1. Chapter 19 Principles of Reactivity: Entropy & Free Energy
Chem 106, Chapter 19
Chapter 19 Principles of Reactivity: Entropy & Free Energy
Copyright 2007, David R. Anderson

2. Spontaneity
Spontaneous processes
Factors affecting spontaneity
Entropy
Entropy changes
The third law of thermodynamics
Calculating entropy changes in chemical reactions
The second law of thermodynamics
Gibbs Free Energy
Definition
Calculating free energy changes in chemical reactions
Standard free energy of formation
Free energy and temperature
Free energy and equilibrium

3. Chemical Thermodynamics
Chapter 6: Thermochemistry
heat of reaction, DH
first law: conservation of energy
Thermodynamics: energy changes
entropy (DS), free energy (DG)
second and third laws
 spontaneity of reaction, position of equilibirum
I. Spontaneity
A. Spontaneous process
= one that, once started, continues on its own without input of energy
nonspontaneous = needs continual input of energy

4. I. Spontaneity B. Factors affecting spontaneity
1. decrease in potential energy (DH < 0)
apple falling
water flowing downhill
2H2 + O2  2H2O
HCl + NaOH  NaCl + H2O
thermite
2. increase in disorder (tendency toward higher probability)
deck of cards (6 x 109 combinations)
gas molecules in a container
salt dissolving in water
ice melting, water evaporating

5. II. Entropy, S = degree of disorder (probability)
Chem 106, Chapter 19
II. Entropy, S = degree of disorder (probability)
A. Entropy changes
Entropy change, DS = Sf - Si
1. examples
expansion of a gas DS
dissolution of a solid DS
H2O(s)  H2O(l)  H2O(g) DS
H2O(g)  H2O(l)  H2O(s) DS
H2CO3  CO2+ H2O DS
2H2 + O2  2H2O DS
CaO + SO2  CaSO3 DS
remove
partition
NaCl(s)
NaCl(aq)
H2O
Copyright 2007, David R. Anderson

6. II. Entropy A. Entropy changes
2. calculating DS for fusion and vaporization
DS  q
increase KE  increase in disorder
DS = q T
10 K  20 K DS1 q
DS  1 T
DS1 > DS2
300 K  310 K DS2 q
e.g., For H2O(s)  H2O(l), DHfus = kJ/mol at 0ºC
and for H2O(l)  H2O(g), DHvap = kJ/mol at 100ºC.
What are DSfus and DSvap?

7. II. Entropy B. The third law of thermodynamics
“For a pure, crystalline substance at 0 K, S = 0”
Increase T  increase in: translational motion
vibrational motion
rotational motion
increase in S
Standard state entropy, Sº
= entropy of a substance at 25ºC, 1 atm (Table 19.1, Appendix L)
(Sº  0 for elements!)

8. II. Entropy C. Calculating entropy changes in chemical reactions
Standard entropy change, DSº
DSºrxn = Sºproducts - Sºreactants
e.g., 2H2 + O2  2H2O
DSºrxn = [2Sº(H2O)] - [2Sº(H2) + Sº(O2)]
= [2(188.8)] - [2(130.6) + (205.0)]
= J/K
(note units)

9. II. Entropy C. Calculating entropy changes in chemical reactions
e.g., Calculate the standard entropy change for the reaction,
2Al(s) + 3Cl2(g)  2AlCl3(s) Sº(Al) = 28.3 J/mol·K
Sº(Cl2) = J/mol·K
Sº(AlCl3) = J/mol·K

10. II. Entropy D. The second law of thermodynamics
“In any spontaneous process, the entropy of the universe must increase (DSuniverse > 0).
system
surroundings
universe
DSuniverse = (DSsystem + DSsurroundings) > 0
e.g., erecting a building
6CO2 + 6H2O  C6H12O6 + 6O2
CFCs

11. III. Gibbs Free Energy A. Definition
DSuniverse = DSsystem + DSsurroundings
DSsurroundings = –
qsystem T
= –
DHsystem
DSuniverse = DSsystem –
DHsystem T
TDSuniverse = DHsystem – TDSsystem
DGsystem = DHsystem – TDSsystem
Gibbs free energy
state function
relates spontaneity to heat of reaction (DH) and entropy (DS)
for a process to be spontaneous, DSuniverse > 0 or DGsystem < 0

12. III. Gibbs Free Energy A. Definition DG < 0: exergonic, spontaneous
DG > 0: endergonic, nonspontaneous
DG = 0: system is at equilibrium (no free energy remains)
G 
Reactants
Products
Equilibrium
DGrxn < 0 “spontaneous”
R  E DG < 0 spontaneous
P  E DG < 0 spontaneous
E  R DG > 0 nonspontaneous
E  P DG > 0 nonspontaneous

13. III. Gibbs Free Energy
B. Calculating free energy changes in chemical reactions
e.g., C(s) + ½O2(g) + 2H2(g)  CH3OH(l) at 25º
DHºrxn = DHfº(products) – DHfº(reactants)
= –238.7 kJ/mol
DSºrxn = Sºproducts – Sºreactants
= [Sº(CH3OH)] – [Sº(C) + ½Sº(O2) + 2Sº(H2)]
= [126.8] – [(5.7) + ½(205.1) + 2(130.7)]
= –242.9 J/mol·K
DGºrxn = DHºrxn – TDSºrxn
= (–238.7 kJ/mol) – (298 K)(–242.9 J/mol·K)
= –166.3 kJ/mol
= DGfº(CH3OH)

14. III. Gibbs Free Energy C. Using standard free energies of formation
DGºrxn = DGfº(products) - DGfº(reactants)
standard free energies of formation
(Table 19.3, Appendix L; note: DGfº(element) = 0)
e.g., CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
DGºrxn = [DGfº(CO2) + 2DGfº(H2O)] - [DGfº(CH4)]
= [(-394.4) + 2(-237.1)] - [(-50.7)]
= kJ

15. III. Gibbs Free Energy C. Using standard free energies of formation
e.g., If DGºrxn for the following reaction is kJ, what is DGºf for TiCl2?
TiCl2(s) + Cl2(g)  TiCl4(s) (DGºf(TiCl4) = kJ/mol)

16. III. Gibbs Free Energy D. Free energy and temperature DG = DH – TDS
DH DS DG
– – at all temperatures
+ – + at all temperatures
– at higher temperatures, + at lower temperatures
– – – at lower temperatures, + at higher temperatures
e.g., C12H22O O2  12CO2 + 11H2O (produces heat)
DH < 0, DS > 0  DG < 0 at any temperature
(for reverse reaction, DG > 0 at any temperature; requires input of energy)
NH4NO3(s)  NH4+(aq) + NO3–(aq) (gets cold)
DH > 0, DS > 0  DG < 0 at higher temperatures
H2O

17. III. Gibbs Free Energy D. Free energy and temperature
e.g., CH3–CH3  CH2=CH2 + H2 DHº = +137 kJ
DSº = J/mol·K
at 25ºC, DGº = kJ nonspontaneous
Above what temperature does the reaction become spontaneous?

18. III. Gibbs Free Energy E. Free energy and equilibrium DG = DGº + RTlnQ
At equilibrium, DG = 0, Q = K
0 = DGº + RTlnK
DGº = –RTlnK
DGº < 0  K > 1 equilibrium lies to the right
DGº > 0  K < 1 equilibrium lies to the left
DGº = 0  K = 1 equilibrium lies in the middle
G  R P
DGº > 0
DGº < 0
DGº = 0

19. III. Gibbs Free Energy E. Free energy and equilibrium DGº = –RTlnK
or K = e–Gº/RT
e.g., 2SO2+ O2
2SO3
What is K at 25ºC? DGºf(SO2) = kJ/mol
DGºf(SO3) = kJ/mol