1. Spontaneity, Entropy, and Free Energy
Chapter 17
Spontaneity, Entropy, and Free Energy

2. Fist Law of Thermodynamics
Energy is not created nor destroyed
Total energy in the universe is always held constant
Energy is transformed from one form to another

3. Spontaneity Spontaneous means without intervention Can be fast OR slow
Chapter 12 = rate laws (talked about rxn rates - activation energy, temperature, concentration, catalysts, etc.)
Chapter 17 = thermodynamics show reaction direction and only initial (reactants) and final (products) - don’t need information about the pathway

4. Examples of Spontaneity
A ball rolling down a hill
Steel rusting exposed to air and moisture
Gas uniformly fills its container
Heat flows from hot to cold
Water freezes <0°C and melts >0°C
All of these processes happen spontaneously in one direction and never in reverse
Gravity? Exothermicity? WHY?

5. Entropy (not to be confused with Enthalpy)
Characteristic common to all spontaneous processes
Represented by “S”
Defined as a measure of molecular randomness or disorder
It is natural for disorder to increase

6. Claudia Witt:
Sock Drawer
Examples
Closely associated with probability (nature more likely to proceed towards states that have the highest probability of existing - called positional probability)

7. Positional Probability
Largest entropy
Increases as a gas expands
Increases as solids change to liquids and gases
When solutions are made, volume, positional probability, and entropy increase

8. Examples
Choose the substance with the higher positional entropy (per mole) at a given temperature
Solid CO2 and gaseous CO2
N2 gas at 1 atm and N2 gas at 1.0 X 10-2 atm
Predict the sign of the entropy change for each of the following processes:
Solid sugar is added to water to form a solution
Iodine vapor condenses on a cold surface to form crystals

9. Second Law of Thermodynamics
In any spontaneous process there is always an increase in the entropy of the universe
(first = energy of universe is constant)
∆Suniv = ∆Ssys + ∆Ssurr
Must consider both the system and surroundings to determine if a given process is spontaneous (∆Suniv must be +)

10. Sign of ∆Ssurr Depends on the direction of the heat flow
Exothermic - heat flows into the surroundings, increasing random motions and entropy of the surroundings --> ∆Ssurr is positive
Endothermic - heat flows into the system (away from the surroundings), decreasing random motions and entropy of the surroundings --> ∆Ssurr is negative

11. Magnitude of ∆Ssurr Depends on temperature
At a high temperature, energy transferred as heat won’t really make a percent change in randomness (entropy)
At a low temperature, energy transferred as heat will make a percent change in randomness (entropy)

12. Entropy vs. Enthalpy
∆Ssurr can be expressed in terms of ∆H (needs a sign - shows direction - and a number - shows quantity of energy) at constant pressure:
∆Ssurr = - ∆H/T
Temperature in Kelvins

13. Example
Calculate ∆Ssurr for each of these reactions at 25°C and 1 atm:
Sb2S3(s) + 3Fe(s) -> 2Sb(s) + 3FeS(s)
∆H = -125 kJ
Sb4O6(s) + 6C(s) -> 4Sb(s) + 6CO(g)
∆H = 778 kJ

14. MEMORIZE

15. Free Energy (G) G = H -TS H = enthalpy T = Kelvin temperature
S = entropy
At a constant temperature, the change in free energy is ∆G = ∆H -T∆S. ∆Suniv = - ∆G/T only at constant temperature and pressure!
**All quantities refer to the system
If you see a degree (°) sign after a substance, it represents that it is in its standard state (pg. 255)

16. ∆H° = 31.0 kJ/mol and ∆S° = 93.0 J/Kmol
Example
At what temperatures is the following process spontaneous at 1 atm?
Br2(l) -> Br2(g)
∆H° = 31.0 kJ/mol and ∆S° = 93.0 J/Kmol
What is the normal boiling point of liquid Br2?

17. MEMORIZE!
FYI: if a reaction is not spontaneous as written, it is spontaneous in reverse

18. Entropy in Chemical Reactions
Compare coefficients of products and reactants in balanced chemical reactions
Fewer molecules means fewer possible configurations (means entropy decreases)
STATES of matter must be taken into account (gases have more entropy than solids)

19. Example Predict the sign of ∆S° for each of the following reactions:
(NH4)2Cr2O7(s) -> Cr2O3(s) + 4H2O(l) + N2(g)
Mg(OH)2(s) -> MgO(s) + H2O(g)
PCl5(g) -> PCl3(g) + Cl2(g)

20. Third Law of Thermodynamics
The entropy of a perfect crystal at 0K is zero
(“perfect crystal” is an unattainable ideal - taken as standard, but never actually observed)
Standard entropy values S° in Appendix 4 in your book
∆S°reaction = ∑npS°products - ∑nrS°reactants
Generally, the more complex the molecule, the higher the standard entropy value
(∆H° is calculated the same way)

21. Example
Calculate ∆S° for each of the following reactions using Appendix 4 in your book
N2O4(g) -> 2NO2(g)
Fe2O3(s) + 2Al(s) -> 2Fe(s) + Al2O3(s)
4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)

22. Free Energy and Chemical Reactions
Standard free energy change (∆G°): the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states
N2(g) + 3H2(g) <-> 2NH3(g) ∆G° = kJ
Can not be directly measured, but it can be calculated from other measured quantities
The more negative ∆G°, the further to the RIGHT the reaction will go to reach equilibrium
Pressure and concentration must be held constant
Three ways to calculate ∆G°

23. Calculating ∆G° ∆G° = ∆H° - T∆S° - Requires constant temperature
2. Similar to Hess’s Law
- Flip equation = negates, multiply by constants…
3. ∆G°rxn = ∑npG°f(prod) - ∑nrG°f(reactants)
- Uses standard free energy of formations (appendix 4)

24. Example: ∆G° = ∆H° - T∆S°
Using data for ∆H° and ∆S°, calculate ∆G° for the following reactions at 25°C and 1 atm.
Cr2O3(s) + 2Al(s) -> Al2O3(s) + 2Cr(s)
Answer: -537 kJ; reaction is spontaneous
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)
Answer: kJ; highly spontaneous

25. Example: Like Hess’s Law
Given the following data, calculate ∆G° for
S(s) + O2(g) -> SO2(g):
Eq 1: S(s) + 3/2O2(g) -> SO3(g) ∆G° = -371 kJ
Eq 2: 2SO2(g) + O2(g) -> 2SO3(g) ∆G° = -142 kJ
Answer: -300 kJ

26. Example: ∆G°rxn = ∑npG°f(prod) - ∑nrG°f(reactants)
Calculate ∆G° for the reaction:
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)
using ∆G° data from Appendix 4. Compare the answer with the answer we calculated previously for this reaction
Answer: kJ

27. Pressure vs. Free Energy
Entropy depends on volume, which inversely relates to pressure via gas laws
Slarge volume > Ssmall volume
Slow pressure > Shigh pressure
Equations used in this chapter can be manipulated (pg. 795) to give:
G = G° + RT lnQ
Q = reaction quotient using initial pressures of gases, R = J/K mol, T = Kelvins

28. Example Calculate ∆G at 700K for the following reaction:
C(s, graphite) + H2O(g) <-> CO(g) + H2(g)
Initial Pressures: PH2O = 0.85 atm,
PCO = 1.0 X 10-4 atm, PH2 = 2.0 X 10-4 atm
Answer: +92kJ = not spontaneous under standard conditions…when temperature and pressure are taken into account, ∆G = -10kJ and is spontaneous

29. Free Energy and Equilibrium
Equilibrium Point occurs at the lowest value of free energy available to the reaction system
Gforward rxn = Greverse rxn
If ∆G < 0, it means that ∆Greactants > ∆Gproducts and rxn will go to the RIGHT until Greactants = Gproducts
If ∆G > 0, it means that Greactants < Gproducts and rxn will go to the LEFT until Greactants = Gproducts

30. Relationship Between ∆G° and K
∆G° = -RT ln K
K = equilibrium constant
Example: C(s, graphite) + H2O(g) <-> CO(g) + H2(g) where T = 700 K and ∆G° = 92 kJ, determine the direction of the reaction with the following initial pressure of each gas is PH2O = 0.67 atm, PCO = 0.23 atm, PH2 = 0.51 atm.
Answer: ∆G = 82kJ and reaction will go to the left

31. Cr2O3(s) + 2Al(s) <-> Al2O3(s) + 2Cr(s)
Example #2
Given ∆G° = -537 kJ, calculate K for the following reaction at 25°C
Cr2O3(s) + 2Al(s) <-> Al2O3(s) + 2Cr(s)
Answer: K = 1.3 X 1094, reaction is spontaneous