CHEM 160 General Chemistry II Lecture Presentation Chemical Thermodynamics

1
CHEM 160 General Chemistry IILecture
PresentationChemical Thermodynamics

• Chapter 19

2
Thermodynamics

• Thermodynamics
• study of energy and its transformations
• heat and energy flow
• Helps determine natural direction of reactions
• Allows us to predict if a chemical process will
  occur under a given set of conditions
• Organized around three fundamental laws of nature
• 1st law of thermodynamics
• 2nd law of thermodynamics
• 3rd law of thermodynamics

3
First Law of Thermodynamics

• 1st Law
• Energy can be neither created nor destroyed
• Energy of the universe is constant
• Important concepts from thermochemistry
• Enthalpy
• Hesss law
• Purpose of 1st Law
• Energy bookkeeping
• How much energy?
• Exothermic or endothernic?
• What type of energy?

4
Spontaneous vs Nonspontaneous

• Spontaneous process
• Occurs without outside assistance in the form of
  energy (occurs naturally)
• Product-favored at equilibrium
• May be fast or slow
• May be influenced by temperature
• Nonspontaneous process
• Does not occur without outside assistance
• Reactant-favored at equilibrium
• All processes which are spontaneous in one
  direction cannot be spontaneous in the reverse
  direction
• Spontaneous processes have a definite direction
• Spontaneous processes are irreversible

5
Spontaneous vs. Nonspontaneous

• Spontaneous Processes
• Gases expand into larger volumes
• H2O(s) melts above 0?C
• H2O(l) freezes below 0?C
• NH4NO3 dissolves spontaneously in H2O
• Steel (iron) rusts in presence of O2 and H2O
• Wood burns to form CO2 and H2O
• CH4 gas burns to form CO2 and H2O
• Reverse processes are not spontaneous!
• nonspontaneous

6
Spontaneous vs. Nonspontaneous

• Spontaneous Processes
• H2O(s) melts above 0?C (endothermic)
• NH4NO3 dissolves spontaneously in H2O
  (endothermic)
• Steel (iron) rusts in presence of O2 and H2O
  (exothermic)
• Wood burns to form CO2 and H2O (exothermic)
• CH4 gas burns to form CO2 and H2O (exothermic)
• Heat change alone is not enough to predict
  spontaneity because energy is conserved
• Many, but not all, spontaneous processes are
  exothermic

7
Factors That Favor Spontaneity

• Two thermodynamic properties of a system are
  considered when determining spontaneity
• Enthalpy, H
• Many, but not all, spontaneous processes tend to
  be exothermic as already noted
• Entropy, S (J/K)
• Measure of the disorder of a system
• Many, but not all, spontaneous processes tend to
  increase disorder of the system

8
Entropy

• Entropy, S (J/K)
• describes of ways the particles in a system can
  be arranged in a given state
• More arrangements greater entropy
• S heat change/T q/T
• ?S Sfinal - Sinitial
• ? S gt 0 represents increased randomness or
  disorder

9
Patterns of Entropy Change

• For the same or similar substances Ssolid lt
  Sliquid lt Sgas

solid
vapor
liquid
10
Patterns of Entropy Change
Particles farther apart, occupy larger volume of
space even more positions available to particles
Rigidly held particles few positions available
to particles
Particles free to flow more positions available
for particles
solid
vapor
liquid
11
Patterns of Entropy Change
least ordered
less ordered
most ordered
solid
vapor
liquid
12
Patterns of Entropy Change

• Solution formation usually leads to increased
  entropy

particles more disordered
solvent
solute
solution
13
Patterns of Entropy Change

• Chemical Reactions
• If more gas molecules produced than consumed, S
  increases. (?Srxn gt 0)
• If only solids, ions and/or liquids involved, S
  increases if total particles increases.

14
fewer arrangements possible so lower entropy
more arrangements possible so higher entropy
15
Patterns of Entropy Change

• Increasing temperature increases entropy

System at T1 (S1)
System at T2 (T2 gt T1) (S2 gt S1)
16
Patterns of Entropy Change
more energetic molecular motions
less energetic molecular motions
System at T1 (S1)
System at T2 (T2 gt T1) (S2 gt S1)
17
Third Law of Thermodynamics

• If increasing T increases S, then the opposite
  should be true also.
• Is it possible to decrease T to the point that S
  is zero?
• At what T does S 0?
• If entropy is zero, what does that mean?

18
Third Law of Thermodynamics

• 3rd Law of Thermodynamics
• Entropy of a perfect crystalline substance at 0 K
  is zero
• No entropy highest order possible
• Why?
• Purpose of 3rd Law
• Allows S to be measured for substances
• S 0 at 0 K
• S heat change/temperature q/T
• S? standard molar entropy

19
50
40
Standard entropy, S (J/K)
30
20
10
0
100
50
250
300
150
200
Temperature (K)
20
50
Gas
Liquid
40
Standard entropy, S (J/K)
30
20
Solid
10
0
100
50
250
300
150
200
Temperature (K)
21
Standard Molar Entropies of Selected Substances
at 298 K
22
Entropy versus Probability

• Systems tend to move spontaneously towards
  increased entropy. Why?
• Entropy is related to probability
• Disordered states are more probable than ordered
  states
• S k(lnW)
• k (Boltzmans constant) 1.38 x 10-23 J/K
• W possible arrangements in system

23
Isothermal Gas Expansion
Consider why gases tend to isothermally expand
into larger volumes.
Gas Container two bulbed flask
Ordered State
Gas Molecules
24
Isothermal Gas Expansion
Gas Container
Ordered State
S k(ln 1) (1.38 x 10-23 J/K)(0) 0 J/K
25
(No Transcript)
26
(No Transcript)
27
(No Transcript)
28
(No Transcript)
29
(No Transcript)
30
Disordered States
31
Disordered States
32
Disordered States
More probable that the gas molecules will
disperse between two halves than remain on one
side
33
Disordered States
Driving force for expansion is entropy
(probability) gas molecules have a tendency to
spread out
34
Disordered States
S k(ln 7) (1.38 x 10-23 J/K)(1.95) 2.7 x
10-23 J/K
35
Total Arrangements
Stotal k(ln 23) k(ln 8) (1.38 x 10-23
J/K)(1.79) 2.9 x 10-23 J/K
36
2nd Law of Thermodynamics

• 2nd Law of Thermodynamics
• The entropy of the universe increases in any
  spontaneous process (?Suniv gt 0)
• Increased disorder in the universe is the driving
  force for spontaneity

37
2nd Law of Thermodynamics

• 2nd Law of Thermodynamics
• ?Suniv ?Ssysyem ?Ssurroundings
• ?Suniv gt 0, process is spontaneous
• ?Suniv 0, process at equilibrium
• ?Suniv lt 0, process is nonspontaneous
• Process is spontaneous in reverse direction

38
2nd Law of Thermodynamics

• Spontaneous Processes
• H2O(s) melts above 0?C (endothermic)
• Steel (iron) rusts in presence of O2 and H2O
  (exothermic)
• 4Fe(s) 3O2(g) ? 2Fe2O3(s)
• CH4 gas burns to form CO2 and H2O (exothermic)
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g)
• Each process increases Suniverse.

39
2nd Law of Thermodynamics

• To determine ?Suniv for a process, both ?Ssystem
  and ?Ssurroundings need to be known
• ?Ssysyem
• related to matter dispersal in system
• ?Ssurroundings
• determined by heat exchange between system and
  surroundings and T at which it occurs
• Sign of ?Ssurroundings depends on whether process
  in system is endothermic or exothermic
• Why?
• Magnitude of ?Ssurroundings depends on T
• ?Ssurroundings ?-?Hsystem/T

40
Entropy Changes in a System

• For any reaction
• ?Srxn ?nS(products) - ?mS(reactants)
• Where n and m are stoichiometric coefficients

41
Example 1(1 on Example Problem Handout)

• Using standard molar entropies, calculate ?Srxn
  for the following reaction at 25C
• 2SO2(g) O2(g) --gt 2SO3(g)
• S 248.1 205.1 256.6
  (J K-1mol-1)
• (Ans. -187.9 J/K)

42
Entropy Changes in the Surroundings
Surroundings
System
Heat
Exothermic Reaction (DHrxn lt 0)
43
Entropy Changes in the Surroundings
Surroundings
DSsurr gt 0
44
Entropy Changes in the Surroundings
Surroundings
System
Heat
Endothermic Reaction (DHrxn gt 0)
45
Entropy Changes in the Surroundings
DSsurr µ -DHrxn
DSsurr lt 0
46
Entropy Changes in the Surroundings
High T DSsurr relatively small
surroundings
Low T DSsurr relatively high
DSsurr -DHrxn/T
DSsurr gt 0
47
2nd Law of Thermodynamics

• DSuniv DSsys DSsurr DSsys - DHsys/T

entropy change due to heat flow
created entropy due to matter dispersal
48
Example 2

• Calculate DSuniv for the synthesis of ammonia at
  25C if DSrxn -199 J/K and DHrxn -92.6 kJ.

49
Gibbs Free Energy, G

• 2nd law ?Suniv ?Ssys ?Ssurr ?Ssys -
  Hsys/T
• Rearrange (multiply by -T)
• -T?Suniv -T?Ssys ?Hsys ?Hsys - T?Ssys
• -T?Suniv ?G (-TSuniv G)
• G Gibbs free energy (J or kJ)
• G H - TS
• and
• ?G ?Hsys - T?Ssys
• ?G ?Hsys - T?Ssy (if at standard state)

50
Gibbs Free Energy

• Summary of Conditions for Spontaneity
• ?G lt 0
• ?G gt 0
• ?G 0

51
Gibbs Free Energy

• Summary of Conditions for Spontaneity
• ?G lt 0
• reaction is spontaneous in the forward direction
• (?Suniv gt 0)
• ?G gt 0
• ?G 0

52
Gibbs Free Energy

• Summary of Conditions for Spontaneity
• ?G lt 0
• reaction is spontaneous in the forward direction
• (?Suniv gt 0)
• ?G gt 0
• reaction is nonspontaneous in the forward
  direction
• (?Suniv lt 0)
• ?G 0

53
Gibbs Free Energy

• Summary of Conditions for Spontaneity
• ?G lt 0
• reaction is spontaneous in the forward direction
• (?Suniv gt 0)
• ?G gt 0
• reaction is nonspontaneous in the forward
  direction
• (?Suniv lt 0)
• ?G 0
• system is at equilibrium
• (?Suniv 0)

54
Example 2(2 in Example Problem Handout)

• For a particular reaction, ?Hrxn 53 kJ and
  ?Srxn 115 J/K. Is this process spontaneous a)
  at 25C, and b) at 250C? (c) At what
  temperature does ?Grxn 0?
• (ans. a) ?G 18.7 kJ, nonspontaneous b) 7.1
  kJ, spontaneous c) 460.9 K or 188?C)

55
Gibbs Free Energy and Temperature

• ?G ?H - T?S
• ?S gt 0 ?H lt 0
• Spontaneous at all T
• ?S lt 0 ?H gt 0
• Not spontaneous at any T
• ?S gt 0 ?H gt 0
• Spontaneous at high T nonspontaneous at low T
• ?S lt 0 ?H lt 0
• Spontaneous at low T nonspontaneous at high T

56
Gibbs Free Energy and Temperature

• Spontaneous Processes
• H2O(s) melts above 0?C (endothermic)
• ?H gt 0, ?S gt 0
• Steel (iron) rusts in presence of O2 and H2O at
  25? C (exothermic)
• 4Fe(s) 3O2(g) ? 2Fe2O3(s)
• ?H lt 0, ?S lt 0
• ?G lt 0 for each process
• T determines sign of ?G ?G ?H -T?S

57
Calculating Free Energy Changes

• For any reaction
• ?Grxn ??Gf(products) - ?m?Gf(reactants)
• Where n and m are stoichiometric coefficients
  and ?Gf standard free energy of formation

58
Example 3(3 on Example Problem Handout)

• Calculate the standard free energy change for the
  reaction at 25C
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
• ??Gf -50.8 0 -394.4
  -237.4 (kJ/mol)

59
Gibbs Free Energy

• What does Gibbs free energy represent?
• For a spontaneous process
• Maximum amount of energy released by the system
  that can do useful work on the surroundings
• Energy available from spontaneous process that
  can be used to drive nonspontaneous process
• For a nonspontaneous process
• Minimum amount of work that must be done to force
  the process to occur
• In actuality, ?Gspont gt ?Gnonspont

60
Gibbs Free Energy

• Conversion of rust to iron
• 2Fe2O3 ? 4Fe 3O2 ??G 1487 kJ (NS)
• To convert iron to rust, G must be provided from
  spontaneous rxn
• 2Fe2O3 ? 4Fe 3O2 ??G 1487 kJ (NS)
• 6CO 3O2 ? 6CO2 ??G -1543 kJ (S)

61
Gibbs Free Energy

• Conversion of rust to iron
• 2Fe2O3 ? 4Fe 3O2 ??G 1487 kJ (NS)
• To convert iron to rust, G must be provided from
  spontaneous rxn
• 2Fe2O3 ? 4Fe 3O2 ?G 1487 kJ (NS)
• 6CO 3O2 ? 6CO2 ?G -1543 kJ (S)
• 2Fe2O3 6CO ? 4Fe 6CO2 ? ?G -56 kJ (S)
• Reactions are coupled

62
Gibbs Free Energy

• Many biological rxns essential for life are NS
• Spontaeous rxns used to drive the NS biological
  rxns
• Example photosynthesis
• 6CO2 6H2O ? C6H12O6 6O2 ? ?G gt 0
• What spontaneous rxns drive photosynthesis?

63
sun
big ball of G
G released
high free energy
DGsurr lt 0
proteins, cells etc., lower S
C6H12O6, O2
ATP
NS
Sp
NS
NS
Sp
CO2, H2O
amino acids, sugars, etc., higher S
ADP
photosynthesis
solar nuclear reactions
low free energy
64
Free Energy and Equilibrium

• ?G (Nonstandard State)
• ?G ?G RTlnQ
• R gas constant (8.314 J/Kmol)
• T temperature (K)
• Q reaction quotient

65
Example 4(4 on Example Problem Handout)

• Calculate ?Grxn for the reaction below
• 2A(aq) B(aq) ? C(aq) D(g)
• if ?Grxn 9.9 x 103 J/mol and (a) A 0.8 M,
  B 0.5 M, C 0.05 M, and PD 0.05 atm, and
  (b) A 0.1 M, B 1 M, C 0.5 M, and PD
  0.5 atm. Is the reaction spontaneous under these
  conditions?
• (ans. a) 2121 J, spontaneous b) 17875 J,
  nonspontaneous)

66
Free Energy and Equilibrium

• At equilibrium
• ?Grxn 0 and Q K
• 0 ?Grxn RTlnK
• ?Grxn -RTlnK

67
Example 5(5 on Example Problem Handout)

• Calculate ?Grxn for the ionization of acetic
  acid, HC2H3O2 (Ka 1.8 x 10-5) at 25C. Is this
  reaction spontaneous under standard state
  conditions?
• (ans. 27 kJ)

68
Example 6(6 on Example Problem Handout)

• Calculate ?G for the neutralization of a strong
  acid with a strong base at 25C. Is this process
  spontaneous under these conditions?
• For the reaction below, K 1.0 x 1014.
• H OH- ? H2O
• (ans. -80 kJ)

69
Example 7

• Calculate Keq for a reaction at (a) 25C and (b)
  250C if ?Hrxn 42.0 kJ and ?Srxn 125 J/K.
  At which temperature is this process product
  favored?
• (ans. a) 0.15 b) 216 250?C)

70
Spontaneous Reaction
DGrxn lt 0
Gproducts
Greactants
extent of reaction
71
Spontaneous Reaction
At equilibrium, any change requires an uphill
climb in energy, DGrxn gt 0
Gproducts
Greactants
Pure reactants
Pure products
extent of reaction
72
NonSpontaneous Reaction
DGrxn gt 0
equilibrium position
Gproducts
Greactants
Pure reactants
Pure products
extent of reaction
73
Calculating ?G for Processes

• ?G may be calculated in one of several ways
  depending on the information known about the
  process of interest
• ?G ?H - T?S ?G ?H - T?S
• ?Grxn ??Gf(products) - ?m?Gf(reactants)
• ?G ?G RTlnQ
• ?Grxn -RTlnK

74
Entropy and Life Processes

• If the 2nd law is valid, how is the existence of
  highly-ordered, sophisticated life forms
  possible?
• growth of a complex life form represents an
  increase in order (less randomness)
• lower entropy

75
Entropy and Life Processes
Organisms pay for their increased order by
increasing Ssurr.
Over lifetime, ?Suniv gt 0.
CO2
H2O
heat
76
sun
big ball of G
G released
high free energy
DGsurr lt 0
proteins, cells etc., lower S
C6H12O6, O2
ATP
NS
Sp
NS
NS
Sp
CO2, H2O
amino acids, sugars, etc., higher S
ADP
photosynthesis
solar nuclear reactions
low free energy