of ncedisappeara of rate
The word “disappearance” implies the negative sign. Because of this, all rates are considered to be
POSITIVE!
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a A + b B c C + d D
Reaction rates are always positive, so we must put negative signs in front of reactant concentration changes!
To account for the stoichiometric relationships and their effect on rate, we must always divide by the stoichiometric coefficient for the chemical.
Δt
DΔ
d
1
Δt
CΔ
c
1
Δt
BΔ
b
1
Δt
AΔ
a
1 ratereaction - - + +
Reaction rates
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Reaction rate
2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)
15
4
4 slidesM 10 x 2.6
2
5 slidesM 10 x 1.3
23
sM 10 x 1.3ratereaction
Δt
SnΔ
1
1
Δt
FeΔ
2
1
Δt
SnΔ
1
1
Δt
FeΔ
2
1 ratereaction
15
15
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Measuring rates
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Measuring rates
1) Average rates
2) Slopes
3) Time = 0
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Problem
At some point in the reaction
2 A + B 3 C
[A] = 0.3629 M. At a time 8.25 minutes later [A] = 0.3187 M. What is the average rate of the disappearance of A in Ms-1, what is the average rate of the formation of C in Ms-1, and what is the average rate of reaction in Ms-1 over that time interval?
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Problem answers
15
14
15
sM 10 x 464ratereaction average
sM 10 x 341C offormation of rate average
sM 10 x 8.93A of ncedisappeara of rate average
.
.
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Instantaneous reaction rates
What’s happening at “this instant in time”?
The initial rate is the instantaneous reaction rate for a
reaction at time zero.
We can use instantaneous reaction rates.
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Rate laws
The rate of a chemical reaction depends on the concentration of some or all of the reactants.
A reactant might not affect the rate, regardless of its
concentration.
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Rate laws
The rate law for a reaction
is the equation showing the
dependence of the reaction rate on
concentrations
of the reactants.
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a A + b B + … products
rate = k [A]m[B]n…
k is a constant for the reaction
AT A GIVEN TEMPERATURE,
and is called the rate constant. The stoichiometry of the balanced reaction equation IS NOT ALWAYS the source of the
rate equation exponents!
m DOES NOT HAVE TO equal a
n DOES NOT HAVE TO equal b
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Reaction order
Reaction order with respect to a given reactant
is the value of the exponent of the rate law equation for the specific reactant only.
The overall reaction order is the
sum of the reaction orders for
all reactants.
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Reaction order example
rate = k [A]2[B]
The reaction order with respect to A is 2 or the reaction is second order in A
The reaction order with respect to B is 1 or the reaction is first order in B
The overall reaction order is 3 (2 + 1 = 3) or the reaction is third order overall
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“Sensitivity” to concentration change
Rate change of the reaction if [A] is
doubled depends on 2m
no change
ratedoubles
ratequadruples
ratehalves
zeroth order in A
first order in A
second order in A
NOTE: negative or non-integer orders
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Method of initial rates
Reaction rate laws are often determined experimentally!
We most commonly carry out a series of experiments in which the
initial rate of the reaction is measured as a function of
different initial concentrations of reactants
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Method of initial rates
If you see a table like this with chemical concentrations or pressures and rate data, chances are good the
question is a method of initial rates problem.
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Method of initial rates
IGNORE THE REACTION with this type of problem. The chemicals in the TABLE
are the interesting ones.You always require at least one more
experimental reaction than your number of chemicals given in your table!
Sometimes we are given a table with an extra experiment which we can use to
check if we’ve done everything correctly.
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2 NO (g) + O2 (g) NO
2 (g)
Since rate laws are always expressed in terms of reactants (and sometimes catalysts – we’ll see these later), lets create a general form of the rate law for this reaction based on what chemicals the TABLE tells us are involved in the rate of the reaction.
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2 NO (g) + O2 (g) NO
2 (g)
rate = k [NO]m[O2]n
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Method of initial rates
For our initial reactant order determination we need to choose a pair of reactions where only one reactant concentration changes. Experiments #1 and #2 fulfill this condition.
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Method of initial rates
rate = k [NO]m[O2]n
Since k is a constant then
k for experiment 1
IS EQUAL TO
k for experiment 2!
k = rate / [NO]m[O2]n
n22
m2
n
12m1
2
1n
22m2
2n
12m1
1
ONO
ONO
rate
rate so
ONO
rate
ONO
rate
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Reaction order w.r.t. NO
n22
m2
n
12m1
2
1
ONO
ONO
rate
rate
0.50 log m0.25 log
(0.50) log0.25 log
(0.50)0.25
M) (0.015M) (0.030
M) (0.015M) (0.015
sM 0.192
sM 0.048
m
m
nm
nm
1-
-1
2m0.301
0.602m
0.50 log
0.25 logm
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Reaction order w.r.t. O2
n32
23
n
1221
3
1
ONO
ONO
rate
rate
0.50 logn 0.50 log
(0.50) log0.50 log
(0.50)0.50
M) (0.030M) (0.015
M) (0.015M) (0.015
sM 0.096
sM 0.048
n
n
n2
n2
1-
-1
1n0.301
0.301n
0.50 log
0.50 logn
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ALTERNATE reaction order w.r.t. O2
n
n2
n2
n2
1-
-1
(0.50)]25.0[0.125
(0.50)(0.50)0.125
M) (0.030M) (0.030
M) (0.015M) (0.015
sM 0.384
sM 0.048
n42
24
n
1221
4
1
ONO
ONO
rate
rate
1n
(0.50)0.50
(0.50)0.25
0.125
n
n
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Our rate law
rate = k [NO]2[O2]1
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Rate constant using experiment 1
k = rate / [NO]2[O2]1
12-4
2
368
1
2
1
22
sM 10 x 1.4k
M 10 x 3.3
sM 0.048
M 0.015M 0.015
sM 0.048
ONO
ratek
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Rate constant using experiment 2
k = rate / [NO]2[O2]1
12-4
2
355
1
2
1
22
sM 10 x 1.4k
M 10 x 1.3
sM 0.192
M 0.015M 0.030
sM 0.192
ONO
ratek
The rate constant is the same, as it should be!
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Check using extra experiment
The rate is the same as the experimentally observed rate (within rounding errors). We MUST have done
everything right!
rate = (1.42 x 104 M-2s-1) [NO]2[O
2]1
11-3
35-12-42
212-42
221-24
2
sM 10 x 3.8rate
M 10 x 2.7sM 10 x 1.4rate
M 0.030M 0.030sM 10 x 1.4rate
ONOsM 10 x 1.4rate
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Units of rate constants
Rate always has units in terms of concentration per time unit.
Usually it is molL-1·s-1 (or M·s-1)
To ensure we get the right units for rate means the rate constant must have different units depending
on the overall reaction order.
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Problem
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) I3- (aq) + 2 H2O (l)
The rate of formation of the red-coloured triiodide ion
[I3-]/t (and therefore the reaction rate – why?) can be
determined by measuring the rate of appearance of the colour.
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Problem
a) What is the rate law for the formation of I3-?
b) What is the value for the rate constant?c) What is the initial rate of formation of triiodide
when the concentrations are [H2O2] = 0.300 M and [I-] = 0.400 M?
Expt Initial [H2O2] (M) Initial [I-] (M) Initial [I3-]/t (Ms-1)
1 0.100 0.100 1.15 x 10-4
2 0.100 0.200 2.30 x 10-4
3 0.200 0.100 2.30 x 10-4
4 0.200 0.200 4.60 x 10-4
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Problem answers
a) rate = k [H2O2] [I-]
b) k = 1.15 x 10-2 M-1s-1
c) rate = 1.38 x 10-3 Ms-1
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Zero order reactions
In a zero order reaction the rate of the reaction does NOT depend on the concentration of ANY of the reactants.
rate = k [NH3]0 = k (1) = k = constant
(g) H 3 (g) N (g) NH 2 22catalystPt
K 11303
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Integrated rate law – 0th order rxn
b
0xmy
t
0t
A tkA
OR
k tA A
0-tk AA
dt kAd
dtk Ad
kdt
Ad
k rate
0t
t
0
A
A
t
0
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Features of zero order reactions
Concentration versus time graph gives a
straight line.
Integrated rate law
Rate constant is the negative slope (UNITS!)
b
0xmy
t A tkA
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First order reactions
In a first order reaction the overall order of the reaction is 1. A common type of first order reaction is the decomposition of a chemical.
rate = k [H2O
2]1
(g) O2
1 (l) OH (aq) OH 2222
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Integrated rate law – 1st order rxn
b
0xm
y
t
0t
Aln tkAln
OR
k tAln Aln
0-tk A
Aln
dt kA
Ad
dtk A
Ad
Akdt
Ad
Ak rate
0
t
t
0
A
A
t
0
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Integrated rate law – 1st order rxn
Natural logarithm of concentration versus
time graph gives a straight line.
Rate constant is the negative slope
(UNITS!)
b
0xm
y
t Aln tkAln
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Problem
The reaction
A 2 B + C
is first order. If the initial [A] = 2.80 M and k = 3.02 x 10-3 s-1, what is the value of [A] after 325 s?
Answer: [A] = 1.05 M
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Half-life of 1st order reactions
From the integrated rate law of a first order reaction we can look at what time [A] is one-half of the initial concentration (½ [A]0).
We call this time the half-life (t½).
k
0.693t
k
0.693-
k2
1lnt
k t21ln
k tA
A21
ln
k tA
Aln
21
21
21
21
0
0
0
t
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Half-life of 1st order reactions
The half-life (t½) of a first order
reaction is constant!
k
0.693t
k
0.693-
k2
1lnt
k t21ln
k tA
A21
ln
k tA
Aln
21
21
21
21
0
0
0
t
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Half-life of 1st order reactions
[A]t = 1/2 [A]0 at t½
[A]t = 1/4 [A]0 at 2 x t½
[A]t = 1/8 [A]0 at 3 x t½
and so on
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Problem
Consider the first order reaction
A products
with k = 2.95 x 10-3 s-1. What percent of A remains after 150 s?
Answer: % A remaining = 64.2%
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Problem
At what time (in minutes) after the start of the reaction is a sample of H2O2 (aq) two-thirds decomposed if k = 7.30 x 10-4 s-1?
Answer: time = 25.1 min
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1st order reactions of gases
Amounts of gases are often measured
by pressure instead of concentration
RT
PA so
RT
P
V
nor
nRTPVbut V
n
volume
A molesA
A
A
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Problem
Start with DTBP at a pressure of 800.0
mmHg at 147 C.
What will the pressure be at t = 125 min if
t½ = 8.0 x 10 min?
k tP
Pln
k t
RTP
RTP
ln
k tA
Aln
0A
tA
0
A
t
A
0
t
Answer:
k = 8.66 x 10-3 min-1
pressure = 271 mmHg
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Radioactive decay as 1st order rxns
Many radioactive decay processes are first order…
half-life of 5730 yrs
half-life of 8.04 days
half-life of 4.51 x 109 yrs
eνNC 147
decay β146
-
eνXeI 13154
decay β13153
-
242
23490
decay α23892 HeThU
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Second order reactions
In a second order reaction the overall order of the reaction is 2. If there is only a single reactant chemical, like in
Then the rate is second order with respect to that chemical
rate = k [NO2]2
(g) O (g) NO 2 (g) NO 2 22
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Integrated rate law – 2nd order rxn
0-tk A
1-
A
1-
dt kA
Ad
dtk A
Ad
Akdt
Ad
Ak rate
0t
t
0
A
A2
2
2
2
t
0
b
0xm
y
t
0t
A
1 tk
A
1
OR
k tA
1
A
1
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Integrated rate law – 2nd order rxn
Inverse of concentration versus
time graph gives a straight line.
Rate constant is the slope (UNITS!)
b
0xm
y
t A
1 tk
A
1
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Half-life of 2nd order reactions
From the integrated rate law of a second order
reaction we can look at what time [A] is one-half of
the initial concentration
(½ [A]0).
We also call this time the
half-life (t½).
21
0
02
1
02
1
21
00
0t
tAk
1
Ak t12
Ak t1
1
211
k tA
1
A21
1
k tA
1
A
1
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Half-life of 2nd order reactions
The half-life (t½) of a second
order reaction is NOT constant!
21
0
02
1
02
1
21
00
0t
tAk
1
Ak t12
Ak t1
1
211
k tA
1
A21
1
k tA
1
A
1
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Half-life of 2nd order reactions
The greater the initial conc. of A
([A]0),
the smaller the half-life!
21
0
02
1
02
1
21
00
0t
tAk
1
Ak t12
Ak t1
1
211
k tA
1
A21
1
k tA
1
A
1
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Half-life of 2nd order reactions
Since
reactant concentration
is halved
over a half-life, the
next half-life
is twice as long
compared to the previous half-life.
21
0
tAk
1
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Pseudo first-order reactions
If we carry out some second (or higher) order reactions under conditions where some reactant concentrations do not change significantly, then the reaction may APPEAR to act like a lower order reaction (pseudo-first order, …)
OHHC COOHCHOH HCOOCCH 5232523
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Kinetics summary - rates
•You can calculate rate with known rate law•If you don’t know a rate law, the rate can be determined from the tangent of a [A] versus time graph or by –[A]/t over a small T
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Kinetics summary – rxn orders
•Method of initial rates•Plot graphs to find a straight line
•First order reactions have constant half-life•Use data in integrated rate laws – k MUST NOT change
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Kinetics summary – rate constants
•Rate constants are related to slope of appropriate straight line graphs•Use data in integrated rate laws to get k. Once you know k you can use the integrated rate law to solve for concentrations or times.•Use half-life of a first order reaction to determine k
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Problem
Without graphing, find the order and the rate constant for the following reaction using the given data
B productsTime (s) [B] (M) Time (s) [B] (M)
0 0.88 100 0.44
25 0.74 150 0.31
50 0.62 200 0.22
75 0.52 250 0.16
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Problem answer
Reaction is first order.
k = 6.93 x 10-3 s-1
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Bumper cars
A gasp of surprise
could be a “reaction”
when riding in a
bumper car
“Reactions” generally occur when the bumps are very hard and when they
come from behind
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Collision theory
A + BC AB + C
At some point in time, the B-C bond starts to break, while the A-B bond starts to form.
At this point, all three nuclei are weakly linked together.
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Collision theory
Molecules tend to repel each other when they get close.
It takes energy to force the molecules close together! This is like forcing together the north poles of two magnets.
This energy is the kinetic energy of the molecules. It converts to potential energy as the molecules get closer.
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Collision theory
A---B---C
has a higher potential energy than either A + B-C or A-B + C
A---B---C is the transition state
or the activated complex
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The difference in energy between
products and reactants is E
The difference in energy between the transition state and the reactants is
Ea – the activation energy
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Temperature is the average kinetic energy of molecules.
Collisions between molecules at higher temperatures are more likely to have
energy GREATER THAN the activation energy.
Higher temperatures mean higher rates of reaction!
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Collisions
1 billion collisions per molecule per second
Every reaction should be almost instantaneous.
This is not the case.
Not every collision breaks the activation energy barrier!
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Collisions
The fraction of collisions that have enough energy to overcome the activation energy barrier is
f = e-Ea/RT
e is approximately 2.7183,
Ea is the activation energy, T is the temperature in Kelvin,
R is the gas law constant (8.3145 JK-1mol-1)
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Bumper cars and energy
bumper car - a more energetic collision is more likely to make us gasp (our “reaction”)molecular collisions -higher energy collisions are more likely to lead to reaction (by overcoming the activation energy)
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Bumper cars and orientation
You are also more likely to gasp if you are hit from behind by another bumper car.
The orientation of how the collision occurs is also important to get a “reaction.”
The same is true for molecules where the fraction of collisions that have the right orientation is p. We call this fraction p the
steric factor.
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General reaction A + BC AB + C
Atom A MUST collide with the B side of BC to form the transition state A---B---C.
If atom A hits the C side, we get a different transition state A---C---B.
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General reaction A + BC AB + C
About half of our collisions won’t give the right reaction EVEN IF THEY BREAK THE ACTIVATION
ENERGY BARRIER!The steric factor p will be about 0.5.
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Collision rate = Z [A] [BC]
Z is a constant related to the collision frequency.
Recall only a fraction (f) of the collisions have a collision energy greater than or equal to
the activation energy.
Of those collisions, only a fraction (p) have the correct orientation to proceed through
the transition state to the products.
General reaction A + BC AB + C
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Reaction rate = p x f x Collision rate Reaction rate = pfZ [A] [BC]
If for our general reaction
rate = k [A] [BC]
k = pfZ = pZ e-Ea/RT = A e-Ea/RT (where A = pZ)
General reaction A + BC AB + C
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Arrhenius Equation
pZ = AAs T increases
k increases
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Problem
AB + CD AC + BD
What is the value of the activation energy for
this reaction? Is the reaction endothermic or
exothermic?
Suggest a plausible
structure for the transition
state.
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Using the Arrhenius Equation
If we know the rate constants for a reaction at two different temperatures, we can then
calculate the activation energy.
k = A e-Ea/RT
ln k = ln (A e-Ea/RT)
ln k = ln (A) + ln (e-Ea/RT)
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ln k = ln (A) – (Ea/RT)
This is the equation for a straight line!
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slope = -Ea/R so Ea = - slope x R
A graph of the natural logarithm
of the rate constant versus
inverse temperature
will give a straight line
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If we know
the rate constants (k1 and k2)
at two temperatures (in Kelvin!) T1 and T2
we will see the rate constant has changed because the temperature has changed.
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One of the things that
HAS NOT CHANGED
with the temperature is the
ACTIVATION ENERGY
-Ea/R (at T1) = -Ea/R (at T2) It is constant just like slope is constant
for a straight line!
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ln k2 = ln (A) – (Ea/RT2)
ln k1 = ln (A) – (Ea/RT1)
Subtract the bottom from the top on both sides
ln k2 – ln k1 = [-(Ea/RT2)] – [-(Ea/RT1)]
Pull out –Ea/R and put it in front
ln k2 – ln k1 = (–Ea/R) (1/T2 – 1/T1)
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Your textbook says
ln (k2/k1) = (Ea/R) (1/T1 – 1/T2)
This is absolutely correct as well! Use whichever form of the relation that
you feel more comfortable with mathematically. There are more forms
given in the text.
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Problem
Rate constants for the decomposition of gaseous dinitrogen pentaoxide are
3.7 x 10-5 s-1 at 25 °C and 1.7 x 10-3 s-1 at 55 °C
2 N2O5 (g) 4 NO2 (g) + O2 (g)
What is the activation energy of this reaction in kJmol-1?What is the rate constant at 35 C?
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Problem answer
Ea = 104 kJmol-1
k35C = 1.4 x 10-4 s-1
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Reaction Mechanisms
A reaction mechanism is the sequence of molecular events (elementary steps or elementary processes) that defines the pathway from the reactants to the
products in the overall reaction.
The elementary processes describe the behaviour of individual molecules while
the overall reaction tells us stoichiometry.
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The reaction actually takes place in two elementary processes!
2 NO2 NO and NO3
NO3 + CO NO2 and CO2
NO2 (g) + CO (g) NO (g) + CO2 (g) (Overall Reaction)
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NO2 (g) + CO (g) NO (g) + CO2 (g) (Overall Reaction)
Elementary processes must add together to give the overall
equation!
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NO2 (g) + CO (g) NO (g) + CO2 (g) (Overall Reaction)
Some of the “crossed-out” chemicals are neither reactants nor products in the overall reaction.
For example, in the above reaction NO3 is formed in one elementary step and
consumed in a later elementary step.
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Reaction intermediate
A reaction intermediate is a species that is formed in an elementary process, that is consumed in a later elementary process.
We never see reaction intermediates in the overall reaction!
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Molecularity
The molecularity of an elementary process is the number of molecules
on the reactant side
of the elementary process reaction.
A one molecule elementary reaction is unimolecular.
A two molecule elementary reaction is bimolecular.
A three molecule elementary reaction is termolecular.
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Molecularity
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Chances for molecularity
The chances of a unimolecular reaction occurring only depends on the one molecule, and are good.
A bimolecular reaction requires that two molecules collide with each other. This isn’t difficult and happens quite often.
A termolecular reaction requires that three molecules collide with each other at the same time. The chances of this happening are not very good.
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Bumper cars
Consider bumper cars. Very often, you will hit one other bumper car. It is a very rare occurrence to have a “bumper car” pile-up where three or more cars hit at exactly the same time.
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Problem
A suggested mechanism for the reaction of nitrogen dioxide and molecular fluorine is shown
a) Give the chemical equation for the overall reaction, and identify any reaction intermediates. b) What is the molecularity of each of the elementary processes?
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Problem answer
2 NO2 (g) + F2 (g) 2 NO2F (g)F is a reaction intermediate (formed in step 1 then consumed in step 2)Step 1 is bimolecular.Step 2 is bimolecular.
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Rate Laws and Reaction Mechanisms
UNLIKE an overall reaction the rate law for an elementary process
follows directly from the molecularity of the elementary
reaction!For a general elementary process
a A + b B products
rate = k [A]a [B]b
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Ozone
Unimolecular decomposition of ozone.
O3 (g) O2 (g) + O (g)
The rate law will be first order with respect to ozone
rate = k [O3]
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Bimolecular reaction
A + B productsReaction depends on collisions between
molecules A and B
Increase A, you increase # collisions
Increase B, you increase # collisions
rate = k [A] [B]
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Bimolecular reactions for one reactant
A + A products
Increase A, you increase # collisions for EACH A
rate = k [A] [A] = k [A]2
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Mechanisms and overall rate law
The mechanism of the overall reaction is predicted through the elementary processes therefore
the elementary processes will determine the rate law of the
overall reaction!
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One step mechanisms
If the overall reaction occurs in one elementary process, then
the two reactions are the same.
The rate law for the overall reaction is the same as the
elementary reaction rate law.
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Single step mechanism
rate = k [CH3Br] [OH-]
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Rate-determining step
For reaction mechanisms of more than one elementary step, one of the elementary step reactions MAY HAVE a much slower rate than any of the other steps.
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Rate-determining step
The rate-determining step of an overall reaction is an elementary step reaction
which has a much slower rate than any other step.
The overall rate law will match the rate law of the rate-determining step
in this case.
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Proposing mechanisms
A proposed mechanism must satisfy two criteria to be considered plausible:
1) the elementary processes must add up to give the appropriate overall reaction
2) the rate law that arises from the proposed mechanism must be consistent with the observed rate law
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H2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g)
Observed rate law
rate = k [H2][ICl]
HCl 2 IICl 2 H
HCl IICl HI :FAST
HCl HIICl H :SLOW
22
2
2
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H2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g)
The rate law of the rate-determining step israte = k [H2][ICl]
It turns out this is the same as the experimentally observed rate law, so this
mechanism is plausible.
HCl 2 IICl 2 H
HCl IICl HI :FAST
HCl HIICl H :SLOW
22
2
2
The elementary reactions DO add up to the overall reaction.
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2 NO (g) + O2 (g) 2 NO2 (g)
Observed rate = k [NO]2 [O2]
The rate law is termolecular. The chances of a single slow termolecular
step reaction occurring are very poor!
We need a plausible mechanism where the slow step cannot be the first step.
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Fast reversible step – slow step
A reversible reaction is one wherereactants can become products
and the products can become reactants
AT THE SAME TIME.
Eventually both reactionshave the same rate
(the system is at equilibrium)
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Fast reversible step – slow step
Rate law of slow step
rate3 = k3 [N2O2] [O2]2
k2
2k
222
22k
k
NO 2O NO 2
NO 2O ON :SLOW
ONNO 2 :FAST
3
2
1
BUT N2O2 is a reaction intermediate (concentration is effectively constant!)
We can’t have it in the rate law!
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Fast reversible step – slow step
Rate law of fast forward step
rate1 = k1 [NO]2
2k
2
2k
222
22k
k
NO 2O NO 2
NO 2O ON :SLOW
ONNO 2 :FAST
3
2
1
Rate law of fast reverse step
rate2 = k2 [N2O2]
At equilibrium rate1 = rate2
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Fast reversible step – slow step
2k
2
2k
222
22k
k
NO 2O NO 2
NO 2O ON :SLOW
ONNO 2 :FAST
3
3
2
1
k1 [NO]2 = k2 [N2O2]so [N2O2] = k1/k2 [NO]2
Let’s say that k1/k2 = K
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Fast reversible step – slow step
Rate law of slow step
rate3 = k3 [N2O2] [O2]2
k2
2k
222
22k
k
NO 2O NO 2
NO 2O ON :SLOW
ONNO 2 :FAST
3
2
1
We can substitute in [N2O2] from fast rxn! rate3 = k3 K [NO]2 [O2]
If we let k = k3 K then we see the rate law of the slow reaction does match the
observed rate law!
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Just because a mechanism is plausible doesn’t mean it is right!
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Problem
In a proposed two-step mechanism for CO + NO2 CO2 + NO, the second, fast step is
NO3 + CO NO2 + CO2.
What must be the slow step? What would you expect the rate law of the rxn to be?
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Problem answer
Slow step
NO2 + NO2 NO3 + NO.
Rate law: rate = k [NO2]2
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Problem
Show that the proposed mechanism is plausible for the reaction
2 NO2 (g) + F2 (g) 2 NO2F (g)
if the rate law is rate = k [NO2][F2]
(g) FNONO (g) F :FAST
(g) F (g) FNO(g)FNO :SLOW
(g)FNO(g) F (g) NO :FAST
22
2 22
2222
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Steady-state approximation
More than one elementary process may determine the rate of an overall
reaction.
No rate-determining step!
Can use the steady state approximation in this case.
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2 NO (g) + O2 (g) 2 NO2 (g)
Observed rate = k [NO]2 [O2]
We’ve already seen this as the fast reversible step – slow step mechanism
example.
Let’s propose a mechanism WITHOUT making any initial assumptions about
the step rates!
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2 NO (g) + O2 (g) 2 NO2 (g)
2k
222
k22
22k
NO 2 O ON :3 Process
NO NO ON :2 Process
ONNO NO :1 Process
3
2
1
This is essentially the same as the fast reversible step – slow step mechanism, but we treat the reversible reaction as
two separate forward reactions.
Each step has its own rate law.
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2 NO (g) + O2 (g) 2 NO2 (g)
2k
222
k22
22k
NO 2 O ON :3 Process
NO NO ON :2 Process
ONNO NO :1 Process
3
2
1
We know a reaction intermediate ALWAYS has a near-zero concentration that does not change much with time (otherwise we would see it build up!)
This means [N2O2]/t 0 (steady state)
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2 NO (g) + O2 (g) 2 NO2 (g)
2k
222
k22
22k
NO 2 O ON :3 Process
NO NO ON :2 Process
ONNO NO :1 Process
3
2
1
Choose our starting point as the step which has a rate law closest to the observed one, while including the reaction intermediate.
rate = k3 [N2O2] [O2]
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2 NO (g) + O2 (g) 2 NO2 (g)
[N2O2]/t 0
2222
2222
ON ncedisappeara of rate- ONformation of rate
0 ON ncedisappeara of rate ONformation of rate
We form the intermediate in process 1 and consume it in processes 2 and 3
2223222
nce!disappearafor ratesreaction positive used webecause negative
22
2122
OONkONk
3 process rate 2 process rate- ON ncedisappeara of rate
NOk 1 process rate ONformation of rate
BE CAREFUL! These are two different negative signs!
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2 NO (g) + O2 (g) 2 NO2 (g)
22232222
1
22232222
1
2222
OONkONk NOk
OONkONk- NOk
ON ncedisappeara of rate- ONformation of rate
Solve for [N2O2]
232
21
22
232222
1
Okk
NOkON
Okk ON NOk
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2 NO (g) + O2 (g) 2 NO2 (g)
Substitute into starting point
22
232
31
2232
21
3
2223
ONOOkk
kk rate
OOkk
NOkk rate
OONk rate
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2 NO (g) + O2 (g) 2 NO2 (g)
22
232
31 ONOOkk
kk rate
Very close to the observed rate law! How do we make the first part constant?
Now we assume something about step rates!
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2 NO (g) + O2 (g) 2 NO2 (g)
Assume rate2 >> rate3
so k2 [N2O2] >> k3 [N2O2] [O2]
k2 >> k3 [O2] SO k2 + k3 [O2] k2
22
2
31 ONO
k
k
kk rate
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Catalysis
Reaction rates are not just affected by reactant concentrations and
temperatures.
A catalyst is a substance that increases the rate of a reaction
WITHOUT undergoing permanent change in the reaction.
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How does a catalyst work?
A catalyst makes available a different reaction pathway that
is more efficient than the uncatalyzed mechanism because
this pathway has a lower activation energy.
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Homogeneous catalysts
A homogeneous catalyst exists in the same phase as the reactants.
I- is a homogeneous catalyst for the decomposition of hydrogen peroxide
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Decomposition of H2O2
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Decomposition of H2O2
Catalysts don’t appear in the
overall balanced equation!
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Heterogeneous catalysts
A heterogeneous catalyst exists in a different phase
(usually solid) than the reactants.
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Heterogeneous catalysts
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Heterogeneous catalysts
Most catalysts used in industry
are heterogeneous
It is much easier to separate a solid from a gas or liquid (for example) than two liquids or gases).
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Enzymes are catalysts
In living beings catalysts are usually called enzymes
Carbonic anhydrase catalyzes the reaction of carbon dioxide with water
CO2 (g) + H2O (l) H+ (aq) + HCO3- (aq)
The enzyme increases the rate of this reaction by a factor of 106. Equivalent to
about a 200 K increase …
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Enzymes
Lock-and-key model