Kinetics
Chapter 12
3 OUT OF 75 M/C QUESTIONS
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Chemical Kinetics
Study of
Reaction rates
Reaction mechanism—series
of steps by which a reaction
takes place
Rate of Reaction

the change in concentration of a reactant
or product per unit time
The square
brackets indicate
concentration of
a substance.

Rate = [A]t2 – [A]t1 / (t2 – t1)
OR Rate = D[A] / Dt
Usually considered positive even if concentration is
decreasing.
Example Problem
Nitrogen dioxide decomposes to nitrogen monoxide and
oxygen gas. At the beginning of the reaction, a pressurized
1-L flask contains 2.0 mol of nitrogen dioxide. At the end
of 20 minutes, the flask contains 1.5 mol of nitrogen
dioxide and 0.5 mol of nitrogen monoxide. What is the
rate of the reaction in terms of M/s of nitrogen dioxide?
Example Problem Solution
• Important information:
2.0 mol/ L nitrogen dioxide initially
After 20 minutes, 1.5 mol/L of nitrogen dioxide
• Change in concentration: 2.0 mol – 1.5 mol = 0.5 mol
• Change in time: 20 min
• Rate: 0.5 mol / 20 min = 0.025 mol/min
Average vs. Instantaneous
Rate
Rate
is usually not constant, but
decreases over time
Average rate—total change over total
time (like the previous example)
Instantaneous rate—rate at a particular
instant in time
Instantaneous Rate
Calculated
by
line’s slope
tangent to the
point in time on
the rate curve
Think about a slope:
change in "y” over
change in “x”—It’s the
same thing as the
average rate if y is the
concentration and x is
the time.
Page
558
Rate Law
A
mathematical expression which
This is a very important
describes the dependence of
concept!
reaction rate on concentration of
The [A] means concentration of
reactants
substance A. The exponent “n” is
known as the order of the reaction. The
n
must be determined
The lower case k is the
Rate = k[A] order
experimentally from data.
rate constant. It must be
lower case because
k = proportionality constant
upper case K is an
equilibrium constant.
Rate Law
[A]
= concentration of reactant(s)
n = order of reactant
Both k and n must be
experimentally determined
Cannot determine from balanced
equation!!!
Write the rate law
2NO2
Rate law always begins with “rate =
k….”
 2NO + O2 if NO2 is Next, put the formula for each
reactant in brackets and raise it to
second order
the power of the order.
Rate = k [NO2]2
You don’t need
to include zero
order reactants
2H2 + 2NO N2 + 2H2O if NO is
since anything
to the zero
second order and H2 is first order raised
power is equal to
one.
Combining Rate Laws
Rate = D[A] / Dt
n
Rate = k[A]
D[A] / Dt =
n
k[A]
Since both of these
are equal to the
rate, they are equal
to each other.
More product terms can be added
Use this
equation to
solve for either
rate, the rate
constant or a
concentration.
Example
Rate =
n
m
p
k[A] [B] [C]
(Greater value of exponent
means greater effect on rate.)
To find exponents, determine
what happens if concentration
of a reactant doubles.
Determining orders of
reactants
Find
two experiments in which
one concentration remains the
same while the other changes.
Rate 2 = (D concentration)n
Rate 1
Example Calculation
Trial
Initial Concentration of
Reactants (M)
[A]
[B]
[C]
Initial
Rate
(M/min)
1
0.10
0.10
0.10
0.01
2
0.10
0.10
0.20
0.01
3
0.10
0.20
0.10
0.02
4
0.20
0.20
0.10
0.08
To determine orders of reactions, by
method of initial rates:
1. Look at the data and find two trials
that start with the same
concentrations in all but one
reactant.
2. See trials 1 and 2. The concentration
of reactant C was doubled while A
and B stayed the same. There was
no effect on the rate, so reactant C
is zero order. Leave it out of the rate
law.
Continued on next slide
Example Calculation
Trial
Initial Concentration of
Reactants (M)
[A]
[B]
[C]
Initial
Rate
(M/min)
1
0.10
0.10
0.10
0.01
2
0.10
0.10
0.20
0.01
3
0.10
0.20
0.10
0.02
4
0.20
0.20
0.10
0.08
3. Compare trials 1 and 3. While the
concentrations of A and C stayed
the same, the concentration of B
was doubled and the rate doubled.
Since the effect was the same, B is a
first order reactant. (2 x B = 2 x rate;
21 = 2)—The exponent is the order.
4. Compare trials 3 and 4. The
concentrations of B and C are the
same while the concentration of A
was doubled. The rate increased by
a factor of 4 so A is a second order
reactant. (2 x A = 4 x rate; 22 = 4)—
The exponent is the order.
5.Rate = k [A]2 [B]1
Example
Overall
Rate:
Rate =
Overall
2
1
0
k[A] [B] [C]
order of reaction:
2+1+0=3
Applying Rate Laws
Once
the rate law is
established, use data to solve
for k and to find rates at
different conditions.
Example
 Calculate
the rate constant k for the previous reaction.
Rate = k[A]2[B]
Trial
1
Initial Concentration of
Reactants (M)
[A]
[B]
[C]
0.10
0.10
Initial
Rate
(M/min)
0.10 0.01
Example
 Since
[C] has no effect on rate, we
can leave it out when solving for k.
To find the rate
constant, k, choose any
of the trials and
substitute the
concentrations and
initial rate into the
equation. Solve for k.
Pay attention to units
and include them for k.
k
Rate = k[A]2[B]
= Rate / [A]2[B]
k = .01M/sec / (.10M)2 (.10M)
k = 10 M-2sec-1
Write the Rate Law Equation
 Look
at data sets from different trials:
+
NH4
+
NO2
 N2 + 2 H2O
Experiment
#
Initial
Conc. Of
NH4+
Initial
Conc. Of
NO2-
Initial Rate
(mol/L *s)
1
0.100 M
0.0050 M
1.35 x 10-7
2
0.100 M
0.010 M
2.70 x 10-7
3
0.200 M
0.010 M
5.40 x 10-7
Answer
Both
reactants are first order, so
Rate = k [NH4+]1[NO2-]1
Solve for k.
Answer
 Rate
= k [NH4+]1[NO2-]1
 Using information from trial #1:
 1.35 x 10-7 M/s = k (0.100 M)(0.0050 M)
 k = 2.7 x 10-4 M-1s-1
2 ClO2 + 2 OH-  ClO3- + H2O
[ClO2] in mol/L
[OH-] in mol/L
0.0500
0.100
0.100
0.100
0.100
0.0500
Initial Rate in
M/s
5.75 x 10-2
2.30 x 10-1
1.15 x 10-1
Write the rate law equation and find the rate constant.
Answer
Rate
= k [ClO2]2 [OH-]
k = 230 M-2 s-1
Types of Rate Laws
Differential
Rate Law—expression of how
rate depends on concentration (previous
problems)
Integrated Rate Law—expression of how
concentration depends on time
When one is determined, the other can be
calculated.
Integrated Rate Law
Shows
how concentration
of A depends on time
Usually given [A]0 and k
Equation (whichever form)
is in the form y = mx + b so it
will be a straight line when
graphed.
Calculating Integrated
Rate Law—First Order
ln[A]
= -kt + ln[A]0
ln = natural logarithm
t = time
[A] = conc. at time t
[A]0 = conc. at time 0
Integrated Rate Law
Equation
Equation
can also be
expressed as a ratio:
ln ([A]0 / [A]) = kt
Half Life
The
time required for
a reactant to reach
half its original
concentration
t1/2
Calculating Half Life
When
t = t1/2, then [A] = [A]0/2
Use equation ln ([A]0 / [A]) = kt
ln
/
([A]0 [A]0/2) = kt1/2
So…ln(2) = kt1/2
Notice, half life does not
depend on concentration.
Calculating Integrated
Rate Law—Second
Order
Rate
= k[A]2
Integrated Rate Law
1/[A] = kt + 1/[A]0
Integrated Second
Order
Plot of 1/[A] versus time is a
straight line with slope = k
Half life for second order:
t1/2 = 1/k[A]0
For 2nd order, half life does
depend on initial
concentration.
Zero Order Rate
Laws
Often encountered when a
reaction rate is limited
because a surface (such as
a platinum catalytic
converter) is required.
When surface is covered,
increasing concentration of
a reactant has no effect.
If the reaction takes place
on a surface, increasing
concentration will not
Zero Order Rate
Laws
0
Rate = k[A] = k(2) = k
Integrated
Rate Law:
[A] = -kt + [A]0
Half
Life:
t1/2 = [A]0 / 2k
More Complex Rate
Laws
Many reactions have several
reactants—each affects rate
To study them, high
concentrations of all but one
reactant are used.
Highly concentrated
reactants stay practically
constant, so the remaining
reactant can be studied.
Pseudo-First-Order
If
conc. of B & C are
large,
n
m
p
Rate = k[A] [B] [C] =
k’[A]
Calculate k’ and solve
for k
Good Summary—Table
12.6
Importance
Helps
determine reaction
mechanism
Find rate determining step
in a series of reactions so
total process can be sped
up
Reaction Mechanisms
Sum
of elementary steps
must give overall
balanced equation
Must agree with
experimentally
determined rate law
Reaction Mechanisms
Most
reactions are much
more complicated that they
appear from their equations.
Intermediate—a species that
is neither a reactant nor a
product but that appears and
then is consumed in the
course of a reaction
Reaction Mechanisms
Each
reaction is called an
elementary step.
Rate of each elementary step
can be written from its
molecularity.
Molecularity = the number of
species that must collide to
cause the reaction
Molecularity
Unimolecular—depends
on
one molecule; Rate = k[A]
Bimolecular—depends on two
molecules; Rate = k[A]2 or
Rate = k[A][B]
Termolecular—depends on 3
molecules; Rate = k[A]2[B]
(etc)
Rate Determining
Step
The slowest step in a
reaction mechanism
Determines overall rate
much like pouring water
through a funnel—
limiting factor
Rate Law
Comes
from the slow step
and every step prior to it.
Only consider species that
are in the overall reaction.
Only consider reactants.
Species Not in Overall
Reaction
 Intermediate—a
species that is
produced and then consumed in a
reaction—appears in mechanism but
not in overall reaction—appears as a
product and then a reactant
 Catalyst—a species that is used during
one step of a mechanism but is
reproduced later—appears as a
reactant and then a product
Determining
Mechanism
See if rates of elementary
steps agree with
observed rate law
If yes, it is an acceptable
mechanism
Never proven—only
possibly correct
Example Problem
2NO2
+ F2  2NO2F
Rate = k[NO2][F2]
Possible Mechanism??
NO2
+ F2  NO2F + F
(slow)
F + NO2  NO2F
(fast)
Example
 Overall
Reaction: I2 + H2  2HI
 Step
1: I2  2I
 Step
2: I + H2  H2I
 Step
3: H2I + I  2HI
 Does
this give the overall equation?
 If
step one is rate determining, what is
the rate law? Step 2?
 Identify
any catalysts or intermediates.
Example
2
NO2 (g) + F2 (g)  2NO2F (g)
 Rate Law: Rate = k [NO2] [F2]
 Possible Mechanism:
NO2
+ F2  NO2F + F (slow)
F + NO2  NO2F
(fast)
Is this a possible mechanism?
Identify any catalysts or
intermediates.
Example
 NO2(g)
+ CO(g)  NO (g) + CO2(g)
 Step 1: NO2 + NO2  NO3 + NO (slow)
 Step 2: CO + NO3  NO2 + CO2 (fast)
 If the mechanism is correct, what is the
rate law?
 Identify any catalysts or intermediates.
Example
Write
the rate law and overall
reaction:
A + A + B  C + D (fast)
C + E  D + B
(slow)
Identify any catalysts or
intermediates.
Collision Model
Molecules
must collide in
order to react
Anything that increases
frequency or energy or
effectiveness of collisions
increases reaction rate.
Effective Collisions
Only
a small fraction of
collisions produce
reactions.
Many ineffective
collisions occur because
energy is too low or
orientation is wrong.
Some orientations for a collision
between BrNO molecules.
Orientations (a) and (b) can lead to
Activation Energy
The
minimum amount
(threshold) of energy
that a system must
have to produce a
reaction
Using k to Calculate Activation
Energy
k=
-E
/RT
a
zpe
k = rate constant
z = collision frequency
p = steric factor (fraction of
collisions with proper
orientation)
e-E /RT = fraction of collisions with
enough energy to produce
reaction
a
Arrhenius Equation
-E /RT
k = Ae
a
zp replaced by A
A is the frequency factor for the
reaction
Ea is activation energy
R is universal gas constant (8.3145)
Taking ln of both sides gives a
y = mx +b form line equation
Arrhenius Equation
ln(k) = -Ea/RT+ lnA
In slope intercept form:
Slope = Ea
Intercept = A
To Find Activation
Energy
Measure
rate
constant (k) at
several temps
Plot ln(k) versus 1/T
To Find Activation
Energy
Use
equation:
ln(k2/k1) = Ea/R (1/T1 –
1/T2)
Example Problem—p. 588
To Speed up a
Reaction:
Increase temperature
2. Increase pressure
3. Increase concentration
4. Increase surface area
5. Add a catalyst
1.
Catalyst
A
substance that speeds up a
reaction without being
consumed in the reaction
Enzymes—biological catalysts
Works by providing the
reaction with a new pathway
with lower activation energy
Effect of a Catalyst
Notice that
energy
difference
between
products and
reactants is
unchanged.
Effect of a Catalyst-
Heterogeneous
Catalysis
Usually
gaseous reactants
adsorbed on a metal surface
Example—hydrogenation of
ethylene
Breaking H-H bond requires
lots of energy, but metal-H
interactions weaken bond
and allow bonds to break at
lower energy
Homogeneous
Catalysis
Catalyst
and reactants exist in
the same phases
Example: Catalysis of ozone
by nitric oxide and freons
(CCl2F2)