1. Chapter 20: Thermodynamics
Entropy, Free Energy, and the Direction of Chemical Reactions
20.1 Predicting Spontaneous Change
The Second Law of Thermodynamics:
20.2 Calculating the Change in Entropy of a Reaction
20.3 Entropy, Free Energy, and Work
20.4 Free Energy, Equilibrium, and Reaction Direction

2. The First Law of Thermodynamics
q = heat w = work
 E = q + w = heat + work =  Energy
Everything that is not part of a system (sys) is the surroundings (surr),
and vice versa:
Euniv = Esys + Esurr
qsys = -qsurr
wsys = -wsurr
The total mass-energy of the universe is constant. Therefore
mass-energy cannot be created or destroyed.
Mass can be converted into energy in nuclear reactions, and energy can be converted into mass in nuclear accelerators.

3. Spontaneous Processes
Thermodynamics is concerned with the question: can a reaction occur?
First Law of Thermodynamics: Energy is conserved.
Any process that occurs without outside intervention is spontaneous.
When two eggs are dropped they spontaneously break.
The reverse reaction (two eggs leaping into your hand with their shells back intact) is not spontaneous.
We can conclude that a spontaneous process has a direction.

4. Spontaneous Processes
A process that is spontaneous in one direction is not spontaneous in the opposite direction.
Example: Ice turns to water spontaneously at T > 0C, Water turns to ice spontaneously at T < 0C.

5. Spontaneous Processes
Chemical systems in equilibrium are reversible and are not spontaneous.
In any spontaneous process, the path between reactants and products is irreversible.
Thermodynamics provides the direction of a process. It cannot predict the speed (rate) at which the process will occur.
Why can both endothermic and exothermic reactions be spontaneous?

6. Heat of reaction (H) & Spontaneous Change
All combustion reactions are spontaneous and exothermic:
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g)
Horxn = -802 kJ
Iron rusts spontaneously and exothermically:
2 Fe(s) O2 (g) Fe2O3 (s) 3 2
 Horxn = -826 kJ
Ionic compounds form spontaneously from their elements with a large
release of heat: 1 2
 Horxn = -411 kJ
Na(s) + Cl2 (g) NaCl(s)
At 1 atm, water freezes below 0°C but melts above 0°C.
Both processes are spontaneous, but the first is exothermic and the
second endothermic.
H2O(l) H2O(s)  Horxn = kJ
(exothermic; spontaneous at T < 0oC)
H2O(s) H2O(l)  Horxn = kJ
(endothermic; spontaneous at T > 0oC)

7. A Spontaneous, Endothermic Chemical Reaction

8. Spontaneous Expansion of a Gas

9. Entropy and the Second Law of Thermodynamics
Entropy (S) is a measure of disorder
A system with little freedom, such as a crystalline solid or a deck of cards in a specific sequence, has relatively small disorder and low entropy.
A system with many degrees of freedom (possible arrangements), such as a gas or a shuffled deck of cards, has relatively large disorder and high entropy.
Sdisorder > Sorder
 Ssys = Sfinal - Sinitial

10. Entropy and the Second Law
The Second Law of Thermodynamics – All spontaneous processes occur in the direction that increases the total entropy of the universe (system plus surroundings).
Suniv = Ssys + Ssurr
Entropy is not conserved: Suniv is increasing
For a reversible process: Suniv = 0.
For a spontaneous process (i.e. irreversible): Suniv > 0.
Note: the second law states that the entropy of the universe must increase in a spontaneous process. It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases.
For an isolated system, Ssys = 0 for a reversible process and Ssys > 0 for a spontaneous process.

11. Temperature Changes and the Increase in Entropy from a Solid to a Liquid to a Gas

12. The Increase in Entropy From Solid to Liquid to Gas

13. The Molecular Interpretation of Entropy
As we heat a substance, the entropy must increase.
The entropy increases markedly at the solid to liquid phase change.
Boiling corresponds to a much greater change in entropy than melting.
Entropy will increase when:
liquids or solutions are formed from solids,
gases are formed from solids or liquids,
the number of gas molecules increase,
the temperature is increased.

15. The Small Increase in Entropy when Methanol Dissolves in Water

16. The Large Decrease in Entropy of a Gas When it Dissolves in a Liquid

17. The Molecular Interpretation of Entropy
There are three atomic modes of motion:
translation (the moving of a molecule from one point in space to another),
vibration (the shortening and lengthening of bonds, including the change in bond angles),
rotation (the spinning of a molecule about some axis).

18. The Molecular Interpretation of Entropy
and the Effects of Temperature on Molecular Motion

19. The Molecular Interpretation of Entropy
Energy is required to get a molecule to translate, vibrate or rotate.
The more energy stored in translation, vibration and rotation, the greater the degrees of freedom and the higher the entropy.
In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this is a state of perfect order.
The Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.

20. Predicting Relative Entropy Values
Problem: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in(f)]:
(a) 1 mol NaCl(s) or 1 mol NaCl(aq)
(b) 1 mol SF6 or 1 mol SCl6
(c) 1 mol CO(g) or 1 mol CO2 (g)
(d) 1 mol S8 or 4 mol S2
(e) 1 mol H2O(s) or 1 mol H2O(g)
(f) Lipton’s noodle soup at 24oC or at 95oC
Plan: The less ordered a system, the greater the entropy. A higher temperature increases entropy. Assume constant temperature in a-e.

21. Solutions:
(a) 1 mol NaCl(aq). The two samples have the same number of ions,
but in the solid they are highly ordered, and in the solution they are
randomly dispersed.
(b) 1 mol SCl6. For similar compounds, entropy increases with molar
mass.
(c) 1 mol CO2 (g) . For equal numbers of moles of substances with the
same types of atoms in the same physical state, the more atoms per
molecule, the more types of motion available to it and, thus, the higher
its entropy.
(d) 4 mol S2. The two samples contain the same number of sulfur atoms,
but different numbers of molecules. Despite the greater complexity of
S8 , the greater number of molecules dominates in this case because
there are many more ways to arrange 4 moles of particles than one mole.
(e) 1 mol H2O(g). For a given substance entropy increases in the
sequence: solid < liquid < gas.
(f) Soup at 95oC. Entropy increases with increasing temperature.

22. Calculations of Entropy Changes
Absolute entropy can be determined from complicated measurements.
Standard molar entropy, S: entropy of a substance in its standard state. Similar in concept to H.
Units: J/mol-K. Note units of H: kJ/mol.
Standard molar entropies of elements are not zero.
For a chemical reaction which produces n products from m reactants:

23. . . The Standard Entropy of Reaction,  Sorxn
For many chemical reactions:
 So = Soproducts - Soreactants > 0
But for reactions in which the total # of moles of products decrease,
particularly gases which have very high entropy, we predict that the
entropy of the products is less than that of the reactants. Therefore,
the entropy will decrease during the reaction:
N2 (g) + 3 H2 (g) NH3 (g)  So = Soproducts - Soreactants< 0
Standard entropy of reaction,  Sorxn:
mSoproducts -
nSoreactants
Sorxn = .
 Sorxn = (2 mol NH3 x So of NH3) - [(1 mol x J/mol K) +
(3 mol x J/mol K)]
 Sorxn = -197 J/K .
As predicted,  So < 0

24. Calculating the Standard Entropy of Reaction, Sorxn
Problem: Calculate the  Sorxn for the oxidation of one mole of S8 to form either SO2 (g), or SO3 (g) at 25oC:
S8 (s) + 8 O2 (g) SO2 (g)
versus
S8 (s) + 12 O2 (g) SO3 (g)
Plan: To determine  Sorxn, apply Equation We predict the sign of  Sorxn from the change in the number of moles of gas: 8 = 8 or 12 = 8, so the entropy will decrease ( Sorxn< 0).

25. Calculating the Standard Entropy of Reaction, Sorxn
Solution: Calculate  Sorxn using Appendix B values,
Rxn #1
Sorxn = ( 8 mol SO2 x So of SO2) - [(1 mol S8 x So of S8) +
( 8 mol O2 x So of O2)]
= ( 8 mol x J/mol K) - [(1 mol x J/mol K)
+ (8 mol x J/mol K)]
= (1,985.6 J/K) - [( J/K) + (1,640.0 J/K)]
= J/K - 2, J/K = J/K

26. Calculating the Standard Entropy of Reaction, Sorxn
Sorxn = ( 8 mol SO3 x So of SO3) - [(1 mol S8 x So of S8) +
( 12 mol O2 x So of O2)]
= ( 8 mol x J/mol K) - [(1 mol x J/mol K)
+ (12 mol x J/mol K)]
= (2, J/K) - [( J/K) + (2,460.0 J/K)]
= 2, J/K - 2, J/K = J/K

27. Calculating the Standard Entropy of Reaction, Sorxn
Summary:
For Reaction #1 (SO2 is product)  Sorxn = J/K
For Reaction #2 (SO3 is the product)  Sorxn = J/K
As predicted they are both negative, but Rxn #1 is close to zero, and we would also predict that it would be close to zero, since the number of moles of gaseous molecules did not change from reactant to product.

28. Entropy Changes in the Surroundings
The second law of thermodynamics restated says:
Decreases in the entropy of the system can occur only if increases in the entropy of the surroundings outweigh them.
The surroundings either add energy to the system or remove energy from it.
Therefore there are two possible types of enthalpy changes in chemical reactions.

29. 1) Can you name the two types of processes: (a)? & (b)?
2) If (b) is a spontaneous process, what is happening to Suniv?

30. Entropy Changes in the Surroundings
Those two possible types of enthalpy changes are:
1. Exothermic. Heat lost by the system is gained by the surroundings.
2. Endothermic. Heat gained by the system is lost by the surroundings.
qsys < 0, qsurr > 0 and Ssurr > 0
qsys > 0, qsurr < 0 and Ssurr < 0
The change in entropy of the surroundings
is directly related to an opposite change in
the energy of the system and inversely related
to the temperature of the surroundings before
heat is transferred.
Hsys
Ssurr = - T

31. Components of Souniv for Spontaneous Reactions