1. Chemical Equilibrium

2. Equilibrium is a state in which there are no observable
changes as time goes by.
Chemical equilibrium is achieved when:
• the rates of the forward and reverse reactions are equal and
• the concentrations of the reactants and products remain
constant
Physical equilibrium
H2O (l)
Chemical equilibrium
N2O4 (g)
14.1
H2O (g)
2NO2 (g)

3. N2O4 (g) 2NO2 (g)
Start with NO2 Start with N2O4 Start with NO2 & N2O4
equilibrium
equilibrium
equilibrium
14.1

4. 14.1
constant

5. N2O4 (g) 2NO2 (g)
= 4.63 x 10-3
K =
[NO2]2
[N2O4]
aA + bB cC + dD
K =
[C]c[D]d
[A]a[B]b
Law of Mass Action
14.1

6. K >> 1
K << 1
Lie to the right Favor products
Lie to the left Favor reactants
Equilibrium Will
K =
[C]c[D]d
[A]a[B]b
aA + bB cC + dD
14.1

7. Homogenous equilibrium applies to reactions in which all
reacting species are in the same phase.
N2O4 (g) 2NO2 (g)
Kc =
[NO2]2
[N2O4]
Kp =
NO2
P2
N2O4
P
aA (g) + bB (g) cC (g) + dD (g)
14.2
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
In most cases
Kc  Kp

8. Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
Kc =
‘
[CH3COO-][H3O+]
[CH3COOH][H2O]
[H2O] = constant
Kc =
[CH3COO-][H3O+]
[CH3COOH]
= Kc [H2O]
‘
General practice not to include units for the
equilibrium constant.
14.2

9. The equilibrium concentrations for the reaction between
carbon monoxide and molecular chlorine to form COCl2 (g)
at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] =
0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)
Kc =
[COCl2]
[CO][Cl2]
=
0.14
0.012 x 0.054
= 220
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
14.2

10. The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if
the PNO = 0.400 atm and PNO = 0.270 atm?
2
2NO2 (g) 2NO (g) + O2 (g)
14.2
Kp =
2
PNO PO
2
PNO
2
2
PO2 = Kp
PNO
2
2
PNO
2
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm

11. Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
Kc =
‘
[CaO][CO2]
[CaCO3]
[CaCO3] = constant
[CaO] = constant
Kc = [CO2] = Kc x
‘
[CaCO3]
[CaO]
Kp = PCO2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
14.2

12. PCO2
= Kp
CaCO3 (s) CaO (s) + CO2 (g)
PCO2
does not depend on the amount of CaCO3 or CaO
14.2

13. Consider the following equilibrium at 295 K:
The partial pressure of each gas is 0.265 atm. Calculate
Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g)
Kp = P
NH3 H2S
P = 0.265 x 0.265 = 0.0702
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
Dn = 2 – 0 = 2 T = 295 K
Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4
14.2

14. A + B C + D
C + D E + F
A + B E + F
Kc =
‘
[C][D]
[A][B]
Kc =
‘‘
[E][F]
[C][D]
[E][F]
[A][B]
Kc =
Kc
‘
Kc‘
‘
Kc
Kc = Kc‘
‘
Kc
‘ x
If a reaction can be expressed as the sum of
two or more reactions, the equilibrium
constant for the overall reaction is given by
the product of the equilibrium constants of
the individual reactions.
14.2

15. N2O4 (g) 2NO2 (g)
= 4.63 x 10-3
K =
[NO2]2
[N2O4]
2NO2 (g) N2O4 (g)
K =
[N2O4]
[NO2]2
‘ =
1
K
= 216
When the equation for a reversible reaction
is written in the opposite direction, the
equilibrium constant becomes the reciprocal
of the original equilibrium constant.
14.2

16. Writing Equilibrium Constant Expressions
1. The concentrations of the reacting species in the
condensed phase are expressed in M. In the gaseous
phase, the concentrations can be expressed in M or in atm.
2. The concentrations of pure solids, pure liquids and solvents
do not appear in the equilibrium constant expressions.
3. The equilibrium constant is a dimensionless quantity.
4. In quoting a value for the equilibrium constant, you must
specify the balanced equation and the temperature.
5. If a reaction can be expressed as a sum of two or more
reactions, the equilibrium constant for the overall reaction is
given by the product of the equilibrium constants of the
individual reactions.
14.2

17. 14.3
Chemical Kinetics and Chemical Equilibrium
A + 2B AB2
kf
kr
ratef = kf [A][B]2
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2]
kf
kr
[AB2]
[A][B]2
= Kc =

18. The reaction quotient (Qc) is calculated by substituting the
initial concentrations of the reactants and products into the
equilibrium constant (Kc) expression.
IF
• Qc > Kc system proceeds from right to left to reach equilibrium
• Qc = Kc the system is at equilibrium
• Qc < Kc system proceeds from left to right to reach equilibrium
14.4

19. Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown x,
which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
14.4

20. At 12800C the equilibrium constant (Kc) for the reaction
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063
M and [Br] = 0.012 M, calculate the concentrations of these
species at equilibrium.
Br2 (g) 2Br (g)
Br2 (g) 2Br (g)
Let x be the change in concentration of Br2
Initial (M)
Change (M)
Equilibrium (M)
0.063 0.012
-x +2x
0.063 - x 0.012 + 2x
[Br]2
[Br2]
Kc = Kc =
(0.012 + 2x)2
0.063 - x
= 1.1 x 10-3 Solve for x
14.4

21. Kc =
(0.012 + 2x)2
0.063 - x
= 1.1 x 10-3
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
ax2 + bx + c =0
-b ± b2 – 4ac

2a
x =
Br2 (g) 2Br (g)
Initial (M)
Change (M)
Equilibrium (M)
0.063 0.012
-x +2x
0.063 - x 0.012 + 2x
x = -0.00178
x = -0.0105
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.062 – x = 0.0648 M
14.4

22. If an external stress is applied to a system at equilibrium, the
system adjusts in such a way that the stress is partially offset
as the system reaches a new equilibrium position.
Le Châtelier’s Principle
• Changes in Concentration
N2 (g) + 3H2 (g) 2NH3 (g)
Add
NH3
Equilibrium
shifts left to
offset stress
14.5

23. Le Châtelier’s Principle
• Changes in Concentration continued
Change Shifts the Equilibrium
Increase concentration of product(s) left
Decrease concentration of product(s) right
Decrease concentration of reactant(s)
Increase concentration of reactant(s) right
left
14.5
aA + bB cC + dD
Add
Add
Remove Remove

24. Le Châtelier’s Principle
• Changes in Volume and Pressure
A (g) + B (g) C (g)
Change Shifts the Equilibrium
Increase pressure Side with fewest moles of gas
Decrease pressure Side with most moles of gas
Decrease volume
Increase volume Side with most moles of gas
Side with fewest moles of gas
14.5

25. Le Châtelier’s Principle
• Changes in Temperature
Change Exothermic Rx
Increase temperature K decreases
Decrease temperature K increases
Endothermic Rx
K increases
K decreases
14.5
colder hotter

26. uncatalyzed catalyzed
14.5
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
Le Châtelier’s Principle

27. Chemistry In Action
Life at High Altitudes and Hemoglobin Production
Kc =
[HbO2]
[Hb][O2]
Hb (aq) + O2 (aq) HbO2 (aq)

28. Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g) 2NH3 (g) DH0 = -92.6 kJ/mol

29. Le Châtelier’s Principle
Change Shift Equilibrium
Change Equilibrium
Constant
Concentration yes no
Pressure yes no
Volume yes no
Temperature yes yes
Catalyst no no
14.5

30. Thermodynamics

31. Energy is the capacity to do work
• Radiant energy comes from the sun and is earth’s
primary energy source
• Thermal energy is the energy associated with the
random motion of atoms and molecules
• Chemical energy is the energy stored within the
bonds of chemical substances
• Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
• Potential energy is the energy available by virtue
of an object’s position
6.1

32. Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Energy Changes in Chemical Reactions
Temperature is a measure of the thermal energy.
900C
400C
greater thermal energy
6.2
Temperature = Thermal Energy

33. Thermochemistry is the study of heat change in chemical
reactions.
The system is the specific part of the universe that is of
interest in the study.
open
mass & energy
Exchange:
closed
energy
isolated
nothing
6.2

34. Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
2H2 (g) + O2 (g) 2H2O (l) + energy
H2O (g) H2O (l) + energy
energy + 2HgO (s) 2Hg (l) + O2 (g)
6.2
energy + H2O (s) H2O (l)

35. Exothermic Endothermic
6.2

36. Thermodynamics is the scientific study of the interconversion
of heat and other kinds of energy.
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
energy, pressure, volume, temperature
6.3
DE = Efinal - Einitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial

37. First law of thermodynamics – energy can
be converted from one form to another, but
cannot be created or destroyed.
DEsystem + DEsurroundings = 0
or
DEsystem = -DEsurroundings
C3H8 + 5O2 3CO2 + 4H2O
Exothermic chemical reaction!
6.3
Chemical energy lost by combustion = Energy gained by the surroundings
system surroundings

38. Another form of the first law for DEsystem
6.3
DE = q + w
DE is the change in internal energy of a system
q is the heat exchange between the system and the surroundings
w is the work done on (or by) the system
w = -PDV when a gas expands against a constant external pressure

39. Work Done On the System
6.3
w = F x d
w = -P DV
P x V = x d3 = F x d = w
F
d2
DV > 0
-PDV < 0
wsys < 0
Work is
not a
state
function!
Dw = wfinal - winitial
initial final

40. A sample of nitrogen gas expands in volume from 1.6 L to
5.4 L at constant temperature. What is the work done in
joules if the gas expands (a) against a vacuum and (b)
against a constant pressure of 3.7 atm?
w = -P DV
(a) DV = 5.4 L – 1.6 L = 3.8 L P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 joules
(b) DV = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm
w = -3.7 atm x 3.8 L = -14.1 L•atm
w = -14.1 L•atm x
101.3 J
1L•atm
= -1430 J
6.3

41. Chemistry in Action: Making Snow
DE = q + w
q = 0
w < 0, DE < 0
DE = CDT
DT < 0, SNOW!

42. Enthalpy and the First Law of Thermodynamics
6.4
DE = q + w
DE = DH - PDV
DH = DE + PDV
q = DH and w = -PDV
At constant pressure:

43. Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0 6.4

44. Thermochemical Equations
H2O (s) H2O (l) DH = 6.01 kJ
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
6.4

45. Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
6.4

46. H2O (s) H2O (l) DH = 6.01 kJ
• The stoichiometric coefficients always refer to the number
of moles of a substance
Thermochemical Equations
• If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s) DH = -6.01 kJ
• If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ
6.4

47. H2O (s) H2O (l) DH = 6.01 kJ
• The physical states of all reactants and products must be
specified in thermochemical equations.
Thermochemical Equations
6.4
H2O (l) H2O (g) DH = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ
266 g P4
1 mol P4
123.9 g P4
x
3013 kJ
1 mol P4
x = 6470 kJ

48. A Comparison of DH and DE
2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) DH = -367.5 kJ/mol
DE = DH - PDV At 25 0C, 1 mole H2 = 24.5 L at 1 atm
PDV = 1 atm x 24.5 L = 2.5 kJ
DE = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
6.4

49. The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the substance
by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q)
required to raise the temperature of a given quantity (m) of the
substance by one degree Celsius.
C = m x s
Heat (q) absorbed or released:
q = m x s x Dt
q = C x Dt
Dt = tfinal - tinitial
6.5

50. How much heat is given off when an 869 g iron bar cools
from 940C to 50C?
s of Fe = 0.444 J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
6.5

51. Constant-Volume Calorimetry
No heat enters or leaves!
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = m x s x Dt
qbomb = Cbomb x Dt
6.5
Reaction at Constant V
DH ~ qrxn
DH = qrxn

52. Constant-Pressure Calorimetry
No heat enters or leaves!
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = m x s x Dt
qcal = Ccal x Dt
6.5
Reaction at Constant P
DH = qrxn

53. 6.5

54. Chemistry in Action:
Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol
1 cal = 4.184 J
1 Cal = 1000 cal = 4184 J
Substance DHcombustion (kJ/g)
Apple -2
Beef -8
Beer -1.5
Gasoline -34

55. Because there is no way to measure the absolute value of
the enthalpy of a substance, must I measure the enthalpy
change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of
formation (DH0) as a reference point for all enthalpy
expressions.
f
Standard enthalpy of formation (DH0) is the heat change
that results when one mole of a compound is formed from
its elements at a pressure of 1 atm.
f
The standard enthalpy of formation of any element in its
most stable form is zero.
DH0 (O2) = 0
f
DH0 (O3) = 142 kJ/mol
f
DH0 (C, graphite) = 0
f
DH0 (C, diamond) = 1.90 kJ/mol
f
6.6

56. 6.6

57. The standard enthalpy of reaction (DH0 ) is the enthalpy of a
reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn dDH0 (D)
f
cDH0 (C)
f
= [ + ] - bDH0 (B)
f
aDH0 (A)
f
[ + ]
DH0
rxn nDH0 (products)
f
= S mDH0 (reactants)
f
S
-
6.6
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)

58. C (graphite) + 1/2O2 (g) CO (g)
CO (g) + 1/2O2 (g) CO2 (g)
C (graphite) + O2 (g) CO2 (g)
6.6

59. Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ
2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJ
rxn
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ
rxn
+
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
rxn
6.6

60. Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0
rxn nDH0 (products)
f
= S mDH0 (reactants)
f
S
-
DH0
rxn 6DH0 (H2O)
f
12DH0 (CO2)
f
= [ + ] - 2DH0 (C6H6)
f
[ ]
DH0
rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
2 mol
= - 2973 kJ/mol C6H6
6.6

61. Chemistry in Action: Bombardier Beetle Defense
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l) DH0 = ?
C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) DH0 = 177 kJ/mol
H2O2 (aq) H2O (l) + ½O2 (g) DH0 = -94.6 kJ/mol
H2 (g) + ½ O2 (g) H2O (l) DH0 = -286 kJ/mol
DH0 = 177 - 94.6 – 286 = -204 kJ/mol
Exothermic!

62. The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hsoln - Hcomponents
6.7
Which substance(s) could be
used for melting ice?
Which substance(s) could be
used for a cold pack?

63. The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
6.7

64. Entropy, Free Energy,
and Equilibrium
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

65. Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• A lump of sugar dissolves in a cup of coffee
• At 1 atm, water freezes below 0 0C and ice melts above 0 0C
• Heat flows from a hotter object to a colder object
• A gas expands in an evacuated bulb
• Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2

66. spontaneous
nonspontaneous
18.2

67. Does a decrease in enthalpy mean a reaction proceeds
spontaneously?
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = -890.4 kJ
H+ (aq) + OH- (aq) H2O (l) DH0 = -56.2 kJ
H2O (s) H2O (l) DH0 = 6.01 kJ
NH4NO3 (s) NH4
+(aq) + NO3
- (aq) DH0 = 25 kJ
H2O
Spontaneous reactions
18.2

68. Entropy (S) is a measure of the randomness or disorder of a
system.
order S
disorder
S
DS = Sf - Si
If the change from initial to final results in an increase in randomness
Sf > Si DS > 0
For any substance, the solid state is more ordered than the
liquid state and the liquid state is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s) H2O (l) DS > 0
18.3

69. W = 1
W = 4
W = 6
W = number of microstates
S = k ln W
DS = Sf - Si
DS = k ln
Wf
Wi
Wf > Wi then DS > 0
Wf < Wi then DS < 0
Entropy
18.3

70. Processes that
lead to an
increase in
entropy (DS > 0)
18.2

71. How does the entropy of a system change for each of the
following processes?
(a) Condensing water vapor
Randomness decreases Entropy decreases (DS < 0)
(b) Forming sucrose crystals from a supersaturated solution
Randomness decreases Entropy decreases (DS < 0)
(c) Heating hydrogen gas from 600C to 800C
Randomness increases Entropy increases (DS > 0)
(d) Subliming dry ice
Randomness increases Entropy increases (DS > 0)
18.3

72. Entropy
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
energy, enthalpy, pressure, volume, temperature, entropy
18.3

73. First Law of Thermodynamics
Energy can be converted from one form to another but
energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous
process and remains unchanged in an equilibrium process.
DSuniv = DSsys + DSsurr > 0
Spontaneous process:
DSuniv = DSsys + DSsurr = 0
Equilibrium process:
18.4

74. Entropy Changes in the System (DSsys)
aA + bB cC + dD
DS0
rxn dS0(D)
cS0(C)
= [ + ] - bS0(B)
aS0(A)
[ + ]
DS0
rxn nS0(products)
= S mS0(reactants)
S
-
The standard entropy of reaction (DS0 ) is the entropy change
for a reaction carried out at 1 atm and 250C.
rxn
18.4
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)
S0(CO) = 197.9 J/K•mol
S0(O2) = 205.0 J/K•mol
S0(CO2) = 213.6 J/K•mol
DS0
rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
DS0
rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol

75. Entropy Changes in the System (DSsys)
18.4
When gases are produced (or consumed)
• If a reaction produces more gas molecules than it
consumes, DS0 > 0.
• If the total number of gas molecules diminishes,
DS0 < 0.
• If there is no net change in the total number of gas
molecules, then DS0 may be positive or negative
BUT DS0 will be a small number.
What is the sign of the entropy change for the following
reaction? 2Zn (s) + O2 (g) 2ZnO (s)
The total number of gas molecules goes down, DS is negative.

76. Entropy Changes in the Surroundings (DSsurr)
Exothermic Process
DSsurr > 0
Endothermic Process
DSsurr < 0
18.4

77. Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the
absolute zero of temperature.
18.3
S = k ln W
W = 1
S = 0

78. DSuniv = DSsys + DSsurr > 0
Spontaneous process:
DSuniv = DSsys + DSsurr = 0
Equilibrium process:
Gibbs Free Energy
For a constant-temperature process:
DG = DHsys -TDSsys
Gibbs free
energy (G)
DG < 0 The reaction is spontaneous in the forward direction.
DG > 0 The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
DG = 0 The reaction is at equilibrium.
18.5

79. 18.5
aA + bB cC + dD
DG0
rxn dDG0 (D)
f
cDG0 (C)
f
= [ + ] - bDG0 (B)
f
aDG0 (A)
f
[ + ]
DG0
rxn nDG0 (products)
f
= S mDG0 (reactants)
f
S
-
The standard free-energy of reaction (DG0 ) is the free-
energy change for a reaction when it occurs under standard-
state conditions.
rxn
Standard free energy of formation
(DG0) is the free-energy change
that occurs when 1 mole of the
compound is formed from its
elements in their standard states.
f
DG0 of any element in its stable
form is zero.
f

80. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DG0
rxn nDG0 (products)
f
= S mDG0 (reactants)
f
S
-
What is the standard free-energy change for the following
reaction at 25 0C?
DG0
rxn 6DG0 (H2O)
f
12DG0 (CO2)
f
= [ + ] - 2DG0 (C6H6)
f
[ ]
DG0
rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is the reaction spontaneous at 25 0C?
DG0 = -6405 kJ < 0
spontaneous
18.5

81. DG = DH - TDS
18.5

82. CaCO3 (s) CaO (s) + CO2 (g)
DH0 = 177.8 kJ
DS0 = 160.5 J/K
DG0 = DH0 – TDS0
At 25 0C, DG0 = 130.0 kJ
DG0 = 0 at 835 0C
18.5
Temperature and Spontaneity of Chemical Reactions
Equilibrium Pressure of CO2

83. Gibbs Free Energy and Phase Transitions
H2O (l) H2O (g)
DG0 = 0 = DH0 – TDS0
DS =
T
DH
=
40.79 kJ
373 K
= 109 J/K
18.5

84. Efficiency = X 100%
Th - Tc
Tc
Chemistry In Action: The Efficiency of Heat Engines
A Simple Heat Engine

85. Gibbs Free Energy and Chemical Equilibrium
DG = DG0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
DG = 0 Q = K
0 = DG0 + RT lnK
DG0 = - RT lnK
18.6

86. 18.6
Free Energy Versus Extent of Reaction
DG0 < 0 DG0 > 0

87. DG0 = - RT lnK
18.6

88. 18.7
ATP + H2O + Alanine + Glycine ADP + H3PO4 + Alanylglycine
Alanine + Glycine Alanylglycine
DG0 = +29 kJ
DG0 = -2 kJ
K < 1
K > 1

89. 18.7
The Structure of ATP and ADP in Ionized Forms

90. High Entropy Low Entropy
TDS = DH - DG
Chemistry In Action: The Thermodynamics of a Rubber Band

91. Electrochemistry

92. 2Mg (s) + O2 (g) 2MgO (s)
2Mg 2Mg2+ + 4e-
O2 + 4e- 2O2-
Oxidation half-reaction (lose e-)
Reduction half-reaction (gain e-)
19.1
Electrochemical processes are oxidation-reduction reactions
in which:
• the energy released by a spontaneous reaction is
converted to electricity or
• electrical energy is used to cause a nonspontaneous
reaction to occur
0 0 2+ 2-

93. Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation
number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to
the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In H2O2
and O2
2- it is –1.
4.4

94. 4. The oxidation number of hydrogen is +1 except when
it is bonded to metals in binary compounds. In these
cases, its oxidation number is –1.
6. The sum of the oxidation numbers of all the atoms in a
molecule or ion is equal to the charge on the
molecule or ion.
5. Group IA metals are +1, IIA metals are +2 and fluorine is
always –1.
HCO3
-
O = -2 H = +1
3x(-2) + 1 + ? = -1
C = +4
Oxidation numbers of all
the atoms in HCO3
- ?
4.4

95. Balancing Redox Equations
19.1
1. Write the unbalanced equation for the reaction ion ionic form.
The oxidation of Fe2+ to Fe3+ by Cr2O7
2- in acid solution?
Fe2+ + Cr2O7
2- Fe3+ + Cr3+
2. Separate the equation into two half-reactions.
Oxidation:
Cr2O7
2- Cr3+
+6 +3
Reduction:
Fe2+ Fe3+
+2 +3
3. Balance the atoms other than O and H in each half-reaction.
Cr2O7
2- 2Cr3+

96. Balancing Redox Equations
4. For reactions in acid, add H2O to balance O atoms and H+ to
balance H atoms.
Cr2O7
2- 2Cr3+ + 7H2O
14H+ + Cr2O7
2- 2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to balance the
charges on the half-reaction.
Fe2+ Fe3+ + 1e-
6e- + 14H+ + Cr2O7
2- 2Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two half-
reactions by multiplying the half-reactions by appropriate
coefficients.
6Fe2+ 6Fe3+ + 6e-
6e- + 14H+ + Cr2O7
2- 2Cr3+ + 7H2O
19.1

97. Balancing Redox Equations
7. Add the two half-reactions together and balance the final
equation by inspection. The number of electrons on both
sides must cancel.
6e- + 14H+ + Cr2O7
2- 2Cr3+ + 7H2O
6Fe2+ 6Fe3+ + 6e-
Oxidation:
Reduction:
14H+ + Cr2O7
2- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O
8. Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
19.1
9. For reactions in basic solutions, add OH- to both sides of the
equation for every H+ that appears in the final equation.

98. Galvanic Cells
19.2
spontaneous
redox reaction
anode
oxidation
cathode
reduction

99. Galvanic Cells
19.2
The difference in electrical
potential between the anode
and cathode is called:
• cell voltage
• electromotive force (emf)
• cell potential
Cell Diagram
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M & [Zn2+] = 1 M
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode cathode

100. Standard Reduction Potentials
19.3
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
2e- + 2H+ (1 M) H2 (1 atm)
Zn (s) Zn2+ (1 M) + 2e-
Anode (oxidation):
Cathode (reduction):
Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

101. Standard Reduction Potentials
19.3
Standard reduction potential (E0) is the voltage associated
with a reduction reaction at an electrode when all solutes
are 1 M and all gases are at 1 atm.
E0 = 0 V
Standard hydrogen electrode (SHE)
2e- + 2H+ (1 M) H2 (1 atm)
Reduction Reaction

102. 19.3
E0 = 0.76 V
cell
Standard emf (E0 )
cell
0.76 V = 0 - EZn /Zn
0 2+
EZn /Zn = -0.76 V
0 2+
Zn2+ (1 M) + 2e- Zn E0 = -0.76 V
E0 = EH /H - EZn /Zn
cell
0 0
+ 2+
2
Standard Reduction Potentials
E0 = Ecathode - Eanode
cell
0 0
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

103. Standard Reduction Potentials
19.3
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) 2H+ (1 M) + 2e-
Anode (oxidation):
Cathode (reduction):
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)
E0 = Ecathode - Eanode
cell
0 0
E0 = 0.34 V
cell
Ecell = ECu /Cu – EH /H
2+ +
2
0 0 0
0.34 = ECu /Cu - 0
0 2+
ECu /Cu = 0.34 V
2+
0

104. 19.3
• E0 is for the reaction as
written
• The more positive E0 the
greater the tendency for the
substance to be reduced
• The half-cell reactions are
reversible
• The sign of E0 changes
when the reaction is
reversed
• Changing the stoichiometric
coefficients of a half-cell
reaction does not change
the value of E0

105. What is the standard emf of an electrochemical cell made
of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr
electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V
Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V
Cd is the stronger oxidizer
Cd will oxidize Cr
2e- + Cd2+ (1 M) Cd (s)
Cr (s) Cr3+ (1 M) + 3e-
Anode (oxidation):
Cathode (reduction):
2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)
x 2
x 3
E0 = Ecathode - Eanode
cell
0 0
E0 = -0.40 – (-0.74)
cell
E0 = 0.34 V
cell
19.3

106. 19.4
Spontaneity of Redox Reactions
DG = -nFEcell
DG0 = -nFEcell
0
n = number of moles of electrons in reaction
F = 96,500
J
V • mol
= 96,500 C/mol
DG0 = -RT ln K = -nFEcell
0
Ecell
0 =
RT
nF
ln K
(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol)
ln K
=
=
0.0257 V
n
ln K
Ecell
0
=
0.0592 V
n
log K
Ecell
0

107. Spontaneity of Redox Reactions
19.4
DG0 = -RT ln K = -nFEcell
0

108. 2e- + Fe2+ Fe
2Ag 2Ag+ + 2e-
Oxidation:
Reduction:
What is the equilibrium constant for the following reaction
at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)
=
0.0257 V
n
ln K
Ecell
0
19.4
E0 = -0.44 – (0.80)
E0 = -1.24 V
0.0257 V
x n
E0
cell
exp
K =
n = 2
0.0257 V
x 2
-1.24 V
= exp
K = 1.23 x 10-42
E0 = EFe /Fe – EAg /Ag
0 0
2+ +

109. The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE 0
-nFE = -nFE0 + RT ln Q
E = E0 - ln Q
RT
nF
Nernst equation
At 298
19.5
-
0.0257 V
n
ln Q
E0
E = -
0.0592 V
n
log Q
E0
E =

110. Will the following reaction occur spontaneously at 250C if
[Fe2+] = 0.60 M and [Cd2+] = 0.010 M?
Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
2e- + Fe2+ 2Fe
Cd Cd2+ + 2e-
Oxidation:
Reduction:
n = 2
E0 = -0.44 – (-0.40)
E0 = -0.04 V
E0 = EFe /Fe – ECd /Cd
0 0
2+ 2+
-
0.0257 V
n
ln Q
E0
E =
-
0.0257 V
2
ln
-0.04 V
E =
0.010
0.60
E = 0.013
E > 0 Spontaneous
19.5

111. Batteries
19.6
Leclanché cell
Dry cell
Zn (s) Zn2+ (aq) + 2e-
Anode:
Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)
+
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

112. Batteries
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-
Anode:
Cathode: HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)
Mercury Battery
19.6

113. Batteries
19.6
Anode:
Cathode:
Lead storage
battery
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e- PbSO4 (s) + 2H2O (l)
4
Pb (s) + SO2- (aq) PbSO4 (s) + 2e-
4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l)
4

114. Batteries
19.6
Solid State Lithium Battery

115. Batteries
19.6
A fuel cell is an
electrochemical cell
that requires a
continuous supply of
reactants to keep
functioning
Anode:
Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)
2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-
2H2 (g) + O2 (g) 2H2O (l)

116. Chemistry In Action: Bacteria Power
CH3COO- + 2O2 + H+ 2CO2 + 2H2O

117. Corrosion
19.7

118. Cathodic Protection of an Iron Storage Tank
19.7

119. 19.8
Electrolysis is the process in which electrical energy is used
to cause a nonspontaneous chemical reaction to occur.

120. Electrolysis of Water
19.8

121. Electrolysis and Mass Changes
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
19.8

122. How much Ca will be produced in an electrolytic cell of
molten CaCl2 if a current of 0.452 A is passed through the
cell for 1.5 hours?
Anode:
Cathode: Ca2+ (l) + 2e- Ca (s)
2Cl- (l) Cl2 (g) + 2e-
Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)
2 mole e- = 1 mole Ca
mol Ca = 0.452
C
s
x 1.5 hr x 3600
s
hr 96,500 C
1 mol e-
x
2 mol e-
1 mol Ca
x
= 0.0126 mol Ca
= 0.50 g Ca
19.8

123. Chemistry In Action: Dental Filling Discomfort
Hg2 /Ag2Hg3 0.85 V
2+
Sn /Ag3Sn -0.05 V
2+
Sn /Ag3Sn -0.05 V
2+

124. Chemical Kinetics

125. Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A B
rate = -
D[A]
Dt
rate =
D[B]
Dt
D[A] = change in concentration of A over
time period Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
13.1

126. A B
13.1
rate = -
D[A]
Dt
rate =
D[B]
Dt
time

127. Reactants & Products over Time

128. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nm
light
Detector
D[Br2] a DAbsorption
13.1

129. Br2 (aq) + HCOOH (aq) 2Br– (aq) + 2H+ (aq) + CO2 (g)
average rate = –
D[Br2]
Dt
= –
[Br2]final – [Br2]initial
tfinal - tinitial
slope of
tangent
slope of
tangent slope of
tangent
instantaneous rate = rate for specific instance in time
13.1

130. rate a [Br2]
rate = k [Br2]
k =
rate
[Br2]
13.1
= rate constant
= 3.50 x 10–3 s–1

131. Reaction Rates and Stoichiometry
13.1
2A B
Two moles of A disappear for each mole of B that is formed.
rate =
D[B]
Dt
rate = –
D[A]
Dt
1
2
aA + bB cC + dD
rate = –
D[A]
Dt
1
a
= –
D[B]
Dt
1
b
=
D[C]
Dt
1
c
=
D[D]
Dt
1
d

132. Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = –
D[CH4]
Dt
= –
D[O2]
Dt
1
2
=
D[H2O]
Dt
1
2
=
D[CO2]
Dt
13.1

133. The Rate Law
13.2
The rate law expresses the relationship of the rate of a reaction
to the rate constant and the concentrations of the reactants
raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y
reaction is xth order in A
reaction is yth order in B
reaction is (x + y)th order overall

134. F2 (g) + 2ClO2 (g) 2FClO2 (g)
Determine x and y in the rate law Rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] held constant:
The rate doubles
Therefore, x = 1
Quadruple [ClO2] with [F2] held constant:
The rate quadruples
Therefore, y = 1 The rate law is
Rate = k [F2]1[ClO2]1
13.2

135. F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2][ClO2]
Rate Laws
• Rate laws are always determined experimentally.
• Reaction order is always defined in terms of reactant
(not product) concentrations.
• The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the balanced
chemical equation.
1
13.2

136. Determine the rate law and calculate the rate constant for
the following reaction from the following data:
S2O8
2– (aq) + 3I– (aq) 2SO4
2– (aq) + I3
– (aq)
Experiment [S2O8
2 – ] [I – ]
Initial Rate
(M/s)
1 0.08 0.034 2.2 x 10–4
2 0.08 0.017 1.1 x 10–4
3 0.16 0.017 2.2 x 10–4
rate = k [S2O8
2–]x[I–]y
Double [I–], rate doubles (experiment 1 & 2)
y = 1
Double [S2O8
2–], rate doubles (experiment 2 & 3)
x = 1
k =
rate
[S2O8
2–][I–]
=
2.2 x 10–4 M/s
(0.08 M)(0.034 M)
= 0.08/M•s
13.2
rate = k [S2O8
2–][I–]

137. First-Order Reactions
13.3
A product rate = -
D[A]
Dt
rate = k [A]
k =
rate
[A]
= 1/s or s-1
M/s
M
=
D[A]
Dt
= k [A]
–
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
[A] = [A]0exp(–kt) ln[A] = ln[A]0 – kt

138. 13.3
2N2O5 4NO2 (g) + O2 (g)
k = 5.714 X 10–4 s–1

139. The reaction 2A B is first order in A with a rate
constant of 2.8 x 10–2 s–1 at 800C. How long will it take for
A to decrease from 0.88 M to 0.14 M ?
ln[A]t = ln[A]0 – kt
kt = ln[A]0 – ln[A]
t =
ln[A]0 – ln[A]
k
= 66 s
[A]0 = 0.88 M
[A]t = 0.14 M
ln
[A]0
[A]
k
=
ln
0.88 M
0.14 M
2.8 x 10–2 s–1
=
13.3

140. Half-Life of First-Order Reactions
13.3
The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln
[A]0
[A]0/2
k
=
t½
ln2
k
=
0.693
k
=
What is the half-life of N2O5 if it decomposes with a rate
constant of 5.7 x 10–4 s–1?
t½
ln2
k
=
0.693
5.7 x 10–4 s–1
= = 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)

141. Half-Life of a First-Order Reaction
The half-life of a first-order reaction stays the same.

142. Comparison of Graphs for a First-Order Reaction
A straight line is obtained from a graph of ln[A] vs. time,
characteristic of a first-order reaction.

143. Second-Order Reactions
13.3
A product rate = -
D[A]
Dt
rate = k [A]2
k =
rate
[A]2
= 1/M•s
M/s
M2
=
D[A]
Dt
= k [A]2
–
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
1
[A]
=
1
[A]0
+ kt
t½ = t when [A] = [A]0/2
t½ =
1
k[A]0

144. Half-Lives of Second-Order Reactions
Each half-life is double the time of the previous half-life.

145. Second-Order Reaction
Comparison of Graphs
The data give a straight line when plotting 1/[A] vs. time,
characteristic of a second-order reaction.

146. Zero-Order Reactions
13.3
A product rate = -
D[A]
Dt
rate = k [A]0 = k
k =
rate
[A]0
= M/s
D[A]
Dt
= k
–
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
t½ =
[A]0
2k
[A] = [A]0 – kt

147. Half-Lives of a Zero-Order Reaction
Each half-life is ½ the time of the previous half-life.

148. Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order Rate Law
Concentration-Time
Equation Half-Life
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
1
[A]
=
1
[A]0
+ kt
[A] = [A]0 - kt
t½
ln2
k
=
t½ =
[A]0
2k
t½ =
1
k[A]0
13.3

149. Comparison of Graphs for H2O2 Decomposition
The reaction is
second order
with rate law
Rate = k[H2O2]2
From www.sparknotes.com

150. Exothermic Reaction Endothermic Reaction
The activation energy (Ea ) is the minimum amount of
energy required to initiate a chemical reaction.
13.4
A + B AB C + D
+
+

151. Activation Energy

152. Temperature Dependence of the Rate Constant
k = A • exp( -Ea / RT )
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
lnk = -
Ea
R
1
T
+ lnA
(Arrhenius equation)
13.4
For Two Temperatures: ln(k1/k2) = Ea/R(1/T2 – 1/T1)

153. 13.4
lnk = -
Ea
R
1
T
+ lnA
Slope = –Ea/R
Y-Intercept = lnA

154. 13.4
Importance of Orientation

155. K + CH3I KI + •CH3

156. 13.5
Reaction Mechanisms
The overall progress of a chemical reaction can be represented
at the molecular level by a series of simple elementary steps
or elementary reactions.
The sequence of elementary steps that leads to product
formation is the reaction mechanism.
2NO (g) + O2 (g) 2NO2 (g)
N2O2 is detected during the reaction!
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+

157. 2NO (g) + O2 (g) 2NO2 (g)
13.5

158. 13.5
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
Intermediates are species that appear in a reaction
mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step
and consumed in a later elementary step.
The molecularity of a reaction is the number of molecules
reacting in an elementary step.
• Unimolecular reaction – elementary step with 1 molecule
• Bimolecular reaction – elementary step with 2 molecules
• Termolecular reaction – elementary step with 3 molecules

159. Unimolecular reaction A products rate = k [A]
Bimolecular reaction A + B products rate = k [A][B]
Bimolecular reaction A + A products rate = k [A]2
Rate Laws and Elementary Steps
13.5
Writing plausible reaction mechanisms:
• The sum of the elementary steps must give the overall
balanced equation for the reaction.
• The rate-determining step should predict the same rate
law that is determined experimentally.
The rate-determining step is the slowest step in the
sequence of steps leading to product formation.

160. 13.5
Sequence of Steps in Studying a Reaction Mechanism

161. The experimental rate law for the reaction between NO2
and CO to produce NO and CO2 is rate = k[NO2]2. The
reaction is believed to occur via two steps:
Step 1: NO2 + NO2 NO + NO3
Step 2: NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
13.5

162. A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
k = A • exp( -Ea / RT ) Ea k
ratecatalyzed > rateuncatalyzed
Ea < Ea
‘ 13.6
Uncatalyzed Catalyzed

163. In heterogeneous catalysis, the reactants and the catalysts
are in different phases.
In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
• Haber synthesis of ammonia
• Ostwald process for the production of nitric acid
• Catalytic converters
• Acid catalysis
• Base catalysis
13.6

164. N2 (g) + 3H2 (g) 2NH3 (g)
Fe/Al2O3/K2O
catalyst
Haber Process
13.6

165. Ostwald Process
Hot Pt wire
over NH3 solution
Pt-Rh catalysts used
in Ostwald process
4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)
Pt catalyst
2NO (g) + O2 (g) 2NO2 (g)
2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)
13.6

166. Catalytic Converters
13.6
CO + Unburned Hydrocarbons + O2 CO2 + H2O
catalytic
converter
2NO + 2NO2 2N2 + 3O2
catalytic
converter

167. Enzyme Catalysis
13.6

168. uncatalyzed
enzyme
catalyzed
13.6
rate =
D[P]
Dt
rate = k [ES]