Chemistry note : Chemical Kinetics grade 11

Chemistry notes

1. Chemical Kinetics
Kinetics = The area of chemistry that is concerned with reaction rates.
What do Kinetics tell us?
Stoichiometry tells us what and how many reactants/products there are
Thermodynamics tells us if the reaction is energetically favorable (-DH)
Kinetics describes how fast or slow a reaction occurs
2 H
2+ O
2 2 H
2O DH = -242 kJ/mol
This spontaneous reaction is so slow, it essentially doesn’t happen!
We need Kinetics to fully describe chemical reactions
Kinetics help us determine how a reaction occurs = Mechanism

2. Reaction Rates= Change in concentration (conc) of a reactant or product
per unit time.
For generic reaction: aA + bB cC + dDt
[D]1
t
[C]1
t
[B]1
t
[A]1
Rate
D
D
=
D
D
=
D
D
−=
D
D
−=
dcba
This is an average rate
over this time period.
If you find the rate for one
reactant or product, you’ve
found them all!
For the general reaction A → B, we measure the concentration of A at t
1and at t
2:
D[A]
Dt
Rate =
change in concentration of A
change in time
conc A
2 -conc A
1
t
2 -t
1
= - = -
Square brackets indicate a concentration in moles per liter.
The negativesign is used because the concentration of A is decreasing.
This gives the ratea positivevalue.

3. Concentration of O
3at Various Times in its Reaction with C
2H
4at
303 K
Time (s)
0.0
20.0
30.0
40.0
50.0
60.0
10.0
Concentration of O
3
(mol/L)
3.20x10
-5
2.42x10
-5
1.95x10
-5
1.63x10
-5
1.40x10
-5
1.23x10
-5
1.10x10
-5
D[C
2H
4]
Dt
rate = -
D[O
3]
Dt
= -

4. Three types of reaction rates for the reaction of O
3and C
2H
4.

5. Plots of [reactant] and [product] vs. time.
C
2H
4+ O
3→ C
2H
4O + O
2
[O
2] increases just as fast as
[C
2H
4] decreases.
Rate = -
D[C
2H
4]
Dt
= -
D[O
3]
Dt
D[C
2H
4O]
Dt
D[O
2]
Dt
= =

6. Rates of Product Formation 2 NO
2 2 NO + O
2
2 NO produced for every 2 NO
2consumed
Rate same at any point in the reaction
Curve is same, only inverted
Rate [NO] at 250 s = 8.6 x 10
-6
mol/L.s
1 O
2produced for every NO
2consumed
Rate = ½ the rate of NO
2consumption at any point
Rate = ½ the rate of NO production at any point
Rate [O
2] at 250 s = 4.3 x 10
-6
mol/L.s
O
2curve has a different shape
We can write an equation of the rates from the balanced equation
We must be specific when we talk about a reaction rate.
Rate depends on which reactant or product
Rate depends on how long the reaction has been goingt
]O[
1
1
t
[NO]
2
1
t
][NO
2
1
Rate
22
D
D
=
D
D
=
D
D
−=

7. The rate of a reaction is dependent on the concentration of
reactants
As reactant concentrations decrease, the reaction rates decrease
The rate of a reaction depends on the following variables:
reactant concentration
temperature
presence and concentration of a catalyst
surface area of solids, liquids or catalysts
The dependence of reaction rate on concentrations is expressed
mathematically by the rate law
Reactant concentrations are raised to exponential powerswhose
product equals the rate of the reaction

8. Rate Laws
•Background
•Chemical reaction are reversible. Products recombine to give reactant.
2 NO
2 2 NO + O
2
When you start with only reactants, the reaction only goes one way
1)D[NO
2] at this point depends on only the forward reaction
2)This makes equations simple
After some products are formed, the reaction goes both directions
•D[NO
2] at this point depends on both forward and reverse reactions
•This makes equations complicated
We will consider reactions only at times and conditions that have only the forward
reaction of any significance

9. Rate Law
aA+ bB→products
Rate = -D[A]/Dt= k[A]
x
[B]
y
[C]
z
Rate Law Basics
•A Rate Law describes how concentrations are changing in a reaction
Rate equation
H2(g) + I2(g) 2HI(g)
•the rate of the reaction is proportional to concentration of the reactants
•(within the limits of experimental error). We can express this mathematically as:
rate of reaction = k . [H2]
m
[I2]
n
•The proportionality constant, k, is called the rate constant.
•The power ‘m and n’iscalled the orderof the reaction
•The overall expression (rate of reaction = k . [H2]
m
[I2]
n
is the rate equation for this
particular reaction.
•Rate equations are generally written without the . sign.
•For example:
rate = k[H2]
m
[I2]
n

10. •Products do not appear in the rate law if we use conditions giving the forward reaction only
Order of Reaction
The order of a reaction shows how the concentration of a reagent affects the rate of reaction.
The order of reaction with respect to a particular reactant is the power to which the concentration of
that reactant is raised in the rate equation.
•The order n must be determined experimentally; it can’tbe written directly from the balanced equation
because the stoichiometric equation does not indicate mechanism of the reaction
•We must specify which species we are using [A] in each rate law
Example : rate = k[H2] [NO]
2
We say that this reaction is:
first-order with respect to H2
(as rate is proportional to [H2]
1
)
second-order with respect to NO
(as rate is proportional to [NO]
2
)
third-order overall (as the sum of the powers is 1 + 2 = 3).
Overall reaction order = sum of exponents in rate equation
Order of RxnPossible Expression of Rate Law
1 k[A]
2 k[A]
2
or k[A][B]
3 k[A]
2
[B] or k[A][B][C]

11. Finding the order
•We can identify the order of a reaction in three ways:
•plot a graph of reaction rate against concentration of reactant
•plot a graph of concentration of reactant against time
•deduce successive half-lives from graphs of concentration
•against time.

12. •A zeroth-order reaction is one whose rate is independent of
concentration;
•its differential rate law is: rate = k.
•We refer to these reactions as zeroth order because we could also
write their rate in a form such that the exponent of the reactant in
the rate law is 0:
•rate=−Δ[A] / Δt= k[reactant]
0
= k(1) = k
•Because rate is independent of reactant concentration, a graph of
the concentration of any reactant as a function of time is a straight
line with a slope of −k.
•The value of k is negative because the concentration of the reactant
decreases with time. Conversely, a graph of the concentration of any
product as a function of time is a straight line with a slope of k, a
positive value.
•The integrated rate law for a zeroth-order reaction also produces a
straight line and has the general form
•[A]=[A]
0−kt
•where [A]
0is the initial concentration of reactant A.
•(the form of the algebraic equation for a straight line,y=mx+b,
with y= [A],m = -k, x= −t, andb= [A]
0.)
•In a zeroth-order reaction, the rate constant must have the same
units as the reaction rate, typically moles per literper second.

13. Zeroth order graph

14. 

16. Second-Order Reactions
•The simplest kind ofsecond-order reactionis one whose rate is proportional to the square
of the concentration of one reactant.
•These generally have the form 2A →products.
•A second kind of second-order reaction has a reaction rate that is proportional to the
product of the concentrations of two reactants. Such reactions generally have the form A +
B →products.
•An example of the former is a dimerization reaction, in which two smaller molecules, each
called a monomer, combine to form a larger molecule (a dimer).
•The differential rate law for the simplest second-order reaction in which 2A →products is
as follows:
rate=−Δ[A]
2/
Δt=k[A]
2
•Consequently, doubling the concentration of A quadruples the reaction rate. For the units
of the reaction rate to be moles per literper second (M/s), the units of a second-order rate
constant must be the inverse (M
−1
·s
−1
). Because the units of molarity are expressed as
mol/L, the unit of the rate constant can also be written as L(mol·s).
•For the reaction 2A →products, the following integrated rate law describes the
concentration of the reactant at a given time:
1/[A]=1/[A]
0+kt
•Becausethe Equationhas the form of an algebraic equation for a straight line,y=mx+b,
withy= 1/[A] andb= 1/[A]
0, a plot of 1/[A] versustfor a simple second-order reaction is
a straight line with a slope ofkand an intercept of 1/[A]
0.
•Second-order reactions generally have the form 2A →products or A + B →products.

17. Graphs for second order reactions

18. Differential Rate Lawtells us how rate changes with concentration
This is what we will call the rate law

Integrated Rate Lawtells us how concentrations depend on time
This can be derived from the differential rate law (and vice versa)
We will look at integrated rate laws later
What do we learn from rate laws?
They help us determine if a reaction is fast enough to be useful
They help us figure out the exact steps (mechanism) of a reaction
Slowest step determines the overall rate
Chemists speed up reactions by changing that stept
]A[
k[A]rate
n
D
D
−==

19. Determining the Form of a Rate Law
•We must do experiments to determine the order of each reactant in the rate law
•Example: 2 N
2O
5 4 NO
2+ O
2(g)
•O
2gas escapes, so there is no reverse reaction
•We can determine rate at various times from the data
[N
2O
5]Time (s)
1.00 0
0.88 200
0.78 400
0.69 600
0.61 800
0.54 1000
0.48 1200
0.43 1400
0.38 1600
0.34 1800
0.30 2000

20. Examine how the rate changes as concentration changes
[N
2O
5] = 0.90 M rate = 5.4 x 10
-4
mol/L.s
[N
2O
5] = 0.45 M rate = 2.7 x 10
-4
mol/L.s
How does the [A] ratio compare to the rate ratio?
•If the ratios are the same, the order of the reactant is 1
•
•When the order is 1, we call this a First Order Reaction
The general expression for a first order reaction is:1
2
45.0
90.0
= 1
2
107.2
104.5
4
4
=


−
− 1
52
52
]Ok[N
t
]O[N
- rate =
D
D
= 1
k[A]
t
[A]
- rate =
D
D
=

21. The Method of Initial Rates
•Initial rate= rate just after a reaction starts (t = 0). This is used because we don’t have to
worry about reverse reaction.
•We start the reaction several time with different initial concentrations of reactants.
•If the rate changes in the same ratio as the concentration, n = 1
•If the rate changes as the square of the concentration ratio, n = 2
•If the rate changes as the cube of the concentration ratio, n = 3, etc…
•If the rate doesn’t change at all, n = 0
Examplewith 2 reactants: NH
4
+
+ NO
2
-
N
2+ 2 H
2O
4)First order in both reactants, and overall second order (n + m = 2)
Exp. #[NH
4
+
]
0[NO
2
-
]
0
Rate
(mol/L.s)
1 0.1000.00501.35 x 10
-7
2 0.1000.01002.70 x 10
-7
3 0.2000.01005.40 x 10
-7
Rate = k[NH
4
+
]
n
[NO
2
-
]
m
Exp1→Exp2 [NH
4
+
] same, [NO
2
-
] doubles
rate doubles so m = 1
Exp2→Exp3 [NH
4
+
] doubles, [NO
2
-
] same
rate doubles so n = 1
Rate = k[NH
4
+
]
1
[NO
2
-
]
1
= k[NH
4
+
][NO
2
-
]

22. Now we can actually find what k is from any of the experiments since we know that: Rate
= k[NH
4
+
][NO
2
-
]
Example: Find rate law, k, and order of the reaction
BrO
3
-
+ 5 Br
-
+ 6 H
+
3 Br
2+ 3 H
2OsL/mol 107.2
]mol/L 005.0][mol/L 100.0[
smol/L 1035.1
]][NO[NH
rate
k
4
7
-
24
=

==
−
−
+
Exp. #[BrO
3
-
][Br
-
] [H
+]
rate
1 0.100.100.108.0 x 10
-4
2 0.200.100.101.6 x 10
-3
3 0.200.200.103.2 x 10
-3
4 0.100.100.203.2 x 10
-3

24. Sample Problem 1
SOLUTION:
Determining Reaction Orders from Rate
Laws
PLAN:We inspect the exponents in the rate law, notthe coefficients
of the balanced equation, to find the individual orders. We add
the individual orders to get the overall reaction order.
(a)The exponent of [NO] is 2 and the exponent of [O
2] is 1, so the
reaction is second order with respect to NO, first order
with respect to O
2and third order overall.
PROBLEM:For each of the following reactions, use the give rate law
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
-
]

25. Sample Problem 1
(b)The reaction is order in CH
3CHOand order overall.
3
2
3
2
(c)The reaction is first order in H
2O
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 with respect to H
+
.

26. Determining Reaction Orders
For the general reaction A + 2B → C + D,
the rate law will have the form
Rate = k[A]
m
[B]
n
To determine the values of mand n, we run a series of
experiments in which one reactant concentration
changes while the other is kept constant, and we
measure the effect on the initial rate in each case.

27. Initial Rates for the Reaction between A and B
Experiment
Initial Rate
(mol/L·s)
Initial [A]
(mol/L)
Initial [B]
(mol/L)
1 1.75x10
-3
2.50x10
-2
3.00x10
-2
2 3.50x10
-3
5.00x10
-2
3.00x10
-2
3 3.50x10
-3
2.50x10
-2
6.00x10
-2
4 7.00x10
-3
5.00x10
-2
6.00x10
-2
[B] is kept constant for experiments 1 and 2, while [A] is doubled.
Then [A] is kept constant while [B] is doubled.

28. Rate 2
Rate 1
=
k[A] [B]
k[A] [B]
m
2
n
2
m
1
n
1
Finding m, the order with respect to A:
We compare experiments 1 and 2, where [B] is kept
constant but [A] doubles:
=
[A]
[A]
m
2
m
1
=
[A]
2
[A]
1
m
3.50x10
-3
mol/L·s
1.75x10
-3
mol/L·s
=
5.00x10
-2
mol/L
2.50x10
-2
mol/L
m
Dividing, we get 2.00 = (2.00)
m
so m= 1

29. Rate 3
Rate 1
=
k[A] [B]
k[A] [B]
m
3
n
3
m
1
n
1
Finding n, the order with respect to B:
We compare experiments 3 and 1, where [A] is kept
constant but [B] doubles:
=
[B]
[B]
n
3
n
1
=
[B]
3
[B]
1
n
3.50x10
-3
mol/L·s
1.75x10
-3
mol/L·s
=
6.00x10
-2
mol/L
3.00x10
-2
mol/L
m
Dividing, we get 2.00 = (2.00)
n
so n= 1

30. Initial Rates for the Reaction between O
2and NO
Initial Reactant
Concentrations (mol/L)
Experiment
Initial Rate
(mol/L·s) [O
2] [NO]
1 3.21x10
-3
1.10x10
-2
1.30x10
-2
2 6.40x10
-3
2.20x10
-2
1.30x10
-2
3 12.48x10
-3
1.10x10
-2
2.60x10
-2
4 9.60x10
-3
3.30x10
-2
1.30x10
-2
5 28.8x10
-3
1.10x10
-2
3.90x10
-2
O
2(g) + 2NO(g) → 2NO
2(g) Rate = k[O
2]
m
[NO]
n

31. Rate 2
Rate 1
=
k[O
2] [NO]
k[O
2] [NO]
m
2
n
2
m
1
n
1
Finding m, the order with respect to O
2:
We compare experiments 1 and 2, where [NO] is kept
constant but [O
2] doubles:
=
[O
2]
[O
2]
m
2
m
1
=
[O
2]
2
[O
2]
1
m
6.40x10
-3
mol/L·s
3.21x10
-3
mol/L·s
=
2.20x10
-2
mol/L
1.10x10
-2
mol/L
m
Dividing, we get 1.99 = (2.00)
m
or 2 = 2
m
, so m= 1
The reaction is first order with respect to O
2.

32. Sometimes the exponent is not easy to find by inspection.
In those cases, we solve for mwith an equation of the form
a= b
m
:
log a
log b
m=
log 1.99
log 2.00
= = 0.993
This confirms that the reaction is first order with respect to O
2.
Reaction orders may be positive integers, zero, negative integers, or
fractions.

33. Finding n, the order with respect to NO:
We compare experiments 1 and 3, where [O
2] is kept
constant but [NO] doubles:
Rate 3
Rate 1
=
[NO]
3
[NO]
1
n
12.8x10
-3
mol/L·s
3.21x10
-3
mol/L·s
=
2.60x10
-2
mol/L
1.30x10
-2
mol/L
n
Dividing, we get 3.99 = (2.00)
n
or 4 = 2
n
, so n= 2.
The reaction is second order with respect to NO.
log a
log b
n=
log 3.99
log 2.00
= = 2.00Alternatively:
The rate law is given by: rate = k[O
2][NO]
2

34. Sample Problem 2 Determining Reaction Orders from Rate Data
PROBLEM:Many gaseous reactions occur in a car engine and exhaust
system. One of these is
NO
2(g) + CO(g) → NO(g) + CO
2(g) rate = k[NO
2]
m
[CO]
n
Use the following data to determine the individual and
overall reaction orders:
Experiment
Initial Rate
(mol/L·s)
Initial [NO
2]
(mol/L)
Initial [CO]
(mol/L)
1 0.0050 0.10 0.10
2 0.080 0.40 0.10
3 0.0050 0.10 0.20

35. Sample Problem 2
PLAN:We need to solve the general rate law for mand for nand
then add those orders to get the overall order. We proceed by
taking the ratio of the rate laws for two experiments in which
only the reactant in question changes concentration.
SOLUTION:
rate 2
rate 1
[NO
2]
2
[NO
2]
1
m
=
k[NO
2]
m
2[CO]
n
2
k[NO
2]
m
1 [CO]
n
1
=
0.080
0.0050
0.40
0.10
=
m
16 = (4.0)
m
so m= 2
To calculate m, the order with respect to NO
2, we compare
experiments 1 and 2:
The reaction is second order in NO
2.

36. k[NO
2]
m
3[CO]
n
3
k[NO
2]
m
1 [CO]
n
1
[CO]
3
[CO]
1
n
=
rate 3
rate 1
=
0.0050
0.0050
=
0.20
0.10
n
1.0 = (2.0)
n
so n= 0
rate = k[NO
2]
2
[CO]
0
or rate = k[NO
2]
2
Sample Problem 3
To calculate n, the order with respect to CO, we compare experiments
1 and 3:
The reaction is zero order in CO.

37. Sample Problem 4 Determining Reaction Orders from Molecular
Scenes
PROBLEM:At a particular temperature and volume, two gases, A (red)
and B (blue), react. The following molecular scenes
represent starting mixtures for four experiments:
(a)What is the reaction order with respect to A? With respect to B?
The overall order?
(b)Write the rate law for the reaction.
(c)Predict the initial rate of experiment 4.
PLAN:We find the individual reaction orders by seeing how a change
in each reactant changes the rate. Instead of using
concentrations we count the number of particles.
Expt no:
Initial rate (mol/L·s)
1
0.50x10
-4
2
1.0x10
-4
3
2.0x10
-4
4
?

38. Sample Problem 4
SOLUTION:
(a)For reactant A (red):
Experiments 1 and 2 have the same number of particles of B, but
the number of particles of A doubles. The rate doubles. Thus the
order with respect to A is 1.
(b)Rate = k[A][B]
2
(c)Between experiments 3 and 4, the number of particles of A
doubles while the number of particles of B does not change. The
rate should double, so rate = 2 x 2.0x10
-4
= 4.0x10
-4
mol/L·s
For reactant B (blue):
Experiments 1 and 3 show that when the number of particles of B
doubles (while A remains constant), the rate quadruples. The
order with respect to B is 2.
The overall order is 1 + 2 = 3.

39. Units of the Rate Constant kfor Several Overall
Reaction Orders
Overall
Reaction
Order
Units of k
(tin seconds)
0 mol/L·s
(or mol L
-1
s
-1
)
1 1/s (or s
-1
)
2 L/mol·s
(or L mol
-1
s
-1
)
3 L
2
/mol
2
·s
(or L
2
mol
-2
s
-1
)
General formula:
L
mol
unit of t
order-1
Units of k=
The value of kis easily determined from experimental rate
data. The unitsof kdepend on the overall reaction order.

40. Information sequence to determine the kinetic
parameters of a reaction.
Series of plots of
concentration vs.
time
Initial rates
Reaction
orders
Rate constant (k)
and actual rate law
Determine slope of tangent at t
0for
each plot.
Compare initial rates when[A]changes and
[B]is held constant (and vice versa).
Substitute initial rates, orders, and
concentrations into rate = k[A]
m
[B]
n
, and
solve for k.

41. Integrated Rate Laws
An integrated rate law includes timeas a variable.
First-order rate equation:
rate = -
D[A]
Dt
= k[A]
ln
[A]
0
[A]
t
= -kt
Second-order rate equation:
rate = -
D[A]
Dt
= k[A]
2
1
[A]
t
1
[A]
0
- = kt
Zero-order rate equation:
rate = -
D[A]
Dt
= k[A]
0 [A]
t-[A]
0= -kt

42. Sample Problem 5 Determining the Reactant Concentration
after a Given Time
PROBLEM:At 1000
o
C, cyclobutane (C
4H
8) decomposes in a first-order
reaction, with the very high rate constant of 87 s
-1
, to two
molecules of ethylene (C
2H
4).
(a)If the initial C
4H
8concentration is 2.00 M, what is the
concentration after 0.010 s?
(b)What fraction of C
4H
8has decomposed in this time?
PLAN:We must find the concentration of cyclobutane at time t, [C
4H
8]
t.
The problem tells us the reaction is first-order, so we use the
integrated first-order rate law:
ln
[C
4H
8 ]
0
[C
4H
8 ]
t
= -kt

43. Sample Problem 5
SOLUTION:
(b) Finding the fraction that has decomposed after 0.010 s:
[C
4H
8]
0-[C
4H
8]
t
[C
4H
8]
0
=
2.00 M -0.87 M
2.00 M
= 0.58
ln
[C
4H
8 ]
0
[C
4H
8 ]
t
= -kt(a) ln
2.00 mol/L
[C
4H
8 ]
t
= (87 s
-1
)(0.010 s) = 0.87
2.00 mol/L
[C
4H
8 ]
t
= e
0.87
= 2.4
= 0.83 mol/L
2.00 mol/L
2.4
[C
2H
4] =

44. A Graphical method for finding the reaction order
from the integrated rate law.
First-order reaction
ln
[A]
0
[A]
t
= -kt
integrated rate law
ln[A]
t= -kt+ ln[A]
0
straight-line form
A plot of ln [A] vs. time gives a straight line for a first-order reaction.

45. B Graphical method for finding the reaction order
from the integrated rate law.
Second-order reaction
integrated rate law
1
[A]
t
1
[A]
0
- = kt
straight-line form
1
[A]
t
1
[A]
0
= kt+
A plot of vs. time gives a straight line for a second-order reaction.
1
[A]

46. C Graphical method for finding the reaction order
from the integrated rate law.
Zero-order reaction
A plot of [A] vs. time gives a straight line for a first-order reaction.
integrated rate law
[A]
t-[A]
0= -kt
straight-line form
[A]
t= -kt+ [A]
0

47. Graphical determination of the reaction order for the
decomposition of N
2O
5.
The concentration data is used to
construct three different plots. Since the
plot of ln [N
2O
5] vs. time gives a straight
line, the reaction is first order.

48. Reaction Half-life
The half-life(t
1/2)for a reaction is the time taken for the
concentration of a reactant to drop to half its initial value.
For a first-orderreaction, t
1/2does not depend on the
starting concentration.
t
1/2 =
ln 2
k
=
0.693
k
The half-life for a first-order reaction is a constant.
Radioactive decay is a first-order process. The half-life for a
radioactive nucleus is a useful indicator of its stability.

49. A plot of [N
2O
5] vs. time for three reaction half-lives.
t
1/2=
for a first-order process
ln 2
k
0.693
k
=

50. Sample Problem 6Using Molecular Scenes to Find Quantities
at Various Times
PROBLEM:Substance A (green) decomposes to two other
substances, B (blue) and C (yellow), in a first-order
gaseous reaction. The molecular scenes below show a
portion of the reaction mixture at two different times:
(a)Draw a similar molecular scene of the reaction mixture at t= 60.0 s.
(b)Find the rate constant of the reaction.
(c)If the total pressure (P
total) of the mixture is 5.00 atm at 90.0 s, what
is the partial pressure of substance B (P
B)?

51. SOLUTION:
Sample Problem 6
(a)After 30.0 s, the number of particles of A has decreased from 8
to 4; since [A] has halved in this time, 30.0 s is the half-life of
the reaction. After 60.0 s another half-life will have passed, and
the number of A particles will have halved again. Each A
particle forms one B and one C particle.
t= 60.0 s

52. Sample Problem 6
(c)After 90.0 s, three half-lives will have passed. The number of A
particles will have halved once again, and each A will produced one
B and one C. There will be 1 A, 7 B and 7 C particles.
mole fraction of B, X
B=
7
1 + 7 + 7
= 0.467
P
B= X
Bx P
total= 0.467 x 5.00 atm= 2.33 atm
(b)The rate constant kis determined using the formula for t
1/2of a
first-order reaction:
t
1/2=
0.693
k
k=
0.693
t
1/2
so =
0.693
30.0 s
= 2.31 x 10
-2
s
-1

53. Sample Problem 7Determining the Half-Life of a First-Order
Reaction
PROBLEM:Cyclopropane is the smallest cyclic hydrocarbon.
Because its 60°bond angles allow poor 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?
PLAN:The reaction is first order, so we find t
1/2using the half-life
equation for a first order reaction. Once we know t
1/2we can
calculate the time taken for the concentration to drop to 0.25 of
its initial value.

54. Sample Problem 7
SOLUTION:
(b)For the initial concentration to drop to one-quarter of its value
requires 2 half-lives to pass.
(a)t
1/2=
0.693
k
=
0.693
9.2 s
-1
= 0.075 s
Time = 2(0.075 s) = 0.15 s

55. Half-life Equations
For a first-orderreaction, t
1/2does not depend on the
initial concentration.
For a second-orderreaction, t
1/2is inverselyproportional
to the initial concentration:
1
k[A]
0
t
1/2= (second order process; rate = k[A]
2
)
For a zero-orderreaction, t
1/2is directlyproportional to
the initial concentration:
[A]
0
2k
0
t
1/2= (zero order process; rate = k)

56. An Overview of Zero-Order, First-Order, and
Simple Second-Order Reactions
Zero Order First Order Second Order
Rate law rate = k rate = k[A] rate = k[A]
2
Units for k mol/L·s 1/s L/mol·s
Half-life
Integrated rate law
in straight-line form
[A]
t= -kt+ [A]
0 ln[A]
t= -kt+ ln[A]
0
Plot for straight line[A]
tvs. t ln[A]
tvs. t vs. t
Slope, yintercept -k, [A]
0 -k, ln[A]
0 k,
[A]
0
2k
ln 2
k
1
k[A]
0
1
[A]
t
= kt +
1
[A]
0
1
[A]
t
1
[A]
0

57. Reaction Mechanisms
The mechanismof a reaction is the sequence of single
reaction steps that make up the overall equation.
The individual steps of the reaction mechanism are called
elementary stepsbecause each one describes a single
molecular event.
Each elementary step is characterized by its
molecularity, the number of particles involved in the
reaction.
The rate law for an elementary step canbe deduced from
the reaction stoichiometry –reaction order equals
molecularity for an elementary step only.

58. Rate Laws for General Elementary Steps
Elementary Step Molecularity Rate Law
A product
2A product
A + B product
2A + B product
Unimolecular
Bimolecular
Bimolecular
Termolecular
Rate = [A]
Rate = k[A]
2
Rate = k[A][B]
Rate = k[A]
2
[B]

59. Sample Problem 10 Determining Molecularity and Rate Laws
for Elementary Steps
PROBLEM:The following elementary steps are proposed for a
reaction mechanism:
(1) NO
2Cl(g) → NO
2(g) + Cl(g)
(2) NO
2Cl(g) + Cl(g) → NO
2(g) + Cl
2(g)
(a)Write the overall balanced equation.
(b)Determine the molecularity of each step.
(c)Write the rate law for each step.
PLAN:We find the overall equation from the sum of the elementary
steps. The molecularity of each step equals the total number
of reactantparticles. We write the rate law for each step
using the molecularities as reaction orders.

60. SOLUTION:
Step(1) isunimolecular.
Step(2) is bimolecular.
(b)
rate
1= k
1[NO
2Cl]
rate
2= k
2[NO
2Cl][Cl]
(c)
Sample Problem 10
(a)Writing the overall balanced equation:
(1) NO
2Cl(g) → NO
2(g) + Cl(g)
(2) NO
2Cl(g) + Cl(g) → NO
2(g) + Cl
2(g)
2NO
2Cl(g) → 2NO
2(g) + Cl
2(g)

61. The Rate-Determining Step of a Reaction
The sloweststep in a reaction is the rate-determining
or rate-limitingstep.
The rate law for the rate-determining step becomes
the rate law for the overall reaction.
The reaction NO
2(g) + CO(g) → NO(g) + CO
2(g)
has been proposed to occur by a two-step mechanism:
(1)NO
2(g) + NO
2(g) → NO
3(g) + NO(g)[slow; rate-determining]
(2)NO
2(g) + CO(g) → NO
2(g) + CO
2(g) [fast]
Observed rate law: rate = k[NO
2]
2

62. Correlating Mechanism with the Rate Law
A valid mechanism must meet three criteria:
The elementary steps must add up to the overall balanced equation.
The elementary steps must be reasonable.
The mechanism must correlate with the observedrate law.
A mechanism is a hypothesis–we cannot proveit is
correct, but if it is consistent with the data, and can be used
to predict results accurately, it is a useful modelfor the
reaction.

63. Mechanisms with a Slow Initial Step
The overall reaction 2NO
2(g) + F
2(g) →2NO
2F(g) has an experimental
rate law Rate = k[NO
2][F
2].
The accepted mechanism is
(1)NO
2(g) + F
2(g) → NO
2F(g) + F(g)[slow; rate determining]
(2)NO
2(g) + F(g) → NO
2F(g) [fast]
The elementary steps sum to the overall balanced equation:
2NO
2(g) + F
2(g) →2NO
2F(g)
Both steps are bimolecular and are therefore reasonable.
rate
1= k
1[NO
2[F
2]
rate
2= k
2[NO
2][F]
Step 1 is the slow step, and rate
1
correlates with the observedrate law.
The mechanism is therefore reasonable.

64. Mechanisms with a Fast Initial Step
The overall reaction 2NO(g) + O
2(g) →2NO
2(g) has an experimental
rate law Rate = k[NO]
2
[O
2].
A proposed mechanism is
(1)NO(g) + O
2(g) DNO
3(g)[fast]
(2)NO
3(g) + NO(g) → 2NO
2(g) [slow; rate determining]
The elementary steps sum to the overall balanced equation:
2NO(g) + O
2(g) →2NO
2(g)
Both steps are bimolecular and are therefore reasonable.
rate
1(fwd)= k
1[NO][O
2]
rate
1(rev)= k
-1[NO
3]
rate
2= k
2[NO
3][NO]
When equilibrium (1) has been established
rate
1(fwd)= rate
1(rev)
k
1[NO][O
2] = k
-1[NO
3]

65. [NO
3] =
k
1
k
-1
[NO][O
2]
rate
2= k
2[NO
3][NO] = k
2
k
1
k
-1
[NO][O
2][NO]
The ratio of rate constants is itself a constant, equal to the overall rate
constant for the reaction, so
rate
2= k[NO]
2
[O
2] which is consistent with the observed rate law.
For any mechanism, only reactants involved up to and
including the slow (rate-determining) step appear in the
overall rate law.

66. Reaction energy diagram for the two-step reaction of
(A) NO
2 and F
2and of (B) NO and O
2.
Each step in a multi-step reaction has its own transition state,
which occurs at the energy maximum for that step.

67. Catalysis: Speeding up a Reaction
A catalystis a substance that increases the reaction rate
without itself being consumed in the reaction.
In general, a catalyst provides an alternative reaction
pathway that has a lower total activation energythan the
uncatalyzed reaction.
A catalyst will speed up both the forward and the reverse
reactions.
A catalyst does not affect either DHor the overall yield for
a reaction.

68. Reaction energy diagram for a catalyzed (green) and
uncatalyzed (red) process.

69. The catalyzed decomposition of H
2O
2.
A small amount of NaBr is
added to a solution of H
2O
2.
Oxygen gas forms quickly as Br
-
(aq)
catalyzes the H
2O
2decomposition; the
intermediate Br
2turns the solution orange.
A homogeneouscatalyst is in the same phase as the reaction mixture.

70. The metal-catalyzed hydrogenation of ethene.
A heterogeneouscatalyst is in a different phase than the reaction mixture.

71. A Two models of enzyme action.
The lock-and-key model visualizes the enzyme active site having
a fixed shape. This shape matches the shape of its substrate(s).
The active site is therefore specific to its substrate.

72. B Two models of enzyme action.
The induced-fit model visualizes the enzyme active site
changing shape in order to bind its substrate(s) more
effectively. The active site is still selective towards certain
substrates.