1. Spontaneity, Entropy and Free Energy

2. SPONTANEOUS PROCESSES AND ENTROPY
First Law
“Energy can neither be created nor destroyed“.
The energy of the universe is constant.
Enthalpy
Heat content – exothermic is favored.
Spontaneous Processes
Processes that occur without outside intervention
Spontaneous processes may be fast or slow
Many forms of combustion are fast.
Conversion of diamond to graphite is slow.

3. ENTROPY (S) A measure of the randomness or disorder.
The driving force for a spontaneous process is an increase in the entropy of the universe.
Entropy is a thermodynamic function describing the number of arrangements that are available to a system.
Nature proceeds toward the states that have the highest probabilities of existing.

4. SECOND LAW OF THERMODYNAMICS
"In any spontaneous process there is always an increase in the entropy of the universe"
"The entropy of the universe is increasing"
For a given change to be spontaneous, Suniverse must be positive
Suniv = Ssys + Ssurr

5. ENTROPY
The Third Law of Thermodynamics:
the entropy of a perfect crystal at 0K is zero. NOTE: not a lot of perfect crystals out there so, entropy values are RARELY ever zero—even elements. This means the absolute entropy of a substance can then be determined at any temperature higher than 0 K.

6. POSITIONAL ENTROPY
The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved
Ssolid < Sliquid << Sgas

7. POSITIONAL ENTROPY
The fewer molecules there are, the fewer microstates there can be. Increasing particles can increase Entropy.

8. PREDICTING ENTROPY
The greater the disorder or randomness in a system, the larger the entropy.
The entropy of a substance always increases as it changes from solid to liquid to gas.
When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases (carbonates are an exception - they interact with water and actually bring MORE order to the system).

9. PREDICTING ENTROPY
When a gas molecule escapes from a solvent, the entropy increases.
Entropy generally increases with increasing molecular complexity (crystal structure: KCl vs CaCl2) since there are more MOVING electrons.
Reactions increasing the number of moles of particles often increase entropy.

10. CALCULATING ENTROPY CHANGE IN A REACTION
Entropy is an extensive property (a function of the number of moles)
Generally, the more complex the molecule, the higher the standard entropy value

11. ΔS°rxn = ΣΔS°(products) - ΣΔS°(reactants)
IN GENERAL
The greater the number of arrangements, the higher the entropy of the system!
ΔS°rxn = ΣΔS°(products) - ΣΔS°(reactants)
S is + when disorder increases (favored)
S is – when disorder decreases
Units are usually J/K∙ mol (not kJ – pay attention!)

12. PRACTICE ONE
Which of the following has the largest increase in entropy?
a) CO2(s) → CO2(g)
b) H2(g) + Cl2(g) → 2 HCl(g)
c) KNO3(s) → KNO3(l)
d) C(diamond) → C(graphite)

13. PRACTICE TWO
Predict the sign of the entropy change for each of the following processes.
A: Solid sugar is added to water to form a solution.
B: Iodine vapor condenses on a cold surface to form crystals.

14. PRACTICE THREE Calculate the entropy change at 25°C, in J/K for:
2 SO2(g) + O2(g) → 2 SO3(g)
Given the following data:
SO2(g) J/K ∙ mol
O2(g) J/K ∙ mol
SO3(g) J/K ∙ mol

15. REVIEW
The driving force of a spontaneous process is an increase in the entropy of the universe.
Entropy can be viewed as a measure of molecular randomness or disorder. It is also a thermodynamic function that describes the numbers of arrangements (positions and/or energy levels) that are available to a system existing at a given state.
Nature spontaneously proceeds towards the states that has the highest probability of existing.

16. REVIEW Dissolving = > disorder S → L → G = > disorder
Evaporation = >disorder
Bigger molecules = > disorder
Reaction with more moles = > disorder

17. POSITIONAL PROBABILITY
Possible arrangements (states) of four molecules in a two-bulbed flask.

18. POSITIONAL PROBABILITY
Highest probability from the activity?
Why?

19. POSITIONAL PROBABILITY
This concept is also illustrated in changes of state.
Positional entropy increases from solid to liquid to gas.
Also applies to solution making. Solutes and solvents in solution have more available volume to move around in. Thus the formation of solutions is favored.

20. PHASE CHANGE AT A CONSTANT TEMPERATURE
ΔSsurr = heat transferred = q temperature at which change occurs T ** where the heat supplied (endothermic) (q > 0) or evolved (exothermic) (q < 0) is divided by the temperature in Kelvins. ** It is important here to note if the reaction is endothermic or exothermic.

21. PHASE CHANGE AT A CONSTANT TEMPERATURE
Taking favored conditions into consideration, the equation above rearranges into:
ΔSsurr = - ΔH T
Give signs to ΔH following exo/endo guidelines! (If reaction is exothermic; entropy of system increases)

22. PRACTICE FOUR
In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores:
Sb2S3(s) +3Fe(s) → 2Sb(s) +3FeS(s) ΔH = -125kJ
Carbon is used as the reducing agent for oxide ores:
Sb4O6(s) +6C(s) →4Sb(s) +6CO(g) ΔH = 778kJ
Calculate ΔSsurr for each of these reactions at 25°C and 1 atm.

23. ΔSsystem + ΔSsurroundings = ΔSuniverse
ENTROPY SUMMARY
ΔS = + MORE DISORDER (FAVORED CONDITION)
ΔS = - MORE ORDER
Whether a reaction will occur spontaneously may be determined by looking at the ΔS of the universe.
ΔSsystem + ΔSsurroundings = ΔSuniverse
IF ΔSuniverse is +, then reaction is spontaneous.
IF ΔSuniverse is -, then reaction is NONspontaneous.

24. EXAMPLE OF THIS Consider 2 H2(g) + O2(g) → 2 H2O(g)
Ignite this and the reaction is fast!
ΔSsystem = -88.9J/K
entropy declines (due mainly to 3→2 moles of gas!). . . to confirm we need to know entropy of surroundings.
ΔSsurr = q surr T
this comes from ΔH calculation: Hsystem = kJ

25. EXAMPLE OF THIS
First law of thermodynamics demands that this energy is transferred from the system to the surroundings so...
-ΔHsystem = ΔHsurr OR -( kJ) = kJ
ΔS°surr = ΔH°surr = kJ = 1620 J/K
T K
Now we can find ΔS°universe
ΔSsystem + ΔSsurr = ΔSuniverse
(-88.9 J/K) + (1620 J/K) = 1530 J/K

26. EXAMPLE OF THIS
Even though the entropy of the system declines, the entropy change for the surroundings is SOOO large that the overall change for the universe is positive.
Bottom line: A process is spontaneous in spite of a negative entropy change as long as it is extremely exothermic. Sufficient exothermicity offsets system ordering.

27. GIBBS FREE ENERGY
Calculation of Gibb’s free energy is what ultimately decides whether a reaction is spontaneous or not.
NEGATIVE ΔG’s are spontaneous.

28. STANDARD FREE ENERGY CHANGE
G° is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.
G° cannot be measured directly.
The more negative the value for G°, the farther to the right the reaction will proceed in order to achieve equilibrium.
Equilibrium is the lowest possible free energy position for a reaction.

29. METHOD # 1 ΔG°rxn = ΣΔG°(products) - ΣΔG°(reactants)
This works the same way as enthalpy and entropy from tables of standard values. Standard molar free energy of formation. ΔG°f = 0 for elements in standard state.

30. METHOD # 2
ΔG = ΔH - TΔS
If G is -, the rxn is spontaneous in the forward direction.
If G = 0, the rxn is at equilibrium.
If G is +, the rxn is nonspontaneous in the forward direction, but the reverse rxn will be spontaneous.
G°f for elements at standard state (pure elements at 25ºC and 1 atm are assigned a value of zero).

31. METHOD # 3 Hess’s law summation
Works same as Hess’s in the enthalpy section - sum up equations using the guidelines as before.

32. METHOD # 4 G = ΔG° + RT ln(Q)
Define terms: ΔG = free energy not at standard conditions
ΔG° = free energy at standard conditions
T = temperature in Kelvin
R = universal gas constant J/mol∙K
ln = natural log
Q = reaction quotient: (for gases this is the partial pressures of the products divided by the partial pressures of the reactants—all raised to the power of their coefficients) Q = [products]÷ [reactants].

33. METHOD # 5 “RatLink”: ΔG°= -RTlnK
Terms: basically the same as above. Here the system is at equilibrium, so ΔG = 0 and K represents the EQ constant under standard conditions. K = [products]÷ [reactants] still raised to power of coefficients.

34. METHOD #6 “nFe”: ΔG°= - nFE° remember this one
Terms: ΔG° = just like above—standard free energy
n = number of moles of electrons transferred (look at ½ reactions)
F = Faraday’s constant 96,485 Coulombs/mole electrons
E° = standard voltage ** one volt = joule/coulomb**

35. PRACTICE FIVE
Given the following information, find the free energy of formation for the oxidation of water to produce hydrogen peroxide: 2 H2O(l) + O2(g) → 2 H2O2(l)
ΔG°f
H2O(l) kcal/mol
O2(g) kcal/mol
H2O2(l) kcal/mol
ΔG°rxn = ΣΔG°(products) - ΣΔG°(reactants)

36. PRACTICE SIX
Consider the reaction 2SO2(g) +O2(g)→2S)3(g) carried out at 25°C and 1 atm. Calculate ΔH°, ΔS°, and ΔG° using the following data:
Substance ΔH°f S° (J/K mol)
SO2(g)
SO3(g)
O2(g)
ΔG = ΔH - TΔS

37. PRACTICE SEVEN Using the following data (at 25°C)
Cdiamond(s) + O2(g) → CO2(g) ΔG°= -397 kJ
Cgraphite(s) + O2(g) → CO2(g) ΔG°= -394 kJ
Calculate ΔG° for the reaction: Cdiamond(s) →Cgraphite(s)
Hess’s law of summation

38. PRACTICE EIGHT
One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases: CO(g) + 2H2(g) →CH3OH(l) Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol.
ΔG°f kJ
CH3OH -166
H2 0
CO -137
ΔG = ΔG° + RT ln (Q)

39. PRACTICE NINE
The overall reaction for the corrosion (rusting) of iron by oxygen is 4 Fe (s) + 3 O2 (g) ↔ 2 Fe2O3 (s). Using the following data, calculate the equilibrium constant for this reaction at 25°C.
Substance ΔH°f S°(J/K mol)
Fe2O3(s)
Fe(s)
O2(g)
ΔG° = -RTlnK

40. ΔG° = ΔH - T ΔS
Gibb’s equation can also be used to calculate the phase change temperature of a substance.
During the phase change, there is an equilibrium between phases so the value of ΔG° is zero.

41. PRACTICE TEN
Find the thermodynamic boiling point of H2O(l) → H2O(g) given the following information:
Hvap = +44kJ Svap = 118.8J/K
ΔG° = ΔH - T ΔS

42. H, S, G AND SPONTANEITY
G = H - TS H is enthalpy, T is Kelvin temperature
Value of H
Value of TS
Value of G
Spontaneity
Negative
Positive
Spontaneous
Nonspontaneous
???
Spontaneous if the absolute value of H is greater than the absolute value of TS (low temperature)
Spontaneous if the absolute value of TS is greater than the absolute value of H (high temperature)

43. SUMMARY OF FREE ENERGY ΔG = + NOT SPONTANEOUS ΔG = - SPONTANEOUS
Conditions of ΔG: ΔH ΔS E
negative
positive
Spontaneous at all temperatures
Spontaneous at high temperatures
Spontaneous at low temperatures
Not spontaneous, ever

44. RELATIONSHIP OF ΔG TO K AND E
Equilibrium point occurs at the lowest value of free energy available to the reaction system
At equilibrium, G = 0 and Q = K ΔG K E
At EQ; K = 1
negative
>1; products favored
positive
<1; reactants favored

45. RELATIONSHIP OF ΔG TO K AND E
Minimum free energy is reached when 75% of A has been changed to B.
Each point on the curve corresponds to the total free energy of the system for a given combination of A and B.

46. FREE ENERGY AND WORK
The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. wmax = ΔG
The amount of work obtained is always less than the maximum.
Henry Bent's First Two Laws of Thermodynamics:
First law - You can't win, you can only break even
Second law - You can't break even.

47. FREE ENERGY AND WORK
When the reaction is spontaneous, ΔG is the energy available to do work such as moving a piston or flowing of electrons.
When ΔG is positive, and thus non-spontaneous, it represents the amount of work needed to make the process occur.

48. FREE ENERGY AND PRESSURE
Enthalpy, H, is NOT pressure dependent
Entropy, S is pressure dependent
entropy depends on volume, so it also depends on pressure
Slarge volume > Ssmall volume
Slow pressure > Shigh pressure

49. TEMPERATURE DEPENDENCE OF K
So, ln(K)  1/T