1. Entropy and Free Energy Chapter 17
How to predict if a reaction can occur, given enough time?
THERMODYNAMICS
Hydrocarbon just vibrates.
Shows paper buring.
How to predict if a reaction can occur at a reasonable rate?
KINETICS

2. Objectives Spontaneity of Reactions Define Entropy, S
Show how entropy affects molecules
Show how entropy affects phase changes
Second Law of Thermodynamics
Enthalpy, Entropy and Free Energy

3. Ch. 17.1 - Spontaneous Processes and Entropy
If the state of a chemical system is such that a rearrangement of its atoms and molecules would decrease the energy of the system---
then this is a product-favored system.
Most product-favored reactions are exothermic —but this is not the only criterion

4. Thermodynamics
Both product- and reactant-favored reactions can proceed to equilibrium in a spontaneous process.
AgCl(s)  Ag+(aq) + Cl–(aq)
Reaction is not product-favored, but it moves spontaneously toward equilibrium.
Spontaneous does not imply anything about time for reaction to occur.
Second bullet Point: Iron will rust, over time, in the elements, just not instantly.
Spontaneous means a system will proceed to products without any other influences.

5. Thermodynamics and Kinetics
Diamond is thermodynamically favored to convert to graphite, but not kinetically favored.
Paper burns — a product-favored reaction. Also kinetically favored once reaction is begun.

6. Spontaneous Reactions
In general, spontaneous reactions are exothermic.
Fe2O3(s) + 2 Al(s)  2 Fe(s) + Al2O3(s)
∆rH = kJ
This might be a good place to talk qualitatively about reversible and irreversible. It’s probably best to note that spontaneous reactions that are exothermic like the above are NOT going to spontaneously happen in reverse. The amount of energy it released would have to be put into the products to change them back.
Can bring up analogy that the salt (NaCl) sitting in their kitchen will not spontaneously go back to being sodium metal and chlorine gas.
For detailed discussion of reversible/irreversible processes, see p. 864 of text.

7. Spontaneous Reactions
But many spontaneous reactions or processes are endothermic or even have ∆H = 0.
∆H = 0
NH4NO3(s) + heat  NH4NO3(aq)

8. Entropy, S Reaction of K with water
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.
The thermodynamic property related to energy dispersal is ENTROPY, S.
2nd Law of Thermo — a spontaneous process results in an increase in the entropy of the universe.
Your book also refers to this a ‘randomness’ or ‘disorder’. While that is true, what ultimately happens is that energy, overall, is more dispersed by a spontaneous (i.e. product-driven) process.
Reaction of K with water

9. Directionality of Reactions
Probability suggests that a spontaneous reaction will result in the dispersal of energy.
Energy Dispersal

10. Directionality of Reactions Energy Dispersal
Exothermic reactions involve a release of stored chemical potential energy to the surroundings.
The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules.
The final state—with energy dispersed—is more probable and makes a reaction spontaneous.

11. Energy Dispersal
Each of the ways to distribute energy is called a microstate. These microstates give a particular arrangement (State).
In other words, the most probable outcome is where the molecules of the gas are spread out (energy dispersed) as far as they can be.

12. Directionality of Reactions
Matter & energy dispersal
As the size of the container increases, the number of microstates accessible to the system increases, and the density of states increases. Entropy increases.

13. The entropy of liquid water is greater than the entropy of solid water (ice) at 0˚ C.
Energy is more dispersed in liquid water than in solid water.

14. Entropy, S S (solids) < S (liquids) < S (gases) So (J/K•mol)
H2O(liq) 69.95
H2O(gas) 188.8
S (solids) < S (liquids) < S (gases)
Energy dispersal

15. Entropy and States of Matter
S˚(Br2 liq) < S˚(Br2 gas)
S˚(H2O sol) < S˚(H2O liq)

16. Ch. 17.2 - 2nd Law of Thermodynamics
A reaction is spontaneous if ∆S for the universe is positive.
∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Suniverse > 0 for spontaneous process
Calculate the entropy created by energy dispersal in the system and surroundings.
If ΔSuniverse < 0, then the process is spontaneous in the opposite direction.
If ΔS = 0, then there is no tendency to occur.

17. Ch. 17.3 – Temperature and Spontaneity
Discussion Questions!
For the process A(l) A(s), which direction involves an increase in energy dispersal? Positional randomness? Explain your answer. As temperature increases/decreases (answer for both), which takes precedence? Why? At what temperature is there a balance between energy randomness and positional randomness?
Since energy is required to melt a solid, the reaction as written is exothermic, i.e. A(l) -> A(s) + energy. Thus, energy randomness favors the right (product; solid). Since a liquid has less order than a solid, positional randomness favors the left (reactant; liquid).
As temperature increases, positional randomness is favored (at higher temperatures the fact that energy is released becomes less important). As temperature decreases, energy randomness is favored.
There is a balance at the melting point.
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18. Entropy and Temperature
S increases slightly with T
S increases a large amount with phase changes

19. ΔSsurr The sign of ΔSsurr depends on the direction of the heat flow.
The magnitude of ΔSsurr depends on the temperature.
From p. 796 Point 2:
The magnitude of ΔSsurr depends upon the temperature. The transfer of a given quantitiy of energy as heat produces a much greater % change in the randomness of the surroundings at a low temperature than a high temperature, which depends directly on the amount of heat transferred and inversely on the temperature. In other words energy becomes a more important driving force at lower temperatures.

20. ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH
The negative sign is there because the ΔH is determined by the system. If energy flows out of the system (-ΔH), then the energy is flowing into the surrounds, so that process would gain energy.

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22. Entropy Changes for Phase Changes
For a phase change, ∆S = q/T
where q = heat transferred in phase change
For H2O (liq)  H2O(g)
∆H = q = +40,700 J/mol
The heat you need to put in for molecular dispersal.

23. Definition of Entropy Practice
Determine whether entropy increases or decreases for the following reactions.
HCl(g) + NH3(g)  NH4Cl(s)
2H2O2(l)  2H2O(l) + O2(g)
Cooling of nitrogen gas from -20⁰C to -50⁰C
NaCl(s)  Na+(aq) + Cl-(aq)
CaCO3(s)  CaO(s) + CO2(g)
decreases  forming a solid from 2 molecules of gas
increases  Gas forming in products side
decreases  Taking heat away, decrease in temperature
Increases  formation of ions, greater dispersal of energy
increases  formation of a gas from a solid.

24. 2nd Law of Thermodynamics
Dissolving NH4NO3 in water—an entropy driven process.
∆Suniverse =
∆Ssystem + ∆Ssurroundings

25. 2nd Law of Thermodynamics
2 H2(g) + O2(g)  2 H2O(liq)
∆Sosystem = J/K
Can calc. that ∆Hro = ∆Hosystem = kJ
∆Sosurroundings = J/K

26. 2nd Law of Thermodynamics
2 H2(g) + O2(g)  2 H2O(liq)
∆Sosystem = J/K
∆Sosurroundings = J/K
∆Souniverse = J/K
The entropy of the universe is increasing, so the reaction is product-favored.

27. ∆Gosys = ∆Hosys - T∆Sosys
Ch Free Energy, G
∆Suniv = ∆Ssurr + ∆Ssys
Multiply through by -T
-T∆Suniv = ∆Hsys - T∆Ssys
-T∆Suniv = change in Gibbs free energy for the system = ∆Gsystem
Under standard conditions —
∆Gosys = ∆Hosys - T∆Sosys
J. Willard Gibbs

28. ∆Go = ∆Ho - T∆So
This means the a process is spontaneous in the direction in which the free energy decreases.
Remember that ΔSuniv = - ΔG/T

29. Free Energy and Spontaneity Example
H2O(s)  H2O(l)
ΔH° = 6.03x103 J/mol and ΔS°=22.1 J/K
At 10°C, the ΔG° = 6030 J – 283K(22.1 J/K)
= 6030 J – 6254 J
= -224 J
So, ice melts spontaneously at 10°C, even though it’s an endothermic reaction.

30. Free Energy and Spontaneity Example
H2O(s)  H2O(l)
ΔH° = 6.03x103 J/mol and ΔS°=22.1 J/K
At -10°C, the ΔG° = 6030 J – 263K(22.1 J/K)
= 6030 J – 5800 J
= 230 J
So, ice does not melt spontaneously at -10°C. However, water would freeze spontaneously at -10°C because opposite is more thermodynamically favorable.

31. Free Energy and Spontaneity Example
H2O(s)  H2O(l)
ΔH° = 6.03x103 J/mol and ΔS°=22.1 J/K
At 0°C, the ΔG° = 6030 J – 273K(22.1 J/K)
= 6030 J – 6030 J
= 0 J
So, neither process is favorable

32. ∆Go = ∆Ho - T∆So • and entropy increases (positive ∆So)
Gibbs free energy change =
total energy change for system
- energy lost in energy dispersal
If reaction is
• exothermic (negative ∆Ho)
• and entropy increases (positive ∆So)
• then ∆Go must be NEGATIVE
• reaction is spontaneous (and product-favored).

33. ∆Go = ∆Ho - T∆So • and entropy decreases (negative ∆So)
Gibbs free energy change =
total energy change for system
- energy lost in energy dispersal
If reaction is
• endothermic (positive ∆Ho)
• and entropy decreases (negative ∆So)
• then ∆Go must be POSITIVE
• reaction is not spontaneous (and is reactant-favored).

34. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So ∆Ho ∆So ∆Go Process
Exo,<0 >0 <0 Spontaneous at all T
End,>0 <0 >0 Never Spontaneous
Exo, <0 <0 ? T dependent, spontaneous
at lower T’s
End, >0 >0- ? T dependent, spontaneous
at higher T’s

35. Spontaneous or Not? Practice
Use the table on the previous slide to classify the following reactions as one of the 4 types.
CH4(g) + 2O2(g)  2H2O(l) + CO2(g)
ΔH = -891 kJ/mol ΔS = -242 J/K
2Fe2O3(s) + 3C(s, graphite)  4Fe(s) + 3CO2(g)
ΔH = +468 kJ/mol ΔS = +561 J/K
3. C(s, graphite) + O2(g)  CO2(g)
ΔH = -394 kJ/mol ΔS = +3 J/K
This is Exercise 19.3 on p. 875
Large –ΔH, negative ΔS  Spontaneous at lower temps, not spontaneous at higher temps
Large +ΔH, large +ΔS  not spontaneous at low temps, spontaneous at higher temps
Large –ΔH, small +ΔS  Spontaneous at all temps

36. Ch. 17.5 – Entropy Changes in Reactions
The 3rd Law of Thermodynamics:
The entropy of a pure crystal at 0 K is 0 (S=0)
Gives us a starting point.
At all other temperatures, entropy must be > 0. (As you raise T, entropy increases)
This is the reason why we have to use K when doing these calculations, because S is 0 at 0 K and the numbers at 298 C reflect the motion of the molecules at room temperature.

37. Standard Molar Entropies

38. Entropy, S Entropy of a substance increases with temperature.
Molecular motions of heptane, C7H16
Molecular motions of heptane at different temps.
Make a model of a hydrocarbon (at least propane). The movies above just show vibrations. Also show translational and rotational movement.

39. Entropy, S
Increase in molecular complexity generally leads to increase in S.

40. Entropy, S Entropies of ionic solids depend on coulombic attractions.
So (J/K•mol)
MgO 26.9
NaF 51.5
Mg2+ & O2-
Na+ & F-

41. Entropy, S
Liquids or solids dissolve in a solvent in a spontaneous process owing to the increase in entropy. Matter (and energy) are more dispersed.

42. Calculating ∆S for a Reaction
∆So =  So (products) -  So (reactants)
Consider 2 H2(g) + O2(g)  2 H2O(liq)
∆So = 2 So (H2O) - [2 So (H2) + So (O2)]
∆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.

43. Entropy of Reaction Practice
Calculate ∆S˚ for the following reaction:
Al2O3(s) + 3H2(g)2Al(s)+3H2O(g) S⁰
∆S⁰rxn = 2(28) + 3(189) – 51 – 3(131)
= – 51 – 393
= J/K
Remember that ∆Sf are in J/K-mol.
Find ∆S’s in text in Appendix 4 on pp. A-19 – A 22.

44. Ch. 17.6 – Free Energy and Chemical Reactions
Really want to determine the STANDARD FREE ENERGY CHANGE, ΔG° for a reaction
This represents the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.
Note: ΔG is not measured directly.
Use it to compare relative tendency for reactions to occur.
The more negative the ΔG of reaction, the greater tendency of product to form

45. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So
Two methods of calculating ∆Go
a) Determine ∆rHo and ∆rSo and use Gibbs equation.
b) Use tabulated values of free energies of formation, ∆Gfo.
∆Gro =  ∆Gfo (products) -  ∆Gfo (reactants)

46. Free Energies of Formation
Note that ∆Gf˚ for an element = 0

47. Calculating ∆Grxno Combustion of acetylene
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
∆Hrxno = kJ
Use standard molar entropies to calculate
∆Srxno = J/K or kJ/K
∆Grxno = kJ - (298 K)( kJ/K)
= kJ
Reaction is product-favored in spite of negative ∆Srxno.
Reaction is “enthalpy driven”

48. Calculating ∆Grxno NH4NO3(s) + heat  NH4NO3(aq)
Is the dissolution of ammonium nitrate product-favored?
If so, is it enthalpy- or entropy-driven?

49. Calculating ∆rGo NH4NO3(s) + heat  NH4NO3(aq)
From tables of thermodynamic data we find
∆rHo = +26 kJ
∆rSo = J/K or kJ/K
∆rGo = +26 kJ - (298 K)( J/K)
= -6 kJ
Reaction is product-favored in spite of positive ∆Hrxno.
Reaction is “entropy driven”

50. Gibbs Free Energy, G ∆Go = ∆Ho - T∆So
Two methods of calculating ∆Go
a) Determine ∆rHo and ∆rSo and use Gibbs equation.
b) Use tabulated values of free energies of formation, ∆fGo.
∆rGo =  ∆fGo (products) -  ∆fGo (reactants)

51. Calculating ∆Gorxn ∆Gro =  ∆Gfo (products) -  ∆Gfo (reactants)
Combustion of carbon
C(graphite) + O2(g)  CO2(g)
∆Gro = ∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)]
∆rGo = kJ - [ ]
Note that free energy of formation of an element in its standard state is 0.
∆Gro = kJ
Reaction is product-favored as expected.

52. Free Energy and Temperature
2 Fe2O3(s) C(s)  4 Fe(s) CO2(g)
∆Hro = kJ ∆Sro = +560 J/K
∆Gro = kJ
Reaction is reactant-favored at 298 K
At what T does ∆Gro just change from being (+) to being (-)?
When ∆Gro = 0 = ∆Hro - T∆Sro

53. 2SO2(g) + O2(g) 2SO3(g) at 227 oC DHo = -196.6 kJ; DSo = -189.6 J/K
Practice - Calculate the ΔG for this Reaction
2SO2(g) + O2(g) 2SO3(g) at 227 oC
DHo = kJ; DSo = J/K
DGo = – 500 ( ) kJ
DGo = kJ
Note: We are not going to do the last part of Ch (Free Energy and the Equilibrium Constant) right now. We will come back to it when we do equilibrium in ~Feb.
2SO2(g) + O2(g) 2SO3(g) at 227 oC
DHo = kJ; DSo = J/K
DGo = – 500 ( ) kJ
DGo = kJ
Also do additional Problems from P #49, 50, 51, 52, 53, 57. (They are already worked out in a WORD document in the Thermodynamics folder.)

54. Free Energy, Enthalpy and Entropy Practice
2000 AP Exam #6a-c
2003 AP Exam #7b-d
2004 AP Exam #2d-e
2005 AP Exam #8a-c
2005B AP Exam #7c
2006 AP Exam #2a-d (Best one)
Note: We are not covering 17.7 – 17.9 now. We’ll come back to this during equilibrium.