1. Chapter 19. Chemical Thermodynamics.
ΔG = ΔH – TΔS

2. Enthalpy (H) and entropy (S)
In Chapter 5 we discussed the role of enthalpy ( ΔH) in driving reactions. We saw that highly exothermic reactions such as the burning of gasoline proceeded very strongly, while the reverse reaction was highly endothermic, and did not proceed spontaneously at all. (CO2 and H2O do not react spontaneously to form gasoline and oxygen). However, enthalpy is not the only driving force in reactions. There is a second contribution, the entropy. Entropy relates to probability, and the increase in disorder of a system.
We can think about probability in terms of a gas moving to occupy two flasks instead of only one:

3. We connect two
flasks as shown at
the right by a closed
stopcock, one of
which is evacuated.
When the stopcock is
opened, the air will
rush into the evacu-
ated flask to equalize
the pressure. What is
the driving force for
this? It is in fact
probability, or entropy.

4. 19.1 Spontaneous Processes.
A spontaneous process is one that proceeds on its own without any outside assistance.
A gas will expand from one flask into a second as a spontaneous process, but the reverse will never occur.
Processes that are spontaneous in one direction are non-spontaneous in the opposite direction.

5. The rusting of nails in air – a spontaneous process:
Fresh nails
rusted nails
spontaneous
not
spontaneous
2 Fe (s) + 3 O2 (g)  Fe2O3 (s) (‘rust’)

6. Examples of spontaneous processes
water
Ice at 25 oC will spontaneously melt:
ice
spontaneous
not
spontaneous
The reverse will never happen. Water at 25 oC
will never freeze.

7. Examples of spontaneous processes
But water at -10 oC will spontaneously freeze:
ice
spontaneous
water
not
spontaneous

8. Equilibrium: A mixture of water and ice at 0 oC is at equilibrium. ice
The relative quantity
of ice and water remains
the same if we don’t add
or remove heat. There is
no spontaneous
process in either
direction. If we remove
some heat, the
temperature remains
at 0 oC, but some
water turns to ice.
ice
water
(at 0 oC)

9. Reversible processes:
These occur with infinitesmal changes in the system.
The melting and freezing of ice at 0 oC is reversible. Thus, if we add some heat, some of the ice melts, and if we remove some heat, some of the water freezes, but the temperature remains at 0 oC. At all other temperatures an irreversible process occurs.
Melting and freezing of a substance at its melting point are reversible.
Boiling and condensation of a substance at its boiling point are also reversible.
Reversible processes are not spontaneous.

10. Example:
When benzene vapor converts to benzene liquid at 80.1 oC (B.Pt. of benzene) and 1 atm is this a spontaneous process?
The condensation of a gas to a liquid at the boiling point of a substance is not a spontaneous process. To move the equilibrium in either direction we have to add or subtract energy.

11. 19.2 Entropy and the second Law of Thermodynamics.
Entropy is associated with randomness. The entropy, S, of a system is a state function just like the internal energy. The change in entropy is given by
ΔS = Sfinal - Sinitial

12. Relating entropy to heat transfer and temperature:
For a process that occurs at constant temperature such as melting of a solid at its melting point:
ΔSsystem = qrev/T
To calculate the entropy change of a system we need to know the heat change, which is qrev, and divide it by the temperature in K. The term qrev is the reversible flow of heat in a reversible process such as enthalpy of melting of a solid at its melting point, or of boiling of a liquid at its boiling point.

13. Example:
The molar enthalpy of fusion of Hg is 2.29 kJ/mol. What is the entropy change when 50 g of Hg freezes at the normal melting point of oC? (fusion = melting)
The important thing to realize here is that qrev = ΔH(fusion)
Hg(s) → Hg(l) ΔH = kJ/mol
So for freezing:
Hg(l) → Hg(s) ΔH = kJ/mol

14. Moles Hg = 50.0 g x 1mole = mol g
kJ = mole x kJ = kJ = -571 J
1 mole
T = – 38.9 = K
ΔS = qrev/T = -571/ = J/K
Note that units of entropy are J/K. A positive change in entropy means an increase in entropy of the system.
ΔH (freezing) Hg(l)

15. The Second Law of Thermodynamics:
The total entropy of the universe increases in any spontaneous process.
The sum of the entropy change of the system and the surroundings is always positive.
Reversible: ΔSuniverse = ΔSsystem + ΔSsurround = 0
Irreversible: ΔSuniverse = ΔSsystem +ΔSsurround > 0

16. Example:
What happens to the entropy of the universe when a lesser spotted thrush begins to do the makarena?
Answer: It increases, ΔSuniverse > 0
For any spontaneous process, and that’s any process where anything actually happens, the entropy of the universe increases. You may get questions like the above, and if you bear in mind that (except for a reversible process) when anything happens, the entropy of the universe increases.

17. 19.3 The Molecular Interpretation of Entropy.
Molecules can undergo translational, rotational, and vibrational motion.
The number of ways a system can be arranged is the number of microstates it has. The entropy of a system is given by
S = k ln W
where W is the number of possible microstates, and k
is Boltsmann’s constant =
1.38 x J/K.
Ludwig Boltzmann
( )
Boltzmann worked out the nature of entropy.

18. Probability and gas molecules in interconnected flasks:
Consider a pair of flasks connected by a stop-cock. The probability that all the molecules will be in one flask is (½)n, where n is the number of molecules.
one molecule,
probability =
(1/2)1 = 1/2
two molecules,
probability =
(1/2)2 = 1/4
four molecules,
probability =
(1/2)4 = 1/16

19. The relationship between probability and entropy:
So if n = 6.02 x 1023, we have a very small probability of finding all
the molecules in one flask!
Entropy relates to probability. The gas
molecules in the apparatus at right are moving randomly, but the probability aspect keeps them evenly distributed between the two flasks.
6.02 x molecules,
probability = (1/2)6.02 x 10 23
There is thus a chance
that all the air molecules
in this room would rush
into one corner, and we
would all die, but it is so
small that it would never
happen.

20. The standard molar entropy of a substance (Sº):
The entropies of substances may be measured experimentally. The entropy of one mole of a substance in its standard state is known as the standard molar entropy (Sº). The units of Sº are J/mol.K. Some values of standard molar entropies are given overleaf. Note that So values increase solid < liquid < gas, e.g. C6H6 (s) < C6H6 (l) < C6H6 (g); increase with the size of molecules, e.g. HF < HCl < HBr < HI or CH4 < C2H6 < C3H8; and with being a solid or liquid dissolved in solution in another liquid e.g. CH3COOH (l) < CH3COOH (aq).

21. Standard molar entropies of selected substances at 298 K: (So, J/mol-K):
Gases:
H2(g) N2(g)
O2(g) C6H6(g)
Liquids:
H2O(l) C6H6(l)
Solids:
Li(s) Fe(s)
FeCl3(s) NaCl(s) 72.3
C(diamond) C(graphite)

22. Making Qualitative predictions about ΔS.
Entropy increases with temperature, since the particles vibrate and move about more rapidly, and so there is greater disorder. It also increases with increasing volume, as with a an expanding gas, and with the number of independ-ently moving particles. Thus, entropy increases solid < liquid < gas.
Figure shows increase
in entropy of a substance
with increasing temp. as
we pass solidliquidgas.

23. Of all phase states, gases have the highest entropy
Ssolid
< Sliquid
< Sgas
solid
liquid
gas
In a gas, molecules are more randomly distributed

24. Larger Molecules generally have a larger entropy
Ssmall
< Smedium
< Slarge
Larger molecules have more internal motion

25. } } } } size of molecules increases Sgas > Sliquid
dissolving a gas in a
liquid is accompanied by
a lowering of the entropy } }
dissolving a liquid in
another liquid is
accompanied
by an increase in entropy

26. more disordered arrangement = higher entropy
Often, dissolving a solid or liquid
will increase the entropy
solution
liquid
dissolves
lower entropy
more disordered
arrangement =
higher entropy
solid

27. overall more disordered
Dissolving a gas in a liquid
decreases the entropy
gas
dissolves
overall more disordered
arrangement:
higher entropy
solution of gas
in liquid, lower S

28. Entropy increases with increasing temperature along the series solid < liquid < gas

29. Note: S increases with increase in number
of moles of gas in a reaction.
What is the sign of DS for the following reactions?
FeCl2 (s) + H2 (g) → Fe (s) HCl (g)
DS = +
solid
gas
solid
gas
1 mol
2 mol
Ba(OH)2 (s) → BaO (s) + H2O (g)
DS = +
solid
solid
gas
2 SO2 (g) + O2 (g) → 2 SO3 (g)
DS = -
gas
gas
gas
2 mol
1 mol
2 mol
Ag+ (aq) + Cl- (aq) → AgCl (s)
DS = -
in solution
insoluble

30. For each of the following pairs of substances,
which substance has a higher molar entropy
at 25oC ?
HCl (l) HCl (s)
Li (s) Cs (s)
C2H2 (g) C2H6 (g)
Pb2+ (aq) Pb (s)
O2 (g) O2 (aq)
HCl (l) HBr (l)
N2 (l) N2 (g)
CH3OH (l) CH3OH (aq)

31. 19.4 Entropy Changes in Chemical Reactions.
This parallels the calculation of standard molar enthalpies of reaction.
ΔSo = ΣnSo(products) ΣmSo(reactants)
Note that So is not zero for elements.
Recall that So is a state function.

32. Standard molar entropies of selected substances at 298 K: (So, J/mol-K):
Gases:
H2(g) N2(g)
O2(g) C6H6(g)
Liquids:
H2O(l) C6H6(l)
Solids:
Li(s) Fe(s)
FeCl3(s) NaCl(s) 72.3
C(diamond) C(graphite)

33. Example: Calculate ΔSo for the following reaction:
C2H4(g) + H2(g) → C2H6(g)
So products = So C2H6(g) = J/mol-K
So reactants = So C2H4(g) + So H2(g)
= J/mol-K
= J/mol-K
So(products) – So(reactants)
= – J/mol-K
= J/mol-K

34. Entropy Changes in the Surroundings.
The change in entropy in the surroundings is caused entirely by heat transferred from the system to the surroundings, and so we have q = ΔH.
ΔS(surroundings) = -qsys /T = -(ΔH/T)
Thus, if asked to calculate the change in entropy of the surroundings, just use the above equation.

35. Section 19.5. Gibbs Free Energy.
We have seen that both enthalpy changes (ΔH) and entropy changes (ΔS) contribute to the overall energy that drives a reaction. Josiah Willard Gibbs ( ) proposed a new state function, now called G in his honor, known as the free energy.
G = H - TS
J. Willard Gibbs
(1839 – 1903)

36. Again, we work with changes in G, so we have:
ΔG = ΔH TΔS
increase in increase in
entropy of entropy of
surroundings system
-ΔG is in fact the increase in entropy in the universe.

37. ΔH is the increase in entropy of the surroundings from ΔS(surroundings) = -q/T.
Note:
1) If ΔG is negative, the reaction is spontaneous.
2) If ΔG is zero, the reaction is at equilibrium.
3) If ΔG is positive, the reaction is non-
spontaneous , but reverse reaction is
spontaneous.

38. 19.6. Free Energy and temperature. (p. 827)
Temperature helps determine whether ΔG will be negative, and the process spontaneous.
ΔH ΔS -TΔS ΔG (= ΔH – TΔS)
__________________________________________
(spontaneous)
(non-spontaneous)
or – depends on T
or - depends on T

39. DG < 0  reaction is spontaneous
(“product favored”)
“exergonic”
DG > 0  reaction is non-spontaneous
“endergonic”
DG = 0  reaction is at equilibrium

40. 19.4 Free Energy Changes (ΔG) in Chemical Reactions.
This parallels the calculation of standard molar enthalpies of reaction.
ΔGo(rxn) = ΣnΔGfo(products) ΣmΔGfo(reactants)
Note that ΔGof is zero for elements.
Recall that ΔGof is a state function.

41. Table of ΔGof , ΔHof, (kJ/mol) and So (J/mol-K) for some substances (p
Substance ΔHof ΔGof So
________________________________________________________
Al(s)
C(s, graphite)
C2H6(g)
O2(g)
Ni(s)
NiCl2(s)
Cl2(g)

42. Alternatively, from ΔG = ΔH – TΔS
Example: Calculate ΔGo, ΔHo, and ΔSo for the following: (problem 19.54)
Ni(s) + Cl2(g) = NiCl2(s)
ΔHo = – [0 + 0] = kJ
ΔSo = – [ ] = J/K
ΔGo = – [0 + 0] = kJ
Alternatively, from ΔG = ΔH – TΔS
ΔGo = kJ – ( J x 1 kJ x 298 K) =
1 K J kJ
Note
units

43. 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g)
The reaction of
sodium metal
with water:
2 Na (s) H2O (l) → 2 NaOH (aq) + H2 (g)
Is the reaction spontaneous?
What is the sign of DG?
What is the sign of DH?
What is the sign of DS?

44. 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g)
The reaction of
sodium metal
with water:
2 Na (s) H2O (l) → 2 NaOH (aq) + H2 (g)
Is the reaction spontaneous?
Yes
What is the sign of DG?
DG = negative
What is the sign of DH?
DH = negative (exothermic!)
What is the sign of DS?
DS = positive