1. Chapter 18: Entropy, Free Energy, and Equilibrium
(2) First Law of Thermodynamics
Energy (E)system = (KE) System + (PE) System
∆Energy = ∑ ∆ Energy final- ∑ ∆ Energy initial

2. Chapter 18: Thermodynamics: Review
(2) First Law of Thermodynamics
For A Chemical Reaction:
∆Energy= ∑ ∆ Energy products- ∑ ∆ Energy reactants

3. Chapter 18: Thermodynamics:Review
For energy changes in a chemical reaction, not all of the energy is available to do work.
Useful Energy = Gross Energy Change - Random/Disorganized Energy

4. Chapter 18: Chemical Thermodynamics, Enthalpy
Enthalpy (H): Term describes Gross Energy Change
Enthalpy (H)
Enthalpy (H◦)
Exothermic: -∆ H reaction
Endothermic: + ∆ H reaction
∆H reaction = ∑ ∆ H products- ∑ ∆ H reactants

5. Chapter 18: Spontaneous Reactions
Spontaneous Reactions proceed to formation of large amounts of products without outside intervention.
Reaction rates will vary.
Fe(s) + O2(g)  2Fe2O3(s)

6. Chapter 18: Spontaneous Reactions: Probability of Outcomes/Product Formation
Spontaneous Reactions have high probability of product formation.
Spontaneous reactions somehow proceed from states of low probability to states of high probability.
High probability states usually result when energy is dispersed.

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8. Chapter 18: Nonspontaneous Reactions
Nonspontaneous reactions do not proceed to formation of large amounts of products without outside intervention.
Second law of Thermodynamics: Reactions that are spontaneous in one direction are nonspontaneous in the reverse direction.
Fe(s) + O2(g)  2Fe2O3(s)

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10. Chapter 18: Chemical Thermodynamics, Entropy: Molecular Explanation
Entropy (S): Random/Disorganized Energy
Useful Energy/Gibbs (G) : Energy actually available to do work
Useful Energy = Gross Energy Change - Random/Disorganized Energy
∆ G reaction = ∆ H reaction - T∆ S reaction

11. Chapter 18: Entropy (S)
Degree of Random energy/disordered energy is temperature related.
Solid
liquid
gas

12. Chapter 18: Entropy Entropy increases if disorder increases.
NaCl(s)  Na+ + Cl-
Ag+ + Cl-  AgCl(s)

13. ENTROPY & PROBABILITY Probable Event: Can Happen in Many Ways
RANDON
DISORDER
Improbable Event: Can only happen in one or two ways
ORDER

14. Entropy: Microstates
Microstates: Different patterns; Different Outcomes
Boltzman: Entropy Related to ln of Number of Microstates
S = k ln W
K = 1.38E-23 Joules/Kelvin

15. Entropy (S): Spontaneous and Nonspontaneous Reactions
Spontaneous reactions somehow proceed from states of low probability to states of high probability.
High probability states usually result when energy is dispersed.

16. Entropy (S): Spontaneous and Nonspontaneous Reactions
Entropy is random or disorganized energy.
Spontaneous reactions always have an increase in Entropy.

17. Entropy (S): Spontaneous and Nonspontaneous Reactions
∆S reaction = ∑ ∆ S products- ∑ ∆ S reactants
∆S◦ reaction = ∑ ∆ S◦ products- ∑ ∆ S◦ reactants
Spontaneous reactions always have an increase in entropy.
+ ∆S◦ reaction

18. Entropy (S): Spontaneous and Nonspontaneous Reactions
∆S◦ reaction = ∑ ∆ S◦ products- ∑ ∆ S◦ reactants
Nonspontaneous reactions usually have a decrease in entropy.
- ∆S◦ reaction

19. Entropy (S): Spontaneous and Nonspontaneous Reactions
+ ∆S◦ reaction Does not mean reaction will be spontaneous!
(a) As temperature increases, entropy (S) usually also increases.
∆Greaction = ∆H reaction -T ∆S reaction

20. Entropy Projections
A process that is spontaneous in one direction is not spontaneous in the opposite direction.
The direction of a spontaneous process can depend on temperature: Ice turning to water is spontaneous at T > 0C, Water turning to ice is spontaneous at T < 0C.

21. Entropy (S): Spontaneous and Nonspontaneous Reactions
+ ∆S◦ reaction Does not mean reaction will be spontaneous!
(b) Vibrational Energy: Energy may be absorbed by bonds.
(c) Under certain conditions, molecules of reactants may become so disordered that the inherent energy is not available to form products.

22. Problem
Predict the sign of the entropy change of the system for each of the following reactions.
(a) 2SO2(g) + O2(g)  2SO3(g)
(b) Ba(OH)2(s) BaO(s) + H2O(g)

23. Problem Using S◦ values from appendix C, calculate
∆S◦ reaction values for each of the following reactions. In each case, account for the sign of ∆S◦ reaction.
(a) C2H4(g) + H2(g)  C2H6(g)
(b) N2O4(g) 2NO2(g)

24. Chapter 18: Useful/Free Energy (G)
Useful/Free energy is that energy which does work.
SPONTANEOUS: - ∆G◦ reaction
NONSPONTANEOUS: + ∆G◦ reaction

25. Chapter 18: Useful Energy and Equilibrium
N2(g) + 3H2(g) ↔ 2NH3(g)
If start with reactants, must have - ∆G for forward reaction to form products.
If start with products, must have - ∆G for reverse reaction to form reactants.

26. Chapter 18: Useful Energy and Equilibrium, ∆Greaction versus ∆G ◦reaction
At equilibrium,
rate of forward = rate reverse
At equilibrium ∆Greaction is zero.
At equilibrium, ∆G ◦reaction is not zero.

27. Gibbs Free Energy

28. CALCULATION OF ∆ G REACTION
∆ E = ∑ (n) (E products) -∑ (n) (E reactants)
∆ H =∑ (n) (H products) -∑(n)(H reactants)
∆ S =∑ (n)(S products) -∑(n)(S reactants)
∆ G =∑(n)(G products) -∑(n)(G reactants)

29. CALCULATION OF ∆ G REACTION
∆Gº =∑ (n)(Gº products) -∑(n)(Gº reactants)
∆G =∑ (n)(G products) -∑(n)(G reactants)
∆G = ∆H - T∆S
∆G º = ∆H º - T∆S º

30. Chapter 18: Useful energy and Signs
Spontaneous: - ∆ H º
Nonspontaneous: + ∆ Hº
Spontaneous: + ∆ S º
Nonspontaneous: - ∆ S º
Spontaneous: - ∆ G º
Nonspontaneous: + ∆ G º

31. Problem
Using data from Appendix, Calculate ∆ Gº for the following reactions.
(a) 2SO2(g) + O2(g)  2SO3(g)
(b) NO2(g) + N2O(g)  3NO(g)

32. Gibbs Free Energy: Temperature
∆ G reaction = ∆ H reaction - T∆ S reaction

33. Table 18.3

34. Calculation of Actual ∆ G
Why is ∆ G not always the same as ∆ G º ?
∆ G = ∆ G º + RT ln Q
R = Joules/(K) (mole)

35. Problem Consider the reaction 2NO2(g)  N2O4(g).
(a) Using appendix C, calculate ∆ G º at 298 K.
(b) Calculate ∆ G at 298 K if the partial pressures of NO2 and N2O4 are 0.40 atm and 1.60 atm respectively.

36. CHAPTER 18: Free Energy and Equilibrium Constant

37. Problem 18.24: Page 830
For the autoionization of pure water at °C, Kw = 1.0E-14. What is ∆ G º for the reaction.
H2O(l)  H+(aq) + OH-(aq)

38. Problem
Use data from Appendix C to calculate Keq at 298 K for each of the following reactions:
(a) H2(g) + I2(g) ↔ 2HI(g)