Chapter 19: Thermodynamics

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Chapter 19: Thermodynamics

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Title: Chapter 19: Thermodynamics

1
Chapter 19 Thermodynamics

Chemistry The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop

2
Thermodynamics

Study of energy changes and flow of energy
Answers several fundamental questions
Is it possible for a given reaction to occur?
Will the reaction occur spontaneously (without
outside interference) at a given T ?
Will reaction release or absorb heat?
Tells us nothing about time frame of reaction
Kinetics
Two major considerations
Enthalpy changes, ?H (heats of reaction)
Heat exchange between system and surroundings
Nature's trend to randomness or disorder
Entropy

3
Review of First Law of Thermodynamics

Internal energy, E
System's total energy
Sum of KE and PE of all particles in system
or for chemical reaction
?E energy into system
?E energy out of system

4
Two Methods of Energy Exchange Between System and
Surroundings

Heat q Work w
?E q w
Conventions of heat and work

q Heat absorbed by system Esystem increases
q Heat released by system Esystem decreases
w Work done on system Esystem increases
w Work done by system Esystem decreases
5
First Law of Thermodynamics

Energy can neither be created nor destroyed
It can only be converted from one form to another
Kinetic ? Potential
Chemical ? Electrical
Electrical ? Mechanical
E is a state function
?E is a change in a state function
Path independent
?E q w

6
Work in Chemical Systems

Electrical
Pressure-volume or P?V
w P?V
Where P external pressure
If P?V only work in chemical system, then

7
Heat at Constant Volume

Reaction done at constant V
?V 0
P?V 0, so
?E qV
Entire energy change due to heat absorbed or lost
Rarely done, not too useful

8
Heat at Constant Pressure

More common
Reactions open to atmosphere
Constant P
Enthalpy
H E PV
Enthalpy change
?H ?E P?V
Substituting in first law for ?E gives
?H (q P?V) P?V qP
?H qP
Heat of reaction at constant pressure

9
Converting Between ?E and ?H For Chemical
Reactions

?H ? ?E
Differ by ?H ?E P?V
Only differ significantly when gases formed or
consumed
Assume gases are ideal
Since P and T are constant

10
Converting Between ?E and ?H For Chemical
Reactions

When reaction occurs
?V caused by ?n of gas
Not all reactants and products are gases
So redefine as ?ngas
Where ?ngas (ngas)products (ngas)reactants
Substituting into ?H ?E P?V gives
or

Ex. 1 What is the difference between ?H and ?E
for the following reaction at 25 C?
2 N2O5(g) ?? 4 NO2(g) O2(g)
What is the difference between ?H and ?E ?
Step 1 Calculate ?H using data (Table
7.2)
Recall
?H (4 mol)(33.8 kJ/mol)
(1 mol)(0.0 kJ/mol) (2 mol)(11 kJ/mol)
?H 113 kJ

12
Ex. 1. ?H and ?E (cont.)

Step 2 Calculate ?ngas
?ngas (ngas)products (ngas)reactants
?ngas (4 1 2) mol 3 mol
Step 3 Calculate ?E using
R 8.31451 J/K mol T 298 K
?E 113 kJ (3 mol)(8.314 J/K mol)(298 K)(1
kJ/1000 J)
?E 113 kJ 7.43 kJ 106 kJ

13
Ex. 1. ?H and ?E (cont.)

Step 4 Calculate percent difference
Bigger than most, but still small
Note Assumes that volumes of solids and liquids
are negligible
Vsolid ? Vliquid ltlt Vgas

14
Is Assumption that Vsolid ? Vliquid ltlt Vgas
Justified?

Consider
CaCO3(s) 2H(aq) ? Ca2(aq) H2O CO2(g)
37.0 mL 218.0 mL 18 mL 18 mL 24.4 L
Volumes assuming each coefficient equal number of
moles
So ?V ?Vprod ?Vreac 24.363 L ? 24.4 L
Yes, assumption is justified
Note If no gases are present reduces to

15
Learning Check

Consider the following reaction for picric acid
8O2(g) 2C6H2(NO2)3OH(l ) ? 3 N2(g) 12CO2(g)
6H2O(l )
Calculate ?? , ??

8O2(g) 2C6H2(NO2)3OH(l ) ? 3 N2(g) 12CO2(g) 6H2O(l ) 8O2(g) 2C6H2(NO2)3OH(l ) ? 3 N2(g) 12CO2(g) 6H2O(l ) 8O2(g) 2C6H2(NO2)3OH(l ) ? 3 N2(g) 12CO2(g) 6H2O(l ) 8O2(g) 2C6H2(NO2)3OH(l ) ? 3 N2(g) 12CO2(g) 6H2O(l ) 8O2(g) 2C6H2(NO2)3OH(l ) ? 3 N2(g) 12CO2(g) 6H2O(l )
?? f (kJ/mol) 0.00 3862.94 0.00 393.5 241.83
?H 12 mol(393.5 kJ/mol) 6 mol(241.83
kJ/mol) 6 mol(0.00 kJ/mol) 8 mol(0.00
kJ/mol) 2 mol(3862.94 kJ/mol)
?H 13,898.9 kJ
?? ?H ?ngasRT ?H (15 8) mol
298 K 8.314 103 kJ/(mol K)
?? 13,898.9 kJ 29.0 kJ 13,927.9 kJ
16
Your Turn!

Given the following
3H2(g) N2(g) ? 2NH3(g) ?? 46.19 kJ
mol1
Determine ?E for the reaction.
A. 51.14 kJ mol1
B. 41.23 kJ mol1
C. 46.19 kJ mol1
D. 46.60 kJ mol1
?? ?E ?nRT ?E ?? ?nRT
?E 46.19 kJ mol
(2 mol)(8.314 J K1mol1)(298 K)(1 kJ/1000 J)
?E 51.14 kJ mol1

17
Enthalpy Changes and Spontaneity

What are relationships among factors that
influence spontaneity?
Spontaneous Change
Occurs by itself
Without outside
assistance until finished
e.g.
Water flowing over waterfall
Melting of ice cubes in glass on warm day

18
Nonspontaneous Change

Occurs only with outside assistance
Never occurs by itself
Room gets straightened up
Pile of bricks turns into a brick wall
Decomposition of H2O by electrolysis
Continues only as long as outside assistance
occurs
Person does work to clean up room
Bricklayer layers mortar and bricks
Electric current passed through H2O

19
Nonspontaneous Change

Occur only when accompanied by some spontaneous
change
You consume food, spontaneous biochemical
reactions occur to supply muscle power
to tidy up room or
to build wall
Spontaneous mechanical or chemical change to
generate electricity

20
Direction of Spontaneous Change

Many reactions which occur spontaneously are
exothermic
Iron rusting
Fuel burning
?H and ?E are negative
Heat given off
Energy leaving system
Thus, ?H is one factor that influences spontaneity

21
Direction of Spontaneous Change

Some endothermic reactions occur spontaneously
Ice melting
Evaporation of water
Expansion of CO2 gas into vacuum
?H and ?E are positive
Heat absorbed
Energy entering system
Clearly other factors influence spontaneity

22
Your Turn!

We can expect the combustion of propane
to be
A. spontaneous
B. non-spontaneous
C. neither

23
Entropy (Symbol S )

Thermodynamic quantity
Describes number of equivalent ways that energy
can be distributed
Quantity that describes randomness of system
Greater statistical probability of particular
state means greater the entropy!
Larger S, means more possible ways to distribute
energy and that it is a more probable result

24
Fig. 19.6 - Entropy Distribution

Low Entropy (a)
A absorbs E in units of 10
Few ways to distribute E
? represent Es of molecules of A

High Entropy (b)
More ways to distribute E
B absorbs E in units of 5
?represent Es of molecules of B

25
Entropy

If Energy money
Entropy (S ) describes number of different ways
of counting it

26
Examples of Spontaneity

Spontaneous reactions
Things get rusty spontaneously
Don't get shiny again
Sugar dissolves in coffee
Stir moreit doesn't undissolve
Ice ?? liquid water at RT
Opposite does NOT occur
Fire burns wood, smoke goes up chimney
Can't regenerate wood
Common factor in all of these
Increase in randomness and disorder of system
Something that brings about randomness more
likely to occur than something that brings order

27
Entropy, S

State function
Independent of path
?S Change in entropy
For chemical reactions or physical processes

28
Effect of Volume on Entropy

For gases, entropy increases as volume increases
Gas separated from vacuum by partition
Partition removed, more ways to distribute energy
Gas expands to achieve more probable particle
distribution
More random, higher probability, more positive S

29
Effect of Temperature on Entropy

As T increases, entropy increases
(a) T 0 K, particles in equilibrium lattice
positions and S relatively low
(b) T gt 0 K, molecules vibrate, S increases
(c) T increases further, more violent vibrations
occur and S higher than in (b)

30
Effect of Physical State on Entropy

Crystalline solid very low entropy
Liquid higher entropy, molecules can move freely
More ways to distribute KE among them
Gas highest entropy, particles randomly
distributed throughout container
Many, many ways to distribute KE

31
Entropy Affected by Number of Particles

Adding particles to system
Increase number of ways energy can be
distributed in system
So all other things being equal
Reaction that produces more particles will have
positive ?S

32
Your Turn!

Which represents an increase in entropy?
A. Water vapor condensing to liquid
B. Carbon dioxide subliming
C. Liquefying helium gas
D. Proteins forming from amino acids

33
Entropy Changes in Chemical Reactions

Reactions without gases
Calculate number of mole molecules
?n nproducts nreactants
If ?n is positive, entropy increases
More molecules, means more disorder
Usually the side with more molecules, has less
complex molecules

Reactions involving gases
Calculate change in number of moles of gas, ?ngas
If ?ngas is positive , ?S is positive
?ngas is more important than ?nmolecules

34
Entropy Changes in Chemical Reactions

Ex. N2(g) 3H2(g) ?? 2NH3(g)
nreactant 4 nproduct 2
?n 2 4 2
Predict ?Srxn lt 0

Lower positional probability
Higher positional probability
35
Ex. 2 Predict Sign of ?S for Following Reactions

CaCO3(s) 2H(aq) ? Ca2(aq) H2O CO2(g)
?ngas 1 mol 0 mol 1 mol
since ?ngas is positive, ?S is positive
2 N2O5(g) ?? 4 NO2(g) O2(g)
?ngas 4 mol 1 mol 2 mol 3 mol
since ?ngas is positive, ?S is positive
OH(aq) H(aq) ?? H2O
?ngas 0 mol
?n 1 mol 2 mol 1 mol
since ?ngas is negative, ?S is negative

36
Predict Sign of ?S in the Following

Dry ice ? carbon dioxide gas
Moisture condenses on a cool window
AB ? A B
A drop of food coloring added to a glass of
water disperses
2Al(s) 3Br2(l ) ? 2AlBr3(s)

CO2(s) ? CO2(g)
positive
H2O(g) ? H2O(l )
negative
positive
positive
negative
37
Your Turn!

Which of the following has the most entropy at
standard conditions?
H2O(l )
NaCl(aq)
AlCl3(s)
Cant tell from the information

38
Your Turn!

Which reaction would have a negative entropy?
A. Ag(aq) Cl(aq) ? AgCl(s)
B. N2O4(g) ? 2NO2(g)
C. C8H18(l ) 25/2 O2(g) ? 8CO2(g) 9H2O(g)
D. CaCO3(s) ? CaO(s) CO2(g)

39
Both Entropy and Enthalpy Affect Reaction
Spontaneity

Sometimes they work together
Building collapses
PE decreases ?H is negative
Stones disordered ?S is positive
Sometimes work against each other
Ice melting (ice/water mix)
Endothermic
?H is positive nonspontaneous
Increase in disorder of molecules
?S is positive spontaneous

40
Which Prevails?

Hard to telldepends on temperature!
At 25 C, ice melts
At 25 C, water freezes
So three factors affect spontaneity
?H
?S
T
Next few slides will develop the relationship
between ?H, ?S, and T that defines a spontaneous
process

41
Second Law of Thermodynamics

When a spontaneous event occurs, total entropy of
universe increases
(?Stotal gt 0)
In a spontaneous process, ?Ssystem can decrease
as long as total entropy of universe increases
?Stotal ?Ssystem ?Ssurroundings
It can be shown that

42
Law of Conservation of Energy
Spontaneous Reactions (cont.)

Says q lost by system must be gained by
surroundings
qsurroundings qsystem
If system at constant P, then
qsystem ?H
So
qsurroundings ?Hsystem
and

43
Thus Entropy for Entire Universe is

Multiplying both sides by T we get
T?Stotal T?Ssystem ?Hsystem
or
T?Stotal (?Hsystem T?Ssystem)
For reaction to be spontaneous
T?Stotal gt 0 (entropy must increase)
So,
(?Hsystem T?Ssystem) lt 0
must be negative for reaction to be spontaneous

44
Gibbs Free Energy

Would like one quantity that includes all three
factors that affect spontaneity of a reaction
Define new state function, G
Gibbs Free Energy
Maximum energy in reaction that is "free" or
available to do useful work
G ? H TS
At constant P and T, changes in free energy ?G
?H T?S

45
Gibbs Free Energy

?G ?H T?S
G ? state function
Made up of T, H and S state functions
Has units of energy
Extensive property
?G Gfinal Ginitial

?G lt 0 Spontaneous process
?G 0 At equilibrium
?G gt 0 Nonspontaneous
46
Criteria for Spontaneity?

At constant P and T, process spontaneous only if
it is accompanied by decrease in free energy of
system

?H ?S Spontaneous?
?G () T () Always, regardless of T
?G () T () Never, regardless of T
?G () T () ? Depends spontaneous at high T, ?G
?G () T () ? Depends spontaneous at low T, ?G
47
Summary

When ?H and ?S have same sign, T determines
whether spontaneous or nonspontaneous
Temperature-controlled reactions are spontaneous
at one temperature and not at another

48
Your Turn!

At what temperature (K) will a reaction become
nonspontaneous when ?H 50.2 kJ mol1 and ?S
20.5 J K1 mol1?
A. 298 K
B. 1200 K
C. 2448 K
D. The reaction cannot become non-spontaneous at
any temperature

49
Third Law of Thermodynamics

At absolute zero (0 K)
Entropy of perfectly ordered, pure crystalline
substance is zero
S 0 at T 0 K
Since S 0 at T 0 K
Define absolute entropy of substance at higher
temperatures
Standard entropy, S
Entropy of 1 mole of substance at 298 K (25 C)
and 1 atm pressure
S ?S for warming substance from 0 K to 298 K
(25 C)

50
Consequences of Third Law

All substances have positive entropies as they
are more disordered than at 0 K
Heating increases randomness
S is biggest for gasesmost disordered
For elements in their standard states
S ? 0 (but ?Hf 0)
Units of S ? J/(mol K)
Standard Entropy Change
To calculate ?S for reaction, do Hess's Law
type calculation
Use S rather than entropies of formation

51
Learning Check

Calculate ?S for the following
CO2(s) ? CO2(g)
S 187.6 213.7 J/mol K
?S (213.7 187.6) J/mol K
?S 26.1 J/mol K
CaCO3(s) ? CO2(g) CaO(s)
S 92.9 213.7 40 J/mol K
?S (213.7 40 92.9) J/mol K
?S 161 J/mol K

Ex. 3. Calculate ?S for reduction of aluminum
oxide by hydrogen gas
Al2O3(s) 3H2(g) ? 2Al(s) 3H2O(g)

Substance S (J/ K mol)
Al(s) 28.3
Al2O3(s) 51.00
H2(g) 130.6
H2O(g) 188.7
53
Ex. 3 Calculate ?S (cont.)

?S 56.5 J/K 566.1 J/K (51.00 J/K
391.8 )
?S 179.9 J/K

54
Your Turn!

What is the entropy change for the following
reaction?
Ag(aq) Cl(aq) ? AgCl(s)
So 72.68 56.5 96.2 J K1 mol1
A. 32.88 J K1 mol1
B. 32.88 J K1 mol1
C. 32.88 J mol1
D. 112.38 J K1 mol1
?S 96.2 (72.68 56.5) J K1 mol1
?S 32.88 J K1 mol1

55
Standard Free Energy Changes

Standard Free Energy Change, ?G
?G measured at 25 C (298 K) and 1 atm
Two ways to calculate, depending on what data is
available
Method 1
?G ?H T?S
Method 2

56
Ex. 4. Calculate ?G Method 1

Calculate ?G for reduction of aluminum oxide by
hydrogen gas

Al2O3(s) 3H2(g) ? 2Al(s)
3H2O(g)
Step 1 Calculate ?H for reaction using heats
of formation below

Substance (kJ/mol)
Al(s) 0.0
Al2O3(s) 1669.8
H2(g) 0.0
H2O(g) 241.8
57
Ex. 4. Calculate ?G Method 1

?H 0.0 kJ 725.4 kJ 0.00 kJ ( 1669.8
kJ)
?H 944.4 kJ

58
Ex. 4. Calculate ?G Method 1

Step 2 Calculate ?S ? see Example 3
?S 179.9 J/K
Step 3 Calculate ?G ?H (298.15 K)?S
?G 944.4 kJ (298 K)(179.9 J/K)(1 kJ/1000 J)
?G 944.4 kJ 53.6 kJ 890.8 kJ
?G is positive
Indicates that the reaction is not spontaneous

59
Ex. 4. Calculate ?G Method 2

Use Standard Free Energies of Formation
Energy to form 1 mole of substance from its
elements in their standard states at 1 atm and 25
C

60
Ex. 4. Calculate ?G Method 2

Calculate ?G for reduction of aluminum oxide by
hydrogen gas.
Al2O3(s) 3H2(g) ? 2Al(s) 3H2O(g)

Substance (kJ/mol)
Al(s) 0.0
Al2O3(s) 1576.4
H2(g) 0.0
H2O(g) 228.6
61
Ex. 4. Calculate ?G Method 2
?G 0.0 kJ 685.8 kJ 0.00 kJ (
1576.4 kJ) ?G 890.6 kJ
Both methods same within experimental error
62
Spontaneous Reactions Produce Useful Work

Fuels burned in engines to power cars or heavy
machinery
Chemical reactions in batteries
Start cars
Run cellular phones, laptop computers, mp3
players
Energy not harnessed if reaction run in an open
dish
All energy lost as heat to surroundings
Engineers seek to capture energy to do work
Maximize efficiency with which chemical energy is
converted to work
Minimize amount of energy transformed to
unproductive heat

63
Thermodynamically Reversible

Process that can be reversed and is always very
close to equilibrium
Change in quantities is infinitesimally small
Example - expansion of gas
Done reversibly, it does most work on surroundings

64
?G Maximum Possible Work

?G is maximum amount of energy produced during
a reaction that can theoretically be harnessed as
work
Amount of work if reaction done under reversible
conditions
Energy that need not be lost to surroundings as
heat
Energy that is free or available to do work

65
Ex. 5 Calculate ?G

Calculate ?G for reaction below at 1 atm and 25
C, given ?H 246.1 kJ/mol, ?S 377.1
J/(mol K).
H2C2O4(s) ½O2(g) ? 2CO2(g) H2O(l )
?G 25 ?H T?S
?G (246.1 112.4) kJ/mol
?G 358.5 kJ/mol

66
Your Turn!

Calculate ?G for the following reaction,
H2O2(l ) ? H2O(l ) O2(g)
given
?H 196.8 kJ mol1 and ?S 125.72 J K1
mol1.
A. 234.3 kJ mol1
B. 234.3 kJ mol1
C. 199.9 kJ mol1
D. 3.7 105 kJ mol1
?G 196.8 kJ mol1 298 K (0.12572 kJ K1
mol-1)
?G 234.3 kJ mol

67
?G and Position of Equilibrium

When ?G gt 0 (positive)
Position of equilibrium lies close to reactants
Little reaction occurs by the time equilibrium is
reached
Reaction appears nonspontaneous
When ?G lt 0 (negative)
Position of equilibrium lies close to products
Mainly products exist by the time equilibrium is
reached
Reaction appears spontaneous

68
?G and Position of Equilibrium

When ?G 0
Position of equilibrium lies halfway between
products and reactants
Significant amount of both reactants and products
present at time equilibrium is reached
Reaction appears spontaneous, whether start with
reactants or products
Can Use ?G to Determine Reaction Outcome
?G large and positive
No observable reaction occurs
?G large and negative
Reaction goes to completion

69
At Equilibrium

?G RT lnK and K e?G /RT
Provides connection between ?G and K
Can estimate K at various temperatures if ?G is
known
Can get ?G if K is known
Relationship between K and ?G

Keq ?G Reaction
gt 1 Spontaneous Favored Energy released
lt 1 non-spontaneous Unfavorable Energy needed
1 0 At Equilibrium
70
Learning Check

Ex. 6 Given that ?H 97.6 kJ/mol, ?S
122 J/(mol K), at 1 atm and 298 K, will the
following reaction occur spontaneously?
MgO(s) 2HCl(g) ? H2O(l ) MgCl2(s)
?G ?H T?S
97.6 kJ/mol 298 K(0.122 kJ/mol K)
?G 97.6 kJ/mol 36.4 kJ/mol
61.2 kJ/mol

71
Effect of Change in Pressure or Concentration on
?G

?G at nonstandard conditions is related to ?G
at standard conditions by an expression that
includes reaction quotient Q
This important expression allows for any
concentration or pressure
Recall

72
Ex. 7 Calculating ?G at Nonstandard Conditions

Calculate ?G at 298 K for the Haber process
N2(g) 3H2(g) ? 2NH3(g) ?G 33.3 kJ
For a reaction mixture that consists of 1.0 atm
N2, 3.0 atm H2 and 0.5 atm NH3
Step 1 Calculate Q

73
Ex. 7 Calculating ?G at Nonstandard conditions

Step 2 Calculate ?G ?G RT lnQ
?G 33.3 kJ/mol (8.314 J/K mol)(1
kJ/1000J)(298K)(ln(9.3 ? 103))
33.3 kJ/mol (2.479 kJ/mol)(ln(9.3 ?
103))
33.3 kJ (11.6 kJ/mol) 44.9 kJ/mol
At standard conditions all gases (N2, H2 and NH3)
are at 1 atm of pressure
?G becomes more negative when we go to 1.0 atm
N2, 3.0 atm H2 and 0.5 atm NH3
Indicates larger driving force to form NH3
Preactants gt Pproducts

74
Free Energy Diagrams

?G is different than ?G
?G interpreted as the slope of the free energy
curve and tells which direction reaction proceeds
to equilibrium
Minimum, at ?G 0, indicates composition of
equilibrium mixture
Because ?G is positive equilibrium lies closer
to reactants

Non-spont. ?G gt 0
N2O4(g) ? 2NO2(g)
Equilibrium occurs here at Ptotal 1 atm with
16.6 N2O4 decomposed
75
Free Energy Diagrams
Spontaneous Rxn ?G lt 0

Reaction proceeds toward equilibrium from either
A or B where slope, ?G , is not zero
The curve minimum, where ?G 0, lies closer to
the product side for spontaneous reactions. This
is determined by ?G

76
How K is related to ?G

Use relation ?G ?G RT lnQ to derive
relationship between K and ?G
At Equilibrium
?G 0 and Q K
So 0 ?G RT lnK
?G RT lnK
Taking antilog (ex) of both sides gives
K e?G /RT

77
Ex. 8 Calculating ?G from K

Ksp for AgCl(s) at 25 C is 1.8 ? 1010 Determine
?G for the process
Ag(aq) Cl(aq) ? AgCl(s)
Reverse of Ksp equation, so
?G RT lnK
(8.3145 J/K mol)(298 K) ln(5.6 ? 109)(1
kJ/1000 J)
?G 56 kJ/mol
Negative ?G indicates precipitation will occur

78
System at Equilibrium

Neither spontaneous nor nonspontaneous
In state of dynamic equilibrium
Gproducts Greactants
?G 0
Consider freezing of water at 0C
H2O(l ) ? H2O(s)
System remains at equilibrium as long as no heat
added or removed
Both phases can exist together indefinitely
Below 0 C, ?G lt 0 freezing spontaneous
Above 0 C, ?G gt 0 freezing nonspontaneous

79
No Work Done at Equilibrium

?G 0
No free energy available to do work
Consider fully charged battery
Initially
All reactants, no products
?G large and negative
Lots of energy available to do work
As battery discharges
Reactants converted to products
?G less negative
Less energy available to do work
At equilibrium
?G Gproducts Greactants 0
No further work can be done
Dead battery

80
Phase Change Equilibrium

H2O(l ) ? H2O(g)
?G 0 ?H T?S
Only one temperature possible for phase change at
equilibrium
Solid-liquid equilibrium
Melting/freezing temperature (point)
Liquid-vapor equilibrium
Boiling temperature (point)
Thus ?H T?S and
or

81
Ex. 9 Calculate Tbp

Calculate Tbp for reaction below at 1 atm and 25
C, given ?H 31.0 kJ/mol, ?S 92.9 J/ mol
K
Br2(l ) ? Br2(g)
For T gt 334 K, ?G lt 0 and reaction is spontaneous
(?S dominates)
For T lt 334 K, ?G gt 0 and reaction is
nonspontaneous (?H dominates)
For T 334 K, ?G 0 and T normal boiling point

82
Learning Check

What is the expected melting point for Cu?

?Hf (kJ/mol) ?Gf (kJ/mol) S (J/mol K)
Cu(l ) 341.1 301.4 166.29
Cu(s) 0 0 33.1
Cu(s) ? Cu(l )
?H 1mol(341.1 kJ/mol 1mol(0 kJ/mol)
?H 341.1kJ
?S 1mol(166.29 J/mol K 1mol(33.1 J/mol K)
?S 133.19 J/K
83
Effect of Temperature on ?G

Reactions often run at temperatures other that
298 K
Position of equilibrium can change as ?G
depends on temperature
?G ?H T?S
For temperatures near 298 K, expect only very
small changes in ?H and ?S
For reaction at temperature, we can write

84
Ex. 10 Determining Effect of Temperature on
Spontaneity

Calculate ?G at 25 C and 500 C for the Haber
process
N2(g) 3H2(g) ? 2NH3(g)
Assume that ?H and ?S do not change with
temperature
Solving strategy
Step 1. Using data in Tables 6.2 and 18.2
calculate ?H and ?S for the reaction at 25 C
?H 92.38 kJ
?S 198.4 J/K

85
Ex. 10 Determining Effect of Temperature on
Spontaneity

Step 2. Calculate ?G for the reaction at 25 C
using ?H and ?S
N2(g) 3H2(g) ? 2NH3(g)
?H 92.38 kJ
?S 198.4 J/K
?G ?H T?S
?G 92.38 kJ (298 K)(198.4 J/K)
?G 92.38 kJ 59.1 kJ 33.3 kJ
So the reaction is spontaneous at 25 C

86
Ex. 10 Determining Effect of Temperature on
Spontaneity

Step 3. Calculate ?G for the reaction at 500
C using ?H and ?S .
T 500 C 273 773 K
?H 92.38 kJ
?S 198.4 J/K
?G ?H T?S
?G 92.38 kJ (773 K)(198.4 J/K)
?G 92.38 kJ 153 kJ 61 kJ
So the reaction is NOT spontaneous at 500 C

87
Ex. 10 Does this answer make sense?

?G ?H T?S
?H 92.38 kJ
?S 198.4 J/K
Since both ?H and ?S are negative
At low temperature
?G will be negative and spontaneous
At high temperature
T?S will become a bigger positive number and
?G will become more positive and thus
eventually, at high enough temperature, will
become nonspontaneous

88
Ex. 11 Calculating K from ?G

Calculate K at 25 C for the Haber process
N2(g) 3H2(g) ? 2NH3(g)
?G 33.3 kJ/mol 33,300 J/mol
Step 1 Solve for exponent
Step 2 Take e x to obtain K
Large K indicates NH3 favored at room temp.

89
Your Turn!

Calculate the equilibrium constant for the
decomposition of hydrogen peroxide at 298 K given
?G 234.3 kJ mol.
A. 8.5 10-42
B. 1.0 10499
C. 3.4 10489
D. 1.17 1041

90
Ex. 12 Calculating K from ?G, First Calculate ?G

Calculate the equilibrium constant at 25 C for
the decarboxylation of liquid pyruvic acid to
form gaseous acetaldehyde and CO2.

91
Ex. 12 First Calculate ?G from ?Gf
Compound ?Gf, kJ/mol
CH3COH 133.30
CH3COCOOH 463.38
CO2 394.36
92
Ex. 12 Next Calculate Equilibrium Constant
K 1.85 ? 1011
93
Temperature Dependence of K

?G RT lnK ?H T?S
Rearranging gives
Equation for line
Slope ?H /RT
Intercept ?S /R
Also way to determine K if you know ?H and ?S

94
Ex. 12 Calculate K given ?H and ?S

Calculate K at 500 C for Haber process
N2(g) 3H2(g) ? 2NH3(g)
Given ?H 92.38 kJ and ?S 198.4 J/K
Assume that ?H and ?S do not change with T
ln K 14.37 23.86 9.49
K e9.49 7.56 105

95
Bond Energy

Amount of energy needed to break chemical bond
into electrically neutral fragments
Useful to know
Within reaction
Bonds of reactants broken
New bonds formed as products appear
Bond breaking
First step in most reactions
One of the factors that determines reaction rate
e.g. N2 very unreactive due to strong N?N bond

96
Bond Energies

Can be determined spectroscopically for simple
diatomic molecules
H2, O2, Cl2
More complex molecules, calculate using
thermochemical data and Hesss Law
Use ?H formation enthalpy of formation
Need to define new term
Enthalpy of atomization or atomization energy,
?Hatom
Energy required to rupture chemical bonds of one
mole of gaseous molecules to give gaseous atoms

97
Determining Bond Energies