Thermodynamics: Entropy, Free Energy, and Equilibrium

1
Chapter 16

• Thermodynamics Entropy,Free Energy,and
  Equilibrium

In this chapter we will determine the direction
of a chemical reaction and calculate equilibrium
constant using thermodynamic values Entropy and
Enthalpy.
2
Spontaneous Processes
Spontaneous Process A process that, once
started, proceeds on its own without a continuous
external influence.
3
Universe System Surroundings
The system is what you observe surroundings are
everything else.
4
Thermodynamics
State functions are properties that are
determined by the state of the system, regardless
of how that condition was achieved.
energy
, pressure, volume, temperature
Potential energy of hiker 1 and hiker 2 is the
same even though they took different paths.
5
Laws of Thermodynamics

• 1) Energy is neither created nor destroyed.
• 2) In any spontaneous process the total Entropy
  of a system and its surrounding always
  increases.
• We will prove later ?G ?Hsys T?Ssys
• ?G lt 0 Reaction is spontaneous in forward
  direction
• 3) The entropy of a perfect crystalline substance
  is zero at the absolute zero ( K0)

6
Enthalpy, Entropy, and Spontaneous Processes
State Function A function or property whose
value depends only on the present state, or
condition, of the system, not on the path used to
arrive at that state. Enthalpy Change (DH) The
heat change in a reaction or process at constant
pressure. Entropy (DS) The amount of Molecular
randomness change in a reaction or process at
constant pressure.
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Enthalpy Change
Exothermic
Endothermic
8
Entropy Change
DS Sfinal - Sinitial
9
Entropy
10
Entropy
11
The sign of entropy change, DS, associated with
the boiling of water is _______.

1. Positive
2. Negative
3. Zero

12
Correct Answer

1. Positive
2. Negative
3. Zero

Vaporization of a liquid to a gas leads to a
large increase in volume and hence entropy DS
must be positive.
13
Entropy and Temperature 02
14
Third Law of Thermodynamics
The entropy of a perfect crystalline substance is
zero at the absolute zero ( K0)
Ssolid lt Sliquid lt Sgas
15
Entropy Changes in the System (DSsys)
When gases are produced (or consumed)

• If a reaction produces more gas molecules than
  it consumes, DS0 gt 0.
• If the total number of gas molecules diminishes,
  DS0 lt 0.
• If there is no net change in the total number of
  gas molecules, then DS0 may be positive or
  negative .

The total number of gas molecules goes down, DS
is negative.
16
Standard Molar Entropies and Standard Entropies
of Reaction
Standard Molar Entropy (S) The entropy of 1
mole of a pure substance at 1 atm pressure and a
specified temperature.
Calculated by using S k ln W , W Accessible
microstates of translational, vibrational and
rotational motions
17
Review of Ch 8

• ?G ?H T?S see page 301 of your book

We will show the proof of this formula later in
this chapter

• Calculating DH for a reaction
• DH DHf (Products) DHf (Reactants)
• For a balanced equation, each heat of formation
  must be multiplied by the stoichiometric
  coefficient.
• aA bB cC dD
• DH cDHf (C) dDHf (D) aDHf (A)
  bDHf (B)

See Appendix B, end of your book
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Entropy Changes in the System (DSsys)
S0(CO) 197.6 J/Kmol
S0(CO2) 213.6 J/Kmol
S0(O2) 205.0 J/Kmol
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Calculating ?S for a Reaction
?So ? So (products) - ? So (reactants)

• Consider 2 H2(g) O2(g) ---gt 2 H2O(liq)
• ?So 2 So (H2O) - 2 So (H2) So (O2)
• ?So 2 mol (69.9 J/Kmol) - 2 mol (130.6
  J/Kmol) 1 mol (205.0 J/Kmol)
• ?So -326.4 J/K
• Note that there is a decrease in Entropy because
  3 mol of gas give 2 mol of liquid.

20
Calculate DS for the equation below using the
standard entropy data given 2 NO(g) O2(g) ? 2
NO2(g) DS values (J/mol-K) NO2(g) 240, NO(g)
211, O2(g) 205.

1. ?147 J/mol
2. ?76 J/mol

1. 176 J/mol
2. 147 J/mol

21
Correct Answer

1. ?147 J/K
2. ?76 J/K

1. 176 J/K
2. 147 J/K

(
)
(
)
?
?
?
?
D
?
D

?
D
reactants
products
S
S
S
DS 2(240) ? 2(211) 205 DS 480 ? 627
DS ? 147
22
Thermite Reaction
23
Spontaneous Processesand Entropy 07

• Consider the gas phase reaction of A2 molecules
  (red) with B atoms (blue).
• (a) Write a balanced equation for the reaction.
• (b) Predict the sign of ?S for the reaction.

24
Second Law Of Thermodynamic
In any spontaneous process the total Entropy of
a system and its surrounding always increases.
Entropy Changes in the Surroundings (DSsurr)
Exothermic Process DSsurr gt 0
Endothermic Process DSsurr lt 0
25
Entropy and the Second Law of Thermodynamics
Dssurr a ( - DH)
Dssurr a (
)
See example of tossing a rock into a calm waters
vs. rough waters Page 661 of your book
26
2nd Law of Thermodynamics 01

• The total entropy increases in a spontaneous
  process and remains unchanged in an equilibrium
  process.
• Spontaneous ?Stotal ?Ssys ?Ssur gt 0
• Equilibrium ?Stotal ?Ssys ?Ssur 0
• The system is what you observe surroundings are
  everything else.

27
Calculate change of entropy of surrounding in the
following reaction

• 2 H2(g) O2(g) ---gt 2 H2O(liq) ?H -571.7 kJ

?Sosurroundings 1917 J/K
28
Spontaneous Reactions 01

• The 2nd law tells us a process will be
  spontaneous if ?Stotal gt 0 which requires a
  knowledge of ?Ssurr.spontaneous ?Stotal
  ?Ssys ?Ssur gt 0

29
Gibbs Free Energy 02

• The expression T?Stotal is equated as Gibbs
  free energy change (?G), or simply free energy
  change
• ?G ?Hsys T?Ssys
• ?G lt 0 Reaction is spontaneous in forward
  direction.?G 0 Reaction is at equilibrium.?G
  gt 0 Reaction is spontaneous in reverse direction.

T?Stotal ?G
Driving force of a reaction, Maximum work you
can get from a system.
30
Calculate ?Go rxn for the following

• C2H2(g) 5/2 O2(g) --gt 2 CO2(g) H2O(g)
• Use enthalpies of formation to calculate
• ?Horxn -1256 kJ
• Use standard molar entropies to calculate ?Sorxn
  ( see page 658, appendix A-10)
• ?Sorxn -97.4 J/K or -0.0974 kJ/K
• ?Gorxn -1256 kJ - (298 K)(-0.0974 kJ/K)
• -1227 kJ
• Reaction is product-favored in spite of negative
  ?Sorxn.
• Reaction is enthalpy driven

31
Calculate DG for the equation below using the
thermodynamic data given N2(g) 3 H2(g) ? 2
NH3(g) DHf (NH3) ?46 kJ S values
(J/mol-K) are NH3 192.5, N2 191.5, H2
130.6.

1. 33 kJ
2. 66 kJ

1. ?66 kJ
2. ?33 kJ

32
Correct Answer

1. 33 kJ
2. 66 kJ
3. ?66 kJ
4. ?33 kJ

DH ?92 kJ DS ?198.3J/mol . K DG DH ?
TDS DG (?92) ? (298)(?0.1983) DG (?92)
(59.2) ?32.9 KJ
33
Gibbs Free Energy 03

• Using ?G ?H T?S, we can predict the sign of
  ?G from the sign of ?H and ?S.
• 1) If both ?H and ?S are positive,
• ?G will be negative only when the temperature
  value is large.
• Therefore, the reaction is spontaneous only at
  high temperature.
• 2) If ?H is positive and ?S is negative,
• ?G will always be positive.
• Therefore, the reaction is not spontaneous

34
Gibbs Free Energy 04
?G ?H T?S

• 3) If ?H is negative and ?S is positive,
• ?G will always be negative.
• Therefore, the reaction is spontaneous
• 4) If both ?H and ?S are negative,
• ?G will be negative only when the temperature
  value is small.
• Therefore, the reaction is spontaneous only at
  Low temperatures.

35
Gibbs Free Energy 04
36
Gibbs Free Energy 06

• What are the signs (, , or 0) of ?H, ?S, and ?G
  for the following spontaneous reaction of A atoms
  (red) and B atoms (blue)?

37
Gibbs Free Energy 02

• The expression T?Stotal is equated as Gibbs
  free energy change (?G), or simply free energy
  change
• ?G ?Hsys T?Ssys
• ?G lt 0 Reaction is spontaneous in forward
  direction.?G 0 Reaction is at equilibrium.?G
  gt 0 Reaction is spontaneous in reverse direction.

T?Stotal ?G
Driving force of a reaction, Maximum work you
can get from a system.
38
Standard Free-Energy Changes for Reactions
Calculate the standard free-energy change at 25
C for the Haber synthesis of ammonia using the
given values for the standard enthalpy and
standard entropy changes
DH -92.2 kJ
DS -198.7 J/K
DG DH - TDS
-92.2 kJ - 298 K
-33.0 kJ
x
x

39
Gibbs Free Energy 05

• Iron metal can be produced by reducing iron(III)
  oxide with hydrogen
• Fe2O3(s) 3 H2(g) ? 2 Fe(s) 3 H2O(g)
• ?H 98.8 kJ ?S 141.5 J/K
• Is this reaction spontaneous at 25C?
• At what temperature will the reaction become
  spontaneous?

40

• Decomposition of CaCO3 has a ?H 178.3 kJ/mol
  and ?S 159 J/mol ? K. At what temperature
  does this become spontaneous?

1. 121C
2. 395C
3. 848C
4. 1121C

41
Correct Answer

1. 121C
2. 395C
3. 848C
4. 1121C

T ?H/?S T 178.3 kJ/mol/0.159 kJ/mol ? K T
1121 K T (C) 1121 273 848
42
Standard Free Energies of Formation
DG DGf (products) - DGf (reactants)
DG cDGf (C) dDGf (D) - aDGf (A)
bDGf (B)
Reactants
Products
43
Standard free energy of formation (DG0f) is the
free-energy change that occurs when1 mole of the
compound is formed from its elements in their
standard ( 1 atm) states.
44
--
45
Calculate DG for the equation below using the
Gibbs free energy data given 2 SO2(g) O2(g) ?
2 SO3(g) DGf values (kJ) SO2(g) ?300.4,
SO3(g) ?370.4

1. 70 kJ
2. 140 kJ
3. ?140 kJ
4. ?70 kJ

46
Correct Answer

1. 70 kJ
2. 140 kJ
3. ?140 kJ
4. ?70 kJ

(
)
(
)
?
??




?
?
?
reactants
products
G
G
G
?
f
f
DG (2 ? ?370.4) ? (2 ? ?300.4) 0 DG
?740.8 ? ?600.8 ?140
47
Gibbs Free Energy 07
48
Calculation of Nonstandard ?G

• The sign of ?G tells the direction of
  spontaneous reaction when both reactants and
  products are present at standard state
  conditions.
• Under nonstandard( condition where pressure is
  not 1atm or concentrations of solutions are not
  1M) conditions, ?G becomes ?G. ?G ?G
  RT lnQ
• The reaction quotient is obtained in the same way
  as an equilibrium expression.

49
Free Energy Changes and the Reaction Mixture
DG DG RT ln Q
Calculate DG for the formation of ethylene (C2H4)
from carbon and hydrogen at 25 C when the
partial pressures are 100 atm H2 and 0.10 atm
C2H4.
Qp
Calculate ln Qp
50
Free Energy Changes and the Reaction Mixture
Calculate DG
DG DG RT ln Q
8.314
68.1 kJ/mol
(298 K)(-11.51)
( see page A-13 of your book, page 667)
DG 39.6 kJ/mol
51
Free Energy and ChemicalEquilibrium 05

• At equilibrium ?G 0 and Q K
• ?G ?G RT lnQ
• ?Grxn RT ln K

52
Free Energy and Chemical Equilibrium
Calculate Kp at 25 C for the following reaction
Calculate DG
DG DGf (CaO(s)) DGf (CO2(g)) - DGf
(CaCO3(s))
Please see appendix B
(1 mol)(-604.0 kJ/mol) (1 mol)(-394.4
kJ/mol)
- (1 mol)(-1128.8 kJ/mol)
DG 130.4 kJ/mol
53
Free Energy and Chemical Equilibrium
Calculate ln K
DG -RT ln K
-130.4 kJ/mol

(298 K)
8.314
ln K -52.63
Calculate K
-52.63
1.4 x 10-23
K e
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Free Energy and ChemicalEquilibrium 04
?Grxn RT ln K
57
Suppose ?G is a large, positive value. What
then will be the value of the equilibrium
constant, K?

1. K 0
2. K 1
3. 0 lt K lt 1
4. K gt 1

58
Correct Answer
?G ?RTlnK Thus, large positive values of DG
lead to large negative values of lnK. The value
of K itself, then, is very small.

1. K 0
2. K 1
3. 0 lt K lt 1
4. K gt 1

59
More thermo?
You betcha!
60
?Gorxn - RT lnK

• Calculate K for the reaction
• N2O4 ---gt2 NO2 ?Gorxn 4.8 kJ/mole ( see page
  A-11)
• ?G ?RT lnK
• ?Gorxn 4800 J - (8.31 J/mol.K)(298 K) ln K

K 0.14 When ?Gorxn gt 0, then K lt 1