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

11

• 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

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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

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Entropy

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

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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

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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

52

• 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

• Ex. CH4(g) ? C(g) 4H(g)
• ?H atom energy needed to break all bonds in
  molecule
• ?H atom /4 average bond CH dissociation energy
  in methane
• D bond dissociation energy
• Average bond energy to required to break all
  bonds in molecule
• How do we calculate this?
• Use ?H f for forming gaseous atoms from
  elements in their standard states
• Hesss Law

98
Determining Bond Energies

• Path 1 Bottom
• Formation of CH4 from its elements ?Hf
• Path 2 Top 3 step path
• Step 1 break HH bonds
• Step 2 break CC bonds
• Step 3 form 4 CH bonds
• 2H2(g) ? 4H(g) ?H 1 4?H f (H,g)
• C(s) ? C(g) ?H 2 ?H f (C,g)
• 4H(g) C(g) ? CH4(g) ?H 3 ?H atom
• 2H2(g) C(s) ? CH4(g) ?H ?H f(CH4,g)

99
Calculating ?H atom and Bond Energy

• ?H f(CH4,g) 4?H f(H,g) ?H f(C,g) ?H atom
  
• Rearranging gives
• ?H atom 4?H f(H,g) ?H f(C,g) ?H f(CH4,g)
• Look these up in Table 18.3, 6.2 or appendix C
• ?H atom 4(217.9kJ/mol) 716.7kJ/mol
  (74.8kJ/mol)
• ?H atom 1663.1 kJ/mol of CH4

415.8 kJ/mol of CH bonds
100
Table 19.4 Some Bond Energies
101
Using Bond Energies to Estimate ?H f

• Calculate ?H f for CH3OH(g) (bottom reaction)
• Use four step path
• Step 1 break CC bonds
• Step 2 break HH bonds
• Step 3 break OO bond
• Step 4 form 4 CH bonds

102
Using Bond Energies

• ?H f(CH3OH,g)
• ?H f (C,g) 4?H f (H,g) ? H f (O,g)
  ?H atom (CH3OH,g)
• ?H f(C,g) 4?H f (H,g) ?H f (O,g) 716.7
  (4 217.9) 249.2 kJ 1837.5 kJ
• ?H atom (CH3OH,g) 3DCH DCO DOH (3
  412) 360 463 2059 kJ
• ?H f(CH3OH,g) 1837.5 kJ 2059 kJ 222 kJ
• Experimentally find ? H f (CH3OH,g) 201
  kJ/mol
• So bond energies give estimate within 10 of
  actual