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```					   Chapter 14

Rates of Reactions
Kinetics
I. Introduction
A) Demonstrations
B) Chemical Kinetics is the study
of the rates (speeds) of chemical
reactions and the mechanisms of
chemical reactions.
C) The rate of a chemical reaction
is a measure of how fast reactants
are consumed and/or how fast
D) The mechanism of a reaction is a
detailed description of the way a
reaction occurs. It is a sequence of
reactants to products.
Mechanisms can be proven
wrong through _____________,
but they can never be called
_________________________
since they are, in general,
educated guesses.
Practical reasons for studying kinetics:
Some reactions we would like to speed
up: drug delivery, paint drying,
destruction of air pollutants in auto
exhaust breakdown of materials in
landfills.
Some reactions we would like to slow
down: food decay, rubber decay, human
aging, destruction of the ozone layer,
rusting, corrosion
E) Some reactions take place in
a fraction of a second and other
take many years. What
variables affect the reaction
rate?
1) The characteristics of the
reactants and the products.
2) The concentration of the
reactants – in some reactions
the rate is unaffected by the
concentration of one of the
reactants as long as it is there
in some amount.
3) The presence of a catalyst, a
substance that …
4)The temperature at which the
reaction occurs. Increasing the
temperature usually increases
the rate. A general rule is …
5) The surface area of a solid
reactant or catalyst affects the
rate.
G) Reaction rate is the increase
in molar concentration of
product of a reaction per unit
time or the decrease in molar
concentration of a reactant per
unit time.
For example, for the reaction:
2 N2O5  4 NO2 + O2
The following table shows the
concentration of N2O5 as a
function of time at 45 oC.
Time in min   [N2O5] mol/L
0             0.01756
20            0.0093
40            0.00531
60            0.00295
80            0.00167
100           0.00095
160           0.00014
1) The rate of reaction can be
written in the form:

2) Usually the rate is a rapidly
changing quantity, as the reaction
proceeds the reactants are used up
and there remains less and less
material to undergo reaction.
3)Generally we obtain for a
reactant a curve which resembles
the one below.
If we take the first 2 points from
the table above, we can find as
avg. rate of decomposition of
N2O5.
You should be able to see that the
avg. rate is decreasing, hence the
curve.
Even though we obtain a
negative value for the rate, since
N2O5 is decreasing, rates are
reported as positive values.
Look at the graph again,

And make the time interval
smaller and smaller, we can
obtain an instantaneous rate.
The instantaneous rate is equal
to the slope of the line at that
point. Calculus??? To what is the
slope of the line at that point
equal?
4) How is data obtained for a
concentration curve?
a) Monitor a color change.
b) Measure pressure if a gas is
produced.
c) Monitor a change in pH if an
acid or base reaction.
5) A look at the change in rate over
time for another reaction:
2 NO2(g)  2 NO(g) + O2(g) at   300 oC
H) What does the balanced
The equation we will look at is:
H2 (g) + I2 (g) ----> 2 HI(g)
H2 and I2 must disappear at the
same rate since 1 molecule of H2
reacts with 1 molecule of I2.
In the same amount of time, 2
molecules of HI must appear.
The rate of appearance of HI
must equal twice the rate of
disappearance of H2 (g) and I2 (g).
The rate of disappearance of
H2(g) = the rate of disappearance
of I2(g) = ½ the rate of
appearance of HI.
In general, for the equation:
aA + bB -----> cC + dD
To obtain a rate equation of the
rates of the substances in relation
to each other we divide through by
the coefficients.
But we usually want the first
reagent in terms of all the others,
so we multiply by the coefficient of
the first reagent.
II. The Rate Law (Rate
expression), Rate Constant
Order of Reaction
A) The following equation has
been studied in the gaseous state
and the data at 250 K may be
summarized as follows:
F2(g) + 2 ClO2(g) ---> 2 ClO2F(g)

B) From this data the answers to
the following questions can be
obtained:
1) What is the rate law of the
reaction?
2) What is the order of the
reaction?
3) What is the value of the rate
constant k?
4) What is the rate of formation
of ClO2F when [F2]is 0.010 mol/L
and[ClO2] is 0.020 mol/L?
C) What is a rate law?
1) In 1864 it was discovered that
the rate of a reaction is
proportional to some power of
the concentration of reactants at
constant temperature.
2) In general, for the equation:
aA + bB  cC + dD
rate law = rate equation = rate
expression

k is the specific rate constant
which is independent of
concentration.
k depends on the nature of the
reactants; fast reactions have
large k's and slow reactions
have small k's ordinarily k
________________ with
temperature.
THE EXPONENTS x AND y
MUST BE EXPERIMENTALLY
DETERMINED.
They are not automatically
obtained from the balanced
equation.
Some experimentally determined
rate laws for equations are as
follows:
a) 2 N2O5 (soln)  2 N2O4 (soln) + O2 (g)
Notice that the exponent is NOT 2, the
coefficient in the balanced equation, but
has been experimentally determined to
be 1.
The reaction order with respect to a given
species equals the exponent of the
concentration of that species in the rate
law as determined experimentally.
The order of the above reaction with
respect to N2O5 is 1.
The order of a reaction is the sum of the
exponents of the reactants in a rate law.
The order of the above reaction, since 1
is the only exponent, is 1 as well. It is a
first order reaction.
This means that when we double the
concentration of N2O5, we double the rate
of reaction OR if we halve the
concentration, we halve the rate.
b) For the reaction: 2 NO + O2  2 NO2

What is the situation here?
D) The answers to the questions then are
obtained in the following manner.
The rate law for

F2(g) + 2 ClO2(g) ---> 2 ClO2F(g)
will look like the following.
Your job is to find the values of x and y
from the experimental data.
You need to do two division problems to
find x and y.
What is the order of the reaction with
respect to F2? _______
What is the order of the reaction with
respect to ClO2? ______
What is the overall order of the
reaction? __________
To find k, the rate constant, we take the
experimentally determined rate law,
put the data in from one of the
experiments and solve for k. The units
of k are important.
k = _________________
To find the rate of formation of ClO2F
when [F2] is 0.010 M and the [ClO2] is
0.020 M, we need to look at the
relationship between the rate of
formation of [ClO2], and the rate of
disappearance of F2.

Rate = __________________________
Transparency – in class problem –
collect it
IV. A graphical method is often used to
show the order of a reaction, or from a
graph we can obtain the order of a
reaction.
A)For the general reaction of A --->
products
B) If the reaction is first order, we can
write
When we divide both sides by [A] and
multiply both sides by dt we obtain the
following:

Those of you who have had calculus
should recognize this as:
Which can be changed to the following
in log to the base 10.
The above equations can be rewritten in
a more familiar form.
THIS MEANS THAT IF WE PLOT ln
[A] vs. t AND OBTAIN A STRAIGHT
LINE THE RATE IS FIRST ORDER.
Or plot log [A] vs. t.
Slope of the line is -k/2.303. The rate
constant k then is = to -slope of the line
times 2.303.
If we plot ln[A] vs.time then the slope of
the line is equal to -k.
C) Half-life - t½ is the time required
for half of the original concentration
of the limiting reactant to be
consumed.
Half-life is _________________ to the
rate constant k.
A fast reaction with a large k has a
_________________ half-life, a slow
reaction with a small k has a
___________________ half-life.

D) For the reaction :
A ----> B + C
We obtain the following data.
Prove that the reaction is first order in
A. Calculate the half-life of the reaction,
t½ .
What is our criterion for a first order
reaction?
We make the following table.

When we plot log [A] vs. t we obtain a
_________________, therefore the
reaction is first order.
The equation for the line is:
V. Second order reaction for a single
reactant
A) A ----> products
B) The rate law is: rate = k[A]2
C) The integrated form of the rate law
gives us the following:
D) A straight line is obtained with slope
= ____.
E) This means that we can tell the
difference between a first order and
second order reaction by
________________________________.
G) The half-life of a second order reaction
depends on the initial concentration of the
reactant as can be seen from the following
derivation:
H) When we compare this to the formula
for the half-life of a first order reaction,
we notice that the t½ for the first order
reaction is __________________of
the initial concentration of the reactant.

The t½ for the second order reaction is
________________________to the initial
concentration of the reactant.
I) The following equations will be given
to you on an exam or quiz for you to
apply to questions in an appropriate
manner.
VI. Collision Theory
A)Consider the elementary process
A2(g) + B2(g) ---> 2AB
B) Collision theory assumes that for gas
molecules to react they must collide.
C) Binary collisions per unit time at
room temperature and 1 atm pressure
are extremely large, approx. 1 x 1031
collisions per Liter-sec.
D) With this many collisions, why isn't
there an immediate explosion when we
put H2(g) and O2(g) together in a balloon?

E) Why isn't every collision an effective
collision; i.e. one that gives product?
F) For a reaction to occur reactant
molecules must collide with an energy
greater than some minimum value and
with the proper orientation.
G) The minimum energy of collision
required for two molecules to react is
called the activation energy, Ea. The
value of Ea depends on a particular
reaction.
H) Effect of temperature
1) The rates of almost all chemical
reactions_________ when T is raised.
Generally speaking for a _________
increase, the reaction rate ________.
2) What is the relationship between
temperature and collision theory?
a) Approx. ___ of the reaction rate
increase due to an increase in
temperature is accounted for by the
increase in the number of___________.
b) Approx. ____of the reaction rate
increase is accounted for by the
increase in the number of high energy
_______________.
3) Recall our graph of the number of
molecules vs. speed in a gas sample.
J) Arrhenius in 1889 proposed the
following equation which relates the rate
constant k to the temperature:

where k is the rate constant from the
rate equation ( rate = k[A])
A is the frequency factor (the collision
factor and the orientation factor) .
e is the base of the natural logarithms
2.718

Ea is the activation energy in Joules per
mole.
R is the molar gas constant, 8.3145 J/K
mol.
T is the temperature in Kelvins.
J) The most valuable use for this
equation is in determining the activation
energy of a reaction from rate
experiments at different temperatures.
L) Ea is best obtained graphically.

This is of the form y = mx + b
If we plot the log of k vs 1/T we obtain
the slope of the line which is equal to
- Ea/ 2.303 R. then Ea = - 2.303 R slope.
L) The equation can be used to compare
rates at various temperatures when Ea is
known, OR the Ea can be obtained from
a comparison of two sets of rate data.
r1 = k1 [A]x at T1
r2 = k2 [A]x at T2
We divide equation 1 by equation 2,
take the log of both sides, and substitute
for log k –Ea/2.303RT. This should give:

We remove the negative sign and we
get the equation which we will use and
which will be given to you on the exam
or quiz.
1) An example: The activation energy
for the reaction involved in the souring
of raw milk is 75 kJ. Milk will sour in
about 8 hours at 21oC (room temp-
70oF). How long will raw milk last in a
refrigerator at 5oC?
2) The gas phase reaction between
methane and diatomic sulfur is given by
the equation:
CH 4(g) + 2 S2(g) ---> CS2(g) + 2 H2S(g)
At 550 oC the rate constant, k1, for the
reaction is 1.1 L/mol sec. At 625 oC the
rate constant, k2, is 6.4 L/mol sec. What
is the activation energy of this reaction?
N) Transition State Theory
1) The hypothetical elementary process
of:
A2 + B2 --> 2 AB
can be represented as the following:
is an activated complex. It is a
transition state species that
__________________________.
In the past, these could not be detected only
postulated. Now with laser technology we
can detect species which exist for only a
femto second(1x10-15), and evidence for the
existence of some intermediates is available.
2) We can graph the potential energy
changes which occur during the course of
a reaction.
3) We have shown reactions in which
DH is - (negative). The reactions are
exothermic.
DHrxn = Eaf - Ear

4) Potential energy diagrams are helpful
in visualizing why it is that highly
exothermic reactions can be very slow.
If Ea is large then __________________ .
VII. The mechanism of a reaction
A) The mechanism of a reaction is the
detailed pathway followed when
reactants are converted to products.
B) It must agree with
1) ____________________ .
2) ____________________ .
3) ____________________ .
C) What do we mean by the rate
determining step? Let's look at a factory
where toasted corn puffs are made.
What is the rate determining step? ___

How fast can the factory produce
packages of corn puffs with a prize?
_________________

What does the factory need to do to
improve the rate of production?
__________________
D) For the reaction:
2 NO + F2 ---> 2 NOF
nitric oxide plus fluorine gives nitrosyl
fluoride
1) The experimentally determined rate
law is: rate = k [NO][F2]
2) A mechanism is proposed on the
basis of the rate equation and other
information gained from a study of
similar reactions over a period of years.
3) Remember that the mechanism is ___ .
4) A suggested mechanism for this
reaction is:
NO + F2 --> NOF + F (slow step)
NO + F --> NOF (fast step)
The sum of these two reactions gives:
2 NO + F2 + F --> 2 NOF + F
Subtracting F from both sides gives the
5) The rate determining step is the _____ .
6) Each step is an elementary reaction
and the rate equation can be written
directly from a _________________
elementary reaction.
7) The rate law of our reaction is rate = k
[NO][F2]. The proposed mechanism
satisfies the requirements ___________
__________________________________ .
E) Let's look at another equation its
experimentally determined rate law and
the proposed mechanism to see if they
are in agreement.
2 O3 -->3 O2
1) The experimentally determined rate
law is:
2) The proposed mechanism is:

The rate of the overall reaction is simply
the rate of step ______ .
Rate = k2[O3][O]
There is a problem. We do NOT report
rate laws with the concentrations of
species which are not in the _________ .
How can we convert Rate = k2[O3][O]
to
3) We need to use a combination of
knowledge from chemistry and algebra to
change one rate law to another.
The chemical knowledge we bring to
the situation is that at equilibrium,
since both step 1 and step 2 are fast and
2 is the reverse of 1 we will make their
rates equal.
k1[O3] = k-1[O2][O]
Then we can use algebra to solve for [O],
since that is what we want to __________

We can then substitute this relationship
of [O] into the rate law we obtained from
____________ to obtain
F) Catalysts
1) The rates of many reactions are
increased by the presence of a catalyst, a
substance which increases the rate of
reaction without being consumed by it.
2) How is this possible?
3) NO (nitric oxide) catalyzes the
decomposition of O3. The following
mechanism is proposed for this reaction:
What should you notice about NO?
4) A catalyst acts by making available a
new reaction mechanism with a lower
activation energy, Ea.
Sometimes is orients the molecules for a
successful collision, for example the
lock and key model mechanism of
enzymes.
Other times it provides an alternate
pathway for the reaction.The depletion of
ozone in the stratosphere by Cl atoms
provides an example of the lowering of
activation energy by a catalyst. The
uncatalyzed reaction has such a large
activation energy as can be seen from the
following diagram that its rate is
extremely slow.
VIII. Review - Factors which affect the
rate of reaction:
A) Nature of the reactants -
B) Concentration -
C) Temperature -
D) Catalysts -

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