CHEMICAL THERMODYNAMICS * Spontaneity and Entropy * Second Law of Thermodynamics * Gibbs Free Energy and Chemical Equilibrium

3. Type of process which does not need the
application of energy to take place.
A spontaneous process is a physical or
chemical change that occurs by itself

4. Type of process which needs the use of
energy to make it happen.
It is one that occurs without outside
intervention.

5. RUSTING
SPONTANEOUS EXOTHERMIC
PROCESS
ICE MELTING
SPONTANEOUS ENDOTHERMIC
PROCESS
REMOVAL OF SUGAR FROM A CUP
OF COFFEE
THIS IS A NON SPONTANEOUS
PROCESS

7. TERMS

9. Molar entropy of the
solid state will be
LOWEST
Molar entropy of the
liquid state will be
INTERMIDIATE
Molar entropy of the
gaseous state will be
HIGHEST

10. ΔS = S(final state) – S(initial state)

11. ΔS = SH O – SH O2 (l)2 (g)
Just like ΔH and ΔU, a positive value
of ΔS means an increase in entropy
while negative value means
decrease in entropy

12. ΔS =
qrev
T

14. Sample Problem: Predict whether the entropy or disorderliness of the system where the
following changes occur is expected to increase or decrease. (p. 136)
KClO3 (s) -> KCl (s) + O2 (g)
2 H2 (g) + O2 (g) -> 2H2 O
Determine if the entropy change will be positive or negative for the following reactions:
1) (NH4)2Cr2O7(s) → Cr2O3(s) + 4 H2O(l) + CO2(g)
2) 2 H2(g) + O2(g) → 2 H2O(g)
3) PCl5 (g)→ PCl3 (l) + Cl2(g)

15. One Mole of water is vaporized at 100°C with the absorption of 40.7 kJ of heat. What is the molar
entropy of vaporization?
Given:
T= 100°C (373 K)
ΔH/ qrev = 40.7 kJ of heat (40,700 J/mol)
Note (conversions): Formula:
1 J/mol = 0.001 kj/mol
1 kj/mol = 1000 J/mol
Solution:
∆𝑺 =
40,700 𝐽/𝑚𝑜𝑙
373𝐾
Answer:
ΔS = 109 J/mol.K

16. The second law of thermodynamics states that the total entropy of the universe
ALWAYS INCREASES FOR A SPONTANEOUS PROCESS.
It is mathematically stated as,
ΔSuniverse = ΔSsystem+ ΔSsurroundings ≥ 0
Entropy change of the UNIVERSE is determined from the sum of entropy change of the
system and surroundings. (IT MUST BE A POSITIVE VALUE – which corresponds to an
increase in entropy)

17. THIS EXPLAINS WHY EXOTHERMIC AND ENDOTHERMIC REACTIONS MAY BE SPONTANEOUS.
In ENDOTHERMIC REACTIONS
The disorder within the system increases sufficiently so
that the change in disorder of the system is greater
than the decrease of the surroundings, then the total
change in disorder can be positive and the reaction
becomes spontaneous.
In EXOTHERMIC REACTIONS
The heat released by the reaction increases the disorder
of the surroundings of the system is decreased, the large
amount of heat released to the surroundings causes
large increase in entropy resulting to a net increase in
entropy of the universe.

18. The exact entropy change for the process can be
determined in the same way as the enthalpy change for the
reaction using standard absolute entropies instead of
enthalpies of formation.
For any chemical reaction:
ΔS°reaction = ΔS°products - ΔS°reactants
Standard Absolute Entropies of
Some Substances at 25°C and 1 atm
SUBSTANCE S° (J/mol-K)
H2(g) 130.68
O2(g) 205.14
N2(g) 191.61
Cl2(g) 223.07
C(s, graphite) 5.740
C (s, diamond) 2.377
H2O(g) 188.83
H2O(l) 69.91
CO2(g) 213.74
CO(g) 197.67
HCl(g) 186.91
NH3(g) 192.45
NO(g) 210.76
NO2(g) 240.06
SO2(g) 248.22
SO3(g) 256.76
CH4(g) 186.26

19. Exercise (Sample Problem)
Calculate the change in entropy at standard state and 25°C that accompanies the reaction:
1. C(s, graphite) + CO2(g)  2CO(g)
2. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

20. Exercise (Sample Problem)
Calculate the change in entropy at standard state and 25°C that accompanies the reaction:
#2 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g
Given:
S°NH3 = 193 J/K·mol
S°O2 = 205 J/K·mol
S°NO = 211 J/K·mol
S°H2O = 189 J/K·mol
Solution:
ΔS°reaction = (4 S°NO + 6 S°H2O) - (4 S°NH3 + 5 S°O2)
ΔS°reaction = (4(211 J/K·K) + 6(189 J/K·mol)) - (4(193 J/K·mol) + 5(205 J/K·mol))
ΔS°reaction = (844 J/K·K + 1134 J/K·mol) - (772 J/K·mol + 1025 J/K·mol)
ΔS°reaction = 1978 J/K·mol - 1797 J/K·mol)
ΔS°reaction = 181 J/K·mol
(Note, in this type of problem you'll either be given the molar entropy values of the reactants and products
or you'll need to look them up in a table.)
Formula: ΔS°reaction = ΔS°products - ΔS°reactants

22. A criterion for spontaneity that applies
only to the system.
It combines enthalpy and entropy.
G = H – TS
is a measure of the spontaneity of a process and of
the amount of energy that can be obtained from it.
It is determined from difference in values of two
different states.
ΔG = Gfinal – Ginitial
The sign ΔG is used as the basis to determine whether a
process is spontaneous.

24. GIBBS FREE ENERGY, ENTHALPY, AND ENTROPY
 The GIBBS FREE ENERGY CHANGE can be determined from the GIBBS EQUATION
ΔG = ΔH – TΔS
by using change in enthalpy and change in entropy at constant temperature and pressure as shown
by the equation.
Free energy and Spontaneity of a Reaction
- Just like other thermodynamic functions, the free energy change for a reaction can be calculated by
using the standard free energy of formation in the equation
∆𝐆° 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 = 𝚺 ∆𝐆° 𝒇,𝒑𝒓𝒐𝒅𝒖𝒕𝒔 − 𝚺 ∆𝐆° 𝒕,𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔

25. Exercise (Sample Problem)
SAMPLE PROBLEM:
Determine ΔG° of the following reaction using: (A) ΣΔG°, and (B) Gibbs equation. (C) is the
reaction spontaneous at 25°C? (D) Is it spontaneous at 2000K?
Standard
Thermodynamic Values
Reactants Products
CO H2O CO2 H2
ΔH°f,25° (KJ/mol) -110.5 -241.83 -393.5 ----
S25°C (J/mol K) 197.5 188.72 213.7 130.6
ΔG°f,25° (KJ/mol) -137.2 -228.6 -394.4 ----

26. A. ΔG°reaction = (ΔG°f,products) - (ΔG°t,reactants)
= ΔG°f, CO2 + ΔG°f, H2 – [ΔG°f, CO + ΔG°f, H2O]
= -394.4 + 0 – [-137.2 – 228.6]
= -28.6 kJ
Standard
Thermodynamic Values
Reactants Products
CO H2O CO2 H2
ΔH°f,25° (KJ/mol) -110.5 -241.83 -393.5 ----
S25°C (J/mol K) 197.5 188.72 213.7 130.6
ΔG°f,25° (KJ/mol) -137.2 -228.6 -394.4 ----

27. (B) ΔH°reaction = ΔH°f, CO2 + ΔH°f, H2 – [ΔH°f, CO + ΔH°f, H2O]
= -393.5 + 0 – [-110.5 – 241.83]
= - 41.17 kJ
ΔS°reaction = ΔS°CO2 + ΔS°H2 – [ΔS°CO + ΔS° H2O]
= 213.7 + 130.6 – [197.5+188.72]
= - 41.92 kJ
ΔG°reaction = ΔH°reaction – TΔS° reaction
= - 41.17 kJ – [298 K x (-0.04192 kJ/K)]
= - 28.68 kJ
Standard Thermodynamic
Values
Reactants Products
CO H2O CO2 H2
ΔH°f,25° (KJ/mol) -110.5 -241.83 -393.5 ----
S25°C (J/mol K) 197.5 188.72 213.7 130.6
ΔG°f,25° (KJ/mol) -137.2 -228.6 -394.4 ----

28. Exercise (Sample Problem)
SAMPLE PROBLEM:
Determine ΔG° of the following reaction using: (A) ΣΔG°, and (B) Gibbs equation. (C) is the
reaction spontaneous at 25°C? (D) Is it spontaneous at 2000K?
(D)ΔG°reaction = ΔH°reaction – TΔS° reaction
= - 41.17 kJ – [298 K x (-0.04192 kJ/K)]
= - 28.68 kJ
(C) Since ΔG°reaction is negative at 298 K. The reaction is spontaneous.

29. https://www.thoughtco.com/entropy-change-problem-609481
https://www.thoughtco.com/entropy-of-reaction-example-problem-609483
https://www.youtube.com/watch?v=YM-uykVfq_E&t=4s
http://web.mit.edu/10.213/oldpages/f99/supp/readings/qdh/index.html
https://www.quora.com/What-does-q-rev-mean-in-the-definition-of-entropy
Ilao et al., (2017). General Chemistry 2; First Edition. Rex Book Store