1. Spontaneity, Entropy, & Free Energy
Chapter 16

2. 1st Law of Thermodynamics
The first law of thermodynamics is a statement of the law of conservation of energy: energy can neither be created nor destroyed. The energy of the universe is constant, but the various forms of energy can be interchanged in physical and chemical processes.

4. Spontaneous Processes and Entropy
Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process.
A spontaneous process is one that occurs without outside intervention.

5.                                                                            < TARGET="display">

6. Kinetics & Thermodynamics
Chemical kinetics focuses on the pathway between reactants and products--the kinetics of a reaction depends on activation energy, temperature, concentration, and catalysts.
Thermodynamics only considers the initial and final states.
To describe a reaction fully, both kinetics and thermodynamics are necessary.

7. The rate of a reaction depends on the pathway from
reactants to products. Thermodynamics tells whether
the reaction is spontaneous and depends on initial
& final states only.

8. What are some examples of spontaneous processes?
A process that does occur under a given set of conditions is considered spontaneous.
For example, a waterfall flows downhill, but never up, spontaneously.
Heat flows from a warmer object to a cooler one, but the reverse never happens spontaneously.
Iron exposed to water and oxygen forms rust, but rust does not spontaneously change back into iron.

9. Entropy
The driving force for a spontaneous process is an increase in the entropy of the universe.
Entropy, S, can be viewed as a measure of randomness, or disorder.
Nature spontaneously proceeds toward the states that have the most “spread out energy”, or the highest probabilities of existing. In other words, towards an arrangement where energy can be contained in the greatest number of ways.

10. The expansion of an ideal gas into an evacuated bulb.

11. As the number of molecules increases, the probability of finding the molecules only in one bulb becomes VERY small!

12. Ssolid < Sliquid << Sgas
Positional Entropy
A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system.
Therefore,
Ssolid < Sliquid << Sgas
Solutions also have high entropy-accounting for the solubility of many solids into water to form ions in solution.

13. Entropy Which of the following has higher entropy?
a) Solid CO2 or gaseous CO2?
b) N2 gas at 1 atm or N2 gas at 1.0 x 10-2 atm?

14. Entropy What is the sign of the entropy change for the following?
a) Solid sugar is added to water to form a solution?
S is positive
b) Iodine vapor condenses on a cold surface to form crystals?
S is negative

15. Predict the sign of the entropy change for each:
Solid sugar is added to water to form a solution
Iodine vapor condenses on a cold surface to form crystals
Water forms H2O vapor
Water freezes
A gas expands
Student breaks a pyrex beaker
+ ΔS
- ΔS
+ ΔS
- ΔS
+ ΔS
+ ΔS

16. Temperature and Entropy
Entropy is directly affected by temperature changes.
Recall that kinetic molecular theory tells us that matter is made up of particles in motion. Temperature is a measurement of the kinetic energy of particles in a system.
Increasing temperature increases particle movement, increasing the disorder in a system, increasing its entropy.

17. Entropy Values We can make generalizations about a reaction’s entropy;
2KClO3(s)  2KCl(s) + 3O2(g)
two mol solids  two mol solids + 3 mol gases
Entropy appears to increase in this reaction.
CaO(s) + CO2(g)  CaCO3(s)
one mol solid + one mol gas  one mol solid
Entropy appears to decrease in this reaction.

18. More factors that influence entropy

19. Although many factors influence the entropy of a system, there is usually a DOMINANT source.

20. 143.7 J/mol*K  82.6 J/mol*K + 205.1 J/mol*K
Entropy for a reaction
We can assign values to the entropy of formation of a substance, called standard entropy, and calculate a reaction’s entropy quantitatively. Standard entropy is determined at 25°C and 1 atm (gas partial pressure) or 1 M (sol’n concentration).
2KClO3(s)  2KCl(s) + 3O2(g)
143.7 J/mol*K  82.6 J/mol*K J/mol*K
Using ΔS = Sproducts – Sreactants, the reaction has a total entropy change of J/mol*K
NOTE that the units are joules/mol K rather than kJ/mol as for enthalpies. Typically smaller energies.
Check out values in Appendix… what are values for elements? Cmpds?

21. The Third Law of Thermodynamics
. . . the entropy of a perfect crystal at 0 K is zero.
Because S is known (= 0) at 0 K,
S values at other temps can be calculated.
See Appendix 4 for values of S0, which is the entropy value of substances at 1 atm when heated to 298K.
Review: What is the “zero” for enthalpy values?

22. Practice problems
Calculate the standard entropy changes for the following reactions at 25۫C:
(a) CaCO3(s) →CaO(s) + CO2(g)
ΔSorxn = [So(CaO) + So(CO2)] – So(CaCO3)
ΔSorxn = [39.8 J/K.mol J/K.mol] – (92.9 J/K.mol)
ΔSorxn = J/K.mol
(b) N2(g) + 3H2(g) → 2NH3(g)
ΔSorxn = 2So(NH3) – [So(N2) + 3So(H2)]
ΔSorxn = (2)(193) J/K.mol – [192 J/K.mol] + (3)(131 J/K.mol)]
ΔSorxn = -199 J/K.mol
+ΔS= favorable
-ΔS= unfavorable

23. More Practice Soreaction
Calculate S at 25 oC for the reaction
2NiS(s) + 3O2(g) ---> 2SO2(g) +2NiO(s)
S = npS(products)  nrS(reactants)
S = [(2 mol SO2)(248 J/Kmol) + (2 mol NiO)(38 J/Kmol)] - [(2 mol NiS)(53 J/Kmol) + (3 mol O2)(205 J/Kmol)]
S = 496 J/K + 76 J/K J/K J/K
S = -149 J/K # gaseous molecules decreases!

24. Comparing So values
Substances with a greater freedom of motion (or number of possible ways to move) have a greater absolute entropy.
example: I2(g) (So = 261 J/K.mol) and
I2(s) (So = 117 J/K.mol)
example: CH4(g) (So = 186 J/K.mol) and
C2H6(g) (So = 230 J/K.mol)

26. The Second Law of Thermodynamics
. . . in any spontaneous process there is always an increase in the entropy of the universe.
Suniv > 0
for a spontaneous process.

27. SUniverse Suniverse is positive -- reaction is spontaneous.
Suniverse is negative -- reaction is spontaneous in the reverse direction.
Suniverse = 0 -- reaction is at equilibrium.

28. How can we predict whether a chemical reaction will be spontaneous?
Thermodynamics can help us determine the direction of a spontaneous reaction, but cannot tell us about the speed of the process (for that we need kinetics…).
Is this chemist observing an an an exothermic or endo thermic reaction?
What is the sign of ΔH?

29. What factors are important in predicting the spontaneity of a process?
We know that exothermic reactions (negative ΔH) are favored since the system moves to a lower energy state.
However, not ALL exothermic reactions will occur spontaneously, and endothermic reactions may also be spontaneous.
Example: ice will spontaneously melt (endothermic) at room temperature.

30. ENTROPY
The other driving force for a spontaneous process is an increase in entropy (S) in the universe.
A reaction tends to be spontaneous if the change in entropy (ΔS) is positive, and change in enthalpy (ΔH) is negative.
Note: -ΔH is really increasing entropy in disguise! Where is entropy increasing?

31. How can we determine spontaneity using ΔS and ΔH?
Two tendencies exist in nature:
tendency toward higher entropy -- S
tendency toward lower energy -- H
If the two processes oppose each other (e.g. melting ice cube), then the direction is decided by the Free Energy, G, and depends upon the temperature.
G -- Free Energy!!!

32. Gsys means +Suniverse
Gibbs Free Energy
G = H  TS (allows us to focus from the standpoint of the system) A process (at constant T, P) is spontaneous in the direction in which free energy decreases:
Gsys means +Suniverse
But, how do exothermic processes increase the entropy of the universe??
An American physicist and founder of thermo-
dynamics at Yale! ( stamp)
went to Hopkins and graduated from Yale at the age of 19. He was praised by Albert Einstein as "the greatest mind in American history".
Entropy changes in the surroundings are primarily determined by the heat flow. An exothermic process in the system increases the entropy of the surroundings.

33. G, H, & S
Spontaneous reactions (shift to RIGHT) are indicated by the following signs:
G = negative
H = negative
S = positive

34. How are you feeling at this point?
The entropy of your brain must be increasing…

35. Temperature Dependence
Ho is not temperature dependent. So is entropy measured at constant 298K.
Go is temperature dependent, and can be calculated for any constant temperature using the equation:
G ◦ = H ◦  TS ◦

36. G ◦ = H ◦  TS ◦
Calculations showing that the melting of ice is temperature
dependent. As temperature increases, the Free Energy
becomes lower (more negative) -the process is spontaneous
above 0oC.

37. Predicting spontaneity…ΔH ΔS ΔG - +
- ALWAYS SPONTANEOUS
+ NEVER SPONTANEOUS
- only spontaneous when TΔS is
greater than ΔH (at high temp.)
- only spontaneous when TΔS is less
than ΔH (at low temp.)

38. Free Energy G G ◦ = H ◦  TS ◦
G = negative – spontaneous (RIGHT shift)
G = positive -- spontaneous in opposite direction (SHIFT left)
G = 0, system is at equilibrium (with products and reactants in std states)

39. Summary of 3 ways to calculate ΔG at std state
1. ΔGrxn = ΔHo - TΔSo
2. G = npGf(products)  nrGf(reactants)
3. Hess’s Law

40. 1. Free Energy Change and Chemical Reactions
(a) Calculate H, S, & G for the following reaction at 25⁰C
2 SO2(g) + O2(g) ----> 2 SO3(g)
H = npHf(products)  nrHf(reactants)
H = [(2 mol SO3)(-396 kJ/mol)]-[(2 mol SO2)(-297 kJ/mol) + (0 kJ/mol)]
H = kJ kJ
H = -198 kJ

41. G Calculations Continued
S = npS(products)  nrS(reactants)
S = [(2 mol SO3)(257 J/Kmol)]-[(2 mol SO2)(248 J/Kmol) + (1 mol O2)(205 J/Kmol)]
S = 514 J/K J/K J/K
S = -187 J/K

42. G Calculations Continued
Go = Ho  TSo
Go = kJ - (298 K)(-187 J/K)(1kJ/1000J)
Go = kJ kJ
Go = kJ
The reaction is spontaneous in forward direction at 25 oC and 1 atm.
Now try this using the second method…

43. (b) Find ΔG for the following reaction at constant pressure (1 atm) and temperature (900˚C):
CaCO3(s) → CaO(s) + CO2(g)
ΔGrxn = ΔHo - TΔSo
ΔSorxn = [So(CaO) + So(CO2)] – So(CaCO3)
ΔSorxn = [39.8 J/K.mol J/K.mol] – (92.9 J/K.mol)
ΔSorxn = J/K.mol
ΔHorxn = [ΔHfo(CaO) + ΔHfo(CO2)] – ΔHfo(CaCO3)
ΔHorxn = [( kJ/mol) + ( kJ/mol)] – ( kJ/mol)
ΔHorxn = kJ/mol

44. Continued… ΔGrxn = ΔHo – TΔS
ΔGrxn = kJ/mol – 1173K( kJ/K.mol)
ΔGrxn = kJ/mol
Try this: at what Temp (in C) will this become nonspontaneous?

45. G = npGf(products)  nrGf(reactants)
2. More G Calculations
G = standard free energy change that occurs if reactants in their standard state are converted to one mole of product in its standard state. Values given in textbook appendix. Note that Gf for elements in their standard states is zero.
G = npGf(products)  nrGf(reactants)
The more negative the value of G, the further a reaction will go to the right to reach equilibrium.
(formation)

46. Practice Problem ΔG is -, SPONTANEOUS
Calculate the standard free energy changes for the following reactions at 1 atm and 25˚C.
(a) CH4 (g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔGorxn = ΔGproducts – ΔG reactants
ΔGorxn = [Gof(CO2) + 2Gof (H2O)] – [Gof(CH4) + 2Gof(O2)]
ΔGorxn = [(-394 kJ/mol) + (2)(-237 kJ/mol)] –
[(-51 kJ/mol) + (2)(0 kJ/mol)]
ΔGorxn = -818 kJ/mol
ΔG is -, SPONTANEOUS

47. ΔG is +, NOT SPONTANEOUS (b) 2MgO(s) → 2Mg(s) + O2(g)
ΔGorxn = ΔGproducts – ΔG reactants
ΔGorxn = [2Gof(Mg) + Gof (O2)] – 2Gof(MgO)
ΔGorxn = [(2)(0 kJ/mol) + (0 kJ/mol)] – (2)(-570 kJ/mol)
ΔGorxn = 1139 kJ/mol
ΔG is +, NOT SPONTANEOUS

48. Cdiamond(s) ---> Cgraphite(s)
3. Hess’s Law & Go
Cdiamond(s) + O2(g) ---> CO2(g) Go = -397 kJ
Cgraphite(s) + O2(g) ---> CO2(g) Go = -394 kJ
Calculate Go for the reaction
Cdiamond(s) ---> Cgraphite(s)
CO2(g) ---> Cgraphite(s) + O2(g) Go = +394 kJ
Cdiamond(s) ---> Cgraphite(s) Go = -3 kJ
Diamond is kinetically stable, but thermodynamically unstable.

49. ΔG◦ =0, Boiling Point Calculations
(a) What is the normal boiling point for liquid Br2?
Br2(l) ---> Br2(g)
Ho = 31.0 kJ/mol & So = 93.0 J/Kmol
At equilibrium (a phase change), Go = 0
Go = Ho  TS0 = 0
Ho = TS0
T = Ho/S0
T = 3.10 x 104 J/mol/(93.0J/Kmol)
T = 333K

50. (b) Find the increase in entropy of the phase transition: H2O(s) → H2O(l)
During a phase transition, ΔG=0 since the system is at equilibrium.
This transition occurs when T=273K
ΔGrxn = ΔH – TΔS
0 = ΔH – TΔS
ΔSrxn = ΔHfusion T
ΔSrxn = 6010 J/mol
273 K
ΔSrxn = 22 J/K.mol

51. Calculating Ssurroundings
Ssurr is positive -- heat flows into the surroundings out of the system.
Ssurr is negative -- heat flows out of the surroundings and into the system.
Ssurr = - Hsystem T

52. Ssurroundings Calculations
Sb2S3(s) + 3Fe(s) ---> 2Sb(s) + 3FeS(s) H = -125 kJ
Sb4O6(s) + 6C(s) ---> 4Sb(s) + 6CO(g) H = 778 kJ
What is Ssurr for these reactions at 250C & 1 atm.
Ssurr = - Hsystem T
Ssurr = -(-125kJ/298K)
Ssurr = 419 J/K
Ssurr = - Hsystem T
Ssurr = -(778kJ/298K)
Ssurr = x 103 J/K

53. Free Energy and partial pressure
For reactions that occur under partial pressures other than standard (1 atm), the G at those pressures is calculated as follows:
G = G + RT ln(Q)
Q = reaction quotient from the law of mass action, using non-standard pressures.
R= J/mol K

55. ΔG and Equilibrium
Although a reaction with a negative ΔG will move forward spontaneously, it does NOT mean that a reaction will go 100% to completion.
Remember, equilibrium occurs when the forward and reverse reaction rates are equal (kinetics).
From a thermodynamic point of view, this occurs when the reaction system is also at the lowest value of ΔG.
The relationship between K and ΔGo:
ΔGo = -RTln(K)

56. Making connections:
At any point in the reaction, the following equation is used: G = G + RT ln(Q) It follows that when all concentrations are standard (1M), Q=1 ln(Q)=0 and G = G At equilibrium, G = 0 and G = -RT ln(Q)
Q=[prod]a
[react]b

57. A system can achieve the lowest possible free energy
by going to equilibrium, not by going to completion.

58. Relationship between ΔG and K
ΔGo = -RTln(K)
Given this equation, it follows that when products and reactants are in their std states and…
ΔGo = 0, K = 1
ΔGo < 0, K > 1
ΔGo > 0, K < 1
We can use ΔGo to calculate K, since K = e -ΔGo/RT
The relationship between K and ΔG at equilibrium:
ΔGo = -RTln(K) = 0 = ΔHo -TΔSo

59. Two cases approaching equilibrium
Q= [products]
[reactants]
G = G + RT ln(Q)
Case 1: A large (-) value for G will make G negative also. Reaction will proceed to the right, creating MORE products (so lnQ goes from (-) with Q<1 to (+) with Q>1 as Q increases). Eventually, the RT ln(Q) term becomes (+) enough that G is canceled out and G is zero (equilib.)
Case 2: A large (+) value for G will make G positive also. Reaction will proceed right to left, creating MORE reactants (so lnQ goes from (+) with Q>1 to (-) with Q<1 as Q decreases). Eventually, the RT ln(Q) term becomes (-) enough that G is canceled out and G is zero (equilib.)

60. Determine the sign of ΔG for each reaction
Determine the sign of ΔG for each reaction. Which reaction favors reactants? Products? Recall G = Gf(prod)  Gf(rcts)

61. But….
If it looks like a free energy diagram then you are in AP Chemistry!

62. Practice Problem
Calculate ΔG at 25˚C for the following reaction where PCO = 5.0 atm and PH2 = 3.0 atm:
CO(g) + 2H2(g) →CH3OH(l)
ΔG = ΔGo + RTln(Q) Q = = = 2.2x10-2
(PCO)(PH2) (5.00)(3.00)2
ΔGo= ΔGo (CH3OH) – [ΔGo (CO) + ΔGo (H2) ]
ΔGo = -166 kJ/mol – [-137 kJ/mol + 2(0 kJ/mol)] = -29 kJ/mol
ΔG = -2.9x [8.31 J/K.mol)(298K)ln(2.2x10-2)
= -2.9x104 J/mol x103 J/mol
= -3.8 x104 J/mol = -38 kJ/mol
Compare this ΔG to ΔGo: a more negative ΔG means this reaction is more spontaneous (shift to right) than at one atm.

63. Equilibrium Calculations
4Fe(s) + 3O2(g) <---> 2Fe2O3(s)
Calculate K for this reaction at 25 oC.
Go = x 106 J *or find this using table, is (-740 KJ/mol)
G = RT ln(K)
K = e - G/RT
K = e 601 or *too big to handle in calculator
try (e200)3
K is very large because G is very negative.

64. Practice Problem #1
AP 1971
Given the following data for graphite and diamond at 298K: S°(diamond) = 2 J/mol K
S°(graphite) = 6 J/mol K ΔHf° CO2(from graphite) = kJ/mol
ΔHf° CO2(from diamond) = kJ/mol
Consider the change: C (graphite) → C(diamond) at 298K and 1 atm.
(a) What are the values of ΔS° and ΔH° for the conversion of graphite to diamond?
ΔS° = S°(dia.) - S°(graph.) = (2 - 6) J/mol K = - 4 J/mol K
CO2  C(dia.) + O ΔH = kJ/mol
C(graph.) + O2 → CO2 ΔH = kJ/mol
C(graph.) → C(dia.) ΔH = kJ/mol

65. 1971 AP (Continued…)
(b) Perform a calculation to show whether it is thermodynamically feasible to produce diamond from graphite at 298K and 1 atmosphere.
G° = ΔH° - TΔS°
= -1.9x103 J/mol - (298K)(-4 J/mol K)
= -708 J/mol;
a ΔG° is negative, indicates feasible conditions
(c) For the reaction, calculate the equilibrium constant
Keq at 298K. ***check this one
Keq = e-ΔG/RT
= e-(-708/(8.314)(298)) e = 1.3

66. Practice Problem #2
AP 1999
Answer the following question in terms of thermodynamic principles and concepts of kinetic molecular theory.
(a) Consider the reaction represented below, which is spontaneous at 298 K.
CO2(g) + 2 NH3(g)  CO(NH2)2(s) + H2O(l) Hº = –134 kJ
(i) For the reaction, indicate whether the standard entropy change, Sº, is positive, negative, or zero. Justify your answer.
(i) Sº is negative because (1) two different gases make a solid and a liquid (both with smaller entropies) and (2) three molecules of reactant make two molecules of product (a decrease in entropy).

67. AP 1999 (Continued…)
(a) Consider the reaction represented below, which is spontaneous at 298 K.
CO2(g) + 2 NH3(g)  CO(NH2)2(s) + H2O(l) Hº = –134 kJ
(ii) Which factor, the change in enthalpy, Hº, or the change in entropy, Sº, provides the principle driving force for the reaction at 298 K? Explain.
(ii) natural tendency is to maximize entropy and since this reaction decreases entropy and is spontaneous (-Gº), then Hº must be negative to overcome the entropy change and drive this reaction.
(iii) For the reaction, how is the value of the standard free energy change, Gº, affected by an increase in temperature? Explain.
(iii) Gº = Hº –TSº ; as T increases, the value of –TSº increases (it becomes more positive since ΔS is negative) and the value of – Gº becomes a smaller negative number (i.e., moves toward zero).

68. AP 1999 (Continued…)
b) Some reactions that are predicted by their sign of Gº to be spontaneous at room temperature do not proceed at a measurable rate at room temperature.
(i) Account for this apparent contradiction.
(i) the sign of Gº (thermodynamics) does not account for activation energy (kinetics); a large activation energy would effectively prevent a reaction even though there is a favorable free energy change.
(ii) A suitable catalyst increases the rate of such a reaction. What effect does the catalyst have on Gº for the reaction? Explain.
(ii) a catalyst changes neither the Hº nor the Sº for a reaction, therefore, it will have no effect on the Gº.

69. EXTRA: Temperature Dependence of K
y = mx + b
(H and S  independent of temperature over a small temperature range)
If the temperature increases, K decreases for exothermic reactions, but increases for endothermic reactions.

70. Summary
Entropy is usually described as a measure of the disorder of a system. Any spontaneous process must lead to an increase in entropy of the universe.
The standard entropy of a chemical reaction can be calculated from the absolute entropies of reactants and products.
Under conditions of constant temperature and pressure, the free-energy change ΔG is less than zero for a spontaneous process and greater than zero for a non-spontaneous process. For an equilibrium process, ΔG = 0.

71. Summary
4. For a chemical and physical change at constant temperature and pressure, ΔG = ΔH – TΔS. This equation can be used to predict the spontaneity of a process.
5. The standard free energy change for a reaction, ΔGo, can be calculated from the standard free energies of formation of reactants and products.
6. The relationship between ΔG and equilibrium position is given by ΔGo = -RTln(K).
7. The value of ΔG for conditions other than standard is calculated from ΔG =ΔGo + RTln(Q).

72. Tutorials
ry/gilbert2/contents/ch13/studyplan.asp
THE END

73. Reversible vs. Irreversible Processes
Reversible: The universe is exactly the same as it was before the cyclic process.
Irreversible: The universe is different after the cyclic process.
All real processes are irreversible -- (some work is changed to heat).  w < G
Work is changed to heat in the surroundings and the entropy of the universe increases.

74. Laws of Thermodynamics
First Law: You can’t win, you can only break even.
Second Law: You can’t break even.

75. Review…. So So increases with: solid ---> liquid ---> gas
greater complexity of molecules (have a greater number of rotations and vibrations)
greater temperature (if volume increases)
lower pressure (if volume increases)