1. Entropy, Free Energy, and Equilibrium
Chapter 17

2. Thermodynamics is the study of heat and other forms of energy involved in chemical or physical processes.

3. First Law of Thermodynamics
The first law of thermodynamics is essentially the law of conservation of energy applied to a thermodynamic system.
Recall that the internal energy, U, is the sum of the kinetic and potential energies of the particles making up the system:
DU = Uf – Ui

4. Exchanges of energy between the system and its surroundings are of two types: heat, q, and work, w.
Putting this in an equation gives us the first law of thermodynamics.
DU = q + w

5. Sign Convention for q
When heat is evolved by the system, q is negative. This decreases the internal energy of the system.
When heat is absorbed by the system, q is positive. This increases the internal energy of the system.

6. Sign Convention for w
Recall from Chapter 6 that w = –PDV.
When the system expands, DV is positive, so w is negative. The system does work on the surroundings, which decreases the internal energy of the system.
When the system contracts, DV is negative, so w is positive. The surroundings do work on the system, which increases the internal energy of the system.

7. DV is positive, so work is negative.
Here the system expands and evolves heat from A to B.
Zn2+(aq) + 2Cl-(aq) + H2(g)
DV is positive, so work is negative.

8. At constant pressure: qP = DH
The first law of thermodynamics can now be expressed as follows:
DU = DH – PDV

9. To understand why a chemical reaction goes in a particular direction, we need to study spontaneous processes.
A spontaneous process is a physical or chemical change that occurs by itself. It does not require any outside force, and it continues until equilibrium is reached.

10. A ball will roll downhill spontaneously
A ball will roll downhill spontaneously. It will eventually reach equilibrium at the bottom.
A ball will not roll uphill spontaneously. It requires work.

11. 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

12. Examining some spontaneous processes will clarify this definition.
First, heat energy from a cup of hot coffee spontaneously flows to its surroundings—the table top, the air around the cup, or your hand holding the cup. The entropy of the system (the cup of hot coffee) and its surroundings has increased.

13. The rock rolling down the hill is a bit more complicated
The rock rolling down the hill is a bit more complicated. As it rolls down, the rock’s potential energy is converted to kinetic energy. As it collides with other rocks on the way down, it transfers energy to them. The entropy of the system (the rock) and its surroundings has increased.

14. Now consider a gas in a flask connected to an equal-sized flask that is evacuated. When the stopcock is open, the gas will flow into the evacuated flask. The kinetic energy has spread out, and the entropy of the system has increased.

15. spontaneous
nonspontaneous
18.2

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

17. The first law of thermodynamics cannot help us to determine whether a reaction is spontaneous as written.
We need a new quantity—entropy.
Entropy, S, is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy.

18. Examining some spontaneous processes will clarify this definition.
First, heat energy from a cup of hot coffee spontaneously flows to its surroundings—the table top, the air around the cup, or your hand holding the cup. The entropy of the system (the cup of hot coffee) and its surroundings has increased.

19. Entropy (S) is a measure of the randomness or disorder of a system.
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

20. W = number of microstates
Entropy
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
18.3

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

22. (a) Condensing water vapor
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

23. In the case of the cup of hot coffee, as heat moves from the hot coffee, molecular motion becomes more disordered. In becoming more disordered, the energy is more dispersed.

24. Likewise, when the gas moves from one container into the evacuated container, molecules become more disordered because they are dispersed over a larger volume. The energy of the system is dispersed over a larger volume.

25. When ice melts, the molecules become more disordered, again dispersing energy more widely.
When one molecule decomposes to give two, as in
N2O4(g)  2NO2(g)
more disorder is created because the two molecules produced can move independently of each other. Energy is more dispersed.

26. In each of these cases, molecular disorder increases, as does entropy.
Note: This understanding of entropy as disorder applies only to molecular situations in which increasing disorder increases the dispersion of energy. It cannot be applied to situations that are not molecular—such as a desk.

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

28. First Law of Thermodynamics
Energy can be converted from one form to another
Second Law of Thermodynamics
The entropy of the universe (i.e system+surrounding) increases in a spontaneous process and remains unchanged in an equilibrium process.
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
18.4

29. Third Law of Thermodynamics
A substance that is perfectly crystalline at zero Kelvin (0 K) has an entropy of zero.

30. The standard entropy of a substance—its absolute entropy, S°—is the entropy value for the standard state of the species. The standard state is indicated with the superscript degree sign.
For a pure substance, its standard state is 1 atm pressure. For a substance in solution, its standard state is a 1 M solution.

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

34. When wine is exposed to air in the presence of certain bacteria, the ethyl alcohol is oxidized to acetic acid, giving vinegar. Calculate the entropy change at 25°C for the following similar reaction:
CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)
The standard entropies, S°, of the substances in J/(K  mol) at 25°C are CH3CH2OH(l),161; O2(g), 205; CH3COOH(l), 160; H2O(l), 69.9.

35. CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)
S° J/(mol  K)
n mol
nS° J/K
DS = J/K – 366 J/K
DS = –136 J/K

36. Entropy Changes in the System (DSsys)
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) ZnO (s)
The total number of gas molecules goes down, DS is negative.
18.4

37. The opening of Chapter 6, on thermo-chemistry, describes the endothermic reaction of solid barium hydroxide octahydrate and solid ammonium nitrate:
Ba(OH)2  8H2O(s) + 2NH4NO3(s) 
2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)
Predict the sign of DS° for this reaction.
3 moles of reactants produces 13 moles of products. Solid reactants produce gaseous, liquid, and aqueous products.
DS° is positive.

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

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

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

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

42. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C?
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DG0
rxn
nDG0 (products) f = S
mDG0 (reactants) -
DG0
rxn
6DG0 (H2O) f
12DG0 (CO2) = [ + ] -
2DG0 (C6H6)
DG0
rxn =
[ 12x– x–237.2 ] – [ 2x124.5 ] = kJ
Is the reaction spontaneous at 25 0C?
DG0 = kJ
< 0
spontaneous
18.5

43. Calculate the free-energy change, DG°, for the oxidation of ethyl alcohol to acetic acid using standard free energies of formation.
CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)

44. CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)
DGf°, kJ/mol – – –237.2
n, mol
nDGf°, kJ – –392.5 –237.2
–174.8 kJ –629.7 kJ
DG° = –629.7 – (–174.8)
DG° = –454.9 kJ

45. DG = DH - TDS
18.5

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

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

48. Using standard enthalpies of formation, DHf° and the value of DS° from the previous problem, calculate DG° for the oxidation of ethyl alcohol to acetic acid.
CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)

49. CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)
DHf° kJ/mol – – –285.8
n mol
nDHf° kJ – – –285.8
–277.6 kJ –772.8 kJ
DH° = –495.2 kJ
DS° = –136 J/K
T = 298 K

50. The reaction is spontaneous.
DH° = –495.2 kJ
DS° = –136 J/K = –0.136 kJ/K
T = 298 K
DG° = DH° – TDS°
DG° = –495.2 kJ – (298 K)(–0.136 kJ/K)
DG° = –495.2 kJ kJ
DG° = –454.7 kJ
The reaction is spontaneous.

51. 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

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

53. DG0 = - RT lnK
18.6

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

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