1. Chapter 19 Lecture presentation
Free Energy
and
Thermodynamics
Catherine E. MacGowan
Armstrong Atlantic State University

2. Thermodynamics and Spontaneity
Thermodynamics predicts whether a process will occur under the given conditions.
Processes that will occur are called spontaneous.
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 of the system after the reaction.
If the system after the reaction has less potential energy than the system before the reaction, the reaction is thermodynamically favorable.
Spontaneity ≠ fast or slow

3. Reversibility of Process
Any spontaneous process is irreversible because there is a net release of energy when it proceeds in that direction.
It will proceed in only one direction.
A reversible process will proceed back and forth between the two end conditions.
Any reversible process is at equilibrium.
This results in no change in free energy.
If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction.

4. Thermodynamics versus Kinetics

5. Spontaneous Processes
Spontaneous processes occur because they release energy from the system.
Most spontaneous processes proceed from a system of higher potential energy to a system at lower potential energy.
Exothermic
But there are some spontaneous processes that proceed from a system of lower potential energy to a system at higher potential energy.
Endothermic

6. Factors Affecting whether a Reaction Is Spontaneous
There are two factors that determine whether a reaction is spontaneous. They are the enthalpy change and the entropy change of the system.
The enthalpy change, ΔH, is the difference between the sum of the internal energy and PV work energy of the reactants and that of the products.
The entropy change, ΔS, is the difference between the randomness of the reactants and that of the products.

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

8. Entropy and the Second Law of Thermodynamics
Entropy (S) is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases.
Entropy (S) has units of J/mol.
S = k ln W
k = Boltzmann constant = 1.38 × 10−23 J/K
W is the number of energetically equivalent ways a system can exist.
Random systems require less energy than ordered systems.

9. Examples of Entropy: Salt Dissolving in Water

10. Water Evaporating: Entropy Increasing

11. Entropy and the Second Law of Thermodynamics
Remember: The first law of thermodynamics says that energy cannot be created or destroyed.
ΔEuniv = 0 = ΔEsys + ΔEsurr
The total energy of the universe cannot change, but energy can be transferred from on place to another.
The second law of thermodynamics says that the total entropy change of the universe must be positive for a process to be spontaneous.
ΔSuniv = ΔSsys + ΔSsurr
For a reversible process, ΔSuniv = 0.
For an irreversible (spontaneous) process, ΔSuniv > 0.
If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount.
When ΔSsys is negative, ΔSsurr must be positive and big for a spontaneous process.

12. Entropy: Expansion of Gas in a Vacuum
Consider the following situation:
When the stopcock is open, the gas atoms distribute themselves in a number of arrangements.
These are energetically equivalent states (A, B, and C) for the expansion of a gas.
It doesn’t matter, in terms of potential energy, whether the molecules are all in one flask or evenly distributed.
But one of these states is more probable than the other two.

13. Macrostates and Probability
There is only one possible arrangement that gives state A and one that gives state B.

14. Macrostates → Microstates Probabilities
Macrostate C can be achieved through several different arrangements of the particles.
These microstates all have the same macrostate.
There are six different particle arrangements that result in the same macrostate. 14

15. Macrostates and Probability
As seen, state A and state B each have only ONE possible arrangement.
Whereas state C has SIX possible arrangements.
The macrostate with the highest entropy also has the greatest dispersal of energy.
Therefore, state C has higher entropy than either state A or state B.

16. Changes in Entropy, ΔS ΔS = Sfinal − Sinitial
Entropy change is favorable when the result is a more random system.
ΔS is positive.
Changes that increase the entropy are as follows:
Reactions whose products are in a more random state
Solid more ordered than liquid; liquid more ordered than gas
Reactions that have larger numbers of product molecules than reactant molecules
Increase in temperature
Solids dissociating into ions upon dissolving

17. Entropy Change in State Change
When materials change state, the number of macrostates they can have changes as well.
The more degrees of freedom the molecules have, the more macrostates are possible.
Solids have fewer macrostates than liquids, which have fewer macrostates than gases.
Ssolid < Sliquid < Sgas

18. Practice Problem: Predicting the Sign of ΔS

19. ΔS: Change in Entropy and Chemical Reactions
For a process in which the final condition is more random than the initial condition,
ΔSsys is positive; and
entropy change is favorable for the process to be spontaneous.
For a process in which the final condition is more orderly than the initial condition,
ΔSsys is negative; and
entropy change is unfavorable for the process to be spontaneous.
ΔSsys = ΔSrxn = Sn(S°products) – Sn(S°reactants)

20. Standard Conditions
The standard state is the state of a material at a defined set of conditions.
Gas = pure gas at exactly 1 atm pressure
Solid or Liquid = pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest
Usually 25 °C
Solution = substance in a solution with concentration 1 M

21. Standard Absolute Entropies (S°)
S° designates standard state conditions.
Entropy is an extensive physical property of matter.
Entropies are for 1 mol of a substance at 298 K for a particular state, a particular allotrope, a particular molecular complexity, a particular molar mass, and a particular degree of dissolution.

22. The Third Law of Thermodynamics: Absolute Entropy
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 is 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 positive.

23. Relative Standard Entropies: States
The gas state has a larger entropy than the liquid state at a particular temperature.
The liquid state has a larger entropy than the solid state at a particular temperature.

24. Standard Absolute Entropies

25. Relative Standard Entropies: Molar Mass
The larger the molar mass, the larger the entropy.
Available energy states are more closely spaced, allowing more dispersal of energy through the states.

26. Relative Standard Entropies: Allotropes
The less constrained the structure of an allotrope is, the larger its entropy.
The fact that the layers in graphite are not bonded together makes graphite less constrained.

27. Relative Standard Entropies: Molecular Complexity
Larger, more complex molecules generally have larger entropy.
More energy states are available, allowing more dispersal of energy through the states.

28. “Places” for Energy in Gaseous NO

29. Relative Standard Entropies: Dissolution
Dissolved solids generally have larger entropy, distributing particles throughout the mixture.

30. The Standard Entropy Change, ΔS°rxn
The standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions.
ΔS°rxn = (∑npS°products) – (∑nrS°reactants)
Remember, although the standard enthalpy of formation, ΔHf°, of an element is 0 kJ/mol, the absolute entropy at 25 °C, S°, is always positive.

31. Practice Problem: Calculating Standard Changes in Entropy (ΔSrxn)

32. Heat Transfer and Changes in Entropy of the Surroundings
The second law demands that the entropy of the universe increase for a spontaneous process.
Yet processes like condensation of water vapor are spontaneous, even though the water vapor is more random than the liquid water.
If a process is spontaneous, yet the entropy change of the process is unfavorable, there must have been a large increase in the entropy of the surroundings.
The entropy increase must come from heat released by the system; the process must be exothermic!

33. Heat Exchange and ΔSsurr
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 the entropy of the surroundings changes depends on the original temperature.
The higher the original temperature, the less effect addition or removal of heat has.

34. Entropy Change in the System and Surroundings
When the entropy change in a system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive) and large in order to allow the process to be spontaneous.

35. Temperature Dependence of ΔSsurr
When heat is added to surroundings that are cool, it has more of an effect on the entropy than it would have if the surroundings were already hot.
Water freezes spontaneously below 0 °C because the heat released on freezing increases the entropy of the surroundings enough to make ΔS positive.
Above 0 °C the increase in entropy of the surroundings is insufficient to make ΔS positive.

36. Quantifying Entropy Changes in Surroundings
The entropy change in the surroundings is proportional to the amount of heat gained or lost.
qsurr = −qsys
The entropy change in the surroundings is also inversely proportional to the temperature.
At constant pressure and temperature, the overall relationship is as follows:
ΔSsurr =
–qsys
Temperature (T) =
(–) ΔHsys T

37. Practice Problem: Calculating Changes in Entropy in the Surroundings

38. Gibbs Free Energy, ΔG
A process will be spontaneous when ΔG is negative.
ΔG will be negative under the following conditions:
ΔH is negative and ΔS is positive.
Exothermic and more random
ΔH is negative and large and ΔS is negative but small.
ΔH is positive but small and ΔS is positive and large.
Or high temperature
• ΔG will be positive under the following conditions:
ΔH is positive and ΔS is negative.
Never spontaneous at any temperature
When ΔG = 0, the reaction is at equilibrium.

39. ΔG, ΔH, and ΔS

40. Gibbs Free Energy and Spontaneity
It can be shown that –TΔSuniv = ΔHsys – TΔSsys.
The Gibbs free energy, G, is the maximum amount of work energy that can be released to the surroundings by a system for a constant temperature and pressure system.
Gibbs free energy is often called the chemical potential because it is analogous to the storing of energy in a mechanical system.
ΔGsys = ΔHsys – TΔSsys
Because ΔSuniv determines whether a process is spontaneous, ΔG also determines spontaneity.
ΔSuniv is positive when spontaneous, so ΔG is negative.

41. Free Energy Change and Spontaneity

42. Practice Problem: Calculating Gibbs Free Energy Changes and Predicting Spontaneity

43. Calculating ΔG°
At 25 °C,
ΔG°rxn = SnΔG°f(products) – SnΔG°f(reactants)
At temperatures other than 25 °C, assuming the change in ΔH°rxn and ΔS°rxn is negligible,
ΔG°rxn = ΔH°rxn – ΔS°rxn Or
ΔG°total = ΔG°rxn1 – ΔG°rxn2 + …

44. Practice Problem: Calculating Standard Change in Gibbs Free Energy Using the Equation ΔG°rxn= ΔH°rxn – ΔS°rxn

45. Practice Problem: Calculating Gibbs Free Energy

46. Standard Free Energies of Formation
The free energy of formation (ΔG°f) is the change in free energy when 1 mol of a compound forms from its constituent elements in their standard states.
ΔG°rxn = SnΔG°f(products) – SnΔG°f(reactants)
The free energy of formation of pure elements in their standard states is zero.

47. Practice Problem: Calculating ΔG°rxn from Standard Free Energies of Formation

48. ΔG Relationships
If a reaction can be expressed as a series of reactions, the sum of the ΔG values of the individual reaction is the ΔG of the total reaction.
ΔG is a state function.
If a reaction is reversed, the sign of its ΔG value reverses.
If the amount of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor.
The value of ΔG of a reaction is extensive.

49. Practice Problem: Calculating ΔG°rxn for a Stepwise Reaction

50. What’s “Free” about Free Energy?
The free energy is the maximum amount of energy released from a system that is available to do work on the surroundings.
For many exothermic reactions, some of the heat released as a result of the enthalpy change goes into increasing the entropy of the surroundings, so it is not available to do work.
And even some of this free energy is generally lost to heating up the surroundings.

51. Free Energy of an Exothermic Reaction
C(s, graphite) + 2 H2(g) → CH4(g)
• ΔH°rxn = −74.6 kJ = exothermic
• ΔS°rxn = −80.8 J/K = unfavorable
• ΔG°rxn = −50.5 kJ = spontaneous
ΔG° is less than ΔH° because some of the released heat energy is lost to increase the entropy of the surroundings.

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

53. Real Reactions
In a real reaction, some (if not most) of the free energy is “lost” as heat.
Therefore, real reactions are irreversible.

54. ΔG under Nonstandard Conditions
ΔG = ΔG° only when the reactants and products are in their standard states.
Their normal state at that temperature
Partial pressure of gas = 1 atm
Concentration = 1 M
Under nonstandard conditions, ΔG = ΔG° + RT ln Q.
Q is the reaction quotient.
At equilibrium, ΔG = 0.
ΔG° = −RT ln K

55. Free Energy versus Pressure for Water

56. Practice Problem: Calculating ΔGrxn under Nonstandard Conditions

57. ΔG° and K ΔG° = −RT ln(K) Because ΔGrxn = 0 at equilibrium,
When K < 1, ΔG° is positive and the reaction is spontaneous in the reverse direction under standard conditions.
Nothing will happen if there are no products yet!
When K > 1, ΔG° is negative and the reaction is spontaneous in the forward direction under standard conditions.
When K = 1, ΔG° is 0 and the reaction is at equilibrium under standard conditions.

59. Practice Problem: The Equilibrium Constant (K) and ΔGrxn

60. Why Is the Equilibrium Constant Temperature Dependent?
Combining these two equations,
ΔG° = ΔH° – TΔS° and ΔG° = -RT ln(K)
it can be shown that
This equation is in the form = mx + b.
The graph of ln(K) versus inverse T is a straight line
with slope and y-intercept
ln(K) =
-ΔH°rxn R
ΔS°rxn 1 T +
ΔH°rxn R
ΔS°rxn R