Rates of Reactions
Reaction Rates
Chapter 15
Macroscopic
observed rate law
catalysts
increase rate
Microsopic
deduced mechanism
Reaction order
concentration
Rate constant
temperature
change mechanism
and/or lower Ea
no effect on equil'm
Elementary
reaction steps
Rate limiting
activation energy
1
Factors affecting rates
• Temperature
– Increasing temperature increases rate.
– More molecules have sufficient energy to react.
• Reactant concentrations
– Dependence on concentration must be determined experimentally.
– Can be used to deduce mechanism.
• Catalysts
– speed up reactions
– heterogeneous (e.g. solids) or homogeneous (same phase)
2
Dependence of concentration
on time, in solution
1
0.8
concentration
For the simple reaction A  C,
starting with [A] = 1.0 and [C] = 0
• Concentration of A decreases.
• Concentration of C increases at
the same rate.
• Reaction slows, but continues
until A runs out, or until
equilibrium is established.
• At completion, or equilibrium,
concentrations of A and C are
constant.
0.6
A
C
0.4
0.2
0
time
3
Dependence of concentration
on time, in solution
Rate of reaction can be expressed
as the rate of disappearance of A or
as the rate of appearance of C.
1.2
concentration
For the reaction A  C,
Rate = -D[A]/D t
= +D[C]/Dt
1.4
1
0.8
A
C
0.6
0.4
Once we see how the rate of reaction
depends on the concentrations, we will
write mathematical expressions for the
concentrations as a function of time,
these are the integrated rate laws.
0.2
0
time
4
Dependence of concentration
on time, in solution
For the reaction A  2C,
The rate of appearance of C is
twice the rate of disappearance of A
In general, for any reaction
aA+bBcC
1.6
concentration
+ D[C]/ D t = 2(- D[A]/ D t )
2
1.2
A
C
0.8
0.4
- D[A] = - D[B] = + D[C]
aDt
bDt cDt
0
time
5
Measuring the rate of a reaction
• The rate is often measured as D[X]/Dt,
where X may be a reactant or product.
• Depending on the nature of X, the change in
concentration may be monitored by a change in
–
–
–
–
–
colour (intensity of some wavelength)
pressure (for gases)
pH (for OH- or H3O+)
conductivity (ions)
radioactivity, etc.
6
Measuring the initial rate
• The rate of the forward reaction depends on the
concentration of reactants, not products.
• The dependence may be linear, quadratic, etc.,
this must be determined experimentally.
• The rate is measured at the beginning of the
reaction (the initial rate), as a function of the
initial reactant concentrations
• This determines the reaction order.
7
Initial reaction rates
2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
[NO] /M [H2] /M Rate /Ms-1 factor
0.210
0.122
0.122
0.420
double NO
0.210
0.0339
baseline
0.136
4 = 22
R  [NO]2
0.0678
0.244
double H2
2 = 21
R [H2]1
Rate = k [NO]2 [H2]1
8
Initial reaction rates
reaction order
[A] /M
[B] /M
Rate /Ms-1 factor
1
2
double A
1
1
1
4
1
2
2
double B
Rate = k [A]2 [B]1
Reaction is second order in A,
first order in B
and third order overall
baseline
4 = 22
[A]2
2 = 21
[B]1
9
CONSIDER THE RATE DATA FOR THE
REACTION:
2NO + O2
2NO2
[NO]
mol/L
0.01
0.02
0.01
[O2]
mol/L
0.01
0.01
0.02
-D[NO]/ Dt
mol/L.s
-5
2.5 x 10
1.0 x 10
-4
5.0 x 10
-5
10
Reaction order & the rate law
• The rate law is: Rate = k[A]x[B]y
• The order of a reaction is equal to the value of the
exponent in the rate law.
• The reaction has an order with respect to each
reactant (and each catalyst).
• The overall reaction order is the sum of the
individual orders.
• k is the rate constant, which depends on T.
11
The rate constant
• Once the form of the rate law is known, we can fill
in the data from any one run of our determination
to find the rate constant.
e.g. 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
Rate = 0.0339 M·s-1 = k (.210 M)2(.122 M)
k = 6.30 M-2 ·s-1
• The units of k depend on the order of the reaction
12
The rate constant
• The value of k depends on the nature of the reactants and
on the temperature.
• Arrhenius found that the temperature dependence could be
expressed as
k = Ae-Ea / RT
• The preexponential factor A, and the activation energy, Ea,
are relatively independent of temperature.
• What are these parameters?
• Why does k have this dependence?
13
Rate Law Determination
•Consider the combination reaction of NO and O2 to produce NO2 :
•2 NO(g) + O2(g)  2 NO2 (g)
•Determination of the Rate Law (via Methods of Initial Rates)
•
•
Experiment
Initial Concentrations
(mol/ L)
[NO]
[O2]
Initial Rate
•1
•2
•3
•4
•5
0.020
0.020
0.020
0.040
0.010
0.028
0.057
0.114
0.227
0.014
0.010
0.020
0.040
0.020
0.020
(mol/L • s)
•Based on these data, what is the rate equation? What is the value of the rate constant k?
14
Rate
Law;
Solving
for
rate
Constant
The general rate law is:
Rate law = k [NO]2 [O2]
the rate constant k is determine by selecting one of
the experiments and solving the equation.
Consider experiment#1:
Rate = 0.028 = k [0.020]2 [0.010]
k = 0.028 / (4•10-4)(0.010) = 7.1•103 M-2 s-1
Rate Law: Rate = 7.1•103 [NO]2 [O2]
15
Microscopic view
• In order to understand our macroscopic observations about
temperature and concentration dependence, we should look
at the reaction microscopically - on the size scale of atoms
and molecules.
• The rates of chemical reactions are explained by collision
theory, which is based on kinetic theory.
• Collision theory views a reaction as the result of a
‘successful’ collision between two or more reactants and/or
catalysts.
• A few reactions occur without any collision.
16
Collision Theory
• The number of collisions between two or more species is
proportional to the product of their concentrations.
• When the reaction is the result of a single collision – an elementary
step – then the concentration dependence is directly related to the
stoichiometry of that collision.
• The probability that A will collide with B is proportional to [A][B].
• The probability that A will collide with A is proportional to [A]2
• For more complicated processes, the rate law is some combination of
these elementary steps.
• In order to react, the molecules must collide in a
favourable orientation and with sufficient energy.
• These factors are accounted for in the rate constant.
17
Molecularity of elementary steps
• For an elementary step (arising from one collision), the
rate law depends on the stoichiometry of the ‘collision’.
• A step involving only one molecule is called unimolecular.
Rate = k[A]
• A step involving two molecules is called bimolecular.
Rate = k[A][B], or Rate = k[A]2
• A step involving three molecules is called termolecular.
Rate = k[A][B][C], etc.
• Very few elementary steps involve more than 3 molecules.
18
Reaction progress & Ea
• For an elementary process we can plot the potential (chemical) energy
of the molecules as they approach each other, collide, react and move
apart.
• For an elementary process which involves only one molecule, we can
plot the potential energy as some internal coordinate, such as bond
length or angle, changes.
• This plot is sometimes called a reaction coordinate diagram, or an
energy plot. There is typically a maximum near the ‘collision’.
• Molecules move along this reaction coordinate with some initial
kinetic energy. K.E. is converted to P. E. to overcome the energy
barrier, the activation energy.
• Those molecular ‘collisions’ starting with enough kinetic
energy can overcome the barrier and react.
19
Arrhenius and Boltzmann
• We saw in chapter 13 that only a certain proportion of molecules had
enough energy to remain in the gas phase. The same type of energy
distribution is at play here.
 The Boltzmann distribution tells us that at any particular
temperature a certain percentage of the molecules are
above some energy cut-off.
• The cut-off of interest in this case is the activation energy.
• The percentage of molecules with energy above Ea is
related to the factor exp(-Ea/RT) in the Arrhenius
expression in the rate.
• As the temperature increases, so does the percentage of
molecules above the cut-off.
20
Calculations with Ea & T
k = Ae-Ea / RT
ln k = ln A – (Ea/RT)
• Increasing the temperature from 300 K to 310 K increases
the rate by a factor of 2. What is the activation energy?
• Given a set of T and k data, a plot of ln k vs. 1/T has a slope of -Ea/R
21
Rate determining step
• When the reaction is a series of elementary steps, rather
than a single step, the rate of reaction is determined by the
slowest step, which is typically the step with the highest
activation barrier.
• This step is called the rate determining step, and the rate
law for a known mechanism can be written in terms of the
rate for this step.
• If the rate determining step is not the first step, the rate
may depend on some species which do not appear as
reactants in the overall reaction equation.
22
Reaction mechanism
• Chemists often study reaction rates in order to deduce or
confirm a reaction mechanism – the stepwise progress of
the reaction.
• A proposed mechanism is written as a sum of elementary
steps, which may be reversible.
• If the rate law derived from the proposed mechanism
matched the observed rate law, then we are more confident
in our proposal, but still unsure.
• If the rate laws do not match, we must come up with a
different proposal.
23
2 NO (g) + Br2 (g)  2 BrNO
(g)
• Step 1
Rate = k1[Br2][NO]
Br2 (g) + NO (g)  NOBr2 (g)
• Step 2
Rate 2 = k2[Br2][NOBr2]
NOBr2 (g) + NO (g)  2 BrNO (g)
• NOBr2 is an intermediate – it is formed and then used up.
• The overall rate will depend on which step is rate
determining, and on whether either step is reversible.
24
2 NO2 (g) + F2 (g)  2 FNO2 (g)
• Step 1
• Step 2
rate = k1[NO2][F2]
NO2 + F2  FNO2 + F
slow
rate = k2[NO2][F]
NO2 + F  FNO2
fast
Overall rate = k1[NO2][F2]
Rate of reaction = rate of the slowest step
k2 >> k1
25
2 NO (g) + O2 (g)  2 NO2 (g)
• Step 1 is reversible
K1 = [NO3] / [NO][O2]
NO + O2
NO3
fast equilibrium
• Step 2
rate = k2[NO3][NO]
NO3 + NO  2 NO2
slow
• Overall rate = k2 [NO3][NO]
• Rate of reaction = rate of the slowest step, but NO is an intermediate –
difficult to determine its concentration. Want to replace [NO] with
known quantities:
K1 = [NO3] / [NO][O2]
[NO3] = K1 [NO] [O2]
Rate = k2(K1 [NO] [O2]) [NO] = k’ [NO]2[O2]
26
Equilibria in reaction
mechanisms
• Note that this topic is not covered in Kotz and Treichel
• In principle all reaction are reversible, but only some are reversible on
a time scale relevant to the overall process.
• A reaction, or step, which is fast in both the forward and reverse
direction will come to equilibrium rapidly.
• Dynamic equilibrium is reached when the rate of the forward reaction
equals the rate of the reverse reaction.
• For an elementary step 2A B + C, at equilibrium
rate forward = k1[A]2 = k-1[B][C] = rate reverse
k1 [B ][C ]

K
2
k 1
[A]
= equilibrium constant.
27