1. Spontaneity, Entropy & Free Energy
Chapter 16

2. Spontaneity Process and Entropy
16.1
Spontaneity Process and Entropy
Spontaneous – happens without outside intervention
Thermodynamics studies the initial and final states of a reaction
Kinetics is concerned with the pathway
Kinetics is concerned with the time or speed of the reaction

3. Spontaneity Process and Entropy
16.1
Spontaneity Process and Entropy
Entropy (S)
The driving force for spontaneous processes of the universe
The measure of molecular randomness – disorder
Describes the number of arrangements available to a system in a given state – probability
Nature spontaneously moves towards the state that has the highest probability of existing.

4. Entropy and Microstates
16.1
Entropy and Microstates
Microstate – a configuration that gives a particular arrangement
Positional probability – depends on the number of configurations in space

5. Entropy and Microstates
16.1
Entropy and Microstates

6. Entropy and Microstates
16.1
Entropy and Microstates
Positional entropy increases from solid to liquid to gas
Positional entropy increases when solute and solvent are mixed
- Favoring the formation of solutions

7. 16.1
Sample Exercises 16.1 & 16.2
For each of the following pairs, choose the substance with the higher positional entropy
Solid CO2 and gaseous CO2
N2 gas at 1 atm and N2 gas at 1.0 x 10-2 atm
Predict the sign of the entropy change for each of the following processes
Solid sugar is added to water to form a solution
Iodine vapor condenses on a cold surface to form crystal
Will entropy increase or decrease for the following state changes
(s) to (aq)
(g) to (aq)
(l) to (aq)
* Mixtures have more entropy
CO2 gas
At 1.0 x 10-2 atm
positive
negative
increase
decrease
increase

8. Entropy and the Second Law of Thermodynamics
16.2
Entropy and the Second Law of Thermodynamics
In any spontaneous process there is an increase in the entropy of the universe
1st Law  Energy is conserved
2nd Law  Entropy is not conserved; it is increasing
Suniverse is always increasing
If Suniverse is  then the forward process is spontaneous
If Suniverse is  then the reverse process is spontaneous then the forward process is not spontaneous
If Suniverse is 0 then the process has no tendency to occur or the system is at equilibrium

9. Interplay of ΔSsys and ΔSsurr in Determining the Sign of Δsuniv
16.3
Pg. 759

10. 16.2
Sample Exercise 16.3
In a living cell, large molecules are assembled from simple ones. In this process consistent with the second law of thermodynamics?
Suniverse must be positive for the process to be spontaneous. A process for with  Ssystem is negative can be spontaneous if the associated  Ssurroundings is both larger and positive.

11. The Effect of Temperature on Spontaneity
16.3
The Effect of Temperature on Spontaneity
If 50 J of energy are transferred to surrounding where the temperature is high there would not be a big difference in the motion of the molecules.
However if 50 J of energy are transferred to surroundings where the temperature is low, there would be a bigger difference in the motion of the molecules.
The impact of the transfer of a given quantity of energy as heat to or from the surroundings will be greater at lower temperatures
Who would feel the greatest effect from a donation of $50?

12. The Effect of Temperature on Spontaneity
16.3
The Effect of Temperature on Spontaneity
Exothermic increases random motion Ssurr = + Endothermic decreases random motion  Ssurr = -
Entropy of surroundings is determined by heat flow.
The impact of energy transfer to or from a system is greater at lower temperatures
The sign of Ssurr depends on the direction of the heat flow
nature tends to seek the lowest energy
The magnitude of Ssurr depends on the temperature
Ssurr is directly proportional to the quantity of heat transferred Ssurr is inversely proportional to the temperature

13. The Effect of Temperature on Spontaneity
16.3
The Effect of Temperature on Spontaneity
+ for endothermic
- for exothermic
Ssurr =
quantity of heat (J)
temperature (K)
The negative changes the  H from the system to the surroundings
Ssurr = H T

14. The Effect of Temperature on Spontaneity
16.3
The Effect of Temperature on Spontaneity
SYSTEM T1
SYSTEM T1
SURROUNDINGS T2
ENERGY
ENERGY
EXOTHERMIC
ENDOTHERMIC
T2 >T1 at the beginning
Energy flows into the system
T2  as T1  until T2 = T1
Decreasing the entropy of the surroundings
H is positive
T1 >T2 at the beginning
Energy flows out of the system
T2  as T1  until T2 = T1
Increases the entropy of the surroundings
H is negative

15. The Effect of Temperature on Spontaneity
16.3
The Effect of Temperature on Spontaneity
SYSTEM T1
SYSTEM T1
SURROUNDINGS T2
ENERGY
ENERGY
EXOTHERMIC
ENDOTHERMIC
 Ssys + Ssurr
H H T T2
H is the same so S is dependent on temperature
T1 < T2
 Ssys Ssurr
Suniverse > 0
 Ssys + Ssurr
-H H T T2
H is the same so S is dependent on temperature
T2 < T1
-  Ssys + Ssurr Suniverse > 0

16. Sb2S3 + 3Fe  2Sb + 3FeS H = -125 kJ 16.3 Sample Exercise 16.4 H
In the metallurgy of Antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores:
Sb2S3 + 3Fe  2Sb + 3FeS H = -125 kJ
Carbon is used as the reducing agent for oxide ores:
Sb4O6 + 6C  4Sb + 6 CO H = 778 kJ
Calculate the  Ssurr for each of these reactions at 25C and 1 atm.
Ssurr = H T

17. 16.3
Exercise #23
Choose the compound with the greatest positional probability in each case.
1 mol H2 (at STP) or 1 mol H2 (at 100C, 0.5 atm)
1 mol N2 (at STP) or 1 mol N2 (at 100K, 2.0 atm)
1 mol H2O (s) (at 0C) or 1 mol H2O(l) (at 20 C)

18. 16.3
Exercise #25
Predict the sign of Ssurr for the following processes.
H2O(l)  H2O(g)
CO2(g)  CO2(s)

19. Goldfish are Held with Tartar Sauce
16.4
Free Energy
G =  H – T  S
G is Gibbs free energy
H is enthalpy
T is temperture in Kelvin
S is entropy
Refers to the system
Proved on p. 760 from S = H/T
Suniv = - G at constant T & P T
Goldfish are Held with Tartar Sauce
The process will be spontaneous if G is negative (because Suniv is positive)

20. ΔH and ΔS and the Dependence of Spontaneity on Temperature
16.5 G
Spontaneity
 G < 0
Spontaneous
Exergonic
 G > 0
Non spontaneous
Endergonic
 G = 0
Equilibrium
MEMORIZE!!!

21. ΔSuniv and ΔG for H2O (s)  H2O (l)
16.5
Is the melting of ice spontaneous?

22. H = 31.0 kJ/mol and S = 93.0 J/mol ∙ K
16.4
Sample Exercise 16.5
At what temperature is the folloiwng process spontaneous at 1 atm?
Br2 (l)  Br2 (g)
H = 31.0 kJ/mol and S = 93.0 J/mol ∙ K
What is the normal boiling point of liquid Br2?

23. 16.5
Exercise #29
Ethanethiol (C2H5SH; also called ethyl mercaptan) is commonly added to natural gas to provide the “rotten egg” smell of a gas leak. The boiling point of ethanethiol is 35C and its heat of vaporization is 27.5 kJ/mol. What is the entropy of vaporization for this substance?

24. 16.5
Exercise #30
For mercury, the enthalpy of vaporization is kJ/mol and the entropy of vaporization is J/k-mol. What is the normal boiling point of mercury?

25. Entropy Changes in Chemical Reactions
16.5
Entropy Changes in Chemical Reactions
Predicting the Sign of S
4NH3 (g) + 5O2 (g)  4NO (g) + 6H2O (g)
Fewer molecules means fewer possible configurations so positional entropy decreases
The change in positional entropy is dominated by the relative numbers of molecules.
H determines if a reaction is endothermic or exothermic
G determines if a reaction is spontaneous (at constant T & P)

26. Entropy Changes in Chemical Reactions
16.5
Entropy Changes in Chemical Reactions
Third Law of Thermodynamics
The entropy of a perfect crystal at 0 K is zero.
Because S is a state function, its value can be summation
Sreaction = np Sproducts - np Sreactants
However S for elements is not zero

27. 16.5
Sample Exercises 16.8
Calculate the S for the reduction of aluminum oxide by hydrogen gas:
Al2O3 (s) + 3H2 (g)  2Al(s) + 3H2O(g)
Will S be positive or negative?
3:3 gas
Look closely at the form H2 vs. H2O
H2O has not just positional entropy but vibrational and rotational
The more complex the molecule, the higher the standard entropy value.
3 Things to Look For
State of matter
Coefficients
Complexity of substance
Substance
S (J/K-mol)
Al2O3 (s) 51
H2 (g)
131
Al(s) 28
H2O(g)
189
179 J/K

28. Free Energy in Chemical Reactions
16.6
Free Energy in Chemical Reactions
Calculating G
Because G is a state function, its value can be calculated by summation
The standard free energy of formation of an element in its standard state is 0.
Greaction = np Gproducts - np Greactants
or using equations as with H or
Calculate H and S then use G = H - TS

29. The Dependence of Free Energy on Pressure
16.7
The Dependence of Free Energy on Pressure
Why is free energy dependent on pressure?
Isn’t H, enthalpy independent of pressure at constant pressure? No work?
G = H - T S
At a constant temperature S large volume > Ssmall volume
S low pressure > S high pressure
G = G + RT ln(P)
G = G + RT ln(Q)
The entropy and free energy of an ideal gas depends on its pressure.
G is free energy at 1 atm
G is free energy at pressure P
R is universal gas constant J/K-mol
T is temp in Kelvin
Q is reaction quotient using partial pressures (p593)
Remember
Q = 0 equilibrium; Q>K shift towards reactants; Q<K shift towards products

30. 16.7
Sample Exercises 16.13
One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases:
CO (g) + 2H2 (g)  CH3OH (l)
Calculate the G at 25 C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol.
Substance
Gf kJ
CH3OH (l)
-166
H2 (g)
CO(g)
-137
-38 kJ/mol G is more negative than G the reaction is more spontaneous at pressures above 1 atm

31. Free Energy and Equilibrium
16.7
Free Energy and Equilibrium
A system will try to achieve the lowest possible free energy Sometime that is equilibrium not a complete reaction
MINIMUM FREE ENERGY
PRODUCTS
REACTANTS
Minimum free energy happens before the completion of products, B, so it is at equilibrium.
A is changing into B, but it never
fully gets there, so equilibrium
Minimum free energy happens at the completion of products so no equilibrium

32. Free Energy and Equilibrium
16.8
Free Energy and Equilibrium
(a) At the beginning of the reaction
(b) As the reaction is occuring
(c) If Greactants = Gproducts then the system is at equilibrium because G = 0 Should happen at minimum free energy

33. Free Energy and Equilibrium
16.8
Free Energy and Equilibrium
At equilibrium
Gproducts = Greactants or G = Gproducts – Greactants = 0
G = G + RT ln (Q)
G = 0 = G + RT ln (Q)
G = - RT ln (Q)
at equilibrium Q = K
G = - RT ln (K)

34. Free Energy and Equilibrium
16.8
Free Energy and Equilibrium
G = Gproducts – Greactants = 0
G = G + RT ln (Q)
G = - RT ln (K)
Case 1: G = 0
Gproducts = Greactants
K = 1
All pressures = 1 atm
The system is at equilibrium.
Case 2: G < 0
G products< Greactants
The system will shift towards products. K > 1 since the pressures of the products at equilibrium will be greater than the pressures of the reactants
Case 3: G > 0
G products> Greactants
The system will shift towards reactants. K < 1 since the pressures of the reactants at equilibrium will be greater than the pressures of the products

35. Relationship between ΔG and K
16.8

36. The Temperature Dependence of K
16.8
The Temperature Dependence of K
G = - RT ln (K) = H - T S
ln (K) = H ̶ S
ln (K) = -H S
R T R
y = m x b
-RT -R
So K is dependent on T because of the G of the system! +
The graph of ln (K) versus 1/T is a straight line graph.
With slope = - H/R and intercept S/R
This assumes that H and S are independent of temperature.
This is a good approximation over a relatively small temperature range.

37. 16.8 Sample Exercises 16.14 Consider the ammonia synthesis reaction
N2 (g) + 3H2 (g)  2NH3 (g)
where G = kJ per mole of N2 consumed at 25  C. For each of the following mixture of reactants and products at 25  C, predict the direction in which the system will shift to reach equilibrium.
PNH3 = 1.00 atm; PN2= 1.47 atm; PH2= 1.00 x 10-2 atm
PNH3 = 1.00 atm; PN2= 1.00 atm; PH2= 1.00 atm

38. 16.8
Sample Exercises 16.15
The overall reaction for the corrosion (rusting) of iron by oxygen is
4Fe (s) + 3O2 (g)  2Fe2O3 (s)
Using the data below, calculate the equilibrium constant for this reaction at 25 C.
Substance
Hf kJ/mol
S J/K-mol
Fe2O3 (s)
-826 90
Fe (s) 27
O2 (g)
205

39. 16.9
Free Energy and Work
The maximum possible useful work obtained from a process at constant T & P is equal to the change in free energy.
wmax = G % efficiency
Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.
Consider a rechargeable battery
When current from the battery flows through a circuit it heats the wires. The heat is wasted energy- lost to the surroundings. Not energy doing useful work.
When recharging the battery more work is required than the battery discharged.
Even though the battery has returned to its original state, the surrounding have not.
Second Law of Thermodynamics (restated)
In any real cyclic process in the system, work is changed to heat in the surroundings, and the entropy of the universe is increased

40. Cell Potential, Electrical Work, and Free Energy
17.3
Cell Potential, Electrical Work, and Free Energy
Work is done when electrons are transferred through a a wire
Depends on the push behind the electrons which is the emf or potential difference (in volts)
1 volt = 1 J/ 1 C
emf = potential difference (V) = work (J) / charge (C)
E = -w / q
When a cell produces current the potential is positive and the current can be used to do work – so the work is negative
Rearranging – w = q E so wmax = - q Emax
In any real spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum.

41. Cell Potential, Electrical Work, and Free Energy
17.3
Cell Potential, Electrical Work, and Free Energy
Now let’s include free energy
wmax = - q Emax = G
q = nF
G = - nF Emax
G  = - nF E
The maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell.
Provides an experimental means to obtain G for a reaction
Galvanic cells will run in the direction that gives a positive value for Ecell and a negative value for G
n = moles of e-
the subscript “max” is usually left out
at standard conditions

42. Cell Potential, Electrical Work, and Free Energy
17.3
Cell Potential, Electrical Work, and Free Energy
Given maximum potential 2.50 V, moles of electrons 1.33 mol e-, and actual potential 2.10 V
Efficiency can be calculated
Calculate q with the Faraday C/ mol e-
Calculate the actual work
Calculate the efficiency

43. 17.3
Sample Exercise 17.3
Using the data in Table 17.1, calculate the G for the reaction
Cu2+(aq) + Fe(s)  Cu (s) + Fe3+(aq)
Is this reaction spontaneous?l

44. 17.3
Sample Exercise 17.4
Using the data from Table 17.1, predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au3+ solution.

45. 17.3
Sample Exercise 17.3
Chlorine dioxide (ClO2), which is produced by the reaction
2NaClO2 (aq) + Cl2 (g)  2ClO2 (g) + 2NaCl (aq)
has been tested as a disinfectant for municipal water treatment. Using the data from Table 17.1, calculate the E and the G at 25 C for the production of ClO2.

46. 17.3
Table 17.1
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