chapter 13 rates of reaction
DESCRIPTION
Chapter 13 Rates of Reaction. Dr. Peter Warburton [email protected] http://www.chem.mun.ca/zcourses/1011.php. Reaction Rates. Reaction rate is concentration change divided by time change Reaction rate = D [X] / D t. Changes! Final minus initial. D [X] = [X] final – [X] initial - PowerPoint PPT PresentationTRANSCRIPT
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Reaction Rates
Reaction rate is concentration change divided by time change
Reaction rate = [X] / t
[X] = [X]final – [X]initial
t = tfinal – tinitial
We most often use molL-1 as units of concentration
This means that rate often has units molL-1s-1
3
The reaction rate is defined either as the increase in the concentration of a product
over time, or the decrease in the concentration of a reactant over time.
Reaction Rate
A + B C + D
rate = -[A] / t = -[B] / t = +[C] / t = +[D] / t
Rate is always positive, so we must put negative signs in front of reactant concentration changes!
4
2 N2O5 (g) 4 NO2 (g) + O2 (g)
5
Rate of decomposition of N2O5 =
- [N2O5] / t = - (0.0101 molL-1 – 0.0120 molL-1) / (400 s – 300
s) = 1.9 x 10-5 mol(L·s)-1
2 N2O5 (g) 4 NO2 (g) + O2 (g)
Between 300 and 400 seconds:
6
Rate of formation of NO2 =
+ [NO2] / t = + (0.0197 molL-1 – 0.0160 molL-1) / (400 s – 300
s) = 3.7 x 10-5 mol(L·s)-1
2 N2O5 (g) 4 NO2 (g) + O2 (g)
Between 300 and 400 seconds:
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Rate of formation of O2 =
+ [O2] / t = + (0.0049 molL-1 – 0.0040 molL-1) / (400 s – 300
s) = 9 x 10-6 mol(L·s)-1
2 N2O5 (g) 4 NO2 (g) + O2 (g)
Between 300 and 400 seconds:
8
1) Average reaction rate
2) Slopes
3) Time = 0
9
2 N2O5 (g) 4 NO2 (g) + O2 (g)
The three values for rate that we calculated are not the same!
Why?We have different molar amounts.
But the relative rates ARE THE SAME!
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2 N2O5 (g) 4 NO2 (g) + O2 (g)
The relative rate of formation of O2 is(1/1) 9 x 10-6 mol(L·s)-1 = 9 x 10-6 mol(L·s)-1
The relative rate of formation of NO2 is(1/4) 3.7 x 10-5 mol(L·s)-1 = 9.3 x 10-6 mol(L·s)-1
The relative rate of decomposition of N2O5 is(1/2) 1.9 x 10-5 mol(L·s)-1 = 9.5 x 10-6 mol(L·s)-1
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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.
12
Problem
Consider the following reaction
3 I- (aq) + H3AsO4 (aq) + 2 H+ (aq)
→ I3– (aq) + H3AsO3 (aq) + H2O (l)
a) If –[I-]/t = 4.8 x 10-4 mol(L·s)-1, what is the value of [I3
-]/t during the same time interval?
b) What is the average rate of consumption of H+ during the same time interval?
13
Rate Laws and Reaction Order
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.
14
Rate laws
The rate law for a reaction is the equation showing the
dependence of the reaction rate on the
concentrations of the reactants.
15
aA + bB products
Rate = k [A]m[B]n
k is a constant for the reaction at a given temperature, and is called the rate constant.
m does not have to equal a
n does not have to equal b
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“Sensitivity” to concentration change
17
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
19
Problem
Consider three reactions with their given rate laws below. What is the order of each reaction in the various reactants, and what is the overall reaction order for each reaction?
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Experimental Determination of a Rate Law
Reaction rate laws can only be 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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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 the units mol(L·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)
[I3-]/t 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 molL-1 and [I-] = 0.400 molL-1?
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Reaction Rates and Temperature
Increasing the temperature increases a chemical reactions
rate.
In general, reaction rates approximately double if you increase
the temperature by 10 °C.
37
Bumper cars
“Reactions” occur ONLY when the bumps are “very hard” and occur
“from behind”.
A gasp of surprisecould be a “reaction”
when riding in a bumper car.
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Collision theory
A + BC AB + C
If this reaction occurs in a single step, then 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.
We must insert energy to force the molecules close together. This is like forcing together the
north poles of two magnets.
This inserted energy is the kinetic energy of the molecules. It becomes 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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Figure 13.11
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Figure 13.11
There are two useful energy
differences in the Figure.
The difference in energy between products and reactants is H
The difference in energy between the transition state and the reactants is
Ea – the activation energy
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Activation energy
The activation energy (Ea) of a reaction is the will always be positive!
The energy of collision between two
molecules must beAT LEAST as big as
Ea otherwise we cannot make it to the
transition state.
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Collisions between molecules at higher temperatures are more likely to have collision energy GREATER THAN the activation
energy.
Higher temperatures mean higher rates of reaction!
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Collisions
An individual molecule collides with other molecules about once every billionth of a
second (one billion collisions per second).
If every collision was successful in creating products, then every reaction would 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 break the activation barrier is given by
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
(8.314 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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Figure 13.9
Cl2 MUST collide with the N side of
NO to form the transition state O=N--Cl--Cl‡.
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Figure 13.9
If Cl2 hits the O side, we get a
different transition state that might
not give the same products or has a higher activation
energy. (molecules “bounce off” each
other)
52
Steric factor
Our steric factor in this case would be p ~ 0.5 since half the collisions lead to the
wrong transition state.
53
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]
Since for our general reaction
Reaction rate = k [A] [BC]
k = pfZ = pZ e-Ea/RT = A e-Ea/RT (where A = pZ)
(frequency factor)
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!
59
If we graph the natural logarithm of the rate constant versus inverse
temperature
ln k (y axis) vs 1/T (x axis)
we get a straight line where the
slope = -Ea/R
So Ea = - slope x R
60
ln k vs 1/T
61
ln k2 – ln k1 = (–Ea/R) (1/T2 – 1/T1)
OR
(ln k) = (–Ea/R) (1/T)
62
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.
63
Problem
Rate constants for the decomposition of gaseous dinitrogen pentaoxide are
4.8 x 10-4 s-1 at 45 °C and 2.8 x 10-3 s-1 at 60 °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 35C?
64
Reaction Mechanisms
A reaction mechanism is the sequence of molecular events (elementary steps
or elementary reactions) that defines the pathway from the reactants to the products in the overall reaction.
The elementary reactions describe the behaviour of individual molecules while
the overall reaction tells us stoichiometry.
65
The reaction actually takes place in two elementary reactions!
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 reactions 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.
68
Reaction intermediate
A reaction intermediate is a species that is formed in an elementary step reaction, that is consumed in a later elementary step reaction.
We never see reaction intermediates in the overall reaction!
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Molecularity
The molecularity of an elementary reaction is the number of molecules
on the reactant side
of the elementary step 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 only depend 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.
The chances of four or more molecules colliding at the same time are almost impossible.
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Bumper cars
Consider bumper cars. Very often, you will hit one other bumper car. Every once and a while, you and another car will hit a third car at the same time. It is a very rare occurrence to have a “bumper car” pile-up where many cars hit a singlecar at exactly the same time.
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Problem
A suggested mechanism for the reaction of nitrogen dioxide and molecular fluorine is
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Problem
a) Give the chemical equation for the overall reaction, and identify any reaction intermediates b) What is the molecularity of each of the elementary reactions?
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Rate Laws and Reaction Mechanisms
Unlike an overall reaction the
rate law for an elementary reaction
follows DIRECTLY from the
molecularity of the step reaction!
For a general elementary step reaction
aA + bB 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]
77
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]
78
79
Elementary reaction rate laws
80
Mechanisms and overall rate law
The mechanism of the overall reaction is predicted through the elementary reactions therefore
the elementary reactions will determine the rate law of the
overall reaction!
81
Mechanisms and overall rate law
If the overall reaction occurs in ONE elementary step, then the elementary reaction and the overall
reaction ARE THE SAME. The rate law for the overall reaction is given by the
rate law for the step reaction
rate = k [CH3Br] [OH-]
82
Rate-determining step
The rate-determining step
of an overall reaction with a mechanism of two or more steps is the
elementary step reaction
which has the slowest rate.
The overall reaction can occur NO FASTER than its SLOWEST elementary reaction.
83
The second step has to wait for the first step to create the NO3, which is then used rapidly
for the second step reaction.
NO2 (g) + CO(g) NO (g) + CO2 (g)
84
Is the proposed mechanism plausible?
The elementary steps MUST ADD UP to give the overall reaction
AND
the mechanism rate law MUST BE CONSISTENT with the observed rate law.
NO2 (g) + CO(g) NO (g) + CO2 (g)
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The elementary reactions DO add up to the overall reaction.
The rate law of the rate-determining step is
rate = k1 [NO2]2
Since this is the same as the experimentally observed rate law, so this mechanism is
plausible.
NO2 (g) + CO(g) NO (g) + CO2 (g)
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Just because a mechanism is plausible doesn’t mean it
is right!
87
Problem
Write the rate law for each of the elementary reactions:
O3 (g) + O (g) 2 O2 (g)
Br (g) + Br (g) + Ar (g) Br2 (g) + Ar (g)
Co(CN)5(H2O)2- (aq) Co(CN)52- (aq) + H2O (l)
88
Problem
The following substitution reaction has a first order rate law:
Co(CN)5(H2O)2- (aq) + I- (aq) Co(CN)5I3- (aq) + H2O (l)
rate = k [Co(CN)5(H2O)2-]
Suggest a possible reaction mechanism, and show that your reaction mechanism is in accord with the observed rate law.
89
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
being consumed in the reaction.
90
How does a catalyst work?
A catalyst makes available a different reaction
mechanism that is more efficient than the
uncatalyzed mechanism.
91
How does a catalyst work?
To get from one side of a mountain to the other we have to climb up to the top (the activation energy), and then down the other side of the
mountain.
If there is a mountain pass partway up the mountain then we can climb up to the pass (a
lower activation energy) and then climb down to the other side.
Going through the pass will be quicker than going to the top!
92
2 H2O2 (aq) 2 H2O (l) + O2 (g)
Ea for this reaction is 76 kJmol-1
At room temperature, the reaction is slow.
In the presence of iodide ion the reaction is faster because a new pathway with a lower
activation energy is made available.
93
This catalyzed overall reaction is faster than the uncatalyzed reaction because it has a
lower activation energy (our “mountain pass”) of 19 kJmol-1.
Because the activation energy is about 3.75 times lower than in the uncatalyzed reaction,
the catalyzed reaction rate will be about 40 times faster than the uncatalyzed rate
constant.
94
Figure 13.13 (O3 + O is catalyzed by Cl atoms)
95
Homogeneous and Heterogeneous Catalysts
A homogeneous catalyst exists in the same phase as the reactants.
A heterogeneous catalyst exists in a different phase (usually solid) than the
reactants. I- is a
homogeneous catalyst here
Pt is a heterogeneous
catalyst here)()()( 62
Pt solid242 gHCgHgHC
96
Figure 13.15
97
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).
98
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