CHEM 102 lecture nhj notes- Kinetics 1.ppt

basic chemistry

1. Kinetics: Rates and Mechanisms of Chemical Reactions

2. 1. Explain the effects of concentration, physical state, and temperature on reaction rates.
2. Describe how reaction rate is expressed in terms of changing reactant and product
concentrations over
time, and distinguish among average, instantaneous, and initial rates.
3. Describe the information needed to determine the rate law and explain how to calculate
reaction orders
and rate constant.
4. Describe integrated rate laws and relate with concentration, reaction order and explain the
meaning of
half-life.
5. Explain the importance of activation energy and the effect of temperature on the rate constant.
6. Explain collision theory and relate with temperature, and concentrations.
7. Draw reaction energy diagram for some reactions.
8. Give reaction mechanisms for certain reactions.
9. Describe the role of catalysts in the rate of a given chemical reaction.
10. Differentiate between between homogeneous and heterogeneous catalysis.
Objectives

3. Chemical kinetics is the study of reaction rates, the changes in concentrations of reactants
(or products) as a function of time
Figure 1.1 Reaction rate: the central focus of chemical kinetics. The rate at which
reactant becomes product is the underlying theme of chemical kinetics. As time
elapses, reactant (purple) decreases and product (pale green) increases.
The rate of reaction-the change in concentration of reactant (or product) per unit time-varies
as the reaction proceeds; it is fastest at the beginning of the reaction and slowest at the
end.

4. •Reactions occur at a wide range of rates.
In a fraction of a second - a neutralization, a precipitation, or an explosive redox process,
seem to be over as soon as the reactants make contact-in a fraction of a second.
A moderate length of time, from minutes to months – rusting.
Much longer-the formation of coal from dead plants take hundreds of millions of years.
1.1. Factors that influence reaction rate
•Under any given set of conditions, each reaction has its own characteristic rate, which
is determined by the chemical nature of the reactants.
H
2
(g) + F
2
(g) 2HF(g) [very fast]
3H
2
(g) + N
2
(g) 2NH
3
(g) [very slow]
At room temperature, for example, hydrogen reacts explosively with fluorine but
extremely slowly with nitrogen:
We can control four factors that affect the rate of a given reaction:
the concentrations of the reactants,
the physical state of the reactants,
the temperature at which the reaction occurs,
the use of a catalyst.

5. a). Concentration: molecules must collide to react.
 A major factor influencing the rate of a given reaction is reactant concentration.
 A reaction can occur only when the reactant molecules collide. The more molecules present in
the container, the more frequently they collide, and the more often a reaction between them
occurs.
Thus, reaction rate is proportional to the concentration of reactants: Rate α collision frequency α
concentration
Concentration affects rate by influencing the frequency of collisions between reactant molecules.
Fig. 1.2 Effect of reactant concentrations on rate of
reaction
(A) Both beakers contain the same amounts
of reactants: Na
2S
2O
3,in acidic solution and
Na
3
AsO
3
. However, the beaker on the right
contains more water and thus lower
concentrations
of reactants.
(B) Na
2S
2O
3 decomposes slowly in acidic solution
to
yield H
2
S, which reacts quickly with Na
3
AsO
3
to
give
a bright yellow precipitate of As
2
S
3
. The time it
takes the precipitate to form depends on the
decomposition rate of Na
2
S
2
O
3
.
(C) Note that the precipitate forms more slowly
in the
solution of lower concentrations (the beaker on
the
right).

6. b). Physical state: molecules must mix to collide.
The frequency of collisions between molecules also depends on the physical states of the
reactants.
When the reactants are in the same phase, as in an aqueous solution, random thermal motion
brings them into contact.
When they are in different phases, contact occurs only at the interface, so vigorous stirring and
grinding may be needed.
In these cases, the more finely divided a solid or liquid reactant, the greater its surface area per
unit volume, the more contact it makes with the other reactant, and the faster the reaction occurs.
Physical state affects rate by determining the surface area per unit volume of reactant(s).

7. c). Temperature: molecules must collide with enough energy to react.
Temperature usually has a major effect on the speed of a reaction.
at a higher temperature,more collisions occur in a given time. Even more important, however, is
the fact that temperature affects the kinetic energy of the molecules, and thus the energy
of the collisions.
Most collisions result in the molecules simply recoiling, somewhat like billiard balls, with no
reaction taking place. However, some collisions occur with sufficient energy for the molecules to
react.
At higher temperatures, more of these sufficiently energetic collisions occur. Thus, raising the
temperature increases the reaction rate by increasing the number and, especially, the energy of
the collisions: Rate α collision energy α temperature
Temperature affects rate by influencing the frequency and, even more importantly, the energy of
the reactant collisions.
d). Catalyst: a catalyst is a substance that increases the rate of reaction without being consumed in the
overall reaction.
E.g. A solution of pure hydrogen peroxide, H
2
O
2
, is stable, but when hydrobromic acid, HBr (aq) is
added, H
2
O
2
decomposes into H
2
O and O
2
2H
2
O (aq) 2H
2
O (l) + O
2
(g)

8. 1.2. Expressing the reaction rate
•A rate is a change in some variable per unit of time.
•The most common examples relate to the rate of motion (speed) of an object, which is the
change in its position (that is, the distance it travels) divided by the change in time.
•In the case of a chemical change, we are concerned with the reaction rate, the changes in
concentrations of reactants or products per unit time: reactant concentrations decrease while
product concentrations increase.
•The reaction rate is the increase in molar concentration of product of a reaction per unit time or
the decrease in molar concentration of reactant per unit time.
•Consider a general reaction, A B . We quickly measure the starting reactant concentration
(conc A
1) at t
1, allow the reaction to proceed, and then quickly measure the reactant
concentration again (conc A
2) at t
2.
The change in concentration divided by the change in time gives the average rate:

9. •The rate has units of moles per liter per second (mol L
-1
s
-1
, or mol/L s), or any time unit
convenient for the particular reaction (minutes, years, and so on) .
Suppose the concentration of A changes from 1 .2 mol/L (conc A
1
) to 0.75 mol/L (conc A
2
) over a
125-s period. The average rate is
Average, Instantaneous, and Initial Reaction Rates
Examining the rate of a real reaction reveals an important point: not only the concentration,
but the rate itself varies with time as the reaction proceeds.
Consider the reversible gas-phase reaction between ethylene and ozone
the concentrations of both reactants decrease at the same rate in this particular reaction

10. Suppose we have a known concentration of O
3 in a closed reaction vessel kept at 30°C (303
K). Table 1.1 shows the concentration of O
3 at various times during the first minute after we
introduce C
2H
4 gas.
The rate over the entire 60.0 s is the total change in concentration divided by the change in
time:
Table 1.1 Concentration of O
3 at various
times in its reaction with C
2H
4 at 303 K
•This calculation gives us the average rate over
that period; that is, during the first 60.0 s of the
reaction, ozone concentration decreases an
average of 3 .50 × 1 0
-7
mol/L each second.
•However, the average rate does not show that
the rate is changing, and it tells us nothing about
how fast the ozone concentration is decreasing
at any given instant.
•The average reaction rate is the change in reactant
(or product) concentration over a change in time,
Δt. The rate slows as reactants are used up.

11. Between the starting time 0.0 s and 10.0 s, the average rate is
•The earlier rate is six times as fast as the later rate. Thus, the rate decreases during the
course of the reaction. This makes perfect sense from a molecular point of view: as O
3

molecules are used up, fewer of them are present to collide with C
2H
4 molecules, so the rate
decreases.
The change in rate can also be seen by plotting the concentrations vs. the times at which they
were measured.
A curve is obtained, which means that the rate changes. The slope of the straight line (Δy/Δx,
that is Δ[O
3]/Δt) joining any two points gives the average rate over that period.

12. •The shorter the time period we choose, the closer we come to the instantaneous rate, the
rate at a particular instant during the reaction. The slope of a line tangent to the curve at a
particular point gives the instantaneous rate at that time.
Plotting the data in Table 1.1 gives a curve because
the rate changes during the reaction.
The average rate over a given period is the slope of a
line joining two points along the curve.
The slope of line b is the average rate over the first
60.0 s of the reaction.
The slopes of lines c and e give the average rate over
the first and last 10.0-s intervals, respectively. Line c is
steeper than line e because the average
rate over the earlier period is higher.
The instantaneous rate at 35.0 s is the slope of line d,
the tangent to the curve at t = 35.0 s.
The initial rate is the slope of line a, the tangent to the
curve at t = 0 s.
•The instantaneous rate at time t is obtained from the slope
of the tangent to a concentration vs. time curve at time t.

13. Expressing Rate in Terms of Reactant and Product Concentrations
•So far, in our discussion of the reaction of C
2
H
4
and O
3
, we've
expressed the rate in terms of the decreasing concentration of
O
3
, The rate is the same in terms of C
2
H
4
, but it is exactly the
opposite in terms of the products because their
concentrations are increasing.
•We can express the rate in terms of any of the four
substances involved:
Fig.1.4 Plots of [C
2H
4] and [O
2] vs. time.
•The curve shapes are identical in this case
because the equation coefficients are identical.

14. •The initial rate, the instantaneous rate at the moment the reactants are mixed. Under these conditions,
the product concentrations are negligible, so the reverse rate is negligible. The initial rate is measured by
determining the slope of the line tangent to the curve at t = 0 s.
Fig. 1.5 The instantaneous rate of reaction.
In the reaction 2N
2
O
5
(g) 4NO
2
(g) + O
2
(g)
the concentration of O
2
increases over time.
Average rate, when the time changes from 600 s to 1200 s, is
2.5 × 10
-6
mol/(L.s). Later, when the time changes from 4200 s
to 4800 s, the average rate has slowed to 5 × 10
-7
mol/(L.s).
•the instantaneous rate is obtained at a given time
from the slope of the tangent at the point on the curve
corresponding to that time. In this diagram, the slope
equals Δ[O
2
]/Δt obtained from the tangent.
Fig. 1.6 Calculation of the average rate. The average
rate of formation of O
2
during the decomposition of N
2
O
5
was calculated during two different time intervals.
•The rate of the reaction
decreases as the reaction
proceeds.

15. 2N
2
O
5
(g) 4NO
2
(g) + O
2
(g)
•Because the amounts of products and reactants are related by stoichiometry, any substance in the
reaction can be used to express the rate of reaction.
•In the case of the decomposition of N
2
O
5
to NO
2
and O
2
, the rate in terms of the rate of formation of
oxygen is, Δ[O
2]/Δt.
•However, it can also express in terms of the rate of decomposition of N
2
O
5
.
•The rate of decomposition of N
2O
5 and the rate of formation of oxygen are easily related.
•Two moles of N
2O
5 decompose for each mole of oxygen formed, so the rate of
decomposition of N
2
O
5
is twice the rate of formation of oxygen.
•To equate the rates, you must divide the rate of decomposition of N
2
O
5
by 2 (its coefficient
in the balanced chemical equation).

16. where a, b, c, and d are coefficients of the balanced equation. In general, the rate
is related to reactant or product concentrations as follows:
For any reaction,
Example 1:
Consider the reaction of nitrogen dioxide with fluorine to give nitryl fluoride, NO
2
F.
2NO
2
(g) + F
2
(g) 2NO
2
F(g)
How is the rate of formation of NO
2
F related to the rate of reaction of fluorine?
Exercise: For the reaction given in the example, how is the rate of formation of NO
2F
related to the rate of reaction of NO
2?

17. Example: 2
Calculating the Average Reaction Rate

18. Exercise
Iodide ion is oxidized by hypochlorite ion in basic solution.
I
-
(aq) + ClO
-
(aq) Cl
-
(aq) + IO
-
(aq)
In 1.00 M NaOH at 25
o
C, the iodide-ion concentration (equal to the ClO
-
concentration) at
Different times was as follows:
Time [I
-
]
2.00 s 0.00169 M
8.00 s 0.00101 M
Calculate the average rate of reaction of I
-
during this time interval.

19. THE RATE LAW AND ITS COMPONENTS
•A rate law is an equation that relates the rate of a reaction to the concentrations of
reactants (and catalyst) raised to various powers.
Here k, called the rate constant, is a proportionality constant in the relationship between rate and
concentrations. It has a fixed value at any given temperature, but it varies with temperature. It is
specific for a given reaction at a given temperature; it does not change as the reaction
proceeds.
•Experimentally, it has been found that a reaction rate depends on the concentrations of certain
reactants as well as the concentration of catalyst, if there is one.
•Any hypothesis we make about how the reaction occurs on the molecular level must conform to
the rate law because it is based on experimental fact.
For a general reaction,
the rate law has the form
•The exponents m and n, called the reaction orders, define how the rate is affected by reactant
concentration.

20. The exponents m and n, called the reaction orders, define how the rate is affected by
reactant concentration.
•Thus, if the rate doubles when [A] doubles, the rate depends on [A] raised to the first
power, [A]
1
, so m = 1 .
• Similarly, if the rate quadruples when [B] doubles, the rate depends on [B] raised to the
second power, [B]
2
, so n =2.
•In another reaction, the rate may not change at all when [A] doubles; in that case, the
rate does not depend on [A] , or, to put it another way, the rate depends on [A] raised to
the zero power, [A]
0
, so m = 0.
•Keep in mind that the coefficients a and b in the general balanced equation are not
necessarily related in any way to these reaction orders m and n.
The components of the rate law-rate, reaction orders, and rate constant-must be found
by experiment; they cannot be deduced from the reaction stoichiometry.
•Chemists take an experimental approach to finding these components by
1) Using concentration measurements to find the initial rate
2) Using initial rates from several experiments to find the reaction orders
3) Using these values to calculate the rate constant

21. •Many experimental techniques have been developed to accomplish the first of these steps,
the measurement of concentrations in order to find initial rates; here are three common
approaches.
spectroscopic methods –colored substances
a change in number of moles of gas (change in pressure)-
a change in conductivity-
Reaction Order: a reaction can be classified by its orders.
•The reaction order with respect to a given reactant species equals the exponent of the
concentration of that species in the rate law, as determined experimentally.
•The overall order of a reaction equals the sum of the orders of the reactant species in the rate law.
•In the simplest case, a reaction with a single reactant A,
the reaction is first order overall if the rate is directly proportional to [A] : Rate = k[A]
it is second order overall if the rate is directly proportional to the square of [A] : Rate = k[A]
2
and it is zero order overall if the rate is not dependent on [A] at all: Rate = k[A]
0
= k(1) = k

22. othe concentration of H
2O does not even appear in the rate law. Thus, the reaction is zero order
with respect to H
2
O ( [H
2
O]
0
) . This means that the rate does not depend on the concentration
of H
2
O.
•Here are some real examples. For the reaction between nitrogen monoxide and ozone,
the rate law has been experimentally determined to be
This reaction is first order with respect to NO. It is also first order with respect to O
3
.
This reaction is second order overall ( 1 + 1 = 2).
The rate law for this reaction has been determined to be
The reaction is second order in NO and first order in H
2
, so it is third order overall.
othe rate law has been found to be

23. •Reaction orders cannot be deduced from the balanced equation. For the reaction between
NO and H
2 and for the hydrolysis of 2-bromo-2-methylpropane, the reaction orders in the rate
laws do not correspond to the coefficients of the balanced equations. Reaction orders must
be determined from rate data.
•Reaction orders are usually positive integers or zero, but they can also be fractional or
negative. For the reaction
a fractional order appears in the rate law:

24. Examples: Determining Reaction Order from Rate Laws
For each of the following reactions, use the given rate Jaw to determine the reaction
order with respect to each reactant and the overall order:
(a) 2NO(g) + O
2
(g) 2NO
2
(g); rate = k[NO]
2
[O
2
]
(b) CH
3CHO(g) CH
4(g) + CO(g); rate = k[CH
3CHO]
3/2
(c) H
2O
2(aq) + 3I
-
(aq) + 2H
+
(aq) I
3
-
(aq) + 2H
2O(l); rate = k[H
2O
2] [I
-
]
We inspect the exponents in the rate law, not the coefficients of the balanced equation,
to find the individual orders, and then take their sum to find the overall reaction order.
Solution
(a) The exponent of [NO] is 2, so the reaction is second order with respect to
NO, first order with respect to O
2
, and third order overall.
(b) The reaction is 3/2 order in CH
3
CHO and 3/2 order overall.
(c) The reaction is first order in H
2
O
2
, first order in I
-
, and second order overall.
The reactant H
+
does not appear in the rate law, so the reaction is zero order in H
+
.
Exercise: Experiment shows that the reaction
5Br
-
(aq) + BrO
3
-
(aq) + 6H
+
(aq) 3Br
2(l)+ 3H
2O(l) obeys this rate law: rate = k[Br
-
][BrO
3
-
][H+]
2
.
What are the reaction orders in each reactant and the overall reaction order?

25. Determining Reaction Orders Experimentally
•The initial rate method is a simple way to obtain reaction orders. It consists of doing a series
of experiments in which the initial, or starting, concentrations of reactants are varied.
Then the initial rates are compared, from which the reaction orders can be deduced.
Consider the reaction between oxygen and nitrogen monoxide,
The rate law, expressed in general form, is
Rate = k[O
2]
m
[NO]
n
To find the reaction orders, we run a series of experiments, starting each one with a
different set of reactant concentrations and obtaining an initial rate in each case.
Table 1.2 shows experiments that change one reactant concentration while keeping the other
constant.

26. Table 1.2: Initial Rates for a Series of Experiments with the Reaction Between O
2 and NO
oIf we compare experiments 1 and 2, we see the effect of doubling [O
2
] on the rate.
First, we take the ratio of their rate laws:
where [O
2]
2 is the O
2 concentration for experiment 2, [NO]
1 is the NO concentration for experiment
1 , and so forth. Because k is a constant and [NO] does not change between these two experiments,
these quantities cancel:
Substituting the values from Table 1.2, we obtain
1.99 = (2.00)
m
, 2.00 = (2.00)
m
, m = 1
The reaction is first order in O
2
: when [O
2
] doubles, the rate doubles.

27. To find the order with respect to NO, we compare experiments 3 and 1 , in which [O
2
] is held
constant and [NO] is doubled:
As before, k is constant, and in this pair of experiments [O
2] does not change, so these quantities
cancel:
The actual values give
Dividing, we obtain, 3.99 = (2.00)
n
4.00 = 2
n
, n = 2
The reaction is second order in NO: when [NO] doubles, the rate quadruples.
Thus, the rate law is, Rate = k[O
2
][NO]
2
Exercise: Find the rate law and the overall reaction order for the reaction H
2 + I
2 2HI from the
following data at 450°C:

28. Determining the Rate Constant
•With the rate, reactant concentrations, and reaction orders known, the sole remaining unknown in the rate law
is the rate constant, k.
•The rate constant is specific for a particular reaction at a particular temperature. The experiments with the
reaction of O
2
and NO were run at the same temperature, so we can use data from any to solve for k.
•From experiment 1 in Table 1.2, for instance, we obtain
Table 1.3 Units of the Rate Constant k
for Several Overall Reaction Orders

29. INTEGRATED RATE LAWS: CONCENTRATION CHANGES OVER TIME
•They tell us the rate or concentration at a given instant, allowing us to answer a critical question, "How fast is
the reaction proceeding at the moment when y moles per liter of A are reacting with z moles per liter of B ?"
•However, by employing different forms of the rate laws, called integrated rate laws, we can consider the time
factor and answer other questions, such as "How long will it take for x moles per liter of A to be used up?"
and "What is the concentration of A after y minutes of reaction?"
Integrated Rate Laws for First-Order, Second-Order, and Zero-Order Reactions
1) First – order reaction rate law
Consider a simple first-order reaction, A B
Using calculus, this expression is integrated over time to obtain the integrated rate law for a
first-order reaction:
the rate can be expressed as the change in the concentration of A divided by the change in
time:
It can also be expressed in terms of the rate law:
Setting these different expressions equal to each other gives
So, for first – order, we have

30. Example:
The decomposition of N
2
O
5
to NO
2
and O
2
is first order, with a rate constant of 4.80 x10
-
4
/s at 45
o
C.
a. If the initial concentration is 1.65 x 10
-2
mol/L, what is the concentration after 825 s?
b. How long would it take for the concentration of N
2
O
5
to decrease to 1.00 x 10
-2
mol/L
from its initial value, given in a?
Solution a. In this case, you need to use the equation relating concentration to time for a first-
order reaction, which is
Substituting the appropriate values, you get
To solve for [N2O5]t, you take the antilogarithm (antiln) of both sides. This removes the ln from the left
and gives antiln(0.396), or e
-0.396
, on the right, which equals 0.673.

31. b. substitute into the same first-order equation relating concentration to
time.
The left side equals -0.501; the right-side equals -4.80 x 10
-4
/s x t. Hence,
Exercise.
a)What would be the concentration of dinitrogen pentoxide in the example above after 6.00 x 10
2
s?
b)How long would it take for the concentration of N
2
O
5
to decrease to 10.0% of its initial value?

32. 2) Second-Order Rate Law
Consider the reaction aA products and suppose it has the second-order rate law
Integrating over time gives the integrated rate law for a second-order reaction involving one
reactant:
An example is the decomposition of nitrogen dioxide at moderately high temperatures (300
o
C to 400
o
C).
Example: At 330
o
C, the rate constant for the decomposition of NO
2
is 0.775 L/(mols). Suppose
the initial concentration is 0.0030 mol/L. What is the concentration of NO
2
after 645 s? By
substituting into the previous equation, you get
Thus, [NO
2]
t 0.0012 mol/L. Thus, after 645 s, the concentration of NO
2 decreased from 0.0030
mol/L to 0.0012 mol/L.

33. 3) Zero-Order Reactions
There are instances where reactions are zero-order. An example includes the decomposition
of ethyl alcohol in the liver in the presence of the enzyme liver alcohol dehydrogenase.
Once again, consider the reaction with the general form
Thus, the expression for zero-order rate law is usually written as Rate = k
Integrating over time gives the integrated rate law for a zero-order reaction:

34. Graphing of Kinetic Data
•Earlier you saw that the order of a reaction can be determined by comparing initial rates for several
experiments in which different initial concentrations are used (initial-rate method).
•It is also possible to determine the order of a reaction by graphical plotting of the data for a
particular experiment.
•The order of the reaction is determined by which graph gives the best fit to the experimental data.
An integrated rate law can be rearranged into the form of an equation for a straight line, y = mx +
b, where m is the slope and b is the y-axis intercept.
For a first-order reaction, we have
Therefore, a plot of In [A]
t
, vs. time gives a straight
line with slope = -k and y intercept = In [A]
0
Fig. 1.7 Plot of In [A]
t vs. time gives a straight line for a reaction
that is first order in A.

35. For a simple second-order reaction, we have
In this case, a plot of 1 / [A]
t
vs. time gives a straight
line with slope = k , y intercept = 1 / [A]
0
Fig. 1.8 Plot of 1 /[A]
t
vs. time gives a straight line
for a reaction that is second order in A
For a zero-order reaction, we have
Therefore, a plot of [A]
t , vs. time gives a straight line
with slope = -k and y intercept = [A]
0 Fig. 1.9 Plot of [A]
t
vs. time gives a straight
line for a reaction that is zero order in A.

36. Fig. 1.10 Graphical determination of the reaction order for the
decomposition of N
2
O
5
A)A plot of [N
2
O
5
] vs. time is curved , indicating that the reaction is
not zero order in [N
2O
5],
B)A plot of In [N
2
O
5
l vs. time gives a straight line, indicating that the
reaction is first order in N
2O
5
C)A plot of 1 /[N
2O
5] vs. time is curved, indicating that the reaction is
not second order in N
2
O
5
.
Plots A and C support the conclusion from plot B.
A table of time and concentration data
for determining reaction order

37. Half-Life of a Reaction
•As a reaction proceeds, the concentration of a reactant decreases, because it is being consumed.
•The half-life, t
1/2, of a reaction is the time it takes for the reactant concentration to decrease to one-
half of its initial value.
•A half-life is expressed in time units appropriate for a given reaction and is characteristic of that
reaction at a given temperature.
•For a first-order reaction, such as the decomposition of dinitrogen pentoxide, the half-life is
independent of the initial concentration.
•At fixed conditions, the half-life of a first-order reaction is a constant, independent of reactant
concentration.
For example, the half-life for the first-order decomposition of N
2
O
5
at 45°C is 24.0 min. The
meaning of this value is that if we start with, say, 0.0600 mol/L of N
2O
5 at 45°C, after 24 min
(one half-life), 0.0300 mol/L has been consumed and 0.0300 mol/L remains; after 48 min (two
half-lives), 0.0150 mol/L remains; after 72 min (three half-lives), 0.0075 mol/L remains, and
so forth.

38. Fig. 1.11 A plot of [N
2
O
5
] vs. time for three half-lives.
During each half-life, the concentration is halved (T =
45°C and [N
2
O
5
]
0
= 0.0600 mol/L). The blow-up
volumes, with N
2
O
5
molecules as colored spheres,
show that after three half-lives, ½ x ½ x ½ = 1/8 of the
original concentration remains.
We can see from the integrated rate law why the
half-life of a first-order reaction is independent of
concentration:
After one half-life, t = t
1/2, and [A]
t, = ½[A]
0.
Substituting, we obtain
Then, solving for t
1/2 we have

39. Determining the Half-Life of a First-Order Reaction
Example
Cyclopropane is the smallest cyclic hydrocarbon. Because its 60° bond angles reduce orbital
overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at
1000°C via the following first-order reaction:
The rate constant is 9.2 s
-1
.
(a)What is the half-life of the reaction?
(b)How long does it take for the concentration of cyclopropane to reach one-quarter of the
initial value?
Solution: (a) We use the equation t
1/2 = 0.693/k and substitute 9.2 s
-1
for k
•It takes 0.075 s for half the
cyclopropane to form propene at this
temperature.
(b) Each half-life decreases the concentration to one half of its initial value, so two half-lives
decrease it to one-quarter.

40. Exercise: Iodine-123 is used to study thyroid gland function. This radioactive isotope breaks
down in a first-order process with a half-life of 13. 1 h. What is the rate constant for the process?
In contrast to the half-life of a first-order reaction, the half-life of a second order reaction does
depend on reactant concentration:
For a second-order reaction with a high initial reactant concentration has a shorter half-life,
and one with a low initial reactant concentration has a longer half-life. Therefore, as a second-
order reaction proceeds, the half-life increases.
The half-life of a zero order reaction is directly proportional to the initial reactant concentration:
Thus, if a zero-order reaction begins with a high reactant concentration, it has a longer half-life
than if it begins with a low reactant concentration.

41. Table: An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions

42. THE EFFECT OF TEMPERATURE ON REACTION RATE
•Temperature often has a major effect on reaction rate.
•a common organic reaction-hydrolysis, or reaction with water,
of an ester-when reactant concentrations are held constant,
the rate nearly doubles with each rise in temperature of 10 K
(or 10°C).
•If we collect concentration and time data for the same reaction
run at different temperatures (T), and then solve each rate
expression for k, we find that k increases as T increases.
•In other words, temperature affects the rate by affecting the
rate constant.
•A plot of k vs. T gives a curve that increases exponentially
Fig. 1.12 A plot of rate constant vs.
temperature for this reaction shows an
exponentially increasing curve.
•The mathematical equation k = Ae
-Ea/RT
, which expresses
the dependence of the rate constant on temperature, is
called the Arrhenius equation, after its formulator, the
Swedish chemist Svante Arrhenius.
where k is the rate constant, e is the base of
natural logarithms, Ea is the activation energy; R is
the universal gas constant and T is the absolute
temperature, and the symbol A is assumed to be a
constant, called the frequency factor..

43. Taking the natural logarithm of both sides of the Arrhenius equation ( ) gives
This shows that if you plot ln k against 1/T, you should get a straight line (below).
The slope of this line is E
a/R, from which you can obtain the activation energy Ea.
write the equation for two different absolute temperatures T
1
and T
2
. You write k
1
for the rate
constant at temperature T
1
and k
2
for the rate constant at temperature T
2
.
Eliminate ln A by subtracting,
•As the Arrhenius equation shows, rate increases with
temperature because a temperature rise increases the
rate constant.
•The activation energy, Ea, the minimum energy needed
for a reaction to occur, can be determined graphically
from k values at different T values.

44. Example:
The rate constant for the formation of hydrogen iodide from the elements,
H
2
(g) + I
2
(g) 2HI(g)
is 2.7 x 10
-4
L/(mol.s) at 600 K and 3.5 x 10
-3
L/(mol.s) at 650 K.
(a)Find the activation energy Ea.
(b) Calculate the rate constant at 700 K.
Solution:
Substitute the data given in the problem
statement into the equation

45. Exercise
The reaction 2NOCl(g) 2NO(g) + CI
2
(g) has an Ea of 1.00 × 10
2
kJ/mol and a rate
constant of 0.286 L/mol·s at 500 K. What is the rate constant at 490 K?