Chemical kinetics University of Zambia University pdf

Rates of chemical kinetics

1. CHEMICAL KINETICS
INTRODUCTION

2. CHEMICAL KINETICS
•Applications of chemistry focus on chemical reactions. Commercial
use of a reaction requires knowledge of its characteristics, e.g.
stoichiometry, energetics, and rate (speed).
•A reaction is defined by its reactants & products. When reactants
and products are known, the equation for the reaction can be
written and balanced, and stoichiometric calculations computed.
•Another very important characteristic of a rxion is its spontaneity.
This is the inherent tendency for a process to occur but says does
not say anything about speed.
•Spontaneous does not mean fast.

3. CHEMICAL KINETICS
•There are many spontaneous reactions that are so slow that no
rxion occurs several weeks or years at normal temperature, e.g.,
there is a strong inherent tendency for gaseous hydrogen and
oxygen to react, that is,
2H
2(g) + O
2(g) → 2H
2O
(l)
•but no reaction occurs at 25ºC. Similarly, the gaseous rxions:
H
2(g) + Cl
2(g) → 2HCl
(g)
N
2(g) + 3H
2(g) → 2NH
3(g)
•are both highly likely to occur from a thermodynamic point of view,
but no rxions occur under normal conditions.
•In addition, the process of changing diamond to graphite is
spontaneous but is so slow that it is not detectable.

4. CHEMICAL KINETICS
•It is not enough to understand the stoichiometry and
thermodynamics of a reaction; it is important to also understand
the factors that govern the rate of reaction. Chemical kinetics is the
chemistry of reaction rates.
•One of the main goals of chemical kinetics is to understand the
steps by which a reaction takes place. This series of steps is called
the reaction mechanism.
•When rxion mechanism is understood, the rate of reaction can be
controlled, i.e., slowed down or increased.
•In this topic, we will consider the main ideas of chemical kinetics,
including: rate laws, reaction mechanisms, and simple models for
chemical reactions.

5. Reaction Rates
•To introduce the concept of the rate of reaction, let’s consider
decomposition of nitrogen dioxide which decomposes to nitric
oxide and oxygen as follows:
2NO
2(g) → 2NO
(g) + O
2(g)
•If in an experiment we start with a flask of nitrogen dioxide at
300ºC and measure the concentrations of NO
2, nitric oxide, and
oxygen as the NO
2 decomposes, results are summarized in Table
12.1, and the data are plotted in Fig. 12.1.
•Note from these results that concentration of the reactant (NO
2)
decreases with time and concentrations of products (NO and O
2)
increase with time (Fig. 12.2).

6. Reaction Rates

7. Reaction Rates

8. Reaction Rates
•Speed, or rate, is the change in a given quantity over a specific
period of time. For chemical reactions, the quantity that changes
is the amount or concentration of a reactant or product. So the
reaction rate of a chemical reaction is change in concentration of a
reactant or product per unit time:
•Rate = conc of A at time t
2 - conc of A at time t
1
t
2 - t
1
= Δ[A]/Δt
•where A = reactant or product being considered,
•square brackets = concentration in mol/L.
•Δ = change in a given quantity.

9. Reaction Rates
•NB: change can be positive (increase) or negative (decrease), thus
leading to a positive or negative reaction rate by this definition,
but rate is always a positive.
•Rate at which the concentration of NO
2 changes over the first 50
seconds of the reaction using the data given in Table 12.1:
Change in [NO
2] /Time elapsed = Δ[NO
2]/Δt
= [NO
2]
t=50 - [NO
2]
t=0
50s - 0s
= 0.0079 mol/L - 0.0100 mol/L
50 s
= -4.2 x 10
-5 mol/L·s

10. Reaction Rates
•Note: since conc of NO
2 is decreasing with time, Δ[NO
2] is a
negative. Because rate of reaction must always be positive:
Rate = -Δ[NO
2] /Δt
•Since concentrations of reactants always decrease with time, rate
involving a reactant always includes a negative sign. Average rate
of this reaction from 0 to 50 seconds is therefore:
Rate = -Δ[NO
2]/Δt
= -(-4.2 x 10
-5 mol/L·s)
= 4.2 x 10
-5 mol/L·s

11. Reaction Rates
•Average rates for this reaction during several other time intervals
are given in Table 12.2.
*Note that the rate decreases with time.
Table 12.2 Average Rate (in mol/L·s) of Decomposition of Nitrogen Dioxide as a Function of Time*
NB: rate is not constant but decreases with time. Rates given in Table
12.2 are average rates over 50-second time intervals.

12. Reaction Rates
•*The value of reaction rate at a particular time (instantaneous
rate) is obtained by determining the slope of a line tangent to the
curve at that point.*
•Figure 12.1 shows a tangent drawn at t = 100 seconds. The slope
of this line gives the reaction rate at t = 100 seconds as follows:
Slope of the tangent line = change in y / change in x
= Δ[NO
2] / Δt
•Since Rate = -Δ[NO
2]/Δt
•therefore, Rate = -(slope of the tangent line)
= -(-0.0026 mol/L /110 s )
= 2.4 x 10
-5 mol/L·s

13. Reaction Rates
•Note: because stoichiometry determines relative rates of
consumption of reactants and generation of products, coefficients
in the balanced equation for a reaction must be considered in the
rate of reaction, e.g.,
2NO
2(g) → 2NO
(g) + O
2(g)
•both, reactant NO
2 and product NO have a coefficient of 2, thus
NO is produced at the same rate as NO
2 is consumed. Thus, the
curve for NO is the same shape as the curve for NO
2, except that it
is inverted.
•This means, at any point in time, the slope of the tangent to the
curve for NO is the negative of the slope of the tangent to the
curve for NO
2.
•In the balanced equation, the product O
2 has a coefficient of 1. This
means, it is produced half as fast as NO, since NO has a coefficient
of 2, i.e., rate of NO production is twice the rate of O
2 production.

14. Reaction Rates
•For example, at t = 250 seconds,
•Slope of the tangent to the NO curve = 6.0 x 10
-4 mol/L / 70 s
= 8.57 x 10
-6 mol/L·s
•Slope of the tangent to the O
2 curve = 3.0 x 10
-4 mol/L / 70 s
= 4.29 x 10
-6 mol/L·s
•Thus, the slope at t = 250 seconds on the NO curve is twice the
slope on the O
2 curve at t = 250s, showing that the rate of
production of NO is twice that of O
2.

15. Reaction Rates
•In summary:
•Rate of consumption = rate of production = 2(rate of production
of NO
2 of NO of O
2)
• -Δ[NO
2] /Δt = Δ[NO] / Δt = 2(Δ[O
2]/Δt)
•Because reaction rate changes with time, and because the rate is
different (by factors that depend on the coefficients in the
balanced equation), reaction rate for a chemical reaction must be
described very specifically.

16. Rate Laws
•Chemical reactions are reversible.
•Decomposition of NO
2 has been so far considered only the forward
reaction: 2NO
2(g) → 2NO
(g) + O
2(g)
•But, as conc of NO and O
2 increase, they react to re-form NO
2:
O
2(g) + 2NO
(g) → 2NO
2(g)
•When gaseous NO
2 is placed in an empty container, initially the
dominant reaction is 2NO
2(g) → 2NO
(g) + O
2(g)
•and change in concentration of NO
2 (Δ[NO
2]) depends only on the
forward reaction. But, after some time, enough products
accumulate, the reverse rxion becomes important and then, Δ[NO
2]
depends on the difference in the rates of the forward and reverse
reactions.

17. Rate Laws
•To avoid this complication, reaction rate of a reversible reaction is
studied under conditions where the reverse reaction is negligible,
i.e., soon after reactants are mixed together, before the products
have had time to form to significant levels.
•If conditions where the reverse reaction is neglible are chosen,
then reaction rate will depend only on the concentrations of the
reactants.
•For the decomposition of nitrogen dioxide, we can write
Rate = k[NO
2]
n → rate law
•An expression, which shows how reaction rate depends on the
concentrations of reactants, is called a rate law.
•k = proportionality constant, called the rate constant
•n = order of the reactant.

18. Rate Laws
•Both, k and n, are determined by experiment. Order of a reactant
can be an integer (including zero) or a fraction.
•Note two important points about the equation Rate = k[NO
2]
n :
•1. Concentrations of products do not appear in the rate law
because the reaction rate is being studied under conditions where
the reverse reaction is not contributing to the overall rate.
•2. The value of the exponent n is determined by experiment; it
cannot be written from the balanced equation.

19. Types of Rate Laws
•1.Differential rate law, often simply called rate law, is the rate law
that shows how rate depends on concentration.
•2.Integrated rate law is rate law that shows how concentration
depends on time.
•NB: A differential rate law is always related to a type of integrated
rate law, and vice versa.
•*That is, when the differential rate law for a given reaction is
determined, the form of the integrated rate law for the reaction is
automatically also known.* This means, once we determine
experimentally either type of rate law for a rxion, we also know the
other one.
•Which rate law we choose to determine by experiment depends on
which type of data are easiest to collect.

20. Types of Rate Laws
•If it is easier to measure how the rate changes with conc, the
differential (rate/conc) rate law can be easily determined.
•If it is easier to measure conc as a function of time, then the
integrated (conc/time) rate law can be easily determined.
•How are rate laws are actually determined? Why are we
interested in determining the rate law for a reaction? How
does it help us?
•It helps us work backward from the rate law to infer the
steps by which the rxion occurs, i.e., rxion mechanism.

21. Determining the Form of a Rate Law
•How rate laws are actually determined.
•1st step in understanding how a chemical reaction occurs is to
determine the rate law, i.e., to determine by experiment the power
to which each reactant concentration must be raised in the rate law.
•Consider the decomposition of dinitrogen pentoxide in carbon
tetrachloride solution:
2N
2O
5(soln) → 4NO
2(soln) + O
2(g)
•Data for this reaction at 45ºC are listed in Table 12.3 and plotted in
Fig. 12.3.

22. Determining the Form of a Rate Law
Table 12.3 | Concentration/Time Figure 12.3 | A plot of the concentration of N
2O
5
Data for the Reaction as a function of time for the reaction
2N
2O
5(soln) → 4NO
2(soln) + O
2(g) (at 45ºC) 2N
2O
5(soln) → 4NO
2(soln) + O
2(g) (at 45ºC). Note that
the reaction rate at [N
2O
5] = 0.90M is twice that
at [N
2O
5] = 0.45M

23. Determining the Form of a Rate Law
•In this rxion, O
2 gas escapes from the solution and thus does not
react with the NO
2. Thus, the reverse rxion is negligible at all times
over the course of this rxion.
•Evaluation of rxion rates at concs of N
2O
5 of 0.90 M and 0.45 M,
by taking the slopes of the tangents to the curve at these points
(see Fig. 12.3), yields the following data:
[N
2O
5] Rate (mol/L·s)
0.90 M 5.4 x 10
-4
0.45 M 2.7 x 10
-4
NB: when [N
2O
5] is halved, rate halves. This means, the rate of this
rxion depends on the conc of N
2O
5 to the first power.

24. Determining the Form of a Rate Law
In other words, the (differential) rate law for this rxion is
Rate = -Δ[N
2O
5]/ Δt = k[N
2O
5]
1
= k[N
2O
5]
Thus the rxion is first order in N
2O
5. Note: for this rxion the order
is not the same as the coefficient of N
2O
5 in the balanced
equation for the rxion.
This shows that, the order of a reactant must be obtained by
observing how the rxion rate depends on the conc of that
reactant.
By determining the instantaneous rate at two different reactant
concs, the rate law for decomposition of N
2O
5 has the form
Rate = -Δ[A]/Δt = k[A] where A = N
2O
5.

25. Method of Initial Rates
Method of initial rates - common method for determining
the rate law for a rxion.
Initial rate of a rxion is the instantaneous rate determined
just after the rxion begins (just after t = 0) before initial
concs of reactants change significantly. Several expts are
carried out using different initial concs. Initial rate is
determined for each run. Results are then compared to see
how the initial rate depends on the initial concs. This allows
the form of the rate law to be determined.
Illustration:
NH
4
+
(aq) + NO
2
-
(aq) → N
2(g) + 2H
2O
(l)

26. Method of Initial Rates
•Table 12.4 gives initial rates obtained from three experiments
involving different initial concs of reactants for the reaction:
NH
4
+
(aq) + NO
2
-
(aq) → N
2(g) + 2H
2O
(l)
Expt Initial Conc of NH
4
+ Initial Conc of NO
2
- Initial Rate(mol/L·s)
1 0.100 M 0.0050 M 1.35 x 10
-7
2 0.100 M 0.010 M 2.70 x 10
-7
3 0.200 M 0.010 M 5.40 x 10
-7
•The form of the rate law for this rxion is:
Rate = -Δ[NH
4
+]/Δt = k[NH
4
+]
n [NO
2
-]
m

27. Method of Initial Rates
•Values of n and m are determined by observing how the initial rate
depends on initial concs of NH
4
+ and NO
2
-.
•In Expts 1 and 2, initial conc of NH
4
+ remains the same but as initial
conc of NO
2
- doubles, observed initial rate also doubles. Since
Rate = k[NH
4
+]
n[NO
2
-]
m
•for Expt1
Rate = 1.35 x 10
-7 mol/L·s = k(0.100 mol/L)
n (0.0050 mol/L)
m
•for Expt 2
Rate = 2.70 x 10
-7 mol/L·s = k(0.100 mol/L)
n (0.010 mol/L)
m

28. Method of Initial Rates
•The ratio of these rates is
•Rate 2 = 2.70 x 10
-7
mol/L·s = k(0.100 mol/L)
n
(0.010mol/L)
m
•Rate 1 1.35 x 10
-7
mol/L·s k(0.100 mol/L)
n
(0.0050 mol/L)
m

= k(0.100 mol/L)
n
(0.010 mol/L)
m

k(0.100 mol/L)
n
(0.0050 mol/L)
m

= (0.010 mol/L)
m
= (2.0)
m

(0.0050 mol/L)
m

•Thus Rate 2/Rate 1 = 2.00 = (2.0)
m
. Thus m = 1

29. Method of Initial Rates
•This means the rate law for this rxion is 1st order in the reactant NO
2
-.
•for Expts 2 and 3, the ratio:
•Rate 3 = 5.40 x 10
-7 mol/L·s = (0.200 mol/L)
n
•Rate 2 2.70 x 10
-7 mol/L·s (0.100 mol/L)
n
• = 2.00 = (0.200/0.100)
n = (2.00)
n The value of n = 1.
•Note: Both n and m = 1, thus, the rate law is:
Rate = k[NH
4
+]/[NO
2
-]
•Rate law is first order in both NO
2
- and NH
4
+. Note: it is a coincidence
that n and m have same values as the coefficients of NH
4
+ and NO
2
-
in the balanced equation for the rxion. Overall rxion order = sum of
powers n & m. For this rxion, n + m = 2. Rxion is 2nd order overall

30. Method of Initial Rates
•Rate constant k can now be calculated using the results of any of
the 3 expts in Table 12.4.
•From data for Expt 1, we know that Rate = k[NH
4
+]/ [NO
2
-]
Thus, 1.35 x 10
-7 mol/L·s = k(0.100 mol/L)/(0.0050 mol/L)
k = 1.35 x 10
-7 mol/L·s = 2.7 x 10
-4 L/mol·s
(0.100 mol/L) (0.0050 mol/L)
•Example 12.1: Determining a Rate Law
•Rxion between bromate ions and bromide ions in acidic aqueous
soln is given by the equation:
BrO
3
-
(aq) + 5Br
-
(aq) + 6H
+
(aq) → 3Br
2(l) + 3H
2O
(l)

31. Method of Initial Rates
•Using data for results from 4 expts in Table 12.5, determine
the orders for all three reactants, the overall rxion order,
and the rate constant.
Expt Initial Conc of Initial Conc of Initial Conc of Measured Initial
BrO
3
-
(mol/L) Br
-
(mol/L) H
+
(mol/L) Rate (mol/L·s)
1 0.10 0.10 0.10 8.0 x 10
-4

2 0.20 0.10 0.10 1.6 x 10
-3

3 0.20 0.20 0.10 3.2 x 10
-3

4 0.10 0.10 0.20 3.2 x 10
-3

32. Method of Initial Rates
•Solution: The general form of the rate law for this reaction is
Rate = k[BrO
3
-]
n[Br
-]
m[H
+]
p
•Values of n, m, and p can be determined by comparing rates from the
expts. To determine n, we use results from Expts 1 and 2, in which
only [BrO
3
-] changes:
•Rate 2 = 1.6 x 10
-3 mol/L·s = k(0.20 mol/L)
n(0.10 mol/L)
m(0.10 mol/L)
p
•Rate 1 8.0 x 10
-4 mol/L·s k(0.10 mol/L)
n(0.10 mol/L)
m(0.10 mol/L)
p
• 2.0 = 0.20 mol/L
n = (2.0)
n Thus n = 1.
0.10mol/L

33. Method of Initial Rates
•To determine the value of m, we use Expts 2 & 3, in which only [Br
-]
changes:
•Rate 3 = 3.2 x 10
-3mol/L·s = k(0.20 mol/L)
n(0.20 mol/L)
m(0.10 mol/L)
p
•Rate 2 1.6 x 10
-3mol/L·s k(0.20 mol/L)
n(0.10 mol/L)
m(0.10 mol/L)
p
2.0 = ( 0.20 mol/L /0.10 mol/L)
m = (2.0)
m Thus m = 1.
•To determine the value of p, we use Expts 1 & 4, in which [BrO
3
-]
and [Br
-] are constant but [H
+] differs:
•Rate 4 = 3.2 x 10
-3mol/L·s = k(0.10 mol/L)
n(0.10 mol/L)
m(0.20 mol/L)
p
•Rate 1 8.0 x 10
-4mol/L·s k(0.10 mol/L)
n(0.10 mol/L)
m(0.10 mol/L)
p
4.0 = (0.20 mol/L /0.10 mol/L)
p
4.0 = (2.0)
p = (2.0)
2 Thus p = 2.

34. Method of Initial Rates
•Rate of this rxion is 1st order in BrO
3
- and Br- and 2nd order in H
+.
Overall rxion order is n + m + p = 4.
•The rate law therefore is:
Rate = k[BrO
3
-][Br
-][H
+]
2
•Rate constant k can be calculated from results of any of the 4 expts.
•For Expt 1, initial rate = 8.0 x 10
-4 mol/L·s; [BrO
3
-] = 0.100 M,
[Br-] = 0.10 M, and [H
+] = 0.10 M.
•Using these values in the rate law gives:
•8.0 x 10
-4 mol/L·s = k(0.10 mol/L) (0.10 mol/L) (0.10 mol/L)
2
•8.0 x 10
-4 mol/L·s = k(1.0 x 10
-4 mol
4/L
4)
•k = (8.0 x 10
-4 mol/L·s) = 8.0 L
3/mol
3·s
(1.0 x 10
-4 mol
4/L
4)

35. The Integrated Rate Law
•Rate laws considered so far express rate as a function of
reactant concs. Reactant concs can also be expressed as a
function of time, given the (differential) rate law for the rxion.
•How is this done?
•Let’s look at rxions involving a single reactant:
•aA → products all of which have a rate law:
Rate = -Δ[A]/Δt = k[A]
n
•We will develop the integrated rate laws individually for the
cases n = 1 (first order), n = 2 (second order), and n = 0 (zero
order).

36. The Integrated Rate Law:-First-Order Rate Law
•For the rxion: 2N
2O
5(soln) → 4NO
2(soln) + O
2(g) the rate law is :
Rate = -Δ[N
2O
5]/Δt = k[N
2O
5]
•Since the rate of this rxion depends on the conc of N
2O
5 to the first
power, it is a first-order rxion, i.e., if conc of N
2O
5 is doubled, the
rate of production of NO
2 and O
2 will also double.
•This rate law can be expressed differently as follows:
ln[N
2O
5] = -kt + ln[N
2O
5]
0
•where ln = natural logarithm, t = time,
•[N
2O
5] = concentration of N
2O
5 at time t,
•[N
2O
5]
0 = initial concentration of N
2O
5 (at t = 0, start of the expt).
•The rate equation written in this format is called integrated rate
law, and it gives the conc of a reactant as a function of time.

37. The Integrated Rate Law:-First-Order Rate Law
•For a chemical reaction of the form
aA → products
•which is first order in [A], the rate law is
Rate = -Δ[A]/Δt = k[A]
•and the integrated first-order rate law is
ln[A] = -kt + ln[A]
0 (12.2)
•Important things to note about Equation (12.2):
•1. Equation shows how conc of A depends on time. If the initial
conc of A and the rate constant, k, are known, the conc of A at any
time can be calculated.

38. The Integrated Rate Law:-First-Order Rate Law
•2. Equation (12.2) is of the form y = mx + b, of which a plot of y
versus x is a straight line with slope m and intercept b.
In Equation (12.2),
y = ln[A] ; x = t ; m = -k ; b = ln[A]
0
•Thus for a first-order rxion, plot of the natural log of concentration
versus time always gives a straight line. This is proof that a reaction
is first order or not. For the reaction
aA → products,
•the rxion is first-order in A if a plot of ln[A] versus t is a straight line.
If this plot is not a straight line, the rxion is not first order in A.
•3. The integrated rate law for a first-order rxion also can be
expressed as a ratio of [A] and [A]
0 as follows: ln ([A]
0/[A]) = kt

39. Example 12.2: First-Order Rate Law
•Decomposition of N
2O
5 in the gas phase was studied at constant
temperature.
2N
2O
5(g) → 4NO
2(g) + O
2(g)
•The following results were collected:
[N
2O
5] (mol/L) Time (s)
0.1000   0
0.0707  50
0.0500 100
0.0250 200
0.0125 300
0.00625 400

40. Example 12.2: First-Order Rate Law
•Using these data, verify that the rate law is first order in [N
2O
5], and
calculate the rate constant, where the rate = -Δ[N
2O
5]/Δt.
•Solution:
•To verify that rate law is first-order in [N
2O
5], construct a plot of
ln[N
2O
5] versus time.
The plot is a straight line.
This confirms that the rxion is
first order in N
2O
5, since it
follows the equation
ln[N
2O
5] = -kt + ln[N
2O
5]
0.

41. Example 12.2: First-Order Rate Law
•Since the rxion is first order, the slope of the line equals -k, where
•Slope = change in y = Δy = Δ(ln[N
2O
5])
change in x Δx Δt
•The first and last points are exactly on the line, thus they can be
used to calculate the slope:
•Slope = -5.075 - (-2.303) = -2.772 = -6.93 x 10
-3 s
-1
400s - 0s 400s
•k = -(slope) = 6.93 x 10
-3 s
-1

42. Half-Life of a First-Order Reaction
•Half-life, t
½, of a reactant is the time required to reach half its
original conc., e.g., we can calculate the half-life of the
decomposition rxion: 2N
2O
5(g) → 4NO
2(g) + O
2(g).
The data plotted in Fig. 12.4 show that the half-life for this rxion is
100 s. We can see this by considering the following numbers:
[N
2O
5] (mol/L) Δt(s) Ratio of Concentrations
0.100 0 [N
2O
5]
t=100 = 0.050 = 1/2
Δt = 100s [N
2O
5]
t=0 0.100
0.0500 100 [N
2O
5]
t=200 = 0.025 = 1/2
Δt = 100s[N
2O
5]
t=100 0.050
0.0250 200 [N
2O
5]
t=300 = 0.0125 = 1/2
Δt = 100s[N
2O
5]
t=200 0.0250
0.0125 300

43. Half-Life of a First-Order Reaction
Note: it’s taking 100 s
for [N
2O
5] to be
halved in this rxion.
A general formula for
half-life of a first-
order rxion can be
derived from the
integrated rate law for
the general rxion:
aA → products
If the rxion is first-
order in [A], then
ln([A]
0/[A] ) = kt

44. Half-Life of a First-Order Reaction
•By definition, when t = t½, [A] = [A]
0/2
•Then, for t = t½, the integrated rate law becomes
In ([A]
0) = kt½ orln(2) = kt½
([A]
0/2)
•Substituting the value of ln(2) and solving for t½ gives
t½ = 0.693/k (12.3)
•This is the general equation for the half-life of a first-order rxion.
Equation (12.3) can be used to calculate t½ if k is known or k if t½
is known. Note that for a first-order reaction, the half-life does not
depend on concentration.

45. Example 12.4:Half-Life of a First-Order Rxion
•A certain first-order rxion has a half-life of 20.0 minutes.
•a. Calculate the rate constant for this rxion.
•b. How much time is required for this rxion to be 75% complete?
•Solution:
•a. Solve Equation (12.3) for k:
k = 0.693/t½ = 0.693/20.0 min = 3.47 x 10
-2
min
-1

•b. Use the integrated rate law in the form:In([A]
0/[A]) = kt
•If the rxion is 75% complete, 75% of reactant has been used, leaving
25%: Thus, [A]
0/[A] x 100% = 25%
•which is thw same as [A]/[A]
0 = 0.25 or[A]
0/[A] = 1/0.25 = 4.0
•Then ln ([A]
0/[A]) = ln(4.0) = kt = (3.47 x 10
-2
/min) t and
t = ln(4.0)/3.47 x 10
-2
/min = 40 min

46. Second-Order Rate Laws
•For a general rxion involving a single reactant, that is, aA →
products that is second order in A, the rate law is:
Rate = -Δ[A]/Δt = k[A]
2 (12.4)
•The integrated second-order rate law is
1/[A] = kt + 1/[A]
0 (12.5)
•Note the following characteristics of Equation (12.5):
•1. Plot of 1/[A] versus t will produce a straight line with a slope
equal to k.
•2. Equation (12.5) shows how [A] depends on time and can be
used to calculate [A] at any time t, provided k and [A]
0 are
known.

47. Second-Order Rate Law
•When one half-life of the second-order reaction has elapsed (t =
t½), by definition, [A] = [A]
0/2
•Equation (12.5) then becomes 1/[A]
0/2 = kt½ + 1/[A]
0
2/[A]
0 - 1/[A]
0 = kt½
1/[A]
0 = kt½
•Solving for t½ gives the expression for half-life of a second-
order rxion: t½ = 1/k[A]
0 (12.6)

48. Zero-Order Rate Law
•Most rxions involving a single reactant show either first-order or
second-order kinetics. However, sometimes such a rxion can be
a zero-order rxion.
•The rate law for a zero-order reaction is
Rate = k[A]
0 = k(1) = k
•For a zero-order rxion, rate is constant. It does not change with
conc. as it does for first-order or second-order rxions.
•The integrated rate law for a zero-order rxion is
[A] = -kt + [A]
0 (12.7)

49. Zero-Order Rate Law
•In this case a plot of [A] versus t gives a straight line of slope -k
(Fig. 12.6). Figure 12.6 A plot of [A] versus t for a zero-order reaction.
•Expression for half-life of a zero-order rxion can be obtained
from the integrated rate law. By definition, [A] = [A]
0/2 when t
= t½, so [A]
0/2 = -kt½ + [A]
0 or
kt½ = [A]
0/2k
•Solving for t½ gives t½ = [A]
0/2k (12.8)

50. Collision Model for Chemical Kinetics
•How do chemical reactions occur? We’ve seen that rates of rxions
depend on concs of reacting species. The initial rate for the rxion:
aA + bB → products can be described by the rate law
Rate = k[A]
n[B]
m
•where the order of each reactant depends on the rxion mechanism.
So, rxion rates depend on concentration.
•What other factors affect reaction rates? How does temperature
affect the speed of a rxion?
•We know wood combusts faster at high temperatures. An egg
cooks much faster at sea level than at high altitude. Thus, chemical
rxions speed up when temperature is increased. All rate constants
increase exponentially with absolute temperature.

51. Collision Model for Chemical Kinetics
•The collision model, says that molecules must collide to react. This
why when conc increases, rxion rates increase. Kinetic theory of
gases says that, increase in temperature increases molecular
vibrations and so increases frequency of collisions.
•This is why rxion rates are greater at higher temperatures. But, the
rate of rxion is much smaller than the calculated collision
frequency in a collection of gas particles. This means, only a small
fraction of collisions produces a rxion. Why is this so?
•In the 1880s, Svante Arrhenius proposed that, there exists a
threshold energy, called activation energy, E
a, that must be
overcome for a rxion to occur, e.g., decomposition of BrNO
(g):
2BrNO
(g) → 2NO
(g) + Br
2(g)

52. Collision Model for Chemical Kinetics
•By the collision model, energy to break the bonds and rearrange
atoms into product molecules comes from kinetic energies of the
reacting molecules before they collide.
•Arrangement of atoms at the top of the potential energy “hill,” or
barrier, is called the activated complex, or transition state.
•Conversion of BrNO to NO and Br
2 is exothermic, because the
products have lower potential energy than the reactants.
•ΔE has no effect on the rate of the rxion. The rate depends on the
magnitude of activation energy E
a.
•The lower the E
a, the faster the rxion. The higher E
a, the slower
the rxion.

53. Collision Model for Chemical Kinetics
Figure 12.10 (a) Change in potential energy as a function of rxion progress for the rxion 2BrNO → 2NO + Br
2.
Activation energy E
a represents energy needed to disrupt the BrNO molecules so that they can form products.
ΔE = net change in energy in going from reactant to products. (b) A molecular representation of the rxion.

54. Collision Model for Chemical Kinetics
•Note: For the two BrNO molecules to “get over the hill” so that
they can react and form products, they need to have a certain
minimum energy. This energy is comes from the energy of the
collision. A collision between two BrNO molecules with small K.E.’s
does not produce enough energy to get over the barrier. At a given
temperature, only a certain fraction of collisions produces enough
energy to result in a rxion and thus, product formation.
•In any sample of gas molecules, there exists a distribution of
velocities. Thus, a distribution of collision energies also exists, as
shown in Fig. 12.11 for two different temperatures. Figure 12.11
also shows the activation energy for the rxion in question.

55. Collision Model for Chemical Kinetics
Figure 12.11 Plot showing the number of collisions with a
particular energy at T
1 and T
2, where T
2 > T
1.

56. Collision Model for Chemical Kinetics
•Only collisions with energy > E
a are able to react (get over the
barrier). At lower temperature, T
1, the fraction of effective collisions
is small. At increased temperature,T
2, the fraction of collisions with
required E
a increases. When temperature is doubled, the fraction of
effective collisions increases exponentially.
•According to Arrhenius, number of collisions with energy greater
than or equal to E
a is given by the expression:
Number of collisions with E
a = (total number of collisions)e
-Ea/RT
where E
a = activation energy, R = universal gas constant,
T = temp in Kelvins.
The factor e
-Ea /RT = fraction of collisions with energy ≥ E
a at
temperature, T.

57. Collision Model for Chemical Kinetics
•Not all collisions are effective in producing chemical rxions
because a minimum energy is required for the rxion to occur.
•Experiments show that, observed rxion rate is much smaller
than rate of effective collisions (i.e. collisions with energy = E
a).
This means, many collisions which have the required energy
still do not produce a rxion. Why?
•This is due to molecular orientations during collisions.
•Some collision orientations lead to rxion, others do not. Thus, a
correction factor to allow for collisions with nonproductive
molecular orientations must be included in the expression.

58. Collision Model for Chemical Kinetics
•To summarize, two requirements must be satisfied for reactants to
collide successfully (i.e., to rearrange to form products):
•1. The collision must generate enough energy to produce a rxion; i.e.,
collision energy must equal or exceed activation energy.
•2. Orientation of reactants must allow formation of new bonds
necessary to produce products.
•Taking all these factors into account, the rate constant becomes:
k = zpe
-Ea /RT
•where z = collision frequency, p = steric factor (always less than 1) =
fraction of collisions with effective orientations, and e
-Ea /RT
= fraction
of collisions with sufficient energy to produce a rxion.
•This equation is often written in form k = Ae
-Ea /RT
(12.9)
called the Arrhenius equation.

59. Collision Model for Chemical Kinetics
•where A = zp = frequency factor for the rxion. Taking the natural
logarithm of each side of the Arrhenius equation gives:
ln(k) = -Ea/R (1/T) + ln(A) (12.10)
•Equation (12.10) is a linear equation of the type y = mx + b,
where y = ln(k), m = -Ea/R = slope, x = 1/T, b = ln(A) = intercept.
•For a rxion where rate constant obeys the Arrhenius equation, a
plot of ln(k) versus 1/T is a straight line. Slope of the line = E
a ;
intercept = A (is characteristic of rxion under consideration).
•Most rate constants obey the Arrhenius equation to a good
approximation. This shows that the collision model for chemical
rxions is physically reasonable

60. •The most common method for finding E
a for a rxion is the
measuring the rate constant k at several temperatures and then
plotting ln(k) versus 1/T. But, E
a can also be calculated from the
values of k at only two temperatures by using a formula derived
from Equation (12.10).
•At temperature T
1, where the rate constant is k
1,
In(k
1) = -Ea/RT
1 + ln(A)
•At temperature T
2, where the rate constant is k
2,
In(k
2) + -Ea/RT
2 + ln(A)
•Subtracting the first equation from the second gives:
ln(k
2) - ln(k
1) = [-Ea/RT
2+ ln(A)] - [-Ea/RT
1+ ln(A)] = -Ea/RT
2+Ea/RT
1
ln(k
2/k
1) = Ea/R (1/T
1 - 1/T
2) (12.11)
•Thus, values of k
1 and k
2 measured at temperatures T
1 and T
2 can
be used to calculate E
a.

61. Example 12.7: Determining Activation Energy
•The rxion 2N
2O
5(g) → 4NO
2(g) + O
2(g)
was studied at several temperatures, and the following values of k
were obtained:
k (s
-1) T (ºC)
2.0 x 10
-5 20
7.3 x 10
-5 30
2.7 x 10
-5 40
9.1 x 10
-5 50
2.9 x 10
-3 60
•Calculate the value of E
a for this rxion.

62. Example 12.7: Determining Activation Energy
•Solution
•To obtain the value of E
a, a plot of ln(k) versus 1/T is done. First, is
to calculate values of ln(k) and 1/T, as shown below:
•T (ºC) T (K) 1/T (K) k (s
-1) ln(k)
20 293 3.41 x 10
-3 2.0 x 10
-5 210.82
•30 303 3.30 x 10
-3 7.3 x 10
-5 29.53
•40 313 3.19 x 10
-3 2.7 x 10
-4 28.22
•50 323 3.10 x 10
-3 9.1 x 10
-4 27.00
•60 333 3.00 x 10
-3 2.9 x 10
-3 25.84

63. Example 12.7: Determining Activation Energy
Fig shows the plot of ln(k) versus
1/T , where the slope
Δln(k)/Δ(1/T) = -1.2 x 10
4 K.
E
a is determined from the
equation: Slope = -Ea/R
Ea = -R(slope) = -(8.3145
J/K·mol) (-1.2 x 10
4 K)
= 1.0 x 10
5 J/mol
Therefore, activation energy for
this rxion = 1.0 x 10
5 J/mol.

64. Example 12.8: Determining Activation Energy
The gas-phase rxion between methane and diatomic sulfur is given
by the equation:CH
4(g) + 2S
2(g) → CS
2(g) + 2H
2S
(g)
•At 550ºC the rate constant for this rxion = 1.1 L/mol·s, and at
625ºC the rate constant is 6.4 L/mol·s. Using these values,
calculate E
a for this rxion.
•Solution
•The relevant data are shown in the following table:
k (L/mol·s) T (ºC) T (K)
1.1 = k
1 550 823 = T
1
6.4 = k
2 625 898 = T
2
•Substituting these values into Equation (12.11) gives
ln (6.4/1.1) = E
a/8.3145 J/K·mol (1/823 K - 1/898 K)
•E
a= (8.3145J/K·mol)ln(6.4/1.1) (1/823K - 1/898K) = 1.4 x 10
5
J/mol