1. Entropy and Free Energy

2. Driving Forces of Reactions
So far we have seen that reactions are spontaneous if they give off heat – exothermic
There is a natural tendency in the universe for systems to get to lowest energy state
Why do endothermic reactions happen?
Demo with barium
Ammonium chloride dissolving in H2O - endothermic

3. Natural direction
Scientists notice that there is a natural direction for processes
Balls roll down hill, not up hill
Ice melts above 0º C, never refreezes above 0 C
A gas fills its container uniformly, never collects in one area
Heat flows from hot to cold; hotter object doesn’t get hotter when exposed to a colder object
Wood burns spontaneously, but CO2 and H2O don’t form wood when heated

4. What makes these processes irreversible?
All of the processes have the following in common:
in all cases you have less information about how the particles are organized than you did before
Analogy: Why don’t we have fire drills during lunch?
Less “information” about where students are during lunch compared to when they are in class
Measure of “information” about a system = ENTROPY

5. Entropy Measure of the amount of randomness or disorder in a system
Entropy was introduced in 1865 by Rudolf J. E. Clausius, a German physicist. Clausius said he derived the term from the Greek words en trope, which means “in the transformation” He used it to describe the dissipation or apparent loss of energy available to do work as energy is transformed in a system.
Measure of the amount of randomness or disorder in a system
Symbol: S
Units: J/K
No matter what the process, entropy (of universe) is always increasing…

6. Entropy Analogies
Throw a card into the air – 2 possible positions (up or down)
Throw a deck of cards into the air – how many possible positions?
One possibility is that they land organized in a stack. How probable is that?
The more cards = the more entropy
MORE MOLES = MORE ENTROPY

7. Given the following reaction, how is entropy changing: N2 (g) + 3 H2(g)  2NH3(g)
Increasing
Decreasing
Stays the same
Need more information 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

8. What causes entropy to increase?
Statistics…
Boltzmann Bucks Demo
At first everyone has $1, we play rock-paper-scissors for awhile. Some people have more money than others, but
no one has all the money and
Not everyone has exactly $1
Why not?
There is only one way for the money to be arranged so that everyone has $1
There are only 9 ways for the money to be arranged so that one person has $9 (assuming class size of 9)
There are many ways for some people to have no money, some to have $1 and some to have $2 or $3.
So there is a higher probability that the $ will be arranged as described in C
Nature spontaneously proceeds to the state that has the highest probability of existing.
Highest probability = most disordered

9. Entropy Changes Increase moles Dissolving and mixing
Increasing temperature
Increase volume
Solid to liquid or liquid to gas (or S to G)
More complicated molecules have higher S than simpler molecles

10. Which one of the following does not generally lead to an increase in entropy of a system?
Increase in total number of moles or particles
Formation of a solution
Formation of a gas
Formation of a solid 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

11. Given the reaction below, how is entropy changing? Br2(l)  Br2(g)
Increasing
Decreasing
Stays the same
Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

12. Given the following reaction, how is entropy changing: Ag+1(aq) + Cl-1(aq)  AgCl(s)
Increasing
Decreasing
Stays the same
Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

13. Given the following reaction, how is entropy changing: 2NO2(g)  N2O4(g)
Increasing
Decreasing
Stays the same
Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

14. Given the following reaction, how is entropy changing: CO(g) + H2O(g)  CO2(g) + H2(g)
Increasing
Decreasing
Stays the same
Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

15. Given the following reaction, how is entropy changing: H2(g) + F2(g)  2HF(g)
Increasing
Decreasing
Stays the same
Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

16. Given the following reaction, what is the sign for ΔS: NaCl(s)  NaCl(aq)
positive
negative
Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

17. Given the following reaction, how is entropy changing: 2OH-(aq) + CO2(g)  H2O(l) + CO32- (aq)
Increasing
Decreasing
Stays the same
Need more info 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

18. Which of the following has the largest increase in entropy?
Pb(NO3)2(s)  Pb(NO3)2(aq)
CaCO3(s)  CaO(s) + CO2(g)
2NH3 (g)  2H2(g) + N2(g)
H2(g) + Br2(g)  2HBr(g) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

19. Spontaneity Spontaneous change: Spontaneous reactions occur when
Occurs w/o continuous input of energy
Spontaneous reactions occur when
Reaction is exothermic ( ΔH < 0)
Increase in entropy for the system (ΔS >0)
But which is more important?
ΔH or ΔS ?

20. Total Entropy System vs. Surroundings vs. Universe
Suniverse = Ssystem + Ssurroundings
Suniv is always increasing.
Two spontaneous processes:
CaCl2(s)  Ca2+(aq) + Cl-1(aq) ΔH = -66kJ
Ssys is increasing
Ssurr is increasing because heat is released to surroundings
NH4Cl(s)  NH4+ + Cl-(aq) ΔH = 15 kJ
Ssurr is decreasing b/c surroundings are losing heat
Non-spotaneous
Na(s)  Na (l) ΔH =2.59 kJ
Ssys = increasing
Ssurr = decreasing

21. NH4Cl(s)  NH4+ + Cl-(aq) ΔH = 15 kJ Na(s)  Na (l) ΔH =2.59 kJ
Na+(g) + Cl-(g)  NaCl(s) ΔH = -771 kJ
Why is A spontaneous, but not B?
Entropy is much greater for A
Why is C spontaneous?
Enthalpy is large

22. How do you know if enthalpy or entropy will make the reaction more spontaneous?
Remember: Suniverse = Ssystem + Ssurroundings
Ssurr depends on temperature
The lower the surrounding temperature, the more significant adding heat is
The higher the surrounding temp, the less significant adding heat is.
Analogy
Imagine you give $1 to someone w/ only $10 to their name?
Imagine the effect of giving $1 to a millionaire.
Who is affected more?

23. Suniverse = Ssystem + Ssurroundings
Exothermic reactions that release heat to the surroundings are a a stronger driver of reactions when the surrounding temperature is low.
At high temperatures, an exothermic reaction isn’t such a strong driving force, entropy is more important.
Negative sign b/c ΔH defined in terms of system: exothermic reaction from system’s perspective causes increase in entropy of surroundings.
Suniverse = Ssystem + Ssurroundings
Entropy of system
Determined by ΔH of system

24. Substitute: - ΔHsys/T for Ssurr Multiply by (-T) Define new quantity
Suniv = Ssys + Ssurr
Substitute: - ΔHsys/T for Ssurr
Multiply by (-T)
Define new quantity
Free energy: ΔG = -T ΔSuniv
Free energy tells whether a rxn will be spontaneous at a given temperature.
Measures the maximum energy available to do useful work
Reactions at equilibrium have ΔG = 0
-T ΔSuniv = ΔHsys - T ΔSsys
ΔG = ΔHsys - T ΔSsys

25. Based on the previous slides and derivation of ΔG, rxns will be spontaneous if ΔG is
Less than 0
Greater than 0
Equal to 0
Spontaneity has nothing to do with ΔG. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

26. Under which conditions will reactions ALWAYS be spontaneous?
ΔH > 0, ΔS >0
ΔH > 0, ΔS < 0
ΔH < 0, ΔS >0
ΔH < 0, ΔS <0
ΔG = ΔHsys - T ΔSsys 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

27. Under which conditions will reactions NEVER be spontaneous?
ΔH > 0, ΔS >0
ΔH > 0, ΔS < 0
ΔH < 0, ΔS >0
ΔH < 0, ΔS <0
ΔG = ΔHsys - T ΔSsys 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

28. Under which conditions will reactions be spontaneous at high temps
Under which conditions will reactions be spontaneous at high temps? (other than when they are always spontaneous)
ΔH > 0, ΔS >0
ΔH > 0, ΔS < 0
ΔH < 0, ΔS >0
ΔH < 0, ΔS <0
ΔG = ΔHsys - T ΔSsys 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

29. ΔH > 0, ΔS >0 ΔG = ΔHsys - T ΔSsys
A large positive value for the term (TΔS) can make ΔG negative if it is bigger than ΔH
Reaction is endothermic so the entropy of surroundings is decreasing. At high temps, this won’t make as big of a difference as it would at lower temps.
ΔG = ΔHsys - T ΔSsys

30. Under which conditions will reactions be spontaneous at low temps
Under which conditions will reactions be spontaneous at low temps? (other than when they are always spontaneous)
ΔH > 0, ΔS >0
ΔH > 0, ΔS < 0
ΔH < 0, ΔS >0
ΔH < 0, ΔS <0
ΔG = ΔHsys - T ΔSsys 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

31. ΔH < 0, ΔS < 0 ΔG = ΔHsys - T ΔSsys
A small negative value for the term (TΔS) can still make ΔG negative if it is smaller than the absolute value of ΔH
Reaction is exothermic so the entropy of surroundings is increasing. At low temps, this will make a bigger difference than it would at higher temps.
ΔG = ΔHsys - T ΔSsys

32. Calculating ΔG
For a reaction at 25 C, ΔH = 100 kJ and ΔS = 80 J/K, determine if the reaction is spontaneous.
For a reaction with a ΔH = 100 kJ and a ΔS of 80 J/K, at what temperature will the reaction become spontaneous?
Watch your units!! Put temps in Kelvin and make sure you aren’t trying to add Joules to Kilojoules!

33. 100 KJ – (298*80/1000) = 76.2 kJ = not spontaneous
0 = (x*80/1000)  100 = x(. 08)
x = 1250 K, reaction becomes spontaneous at temperatures above 1250 K

34. Calculations of ΔS°rxn, ΔH°rxn and ΔG°rxn
Standard Entropy of Formation Tables (ΔS°f )
Σ n(ΔS°f )products - Σ n(ΔS°f )reactants
Standard Gibbs Free Energy of Formation Tables (ΔG°f )
Σ n(ΔG°f )products - Σ n(ΔG°f )reactants
Only good for standard conditions!
For ΔG at non-standard conditions, use ΔG = ΔH - TΔS

35. Calculate the free-energy change, DG°, for the oxidation of ethyl alcohol to acetic acid using standard free energies of formation.
CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)

36. CH3CH2OH(l) + O2(g)  CH3COOH(l) + H2O(l)
DGf°, kJ/mol – – –237.2
n, mol
nDGf°, kJ – –392.5 –237.2
–174.8 kJ –629.7 kJ
DG° = –629.7 – (–174.8)
DG° = –454.9 kJ

37. 2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
Sodium carbonate, Na2CO3, can be prepared by heating sodium hydrogen carbonate, NaHCO3:
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
Estimate the temperature at which the reaction proceeds spontaneously at 1 atm. See Appendix C for data.

38. 2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO(g)
DHf°, kJ/mol –947.7 – –241.8 –393.5
n, mol
nDHf°, kJ – – –241.8 –393.5
– kJ – kJ
DH° = kJ
Sf°, J/mol  K
nSf°, J/K
204 J/K J/K
DS° = J/K

40. Use ΔG to get K
Equilibrium position represents the lowest free energy value available to a particular reaction system

41. ΔG and K
Standard free energy change is related to the thermodynamic equilibrium constant, K, at equilibrium.
IF a reaction is NOT at equilibrium, it is proceeding in some direction (forward or reverse) depending on Q, reaction quotient.
That means there exists energy to do work (make reaction proceed)
ΔG = Δ G° + RT ln Q
At equilibrium:
Δ G = 0, because there is no ability to do any more useful work
and Q = K
So we get:
Δ G° = –RT ln K

42. Calculate the value of the thermodynamic equilibrium constant at 25°C for the reaction
N2O4(g) NO2(g)
The standard free energy of formation at 25°C is kJ/mol for NO2 and kJ/mol for N2O4(g).

43. DG° = 2 mol(51.30 kJ/mol) – 1 mol(97.82 kJ/mol)
DG° = kJ – kJ
DG° = 4.78 kJ