1. Chapter 19 Chemical Thermodynamics
Lecture Presentation
Chapter 19 Chemical Thermodynamics
James F. Kirby
Quinnipiac University
Hamden, CT
© 2015 Pearson Education, Inc.

2. 19.1 Spontaneous Processes Spontaneous Physical and Chemical Processes
Thermodynamics: study of energy conversions and spontaneity of processes
spontaneous
Spontaneous Physical and Chemical Processes
At 1 atm, water freezes below 0 oC and ice melts above 0 oC
Heat flows from a hotter object to a colder object
Iron exposed to oxygen and water forms rust
A gas expands in an evacuated bulb
nonspontaneous

3. 19.1 Spontaneous Processes
Does a decrease in enthalpy mean a reaction proceeds spontaneously?
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = −890.4 kJ/mol
H+ (aq) + OH− (aq) H2O (l) DH0 = −56.2 kJ/mol
H2O (s) H2O (l) DH0 = 6.01 kJ/mol
NH4NO3 (s) NH4+(aq) + NO3− (aq) DH0 = 25 kJ/mol
H2O
Enthaply alone cannot determine spontaneity of a process.

4. 19.2 Entropy and the Second Law of Thermodynamics
Entropy (S) is a measure of the randomness or disorder of a system.
order S
disorder S
DS = Sf – Si
Sf > Si
DS > 0
Processes that lead to an increase in entropy (DS > 0):

5. 19.2 Entropy and the Second Law of Thermodynamics
For a substance in different phases:
Ssolid < Sliquid << Sgas
H2O (s) H2O (l)
DS > 0
For substances in the same phase:
He(g) < Ne(g) < Ar(g)
Molar mass
Molecular complexity
F2(g) < Cl2(g) < Br2(g)
He(g) < H2(g) isoelectronic
Ar(g) < F2(g) isoelectronic
NO(g) < NO2(g) < N2O4(g)

6. Ne(g) SO2(g) Na(s) NaCl(s) H2(g)
EXERCISE #1
Arrange the following substances in order of increasing standard molar entropy at 25oC:
Ne(g) SO2(g) Na(s) NaCl(s) H2(g)
20.2 g/mol
64.1 g/mol
23.0 g/mol
58.5 g/mol
2.0 g/mol
Order: 4 5 1 2 3
Ssolid < Sliquid << Sgas
Molar mass
Molecular complexity

7. 19.2 Entropy and the Second Law of Thermodynamics
First Law of Thermodynamics
Energy can be converted from one form to another but energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0

8. 19.3 Entropy and the Third Law of Thermodynamics
The entropy of a perfect crystal at 0 K is zero.
The entropy of a substance increases with temperature.
EXERCISE #2
Which of the following obeys the third law of thermodynamics?
a) KBr, solid, amorphous, 0 K
b) KBr, solid, perfect crystal, 25 K
c) KBr, solid, perfect crystal, 0 K

9. 19.4 Entropy Changes in Chemical Reactions
Standard entropy of reaction (∆S0)
The entropy change for a reaction carried out at 1 atm and 25oC.
aA + bB cC + dD
∆S0 =
[cS0(C) +
dS0(D)] –
[aS0(A) +
bS0(B)]
∆S0 =
SnS0(products) −
SmS0(reactants)
Hess’s law
S0 is absolute entropy of a substance at 25oC and 1 atm
When gases are produced (or consumed):
Dn > 0
DS0 > 0
Dn < 0
DS0 < 0
Dn = 0
DS0 may be positive or negative BUT DS0 will be a small number.

10. EXERCISE #3
Calculate ΔS0 for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) S0 (J/K·mol) Na(s) 51 H2O(l) 70 NaOH(aq) 50 H2(g) 131
x 2
x 2
x 2
ΔS0 =
[2S0(NaOH, aq) +
S0(H2, g)] –
[2S0(Na, s) +
2S0(H2O, l)] =
[2(50) +
131] –
[2(51) +
2(70)] =
–11 J/K

11. Enthalpy Changes in Chemical Reactions
Standard enthalpy of reaction ( ∆𝐇 𝐫𝐱𝐧 𝟎 )
The enthalpy change for a reaction when it occurs under standard-state conditions.
aA + bB cC + dD
∆H0 =
[c ∆H f 0 (C) +
d ∆H f 0 (D)] –
[a ∆H f 0 (A) +
b ∆H f 0 (B)]
∆H0 =
Sn ∆H f 0 (products) −
Sm ∆H f 0 (reactants)
Hess’s law
∆𝑯 𝐟 𝟎 is standard enthalpy of formation
The enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
∆H f 0 of any element in its stable form is zero.

12. Gibbs Free Energy (G) and Spontaneity For a constant P and T process:
G = Gibbs free-energy
H = enthalpy
T = temperature (in Kelvin)
S = entropy
G = H – TS
For a constant P and T process:
DG = DH − TDS
DG < 0
The reaction is spontaneous in the forward direction.
DG > 0
The reaction is nonspontaneous as written.
The reaction is spontaneous in the reverse direction.
DG = 0
The reaction is at equilibrium.

13. Standard free-energy of reaction (∆G0)
19.5 Gibbs Free Energy
Standard free-energy of reaction (∆G0)
The free-energy change for a reaction when it occurs under standard-state conditions.
aA + bB cC + dD
∆G0 =
[c ∆G f 0 (C) +
d ∆G f 0 (D)] –
[a ∆G f 0 (A) +
b ∆G f 0 (B)]
∆G0 =
Sn ∆G f 0 (products)
Hess’s law −
Sm ∆G f 0 (reactants)
∆𝐆 𝐟 𝟎 is standard free-energy of formation
The free-energy change when 1 mole of a compound is formed from its elements in their standard states.
∆G f 0 of any element in its stable form is zero.

14. CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
EXERCISE #4
Given the data below, calculate the standard free-energy change for the following reaction at 25oC. Is the reaction spontaneous?
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
∆𝐺 𝑓 𝑜 (kJ/mol)
CO2(g) −394.4
H2O(l) −237.2
CH4(g) −50.8
O2(g) 0
x 2
x 2
∆G0 =
[ ∆G f o (CO2, g) +
2 ∆G f o (H2O, l)] –
[ ∆G f o (CH4, g) +
2 ∆G f o (O2, g)]
∆G0 =
[−394.4 kJ/mol +
2(−237.2 kJ/mol)] −
[−50.8 kJ/mol +
2(0 kJ/mol)]
= −818.0 kJ
ΔG0 < 0
yes

15. 19.6 Free Energy and Temperature
Factors affecting the sign of ∆G in the relationship ∆G = ∆H – T∆S ∆H ∆S ∆G
Spontaneity + +
∆G < 0 at high T.
Spontaneous in the forward direction at high T. + –
∆G > 0 at all T.
Nonspontaneous in the forward direction at all T. – +
∆G < 0 at all T.
Spontaneous in the forward direction at all T. – –
∆G < 0 at low T.
Spontaneous in the forward direction at low T.

16. 19.6 Free Energy and Temperature
Spontaneity of Physical Processes at 1 atm:
DG0 = ∆H vap 0 – Tb ∆S vap 0 = 0
EXERCISE #5
The enthalpy of vaporization and the entropy change of ammonia at 1 atm is kJ/mol and 9.74 x 101 J/K∙mol, respectively. Calculate the boiling point of ammonia.
NH3(l) NH3(g)
DG0 = ∆H vap 0 – Tb ∆S vap 0 = 0
∆H vap 0 = Tb ∆S vap 0
∆H vap 0 ∆S vap 0 =
23.35 x J/mol 97.4 J/K∙mol
Tb =
= K
= −33.4oC

17. 19.6 Free Energy and Temperature
Spontaneity of Chemical Reactions at 1 atm:
DG0 = DH0 – TDS0 < 0
EXERCISE #6
The standard enthalpy and the standard entropy change of the reaction
at 1 atm are kJ/mol and J/K·mol, respectively.
CaCO3 (s) CaO (s) + CO2 (g)
a) Calculate the standard free energy of the reaction at 25oC.
DG0 =
DH0 – TDS0 =
(177.8 x 103 J/mol) –
(298 K)
(160.5 J/K·mol)
DG0 =
J/mol
= kJ/mol
DG0 > 0 nonspontaneous at 25oC
b) At what temperatures will the reaction be spontaneous?
∆H0 ∆S0
= x J/mol J/K∙mol
DG0 = DH0 – TDS0 = 0
Tb =
= 1108 K
= 835oC
DG0 = 0 at 835 oC
DG0 > 0 at 25 oC
DG < 0 when T > 835oC
Spontaneous at T > 835oC

18. 19.7 Free Energy and the Equilibrium Constant
Free Energy and Spontaneity of Chemical Equilibrium
DG0 is the standard Gibbs free-energy (kJ/mol)
R is the gas constant (8.314 J/K∙mol)
DG = DG0 + RT lnQ
T is the absolute temperature (K)
Q is the reaction quotient
DG < 0
The reaction is spontaneous in the forward direction.
DG > 0
The reaction is nonspontaneous as written.
The reaction is spontaneous in the reverse direction.
DG = 0
The reaction is at equilibrium.

19. 19.7 Free Energy and the Equilibrium Constant
Free Energy and Equilibrium Constant
DG = DG0 + RT ln Q
DG = 0
Q = K
0 = DG0 + RT ln K
DG0 = −RT ln K
Relationship between ∆Go and K as predicted by ∆Go = −RT ln K K
ln K
∆Go
At equilibrium
> 1 + −
Products favored over reactants.
= 1
Products and reactants equally favored.
< 1 – +
Reactants favored over products.

20. EXERCISE #7
Consider the decomposition of water 2H2O(l) H2(g) + O2(g), and the thermodynamic data below for the reaction at 25oC.
∆H f o (kJ/mol) So(J/K∙mol)
H2O(l) −
H2(g)
O2(g)
x 2
x 2
x 2
x 2
a) Calculate the standard enthalpy of the reaction.
∆H0 =
[ ∆H f o (O2,g) +
2 ∆H f o (H2, g)] –
[2 ∆H f o (H2O, l)]
∆H0 = [0 +
2(0)] –
[−2(286 kJ/mol)]
= 572 kJ
b) Calculate the standard entropy of the reaction.
∆S0 =
[So(O2, g) +
2So(H2, g)] –
[2So(H2O, l)]
∆S0 =
[205 J/K∙mol +
2(131 J/K∙mol)] –
[2(70 J/K∙mol)]
= 327 J/K

21. EXERCISE #7
Continued
c) Calculate the standard Gibbs energy of the reaction.
DH0 = 572 kJ
DS0 = 327 J/K
T = 25oC
= 298 K
∆G0 =
∆H0 – T∆S0 =
(572 x 103 J) –
(298 K)
(327 J/K)
= J
= kJ
d) Calculate the equilibrium constant Kp of the reaction.
∆G0 = −RT ln Kp
R = J/K∙mol
ln Kp = ∆G 0 −RT
= x J − 8.314J/K∙mol)(298 K
= −191.6
Kp = e−191.6
= 6 x 10−84
e) What does DG0 and Kp tell us about the conditions at equilibrium?
∆G0 >> 0 and Kp << 0
reactants are favored over products
f) Assuming DH0 and DS0 are constant, estimate the value of K at 100oC.
ln K = ∆G 0 −RT
= x J − 8.314J/K∙mol)(373 K
= −153.0
K = e−153.0
= 4 x 10−67

22. EXERCISE #8
The standard free-energy change for the reaction
is 5.40 kJ/mol at 298 K. In a certain experiment, the initial pressures are PNO2 = atm and PN2O4 = atm. Calculate ΔG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium.
N2O4(g) NO2(g)
(P NO 2 )2 P N 2 O 4 =
∆G = ∆G0 + RT lnQp
R = J/K∙mol
Qp =
(ln ) ∆G =
5.40 x 103 J/mol +
(8.314 J/K∙mol)
(298 K) ∆G
= −3.06 kJ/mol
= −3.06 x 103 J/mol
∆G < 0
spontaneous in the forward direction

23. End of Chapter 19