CHAPTER 14
Chemical kinetics
Introduction
Spontaneity refers to the inherent tendency for
the process to occur; however, it implies nothing
about speed.
Who might be interested in the rate of a chemical
reaction?
• Would the grocer be interested in fast or slow
chemical reactions?
Introduction
Spontaneity refers to the inherent tendency for
the process to occur; however, it implies nothing
about speed.
Who might be interested in the rate of a chemical
reaction?
• Would the grocer be interested in fast or slow
chemical reactions? Slow reactions; to preserve
food.
Introduction
• How about the doctor and the patient?
Introduction
• How about the doctor and the patient?
Fast or slow-depends? It would be nice to take an
aspirin and have the head ache go away in a second
or two, but if the issue is cancer, then slow!
Introduction
What would be a word to describe extremely fast
chemical reactions that occur in a nanosecond?
Introduction
What would be a word to describe extremely fast
chemical reactions?
Explosion!
Introduction
What would be a word to describe extremely fast
chemical reactions that occur in a nanosecond?
Explosion!
There really is no word for very slow reactions, but some
examples are in order.
• Rusting of Iron
• Rotting of wood
Factors Affecting Rate
During our study of Kinetics the following will be
shown to affect the rate of chemical reactions
• Reactant concentration: The closer reactants are to
each other the higher the probability of combining to
make products
• Reaction temperature: Increased temperature
increases kinetic energy to break reactant bonds to
form products
• Catalyst: Lowers the activation energy.
• Surface Area: Exposes more reactants to the other
reactant.
• Nature of Reactants
How is Rate Measured
All rates are inversely proportional to time, which
means time must be in the denominator. In a
chemical reaction a change in concentration of
reactants or products is measured per unit time.
To measure rate, then concentration and time
must be measured.
Stoichiometry and Rate
Consider the Stoichiometry of reaction:
2 H2 + O2 → 2 H2O
• Hydrogen disappears twice as fast as oxygen.
• Water appears at the same rate as hydrogen
disappears.
• Oxygen disappears ½ as fast as hydrogen appears
General Rate Relations
Note: - means decreasing, while + means increasing.
Consider the equation:
aA + bB → cC + dD
Δ[A] = - 1 Δ[B] = 1 Δ[C] = 1 [∆D]
a Δt
b Δt
c Δt
d ∆t
Rate = - 1
Special Note about Δ
Δ means change. To be
consistent change in chemistry
is always final - initial
Sample Problem
Consider the decomposition of dinitrogenpentoxide
2 N2O5 (g)
→
4 NO2 (g) + O2 (g)
If the rate of appearance of O2 is 0.23 M/min, then what
would the rate of appearance of NO2?
¼ Δ[NO2]/Δt = Δ[O2]/Δt
Δ[NO2]/Δt = 4 Δ[O2]/Δt = 4 [0.23M/min] = 0.92 M/min
Average vs. Instantaneous Rate
Consider a Road Trip:
What does the speedometer tell us?
Using the road signs how can the average rate
be calculated?
Average vs. Instantaneous Rate
Consider a Road Trip:
What does the speedometer tell us?
Instantaneous Rate
Using the road signs how can the average rate
be calculated?
Average vs. Instantaneous Rate
Consider a Road Trip:
What does the speedometer tell us?
Instantaneous Rate
Using the road signs how can the average rate
be calculated? Rateave = ΔD/ Δt
Average vs. Instantaneous Rate
Consider a the following chemical reaction:
CV+ (aq) + OH-(aq) → CVOH (aq)
Where CV is crystal violet calculate the average rate
for the given time frames from the data on the next
slide.
Crystal Violet Data
1. Graph [CV+] vs. time
2. Calculate average rate for
a. t=0.0 to 10.0 seconds
b. t=10 to 20 seconds
3. Calculate the instantaneous
rate at t= 10.0 seconds
Rate =
Rate =
∆[CV+]
∆t
=
(3.68-5.00)(10-5)
10.0-0.00
-1.32X10-5 M/s
Time (s)
0
10.0
20.0
30.0
40.0
50.0
60.0
80.0
90.0
[CV+]
5.00 X 10-5
3.68 X 10-5
2.71 X 10-5
1.99 X 10-5
1.46 X 10-5
1.08 X 10-5
0.793 X 10-5
0.429 X 10-5
0.232 X 10-5
About Graphs
Central Graphing Guidelines
• Graph done on graph paper; or Excell (possibly Logger-Pro)
• Scales clearly and axes clearly labeled
• Allowed scales
– 1,2,3,4 etc.; 10, 20, 30,40 etc.; 100, 200, 300 etc.
– 2,4,6 etc; 20,40, 60 etc.
– 5, 10, 15, 20, etc; 50, 100, 150, 200, etc
• Smoothest curve between data points, or best fit line
• If drawing the line, then the same number of data points should
be above and below the smoothest curve
• Large graphs required; full sheet of graph paper, no smaller
than half sheet
• Design graph to maximize the number of significant figures.
Graphing With Excel
1.
2.
3.
4.
Place ordered pairs in Excell
Highlight the ordered pairs
Go to insert, chart, scatter
Right click on line and choose trend line,
options display equation.
2NO2 → 2NO + O2
Rate Laws
• Most all reactions are reversible, this means that
products can make reactants.
• The overall rate would be the rate of the product
formation minus rate of reactant formation.
• In order not to deal with this difference we determine
rates shortly after the reaction starts, so that the rate
of the reverse reaction is negligible.
• The previous graphs were not linear, but exponential
in nature, due to reactant concentration decreasing.
• The rate of a reaction can be more easily stated as
Rate = k[A]n , where k is the rate constant, [A] the
initial concentration of the reactants, and n the order of
the reaction. This expression is called the rate law.
Rate Law cont.
• The value of n cannot be determined from the
balanced equation, but must be determined
experimentally.
• Concentrations of products can be ignored, since they
are minimal, thus producing a negligible rate in the
reverse direction.
• There are two different types of rate laws
a. The differential rate law (rate law), describes
how the rate depends on concentration.
b. The integrated rate law, which shows how the
concentrations of species depend on time.
The Differential Rate Law
Consider the following reaction:
CH3CO2CH3 + OH- → CH3CO2- + CH3OH
Exp#
1
[CH3CO2CH3]
0.040
[OH-]
0.040
Rate M/s
0.00022
2
0.040
0.080
0.00048
3
0.080
0.080
0.00090
Simplify the Rate Data
Exp# [CH3CO2CH3] [OH-] Rate M/s
assume k =1
Method of initial rates
1
2
3
R = k[CH3CO2CH3]X[OH-]y
R = 1[1]1[1]1 = 1 for exp #1
R = 1[1]1[2]1 = 2 for exp #2
R = 1[1]1[2]1 = 4 for exp #3
Conclusion X=1, y=1
1
1
2
1
2
2
1
2
4
The reaction must be first
order in methylacetate
and first order in
hydroxide, with an overall
order of two
NH4+ + NO2- → N2 + 2 HOH
R1/R2 =?
R2/R3 =?
R1 = k[NH4+]x [NH2-]y
R1/R2 =
R2/R3 =
R2 = k[NH4+]x[NH2-]y
x
y
1.35X10-7 = [0.100] [0.005]
2.70x10-7
[0.100]x[0.010]y
2.70X10-7
5.40X10-7
=
[0.100]x[0.010]y
[0.200]x[0.010]y
0.500 = [0.500]y
y=1
0.500 = [0.500]x
x=1
Log Review
Log(X)(Y) = Log(X) + Log(Y)
Log
X
Y
= Log(X) - Log(Y)
Log (Y)x = X Log (Y)
Example:
1.37 = (4.39)x
Log 1.37 = X Log 4.39
0.136 = X (4.39)
0.136
X=
= 0.212
4.39
Checking on Calculator:
[4.39]
0.212
= 1.37
Reaction Rate and Order
Order is predicted from the exponents in the rate equation.
• Zero order in the reactant, means that there is no
rate effect on the initial concentration of that reactant,
even if it is increased.
• First order in the reactant, means that doubling the
initial concentration will double the rate.
• Second order in a reactant, means that doubling the
initial concentration will make the rate quadruple.
• Third order in a reactant, means that doubling the
initial concentration increases the rate eightfold.
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. What is the order with respect to N2?
2. What is the order with respect to H2?
3. What is the overall order?
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. What is the order with respect to N2? First
2. What is the order with respect to H2?
3. What is the overall order?
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. What is the order with respect to N2? First
2. What is the order with respect to H2? Third
3. What is the overall order?
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. What is the order with respect to N2? First
2. What is the order with respect to H2? Third
3. What is the overall order? Fourth
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. How many times faster if N2 is doubled?
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. How many times faster if N2 is doubled? 2
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. How many times faster if N2 is doubled? 2
2. How many times faster if H2 is doubled?
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. How many times faster if N2 is doubled? 2
2. How many times faster if H2 is doubled? 8
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. How many times faster if N2 is doubled? 2
2. How many times faster if H2 is doubled? 8
3. How many times faster if both reactants are
doubled?
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
Experiments show R = k[N2][H2]3
1. How many times faster if N2 is doubled? 2
2. How many times faster if H2 is doubled? 8
3. How many times faster if both reactants are
doubled? 16
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
1. Experiments show R = k[N2][H2]3
2. What is the overall order? 4
3. How many times faster if N2 is doubled? 2
4. How many times faster if H2 is doubled? 8
5. How many times faster if both are doubled? 16
6. How many times faster if H2 is reduced by half of its
original concentration?
Sample Problems
Consider the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g)
1. Experiments show R = k[N2][H2]3
2. What is the overall order? 4
3. How many times faster if N2 is doubled? 2
4. How many times faster if H2 is doubled? 8
5. How many times faster if both are doubled? 16
6. How many times faster if H2 is reduced by half of its
original concentration? 1/8
Zero-Order Reactions
A →B
• R = k[A]0
• Reactant concentration has no effect on initial rate of
reaction.
• [A] vs. time is linear with a negative slope.
• The rate is equal to k, since [A]0 =1
• The units of k are M/s
• Equation y = mx + B → [A]t = -kt + [A]0
• This is called the integrated rate law (derived on the
next slid), which shows how concentration is related to
time.
• The integrated rate law can give instantaneous rates.
Kinetics and Integral Calculus
Since rate is the change in concentration per unit time
(Δ[A]/Δt) then we can apply calculus principles of
extremely small changes of concentrations per
extremely small changes of time d[A]/dt.
Consider a zeroth order reaction
R = d [A] 0/dt = - k[A0]0 where A0 = initial concentration at time zero
Rearranging this expression gives:
d [A]0 = -k[A0]0 dt
Integrating from t=0 to t=t gives the integrated rate law.
t
t
∫ d [A]0 = ∫ -k[A0]0 dt
0
0
gives [A]t = -kt + [A]0
First-Order Reactions
A →B
• R = k[A]1
• Reactant concentration is directly proportional
to the rate of a reaction.
• [A] vs. time not linear with a negative slope.
• A graph of the ln[A] vs. time is linear.
• The units of k are 1/s
Calculus and Kinetics; First-Order
Integrated Rate Law
Since R = Δ[A]/Δt = -k[A]1 (first order) and
integral calculus uses d for Δ, the equation
has a calculus form of d[A]/[A] 1 dt = -k
Applying the integration process we get;
 d[A] /[A] = -k  dt
ln[A]t / [A]0 = - kt → ln[A]t = - kt + ln[A]0
Second-Order Reactions
A →B
• R = k[A]2
• Reactant concentration is exponentially
proportional to the rate of a reaction.
• [A] vs. time not linear with a negative slope.
• A graph of the 1/[A] vs. time is linear.
• The units of k are 1/Ms
Second-Order
Integrated Rate Law
Since R = Δ[A]/Δt = -k[A]1 (first order) and
integral calculus uses d for Δ, the equation
has a calculus form of d[A]/dt = -k[A]2, or
rearranging d[A]/ [A]2= -k dt
Applying the integration process we get;
 d[A]/[A]2 = -k  dt
1/[A]t = kt + 1/[A]0
Reaction Half Lives
• The half life of a reaction is the amount of time required
for one half of the initial reactant concentration to
disappear.
•The half life for each type of integrated equation can be
calculated from the equation itself.
Zero-Order Half Life
[A]t = ½[A]0 (half life definition)
[A]t = - kt + [A]0 → ½[A]0 = -kt1/2 + [A]0
- [A]0/2 = - kt1/2 → t1/2 = [A]0/2k
Reaction Half Lives
First Order Half Life
ln[A]t = -kt + ln[A]0
ln[A]t - ln[A]0 = - kt1/2 → ln[A]0 /[A]t = - kt
ln[A]t /[A]0 = - kt → ln[A]0 /[A]t = kt
[A]t = ½[A]0 (definition)
ln[A]0 /1/2[A]0 = kt1/2 → ln2[A]0 / [A]0 = kt1/2
ln2 = kt1/2 → t1/2 = 0.693/k
Note: the half life does not depend on initial concentration
Reaction Half Lives
Second-Order Half Life
1/[A]t = kt + 1/[A]0 → 1/[A]t - 1/[A]0 = kt
[A]t = ½[A]0 (definition) → 2/[A]0 - 1/[A]0 = kt1/2
1/[A]0 = kt1/2 → t1/2 = 1/k[A]0
Decomposition of N2O5
0.0500
0.0250
0.0125
Is this a zero order reaction?
Decomposition of N2O5
No, the graph should be linear!
Decomposition of N2O5
Is this a first order reaction?
Decomposition of N2O5
Note the half times are the same
Decomposition of N2O5
T1/2=0.693/k ( a constant divided by a constant
would always give same time)
Consider a Second Order
t1/2= 1/k[A]0 for a first order reaction.
The first half life would be [A]0  [A]0/2
The second half life would be [A]0/2  [A]0/4
Consider a Second Order
t1/2= 1/k[A]0 for a first order reaction.
The first half life would be [A]0  [A]0/2
The second half life would be [A]0/2  [A]0/4
Now do the math:
Consider a Second Order
t1/2= 1/k[A]0 for a first order reaction.
The first half life would be [A]0  [A]0/2
The second half life would be [A]0/2  [A]0/4
Now do the math: Assume [A]0=1 and k=1
t1/2=1/(1)(1)=1 for the first half life
Consider a Second Order
t1/2= 1/k[A]0 for a first order reaction.
The first half life would be [A]0  [A]0/2
The second half life would be [A]0/2  [A]0/4
Now do the math: Assume [A]0=1 and k=1
t1/2=1/(1)(1)=1 first half life
t1/2=1/(1)(1/2)=2 second half life twice as long
Consider a Second Order
t1/2= 1/k[A]0 for a first order reaction.
The first half life would be [A]0  [A]0/2
The second half life would be [A]0/2  [A]0/4
Now do the math: Assume [A]0=1 and k=1
t1/2=1/(1)(1)=1 first half life
t1/2=1/(1)(1/2)=2 second half life twice as long
The third half life would be 4 times longer than the
first.
So the graph must be for a first order reaction since the
time increments for the half-lives are the same.
Order Determination
From the information below, determine the order of the reaction.
Time
(s)
[A]
ln[A]
1/[A]
0
0.01000
-4.605
100
1000
0.00625
-5.075
160
1800
0.00476
-5.348
210
2800
0.00370
-5.599
270
3600
0.00313
-5.767
320
4400
0.00270
-5.915
370
5200
0.00241
-6.028
415
6200
0.00208
-6.175
481
Graphs of the Data
Graphs of the Data
ln[A]t = - kt + ln[A]0
1/[A]t = kt + 1/[A]0
Graphs of the Data
ln[A]t = - kt + ln[A]0
1/[A]t = kt + 1/[A]0
Second Order
Pseudo-Ordered Reactions
If a reaction is first order in each reactant and
second order over all how can we calculate the
rate constant?
Pseudo-Ordered Reactions
If a reaction is first order in each reactant and
second order over all how can we calculate the
rate constant?
Do we use the first order rate law, the second order rate
law, or neither of these?
Pseudo-Ordered Reactions
If a reaction is first order in each reactant and
second order over all how can we calculate the
rate constant?
Do we use the first order rate law, the second order rate
law, or neither of these? Neither; use the pseudoorder technique
Pseudo-Ordered Reactions
The pseduo-order technique uses very large
concentration differences between the reactants.
Always keep one reactant at low concentrations and
the other reactants at higher concentrations by 4-5
orders of magnitude.
Pseudo-Ordered Reactions
The pseudo-order technique uses very large
concentration differences between the reactants.
Always keep one reactant at low concentrations and
the other reactants at higher concentrations by 4-5
orders of magnitude.
Consider the following reaction:
A + B  C
1.0 M 1.0X10-4 M
When half of B disappears its concentration will
be 5.0X10-5. The concentration of A should then
be what?
Pseudo-Ordered Reactions
The pseudo-order technique uses very large
concentration differences between the reactants.
Always keep one reactant at low concentrations and
the other reactants at higher concentrations by 4-5
orders of magnitude.
Consider the following reaction:
A + B  C
1.0 M 1.0X10-4 M
When half of B disappears its concentration will
be 5.0X10-5. The concentration of A should then
be what? 1.0 – 0.00005 = ?
Pseudo-Ordered Reactions
The pseudo-order technique uses very large concentration
differences between the reactants. Always keep one reactant at
low concentrations and the other reactants at higher
concentrations by 4-5 orders of magnitude.
Consider the following reaction:
A + B  C
1.0 M 1.0X10-4 M
When half of B disappears its concentration will
be 5.0X10-5. The concentration of A should then
be what? 1.0 – 0.00005 = 1.0, right?
Rounded to the correct
number of significant figures.
Pseudo-Ordered Reactions
Since the concentration did not change for A it can be
considered a constant and incorporated into the
constant (that is why we use one, since 1 X k = k),
then any rate information is relative to B, thus allowing
calculation of the B order. After the order of B is
determined, then an experiment with [B]=1.0 and
[A]=1.0X10-4 can be done to calculate the order of [A],
thus giving the rate law. Then the experimental data
can be substituted into the rate equation to provide the
value of k.
Pseudo-Ordered Example
Experiment
[A]
[B]
Rate
1
0.1000
0.0050
1.35X10-7
2
0.1000
0.0100
2.70X10-7
3
0.0050
0.1000
2.7X10-7
4
0.0100
0.1000
1.08X10-6
Rate = k[A]x[B]y
R1/R2 =
R3/R4 =
1.35X10-7
2.70x10-7
[0.100]x[0.005]y
=
[0.100]x[0.010]y
y
x
2.70X10-7 = [0.0050] [0.1000]
1.08X10-6
[0.0100]y[0.1000]x
0.500 = [0.500]y
y=1
0.250 = [0.500]y
x=2
Pseudo-Ordered Example
Experiment
[A]
[B]
Rate
1
0.1000
0.0050
1.35X10-7
2
0.1000
0.0100
2.70X10-7
3
0.0050
0.1000
2.7X10-7
4
0.0100
0.1000
1.08X10-6
Rate = k[A]x[B]y
R1/R2 =
R3/R4 =
1.35X10-7
2.70x10-7
[0.100]x[0.005]y
=
[0.100]x[0.010]y
y
x
2.70X10-7 = [0.0050] [0.1000]
1.08X10-6
[0.0100]y[0.1000]x
Rate = k[A]1[B]2
0.500 = [0.500]y
y=1
0.250 = [0.500]y
x=2
Multiple Reactants
Use pseudo-order technique
A + B + C → D
R = k[A]x[B]y[C]z
The experimental conditions, will use high
concentrations of two of the three reactants relative to
the third reactant. This will cause relatively little change
in concentration of the large ones and significant
changes in concentration of the more dilute reactant.
Thus the order to the dilute reactant can be determined
by the method of initial concentrations. Then another
set of experiments will be carried out by keeping two
reactants at high concentrations and the other at dilute
concentration, to determine its order.
A Sample Problem
•
Consider the reaction: 3A + B + C → D + E,
where the rate law is defined as
• Δ[A]/Δt = – k[A]2[B][C] experiment is carried out
where [B]0 = [C]0 = 1.00 M and [A]0 = 1.00 x 10–4
M. After 3.00 minutes,
• [A] = 3.26 x 10–5 M.
– Calculate the value of k using integrated
second order equation.
1
1
= kt +
[A]0
[A]t
1
1
=
kt + 1.00X10-4
-5
3.26X10
30675 = k (3.00 min) + 1.00 X 104
K = 6.89 x 103 min-1
A Sample Problem
•
•
Consider the reaction: 3A + B + C → D + E,
where the rate law is defined as
Δ[A]/Δt = – k[A]2[B][C] experiment is
carried out where [B]0 = [C]0 = 1.00 M and
[A]0 = 1.00 x 10–4 M. After 3.00 minutes,
[A] = 3.26 x 10–5 M.
– Find the concentration of [B] and [C] after 3.00
minutes
[B] = [C] = 1.00 – 0.0000225 = 1.00 M
A Sample Problem
•
•
•
Consider the reaction: 3A + B + C → D + E,
where the rate law is defined as
Δ[A]/Δt = – k[A]2[B][C] experiment is
carried out where [B]0 = [C]0 = 1.00 M and
[A]0 = 1.00 x 10–4 M. After 3.00 minutes,
[A] = 3.26 x 10–5 M.
– Describe, in your own words, an experiment
that could be carried out to determine the order
of reactants B and C.
A Sample Problem
•
•
•
Consider the reaction: 3A + B + C → D + E,
where the rate law is defined as
Δ[A]/Δt = – k[A]2[B][C] experiment is
carried out where [B]0 = [C]0 = 1.00 M and
[A]0 = 1.00 x 10–4 M. After 3.00 minutes,
[A] = 3.26 x 10–5 M.
– Describe, in your own words, an experiment
that could be carried out to determine the order
of reactants B and C. Same way, keep A and
C at 1.00 molar and do rate expt. with B to find
order
Reaction Kinetics: A Summary
1. To calculate rate of reaction when the rate law is
known, use this expression: R = k[A]x[B]y
2. To determine a rate of reaction when the rate law is
not given, use
•
•
The slope of an appropriate tangent line to the graph of [A]
versus t
The expression –Δ[A]/ Δt, with a short time interval Δt
3. To determine the order of a reaction, use one of the
following methods.
•
•
•
Find the graph of rate data that yields a straight line
Test for constancy of the half-life (only first order)
Substitute rate data into integrated rate laws to find the
one that gives a constant value of k.
Reaction Kinetics: A Summary
1. To find the rate constant k for a reaction, use
one of the following methods.
•
•
•
Obtain k from the slope of a straight-line graph.
Substitute concentration-time data into the
appropriate integrated rate law.
Obtain k from the half-life of the reaction (first order
only) t1/2 = 0.693/k
2. To relate reactant concentrations and times,
use the appropriate integrated rate law after
first determining k.
Reaction Mechanisms
1. A reaction mechanism is a step-by-step description
of how chemical changes occur.
2. Each step is called an elementary process
3. A reaction mechanism must meet the following:
• Be consistent with the overall Stoichiometry
• The slowest step must account for the
experimentally determined rate law
• The individual equations must add up to equal the
total balanced equation. Like Hess’s Law
Elementary Processes
1. Are either unimolecular or bimolecular collisions.
termolecular are not probable since they require
simultaneous collision of three particles.
2. The exponents of concentration terms for the rate
law of an elementary step are the coefficients of the
equation. This is not the case for the overall
equation for the chemical change.
3. Elementary processes are reversible and some may
have the forward and reverse reaction rates equal
(equillibrium)
Elementary Processes
4. Certain species are produced in one
elementary process and consumed in another
are called reactive intermediates.
5. Of all of the elementary steps, only one is the
slowest and is called the rate determining
step
Mechanism Examples
Slow Step Followed by a Fast Step.
H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)
Mechanism: (Note: HI is a reactive intermediate)
H2 + ICl → HI + HCl
slow
HI + ICl → I2 + HCl
fast
H2 (g) + 2 ICl (g) → I2 (g) + 2 HCl (g)
Overall Rx
bimolecular
bimolecular
The rate law for the overall reaction is just the rate of the slowest
step.
R = k[H2][ICl]
A reactive intermediate is produced in one reaction
and consumed in the next reaction
Reaction Profile HCl + ICl
Mechanism Examples
Fast followed by slow step
2 NO (g) + O2 (g)
2 NO (g)
N2O2 (g) + O2 (g)
N2O4 (g) overall
k1
k-1
K2
N2O2 (g) fast
2 NO (g) + O2 (g)
R=
2
k[NO] [O
N2O4 (g) slow
N2O4 (g)
experiment
]
2
The Fast Slow Mechanism
• In this mechanism there is a rapid equilibrium
in the first step.
• Some of the product of the first step is slowly
drawn off and consumed in the second step.
• Since N2O2 is a reactive intermediate it
cannot be in the rate law. Rf = K[N2O2][O2]
• Rf = k1[N O]2 and Rr = k-1[N2O2]
• Equilibrium means that Rf = Rr ; k1[NO]2= k-1[N2 O2]
• [N2O2] = k1/k-1[NO]2
The Fast Slow Mechanism
•
•
•
•
•
Rf = K[N2O2][O2] but [N2O2] = k1/k-1[NO2]2
Substituting Rf =Kk1/k-1[NO2]2[O2]
Kk1/k-1 = K
Rf = K[NO2]2[O2] (The rate law from experiment)
This is a very common mechanism when the overall
reaction suggests a termolecular collision
Hydration Mechanism
CH2=CH2 + HOH
Step #1
H+
CH3CH2OH
CH2=CH2 + H+
CH3CH2+ fast
CH3CH2+ + HOH
CH3CH2OH2+ slow
CH3CH2OH2+
CH3CH2OH + H+ slow
Effect of Temperature on Rates
• In 1889 Arrhenius demonstrated that the rate constants of
many chemical reactions vary with temperature according
to the Arrhenius equation below:
K = Ae-Ea/RT
• This form is not useful, since it is exponential.
• A more useful form is found by taking the natural
logarithm of both sides, also below:
-Ea
+ ln A
ln k =
RT
An equation of a straight line where –Ea/R is the slope and ln
A is the y-intercept
About the Arrhenius Equation
K = A e-Ea/RT
1. A is the pre-exponential factor having the same
units as k
a
For a first order reaction the units are s-1, since these are
units of frequency, then sometimes A is called the
frequency factor
2. Ea is the energy of activation
3. T is the absolute temperature
4. R is the ideal gas constant, 8.314 j/mole-K
Rearrangement of Arrhenius Equation
1.Taking the natural logarithm of both sides of the equation
lnK = -Ea/RT + lnA
Fits the equation of a straight line, where –Ea/R is the slope
of the line
Two Point Equation
K1 is constant at a constant T1 temperature
K2 is constant at a constant T2 temperature
lnK1 = -Ea/RT1 + lnA
lnK2 = -Ea/RT2 + lnA
On the next slide the second equation will be subtracted from
the first equation, thus eliminating the y-intercept lnA
The Two Point Equation
Consider the the equation for different
temperatures
ln k1 =
ln k2 =
-Ea
RT1
-Ea
RT2
Lnk1 - ln k2 =
+ ln A
+ ln A
-Ea
RT1
-
Subtract the bottom
equation from the top
equation
-Ea
RT2
Lnk1/lnk2 = Ea/R(1/T2 – 1/T1)
+ ln k1 – lnk2
Collision Theory
• The number of collisions per second is called
the collision frequency.
• Typically gases have a frequency of 1030
collisions per second.
• If each collision made product then the rate
would be 106M/s. (explosive)
• Gas reactions are normally around 10-4M/s
• We must conclude then, that most collisions do
not end up as products.
Collision Theory
1. Collision theory views reaction rate as a process of
reactants colliding with a certain frequency and minimum
energy.
2. The number of collisions per unit time is the upper limit of
reaction rate.
3. This model is restricted to A + B →C
a. If there are 2 A and 2 B reactants then there are possible 4
collisions, which both multiplication and addition will predict.
b. Add another A particle, thus 3 A particles and two B particles,
each a can collide with either of the two B particles, thus a total
of 6 collisions, not 5 that addition would indicate.
c. Add another B particle, so that there are 3 A particles and 3 B
particles; each A can collide with three B particles, for a total of 9
collisions and not 6.
Collision Theory
Increasing the temperature of the reaction
increases the speed of the reacting particles and
therefore their collision frequency.
a.Collision frequency cannot be the only factor
affecting rate, or every gaseous reaction would be
over instantaneously.
b.Since at 1 atm at 20°C the molecules in a milliliter
of gas experience about 1027 collisions per second
(Total collisions)
c. Each molecule, on the other hand, experiences
4X107 collisions
Collision Theory
d. This means that if every collision resulted in a
reaction the reaction would be over in 10-7
seconds
e. Obviously most collisions do not result in a
reaction.
Arrhenius suggested that every reaction had an
energy threshold (Ea) that reactants must meet or
exceed in order to react. This is the energy
require to break the bonds, so that the atoms
can rearrange into other molecules
Collision Theory
Usually a ten degree increase in reaction temperature
doubles or triples the reaction rate
a. Is this increase due to increased frequency?
b. An increase of 10°only increases the speed by about
2%, which we would expect a 4% increase in reaction
rate, which does not explain the two or three fold
increase in reaction rate.
c. Far more important is the increase of the fraction of
particles having the correct kinetic energy to form
products. This is shown on the next slide.
Kinetic Energy Distribution
Energy of activation
About Colliding Particles
Orientation
a Colliding particles must collide with a
particle that can make a product, called
effective collisions
b In Arrhenius equation the frequency factor A
contains information on orientation
1. A = pz
2. P is the steric factor (always less than one) and
represents the fraction of collisions with
effective orientations.
3. Z is the collision frequency
4. e-Ea/RT is the fraction of collisions with sufficient
energy to produce products
About Colliding Particles
As molecules become more complex the
probability factor p (steric factor) decreases.
– When NO collides with NO3 to make NO2, the
probability factor is 0.006, or six out of a
thousand are effective
– In nucleic acids the factor is about 10-6
Transition State Theory
• Transition State was first proposed by Henry
Eyring (1901-19810).
• The hypothetical species that exists between
the reactants and products is called the
activated complex.
• The state where the activated complex is
found is called the transition state.
• The activated complex is formed from
collisions between molecules.
Transition State Theory
• Once the reactants have collided, they can go
back to reactants, or produce products.
• Meaning they roll, like a ball, down one side of
the hill or the other.
• Consider the collision of dinitrogen oxide with
nitrogen monoxide as illustrated below:
N
N O
+
N=O
O
N=O
N N
activated complex
Transition State Theory
Activated Complex
The Role of a Catalyst
– A substance that speeds up a reaction with out
being consumed by the reaction.
– Biological catalyst is called an enzyme.
– Catalysts work by lowering the energy of
activation.
– The enthalpy does not change.
– Homogeneous catalyst, same phase as the
reactants
– Heterogeneous catalyst, not the same phase as
the reactants
Heterogeneous Catalyst
•
Has a different phase than the rest of the reaction
mixture
Common example is an adsorption mechanism
Difference between adsorption and absorption
•
•
–
–
–
–
Adsorption means sticks to the surface like adhesive tape
Absorption means penetrates into the substance itself.
Metals are common choice for non-metal reactions
When a gas for example is adsorbed on to the surface of a
metal, the bonds of the gas are weakened, thus allowing
the other gas, also with weaker bonds to react.
Heterogeneous Catalyst
•
Hydrogenation of an unsaturated hydrocarbon
–
For ethylene reacting with hydrogen, they are
both adsorbed on the surface of a metal such as
Pt, Pd, Ni, Cu, etc.
The metal hydrogen interactions weaken the
covalent bond between the two hydrogen atoms.
The steps are typically
–
–
•
•
•
•
Adsorption and activation of the reactants
Migration of the absorbed reactants on the surface
Reaction among the absorbed substances
Escape, or desorption of the products
Homogeneous Catalysis
•The catalyst is in the same phase as the reaction
mixture
•Homogeneous catalysts have advantages over
heterogeneous catalysts
•Cheaper than transition metals
•Can catalyze a specific portion of a molecule
•Reactions can be carried out under
atmospheric conditions
Heterogeneous Catalyst Example
The hydrogenation of ethylene requires the
use of a heterogeneous catalyst
Pt
CH2=CH2 + H2  CH3CH3
The hydrogen molecules are adsorbed on
the surface of the platinum metal, thus
weakening the bonds holding the hydrogen
atoms together. The exposed protons are
then attracted to the pi bonding electrons
from the ethylene molecule forming new
sigma bonds between the hydrogen atoms
and the carbon atoms of the ethylene.
Homogeneous Catalysis
1. The catalyst is in the same phase as the reaction mixture
2. Homogeneous catalysts have advantages over
heterogeneous catalysts
a. Cheaper than transition metals
b. Can catalyze a specific portion of a molecule
c. Reactions can be carried out under atmospheric conditions
3. Example the hydration of ethene into ethyl alcohol
•
R = K [CH3COOC2H5] uncatalyzed
•
R = Kc[CH3COOC2H5][H+] catalyzed
Review
Predict the Order
N2O 

 N2 + 1/2O2
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
The radioactive isotope 32P decays and has
a half-life of 14.3 days.
a. Predict the order. Since the half-life of 14.3
days is given, and the concentration of 32P
is constant, then the rate cannot be
dependent on the initial concentration,
leaving first order rate law as the only
choice.
The radioactive isotope 32P decays and has
a half-life of 14.3 days.
a. Predict the order. Since the half-life of 14.3
days is given, and the concentration of 32P
is constant, then the rate cannot be
dependent on the initial concentration,
leaving first order rate law as the only
choice.
Why is the concentration constant?
The radioactive isotope 32P decays and has
a half-life of 14.3 days.
a. Predict the order. Since the half-life of 14.3
days is given, and the concentration of 32P
is constant, then the rate cannot be
dependent on the initial concentration,
leaving first order rate law as the only
choice.
Why is the concentration constant, because
the density is constant, right?
About Phosphorus
• Phosphorus, P, is a non-metallic main group
element belonging to Group 15 of the periodic
table. Atomic Number : 15
• Relative Atomic Mass : 31
• Melting Point : 41 degC
• Boiling Point : 280 degC
• Relative Density : 1.8 g/mL (White Phosphorus)
• Relative Density : 2.2 g/mL (Red Phosphorus)
• White phosphorus is the most common allotrope.
Density to Molarity
1.8 g mole
mL 31 g
Density to Molarity
1.8 g mole mL
mL 31 g 10-3 L
Density to Molarity
1.8 g mole mL
= 58 M
mL 31 g 10-3 L
Sample Problem
The radioactive isotope 32P decays and has
a half-life of 14.3 days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay?
Sample Problem
The radioactive isotope 32P decays and has
a half-life of 14.3 days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? Since this is first order, we can
use the first order half-life equation to calculate
the rate constant k.
Sample Problem
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? 95% means [A] =100.0 [A] =5.0
0
t
t1/ 2 =
0.693
k
[A]t
ln [A] = -kt
0
Sample Problem
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? [A] =100.0 [A] =5.0
0
t
t1/ 2 =
0.693
k
T1/2 =
0.693 0.693
=
k
14.3
Sample Problem
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? [A] =100.0 [A] =5.0
0
t
t1/ 2 =
0.693
k
0.693
0.693
=
k
14.3
= 0.04846 1/day
T1/2 =
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? [A] =100.0 [A] =5.0
0
t
t1/ 2 =
0.693
k
[A]t
ln [A] = -kt
0
0.693 0.693
=
k
14.3
= 0.04846 1/day
T1/2 =
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? [A] =100.0 [A] =5.0
0
t
t1/ 2 =
0.693
k
[A]t
ln [A] = -kt
0
0.693 0.693
=
k
14.3
= 0.04846 1/day
T1/2 =
5.0
ln 100.0
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? [A] =100.0 [A] =5.0
0
t
t1/ 2 =
0.693
k
[A]t
ln [A] = -kt
0
0.693 0.693
=
k
14.3
= 0.04846 1/day
T1/2 =
5.0
ln 100.0 = - 0.04846t
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? [A] =100.0 [A] =5.0
0
t
t1/ 2 =
0.693
k
[A]t
ln [A] = -kt
0
0.693 0.693
=
k
14.3
= 0.04846 1/day
T1/2 =
5.0
ln 100.0 = - 0.04846t
- 2.996 = - 0.04846t
The radioactive isotope 32P decays by firstorder kinetics and has a half-life of 14.3
days.
a. Predict the order.
b. How long does it take for 95.0% of a sample of
32P to decay? [A] =100.0 [A] =5.0
0
t
t1/ 2 =
0.693
k
[A]t
ln [A] = -kt
0
0.693 0.693
=
k
14.3
= 0.04846 1/day
T1/2 =
5.0
ln 100.0 = - 0.04846t
- 2.996 = - 0.04846t
t = 61.8 days
Sample Problem
The gas-phase reaction between methane and diatomic
sulfur is given by the equation. Calculate Ea
CH4 (g) 2S2 (g)
K (L/mole-s)
K1 = 1.1
K2 = 6.4
k2
ln
k1
=
Ea
R
[
CS2 (g) + 2 H2S (g)
T (°C)
550
T (K)
823
625
898
1 - 1
]
T1
T2
Sample Problem
The gas-phase reaction between methane and diatomic
sulfur is given by the equation. Calculate Ea
CH4 (g) 2S2 (g)
K (L/mole-s)
K1 = 1.1
K2 = 6.4
k1
ln
k2
6.4
ln
1.1
=
=
Ea
R
Ea
8.314
CS2 (g) + 2 H2S (g)
T (°C)
550
T (K)
823
625
898
[
1 - 1
]
T1
T2
[
1 - 1
823 898
]
Sample Problem
The gas-phase reaction between methane and diatomic
sulfur is given by the equation. Calculate Ea
CH4 (g) 2S2 (g)
K (L/mole-s)
K1 = 1.1
K2 = 6.4
k1
ln
k2
6.4
ln
1.1
=
=
Ea
R
Ea
8.314
[
[
CS2 (g) + 2 H2S (g)
T (°C)
550
T (K)
823
625
898
1 - 1
]
T1
T2
1 - 1
823 898
]
Ea= 1.4 X 102 Kj/mole
Sample Problem
The decomposition of ammonia to nitrogen and hydrogen
was studied on two surfaces. Ea with tungsten is 163
kj/mole and 197 kj/mole using Os. Without catalyst the
activation energy is 335 kj/mole.
a. Which surface is the better catalyst?
b. How many times faster is the reaction at
298 K on tungsten compared to no catalyst?
Sample Problem
The decomposition of ammonia to nitrogen and hydrogen
was studied on two surfaces. Ea with tungsten is 163
kj/mole and 197 kj/mole using Os. Without catalyst the
activation energy is 335 kj/mole.
a. Which surface is the better catalyst? Tungsten
b. How many times faster is the reaction at
298 K on tungsten compared to no catalyst?
Sample Problem
The decomposition of ammonia to nitrogen and hydrogen
was studied on two surfaces. Ea with tungsten is 163
kj/mole and 197 kj/mole using Os. Without catalyst the
activation energy is 335 kj/mole.
a. Which surface is the better catalyst?
b. How many times faster is the reaction at
298 K on tungsten compared to no catalyst?
335 kj 1000 j mole-K
Ea/RT = mole
=135.2
8.3145
j
kj
298 K
K = e-135.2
Sample Problem
The decomposition of ammonia to nitrogen and hydrogen
was studied on two surfaces. Ea with tungsten is 163
kj/mole and 197 kj/mole using Os. Without catalyst the
activation energy is 335 kj/mole.
a. Which surface is the better catalyst?
b. How many times faster is the reaction at
298 K on tungsten compared to no catalyst?
335 kj 1000 j mole-K
Ea/RT = mole
=135.2
8.3145
j
kj
298 K
K = e-135.2
K = 1.920X10-59
Sample Problem
The decomposition of ammonia to nitrogen and hydrogen
was studied on two surfaces. Ea with tungsten is 163
kj/mole and 197 kj/mole using Os. Without catalyst the
activation energy is 335 kj/mole.
a. Which surface is the better catalyst?
b. How many times faster is the reaction at
298 K on tungsten compared to no catalyst?
335 kj 1000 j mole-K
Ea/RT = mole
=135.2
8.3145
j
kj
298 K
163 kj 1000 j mole-K
Ea/RT= mole
=65.78
kj
8.3145j 298 K
K = e-135.2
K = 1.920X10-59
Sample Problem
The decomposition of ammonia to nitrogen and hydrogen
was studied on two surfaces. Ea with tungsten is 163
kj/mole and 197 kj/mole using Os. Without catalyst the
activation energy is 335 kj/mole.
a. Which surface is the better catalyst?
b. How many times faster is the reaction at
298 K on tungsten compared to no catalyst?
335 kj 1000 j mole-K
Ea/RT = mole
=135.2
8.3145
j
kj
298 K
K = e-135.2
K = 1.920X10-59
-29
K
=
2.705X10
1000
j
163 kj
mole-K
Ea/RT= mole
=65.78
kj
8.3145j 298 K
Sample Problem
The decomposition of ammonia to nitrogen and hydrogen
was studied on two surfaces. Ea with tungsten is 163
kj/mole and 197 kj/mole using Os. Without catalyst the
activation energy is 335 kj/mole.
a. Which surface is the better catalyst?
b. How many times faster is the reaction at
298 K on tungsten compared to no catalyst?
335 kj 1000 j mole-K
Ea/RT = mole
=135.2
8.3145
j
kj
298 K
K = e-135.2
K = 1.920X10-59
-29
K
=
2.705X10
1000
j
163 kj
mole-K
Ea/RT= mole
=65.78
kj
8.3145j 298 K
K/K = 1.409X10-30
Sample Problem
The decomposition of ammonia to nitrogen and hydrogen
was studied on two surfaces. Ea with tungsten is 163
kj/mole and 197 kj/mole using Os. Without catalyst the
activation energy is 335 kj/mole.
a. Which surface is the better catalyst?
b. How many times faster is the reaction at
298 K on tungsten compared to no catalyst?
335 kj 1000 j mole-K
Ea/RT = mole
=135.2
8.3145
j
kj
298 K
K = e-135.2
K = 1.920X10-59
-29
K
=
2.705X10
1000
j
163 kj
mole-K
Ea/RT= mole
=65.78
kj
8.3145j 298 K
K/K = 1.409X10-30
1.409X1030 times faster
Sample Problem
The experimental rate law for ozone decomposition has
2
[O
]
3
been determined to be rate = k
Propose a two
[O2]
step mechanism consisting of a fast reversible first step
and a slow second step. 2 O3
3 O2
Remember mechanism steps must add to give overall
reaction.
O3
k1
K-1
O + O3
O2 + O fast
2 O2 slow
Sample Problem
The experimental rate law for ozone decompositin has
2
[O
]
3
been determined to be rate = k
Propose a two
[O2]
step mechanism consisting of a fast reversible first step
and a slow second step. 2 O3
3 O2
Remember mechanism steps must add to give overall
reaction.
O3
k1
O + O3
2 O3
O2 + O fast
K-1
k2
2 O2 slow
3 O2
Sample Problem
The experimental rate law for ozone decompositin has
2
[O
]
3
been determined to be rate = k
Propose a two
[O2]
step mechanism consisting of a fast reversible first step
and a slow second step. 2 O3
3 O2
Remember mechanism steps must add to give overall
reaction.
O3
k1
O + O3
2 O3
O2 + O fast
K-1
k2
2 O2 slow
3 O2
Sample Problem
The experimental rate law for ozone decomposition has
2
[O
]
3
been determined to be rate = k
Propose a two
[O2]
step mechanism consisting of a fast reversible first step
and a slow second step. 2 O3
3 O2
Remember mechanism steps must add to give overall
reaction.
R =R
O3
k1
K-1
O + O3
2 O3
f
O2 + O fast
k2
2 O2 slow
3 O2
r
K1[O3] = k-1[O2][O]
[O] = k1/k-1[O3]/O2
Sample Problem
O3
Rf = Rr
k1
O2 + O fast
K-1
O + O3
k2
2 O3
2 O2 slow
[O] = k1/k-1[O3]/O2
3 O2
R = k2[O][O3]
R = k2 k1/k-1[O3]/O2 [O3]
R=
K1[O3] = k-1[O2][O]
K[O3]2
[O2]
[O] = k1/k-1[O3]/O2
Sample Problem
O3
Rf = Rr
k1
O2 + O fast
K-1
O + O3
k2
2 O3
2 O2 slow
3 O2
R = k2[O][O3]
R = k2 k1/k-1[O3]/O2 [O3]
R=
K[O3]2
[O2]
K1[O3] = k-1[O2][O]
[O] = k1/k-1[O3]/O2
Since the elementary steps add
up to equal the overall reaction
and the slowest step of the
mechanism yields a rate law
consistent with experimental
evidence this must be an
acceptable mechanism. Other
mechanisms are possible.
Review Continued
Conceptual Questions
If the rate of a reaction is directly proportional to
the concentration of a reactant, what does this tell
you about?
a. The order of the reaction with respect to the
reactant?
b. The overall order?
Conceptual Questions
If the rate of a reaction is directly proportional to
the concentration of a reactant, what does this tell
you about?
a. The order of the reaction with respect to the
reactant? The reaction is first order in that
reactant.
b. The overall order?
Conceptual Questions
If the rate of a reaction is directly proportional to
the concentration of a reactant, what does this tell
you about?
a. The order of the reaction with respect to the
reactant? The reaction is first order in that
reactant.
b. The overall order? The overall order is ≥ 1
Conceptual Questions
What is the relationship between the order
of a rate law and the molecularity of a
reaction?
Conceptual Questions
What is the relationship between the order
of a rate law and the molecularity of a
reaction?
The order of a reaction depends upon the
molecularity of the rate determining step
Conceptual Questions
What is the relationship between the rate
law and the balanced chemical equation?
Conceptual Questions
What is the relationship between the rate
law and the balanced chemical equation?
None!
Conceptual Questions
For any given reaction, what is the relationship
between the energy of activation and
a. electrostatic repulsions?
Conceptual Questions
For any given reaction, what is the relationship
between the energy of activation and
a. electrostatic repulsions?
An increase in repulsions between
reactants will increase the activation
energy.
Conceptual Questions
For any given reaction, what is the relationship
between the energy of activation and
a. electrostatic repulsions?
An increase in repulsions between
reactants will increase the activation
energy.
b. bond formation?
Conceptual Questions
For any given reaction, what is the relationship
between the energy of activation and
a. electrostatic repulsions?
An increase in repulsions between
reactants will increase the activation
energy.
b. bond formation?
Bond formation tends to tends to
lower the activation energy
Conceptual Questions
For any given reaction, what is the relationship
between the energy of activation and
a. electrostatic repulsions?
An increase in repulsions between
reactants will increase the activation
energy.
b. bond formation?
Bond formation tends to tends to
lower the activation energy
c. The nature of the activated complex?
Conceptual Questions
For any given reaction, what is the relationship
between the energy of activation and
a. electrostatic repulsions?
An increase in repulsions between
reactants will increase the activation
energy.
b. bond formation?
Bond formation tends to tends to
lower the activation energy
c. The nature of the activated complex?
An increase in the energy of the activated
complex increases the activation energy of
the reaction.
Conceptual Questions
What is the relationship between A, activation
energy, and temperature?
Conceptual Questions
What is the relationship between the frequency
factor (A), activation energy, and temperature?
Conceptual Questions
What is the relationship between the frequency
factor (A), activation energy, and temperature?
The activation energy is not related to either the
temperature or the frequency factor A, but the
frequency factor is usually proportional to the
square root of the temperature.
How does an increase in frequency factor affect the
rate of reaction?
Conceptual Questions
What is the relationship between the frequency
factor (A), activation energy, and temperature?
The activation energy is not related to either the
temperature or the frequency factor A, but the
frequency factor is usually proportional to the
square root of the temperature.
How does an increase in frequency factor affect the
rate of reaction?
An increase in frequency factor increases the
rate.
Conceptual Questions
What effect does a catalyst have on the energy of
activation of a reaction?
Conceptual Questions
What effect does a catalyst have on the energy of
activation of a reaction?
A catalyst lowers the activation energy of a reaction,
thus increasing the fraction of particles having the
necessary activation energy to react.
Conceptual Questions
What effect does a catalyst have on the energy of
activation of a reaction?
A catalyst lowers the activation energy of a reaction,
thus increasing the fraction of particles having the
necessary activation energy to react.
What effect does it have on the frequency factor, A?
Conceptual Questions
What effect does a catalyst have on the energy of
activation of a reaction?
A catalyst lowers the activation energy of a reaction,
thus increasing the fraction of particles having the
necessary activation energy to react.
What effect does it have on the frequency factor, A?
Some catalysts can also orient the reactants and
thereby increase the frequency factor. Catalysts
have no effect on the change in potential energy
for the reaction, ΔH.
Conceptual Questions
Consider the reaction between cerium(IV) and
thallium(I).
2Ce4+ + Tl+
2Ce3+ + Tl3+
This reaction is slow, but it is catalyzed by Mn2+, as
shown in the following mechanism:
Ce4+ + Mn2+
Ce3+ + Mn3+
Ce4+ + Mn3+
Mn4+ + Tl+
Ce3+ + Mn2+
Tl3+ + Mn2+
In what way does Mn2+ increase the reaction rate?
Conceptual Questions
Consider the reaction between cerium(IV) and
thallium(I).
2Ce4+ + Tl+
2Ce3+ + Tl3+
This reaction is slow, but it is catalyzed by Mn2+, as
shown in the following mechanism:
Ce4+ + Mn2+
Ce3+ + Mn3+
Ce4+ + Mn3+
Mn4+ + Tl+
Ce3+ + Mn4+
Tl3+ + Mn2+
In what way does Mn2+ increase the reaction rate?
The Mn2+ ion donates an electron to Ce4+ and then
accepts and electron from Tl+. It functions as an
electron conduit.
Numerical Questions
Half-lives for the reaction A + B
C were
calculated at three values of [A]0, and the
concentrations of B was the same in all cases.
The data are listed bellow.
[A]0, M
t1/2, s
0.50
420
0.75
280
1.0
210
Does this reaction follow first-order kinetics?
Numerical Questions
Half-lives for the reaction A + B
C were
calculated at three values of [A]0, and the
concentrations of B was the same in all cases.
The data are listed bellow.
[A]0, M
t1/2, s
0.50
420
0.75
280
1.0
210
Does this reaction follow first-order kinetics?
No. The reaction is second order: the half-life decreases
with increasing reactant concentration according to t1/2 =
1/k[A]0
Numerical Questions
Nitramide (O2NNH2) decomposes in solution to
N2O and H2O . Give the rate law for this process.
O2NNH2
O2NNH H+
+ OH -
K1
K-1
K2
K3
O2NNH2
[O2NNH2]
Rate = K
[H+]
O2NNH - + H+
fast
N2O + OH -
slow
H2O
fast
N2O + H2O
Note: H+ is not a reactive
intermediate, since it is not
consumed in the next step
The End