1. Chapter 17 Free Energy and
Lecture Presentation
Chapter 17 Free Energy and
Thermodynamics
Catherine MacGowan
Armstrong Atlantic State University
© 2013 Pearson Education, Inc.

2. Kinetics vs. Thermodynamics
How to predict if a reaction can occur at a reasonable rate (speed, intermediate states)
KINETICS
Potential Energy (kJ)
How to predict if a reaction can occur, given enough time (initial and final states, spontaneity)
THERMODYNAMICS
Reaction progress 

3. The Laws of Thermodynamics
Two bodies at the same temperature have or are at thermal equilibrium.
Energy (E) in the universe (system + surroundings) is conserved.
DE = heat (q) + work (w or PDV)
DH = q (at constant P)
Total entropy (S) of the universe (total) MUST increase in a spontaneous process.
DSuniverse = DSsystem + DSsurroundings > 0
Order  disorder
Explains why heat NEVER flows from a cold body to a hot body.
Entropy of a substance at absolute zero (0K) for a pure substance is zero.

4. First Law of Thermodynamics: That’s all there is!
Energy cannot be created or destroyed.
The total energy of the universe cannot change.
However, it can transfer (flow) from one place to another.
DEuniverse = 0 = DEsystem + DEsurroundings
Energy conservation:
Example:
In an exothermic reaction, “lost” heat from the system goes into the surroundings.
Two ways energy is “lost” from a system
Converted to heat, q
Used to do work, w
DE = q + w or DH + PDV, where DH (enthalpy) ~ DE at constant pressure 4

5. You Can’t Break Even: The Energy Tax
Every energy transition results in a “loss” of energy.
Conversion of energy to heat, which is “lost” by heating up the surroundings
Example:
Recharging a battery with 100 kJ of useful energy will require more than 100 kJ.

6. Potential Energy: Mechanical vs. Chemical
The direction of spontaneity for any process can be determined by comparing the potential energy of the system at the start and the end.

7. DH is measured in kJ/mol.
Review: Enthalpy (DH)
DH is measured in kJ/mol.
A reaction is generally exothermic if the bonds in the products are stronger than the bonds in the reactants.
Exothermic = energy released, DH is negative
Stronger bonds = more stable molecules
A reaction is generally endothermic if the bonds in the products are weaker than the bonds in the reactants.
Endothermic = energy absorbed, DH is positive
The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions.

8. Entropy (S) 2nd Law of Thermodynamics
It is a thermochemical physical property related to energy dispersal.
A spontaneous process results in an increase in the entropy of the universe.
One property common to spontaneous processes is that the energy of the final state is more dispersed.
In a spontaneous process energy goes from being more concentrated to being more dispersed.
Entropy change (DS) is favorable when the result is a more random system.
Have a positive (+) value for DSsys

9. Changes That Increase a System’s Entropy
Molecular complexity
The larger the molecule, the more spatial movements.
Temperature elevation/ energy flow
Hot flows to cold
Reactions whose products are in more random state
Ssolid < Sliquid < Sgas
Reactions that have greater number of product molecules than reactant molecules

10. Entropy and Phase Changes
H2O(s)  H2O(l)
Lower entropy
H2O(l)  H2O(g)
Greater entropy

11. Changes That Increase a System’s Entropy
Dissolution of solid into liquid
Liquids or solids dissolve in a solvent in a spontaneous process owing to the increase in entropy.
Matter and energy are more dispersed.
Entropies of ionic solids depend on coulombic attractions.
Ion size
Small more compact; more ions per volume
Example: NaF (52 J/mol) vs. MgO (27 J/mol)

12. Figure 17.5 “Places” for Energy

13. Chapter 17, Unnumbered Figure 2, Page 668

14. Thermodynamics and Spontaneity
Thermodynamics predicts whether a process will proceed under the given conditions.
Spontaneous process
Nonspontaneous processes require energy input to go.
Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy (DG) of the system after the reaction.
If the system after the reaction has less potential energy than before the reaction, the reaction is thermodynamically favorable.
Spontaneity DOES NOT imply fast or slow or length of time.
NOTE: Graphite is more stable than diamond, so the conversion of diamond into graphite is a spontaneous reaction.

15. Directionality of Reactions Energy Dispersal
Probability suggests that a spontaneous reaction/process will result in the dispersal of energy.
More disorder
More places to occupy
As the size of the container increases:
the number of microstates accessible to the system increases
the density of states increases
entropy increases
When a substances changes state, the number of macro states it can have changes as well.
The more degrees of freedom the particles have, the more macro states are possible.
Solids have fewer macro states than liquids, which have fewer macro states than gases.

17. Reversibility of Process
Spontaneous processes are irreversible.
They will proceed in only one direction.
A reversible process will proceed back and forth between the two end conditions.
Equilibrium
Results in no change in free energy
If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction. 17

18. Factors Affecting Whether a Reaction Is Spontaneous
Reactions are spontaneous in the direction of lower chemical potential energy.
There are two factors that determine the thermodynamic favorability:
1) change in enthalpy (DH)
2) change in entropy (DS)
In general, EXOTHERMIC chemical reactions are spontaneous.
The entropy change, DS, is the
difference in randomness of the
reactants compared to the products. 18

19. Problem: Predict whether DSsystem is (+) or (−) for each of the following.
Heating air in a balloon
Water vapor condensing
Separation of oil and vinegar salad dressing
Dissolving sugar in tea
2 HgO(s)  2 Hg(l) + O2(g)
2 NH3(g)  N2(g) + 3 H2(g)
Ag+(aq) + Cl−(aq)  AgCl(s)

20. Heating air in a balloon (+)
Water vapor condensing (-)
Separation of oil and vinegar salad dressing (-)
Dissolving sugar in tea (+)
2 HgO(s)  2 Hg(l) + O2(g) (+)
2 NH3(g)  N2(g) + 3 H2(g) (+)
Ag+(aq) + Cl−(aq)  AgCl(s) (-)

21. Entropy Changes for Phase Changes
For a phase change: ∆S = q/T
where q = heat transferred in phase change
For H2O(l)  H2O(g)
∆Hvap = q = +40,700 J/mol
T = oC
∆S = q/T
∆S = (40,700 J/mol /373 K)
∆S = 1.09 x 102 J/mol K, or 109 J/mol K
NOTE: Entropy units are J/mol K.

22. The Second Law of Thermodynamics
The total entropy change of the universe must be positive for a process to be spontaneous.
For reversible process, DSuniv = 0.
For irreversible (spontaneous) process, DSuniv >0.
DSuniverse = DSsystem + DSsurroundings
If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount.
If DSsystem is negative, then DSsurroundings is positive.

23. DSsystem DSsystem = DSreaction = S(S°prod) − S(S°react)
If the final condition is more random than the initial condition, DSsystem is positive.
Favorable entropy
If the final condition is more orderly than the initial condition, DSsystem is negative.
Unfavorable entropy
DSsystem = DSreaction = S(S°prod) − S(S°react)

24. PROBLEM:
Determine the DSo for the chemical reaction (system):
2 H2(g) + O2(g)  2 H2O(l)
DSsystem = DSreaction = S(S°prod) − S(S°react)
∆So = [2 So (H2O(l))] - [(2 So (H2)(g)) + (So (O2)(g))]
∆So = [2 mol (69.9 J/K·mol)] - [2 mol (130.7 J/K·mol) mol (205.3 J/K·mol)]
∆So = J/K
Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid.

25. Temperature Dependence of DSsurroundings
When a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings.
When a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings.
The amount of entropy of the surroundings changes depending on its initial temperature.
The higher the original temperature, the less effect addition or removal of heat has.
DSsurroundings = -(DHsystem) / T

26. Problem: Given that DSosys = - 326.9 J/K and under
standard conditions, calculate DSosurr and DSuniv
for the chemical reaction 2 H2(g) + O2(g)  2 H2O(l).
Strategy:
Calculate DSosurroundings using
DSsurroundings = - (DHsystem) / T
Calculate DSuniv using
DSuniverse = DSsystem + DSsurroundings

27. DSosurr Answer to Problem: Given DSosys = - 326.9 J/K, calculate the
DSosurr and DSuniv for 2 H2(g) + O2(g)  2 H2O(l).
DSosurr
∆Ho = ∆Hosystem = kJ (note the kJ)
DSosurr = - [( kJ)(1000 J/1 kJ)/ K)]
DSosurr = J/K
* K is room temperature
DSuniv
DSuniv = DSsys + DSsurr
DSuniv = J/K J/K
DSuniv = J/K
Reaction/process is SPONTANEOUS, for DSuniv > 0.

28. Spontaneous at all temperatures Not spontaneous at all temperatures
Spontaneous or Not?
Reaction
Type
DHo (system)
DSo (system)
Spontaneous Process 1
Exothermic < 0
Positive > 0
Spontaneous at all temperatures
DSo (universe) > 0 2
Negative < 0
Depends on relative magnitude of DHo and DSo. Spontaneous at low temperatures. 3
Endothermic > 0
Depends on relative magnitude of DHo and DSo. Spontaneous at high temperatures. 4
Not spontaneous at all temperatures
DSo (universe) < 0

29. Problem: The reaction below has DHrxn = + 66. 4 kJ at 25 °C
Problem: The reaction below has DHrxn = kJ at 25 °C. Determine : 1. DSsurr 2. The sign of DSsys 3. Whether the process is spontaneous Reaction: 2 O2(g) + N2(g)  2 NO2(g) Given: DHsys = kJ, T = 25.0 ºC = 298 K Find: DSsurr (J/K), DSsys (+) or (−), DSuniv (+) or (−)

30. DSsurr is (−); it is unfavorable.
Answer to Problem: Reaction: 2 O2(g) + N2(g)  2 NO2(g) Given: DHsys = kJ, T = 25 ºC = 298 K Find: DSsurr (J/K), DSsys (+) or (−), DSuniverse (+) or (−) Answer: DSsurr = - DHsys / T DSsurr = - (66.4 kJ / 298 K) DSsurr = kJ/K, or – 223 J/K
DSsurr is (−); it is unfavorable.
The major difference is that there are fewer product molecules than reactant molecules; DSsys is unfavorable and (−).
Since both DSsurr and DSsys are (−), DSuniverse is (−).
The process is nonspontaneous.

31. Absolute Entropy The Third Law of Thermodynamics
The absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles.
The third law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol K.
Therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy.
Therefore, the absolute entropy of substances is always +.

32. Gibbs Free Energy (G) The Energy of Work

33. What Is Gibbs Free Energy?
A thermodynamic quantity that combines enthalpy (H) and entropy (S) into a single quantity
G = H – TS
The energy a chemical reaction uses to do work
Energy available after bonds are broken and formed to do chemical or biochemical work
The maximum amount of work energy that can be released to the surroundings by a system that is at a constant temperature and pressure
Often called the chemical potential energy because it is analogous to the storing of energy in a mechanical system

34. Free Energy and Reversible Reactions
The change in free energy is the theoretical limit as to the amount of work that can be done.
If the reaction achieves its theoretical limit, it is a reversible reaction.

35. Real Reactions
In a real reaction (nonideal conditions), some of the free energy is “lost” as heat.
Real reactions are irreversible.

36. Gibbs Free Energy, DG DG and Spontaneity
Gibbs free energy value is a combination of enthalpy (H) and entropy (S) at a given temperature for a system.
DGsys = DHsys−TDSsys
DSuniv determines whether a process is spontaneous.
DG also determines spontaneity.
DSuniv is (+) when spontaneous, so DG must be (−). DG
Its value predicts the direction of a chemical reaction.
Positive (+) DG: the reaction is nonspontaneous
Negative (-) DG: the reaction is spontaneous
DG = 0: reaction is at equilibrium

37. Gibbs Free Energy, DG DG will be negative when:
DH is negative (-) and DS is positive (+)
Exothermic and more random
DH is negative (-) plus large and DS is negative (-) but small
DH is positive (+) but small and DS is positive (+) plus large
Or at high temperature
DG will be positive when:
DH is positive (+) and DS is negative (−)
Never spontaneous at any temperature
DG = 0 when:
the reaction is at equilibrium.

38. Gibbs Free Energy and the Other Thermodynamic Properties
Given the following relationships:
∆Suniv = ∆Ssurr + ∆Ssys and ∆Suniv = - ∆Hsys + ∆Ssys T
With substitutions and multiplying through by (-T), then
-T∆Suniv = ∆Hsys - T∆Ssys
Where (-T∆Suniv) equals the change in Gibbs free energy for the system (∆Gsys)
So, under standard conditions:
∆Gosys = ∆Hosys - T∆Sosys

39. ∆Go = ∆Ho - T∆So ∆Go = ∆Ho - T∆So
Gibbs free = total energy change - energy lost due
energy change for the system to its dispersal
at specific T
If reaction is
• exothermic (negative ∆Ho)
• and entropy increases (positive ∆So)
• then ∆Go must be NEGATIVE
The reaction is spontaneous (and product favored).
• endothermic (positive ∆Ho)
• and entropy decreases (negative ∆So)
• then ∆Go must be POSITIVE
The reaction is not spontaneous (and is reactant favored).

40. DGo Value Summary Table
∆Go = ∆Ho - T∆So
Temperature INDEPENDENT
∆Ho ∆So ∆Go Reaction Spontaneity
exo(–) increase(+) – Product favored Spontaneous
endo(+) decrease(-) Reactant favored Not spontaneous
Temperature DEPENDENT
exo(–) decrease(-) High temperature Not spontaneous
exo(–) decrease(-) Low temperature Spontaneous
endo(+) increase(+) High temperature Spontaneous
endo(+) increase(+) Low temperature Not spontaneous

41. Gibbs Free Energy Relationships
Two methods of calculating ∆Go
Use known values of ∆Ho and ∆So and the Gibbs equation.
∆Go = ∆Ho - T∆So
b) Use tabulated values of free energies of formation, ∆Go.
∆Go = S∆Go (Products) - S∆Go (Reactants)
Note: ∆Go formation values for elements = ZERO

42. DG Relationships
If a reaction can be expressed as a series of reactions, the sum of the DG values of the individual reaction is the DG of the total reaction.
∆Go = S∆Go (Products) - S∆Go (Reactants)
DG is a state function; therefore:
If a reaction is reversed, the sign of its DG value reverses
the DG value is multiplied by the reaction’s stoichiometric factors
The value of DG of a reaction is:
extensive
phase specific

43. Calculating ∆Gorxn Problem: Combustion of carbon:
C(graphite) + O2(g)  CO2(g)
Determine the DGo and spontaneity of the reaction.
Answer: Use DGo table values and summation formula.
∆Go = S∆Go (Products) - S∆Go (Reactants)
∆Go = ∆Go(CO2) - [∆Go(graphite) + ∆Go(O2)]
∆Go = kJ - [ ]
∆Go = kJ
Reaction is product favored and spontaneous.

44. Problem: Calculate DG at 25 C for CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g).
Substance
DG kJ/mol
CH4(g)
−50.5
O2(g)
0.0
CO2(g)
−394.4
H2O(g)
−228.6
O3(g)
163.2
Given:
Solution:
DGo = (SDGproductso – SDGreactantso)
DGo = [( kJ/mol x 1 mol) + ( kJ/mol x 2 mol)] - [(-50.5 kJ/mol x 1 mol) + 0 kJ]
DGo = kJ

45. Problem:. Calculate DG and determine whether the reaction below
Problem: Calculate DG and determine whether the reaction below is spontaneous. CCl4(g)  C (s, graphite) + 2 Cl2(g) Given: DH = kJ, DS = J/K, T = 298 K DG = DH - TDS
Solution:
DG = kJ – {(298 K)( J/K)}
DG = 53.5 kJ
Since DG is +, the reaction is not spontaneous at this temperature.
To make it spontaneous, we need to increase the temperature.

46. C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DGº = −2074 kJ
Problem: Using the given equations, find DGº for the following reaction: C(s) + 4 H2(g)  C3H8(g)
Given:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DGº = −2074 kJ
C(s) + O2(g)  CO2(g) DGº = −394.4 kJ
2 H2(g) + O2(g)  2 H2O(g) DGº = −457.1 kJ
Solution:
3 CO2(g) + 4 H2O(g)  C3H8(g) + 5 O2(g) DGº = kJ
3 x [C(s) + O2(g)  CO2(g) ] DGº = 3(−394.4 kJ)
2 x [2 H2(g) + O2(g)  2 H2O(g)] DGº = 2(−457.1 kJ)
3 C(s) + 4 H2(g)  C3H8(g) DGº = -23 kJ

47. Problem: The combustion of acetylene:
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g)
Find DHo, DSsyso, DSuniv, Dssurr, and DGo for the above reaction at 25.0 oC.
Determine whether the reaction is product favored or reactant favored and why.
What is its spontaneity?
Strategy: Use enthalpies of formation to calculate ∆Ho.
Use standard molar entropies to calculate ∆Ssyso.
Use the relationship of ∆Go = ∆Ho - T∆So.
Use the relationship of ∆Ssurr = -(DHo)/T.
Use the relationship DSuniv = DSsys + DSsurr.

48. Answer:
Use enthalpies of formation to calculate ∆Ho.
Hess’s law: ∆Ho = S∆Hproducts - S∆Hrecatants ∆Ho = kJ
Use standard molar entropies to calculate ∆Ssyso. ∆Ssyso = J/K
∆Ssyso = S∆Sproducts - S∆Srecatants or kJ/K
Use the relationship of ∆Go = ∆Ho - T∆So.
∆Go = kJ - (298 K)( kJ/K) ∆Go = kJ
Use the relationship of ∆Ssurr = -(DHo)/T. ∆Ssurr = 4155 J/K
∆Ssurr = {-(-1238 kJ)(1000 J/kJ)}/298 K
Use the relationship DSuniv = DSsys + DSsurr. DSuniv = 4058 J/K
DSuniv = J/K J/K
Spontaneous and product favored
Driven by negative ∆Ho and large positive DSuniv

49. Thermodynamics and Keq
∆Go is the change in free energy when pure reactants convert COMPLETELY to pure products.
Keq is related to reaction favorability.
Product-favored systems have Keq > 1.
Therefore, both ∆G˚ and Keq are related to reaction favorability.
The larger the value of K, the more negative the value of ∆Go.
∆Go = - RT lnKeq
where R = 8.31 J/K•mol

50. The relation of ∆G, ∆G˚, Q and K, reaction spontaneity, and product or reactant favorabilityDG
Spontaneous?
Q < K
DG < 0
Spontaneous towards products
Q = K
DG = 0
Reaction is at equilibrium
Q > K
DG > 0
Not spontaneous to product formation; favors reactants

52. The relation of ∆G, ∆G˚, Q and K, reaction spontaneity, and product - or reactant favorability.
DGo
Reactant-Favored or Product Favored at Equilibrium
Spontaneous under Standard Conditions?
K >> 1
DGo < 0
Product-Favored
Yes: Spontaneous
K = 1
DGo = 0
[Products]eq = [Reactants]eq
Yes: At equilibrium
K << 1
DGo > 0
Reactant-Favored
No: Not Spontaneous

54. Problem: The ∆Go for the reaction N2O4  2 NO2 is +4. 8 kJ
Problem: The ∆Go for the reaction N2O4  2 NO2 is +4.8 kJ Calculate K for this reaction.
∆Go = -RT lnK
Answer: ∆Go = J = - (8.31 J/K)(298 K) ln K
ln K = - (4800 J/{(8.31 J/K(298 K)}
ln K =
Take the antilog of both sides :
K = e-1.94
K = 0.14
When ∆Go > 0, K < 1.
When ∆Go > 0, the reaction is not spontaneous in that direction; reaction is reactant favored according to K value.

55. ∆G, ∆G˚, and Keq DG = DG + RT lnQ where Q is the reaction quotient
∆G is change in free energy at nonstandard conditions.
DG is change in free energy at standard conditions
DG = DG only when the reactants and products are in their standard states.
Under nonstandard conditions,
DG = DG + RT lnQ where Q is the reaction quotient
When Q < K or Q > K, reaction is spontaneous.
When Q = K, reaction is at equilibrium.
When ∆G = 0, the reaction is at equilibrium.
With DG = 0 at equilibrium, the above equation becomes
DG = −RT lnK

56. Problem: Calculate DG at 298 K for the following reaction under the given conditions: 2 NO(g) + O2(g)  2 NO2(g) DGº = −71.2 kJ DG = DGo + RT lnQ Answer: Q = (PNO2)2 / {(PNO)2(PO2)} Q = (2.00)2 / {(0.100)2(0.100)} = 4.00 x DG = DGo + RT lnQ DG = (-7.12 x 104 J) + (8.314 J/K mol)(298 K)(ln 4.00 x 103) DG = x 104 J, or – 50.7 kJ
Substance
P, atm
NO(g)
0.100
O2(g)
NO2(g)
2.00

57. Problem: Calculate DGrxn for the following reaction under the given conditions: DGº = kJ T = 700 K N2(g) + 3 H2(g)  NH3(g) DG = DGo + RT lnQ Answer: Q = (PNH3)2 / {(PN2)(PH2)2} Q = (2.00)2 / {33)(99)2} = 1.20 x DG = DGo + RT lnQ DG = (+4.61 x 104 J) + (8.314 J/K mol)(700 K)(ln 1.20 x 10-7) DG = x 104 J, or – 46.1 kJ
Substance
P, atm
N2(g) 33
H2(g) 99
NH3(g)
2.0