1. AP Chem Get Thermo Practice WS stamped off Today: Unit 4 Quest Th 11/8
Entropy, Gibbs Free Energy
Spontaneous reactions
Unit 4 Quest Th 11/8

2. First Law of Thermodynamics
Energy cannot be created nor destroyed.
Therefore, the total energy of the universe is a constant.
Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa

3. Spontaneous Process
Spontaneous processes can occur without any outside intervention
** the spontaneous process may be slow or fast

4. 1. Directionality
If one direction of a process is spontaneous the other direction is not
A nail will spontaneously rust but not spontaneously unrust

5. 2. Temperature/Pressure
Pressure and temperature affect spontaneity
Above 0 C it is spontaneous for ice to melt.
Below 0 C the reverse process (freezing) is spontaneous.
At 0 oC both processes are in equilibrium and neither reaction is favored

6. Example Determine whether the following processes are spontaneous:
a) when a piece of metal heated to 150oC is added to water at 40oC, the water gets hotter
b) water at room temperature decomposes into H2(g) and O2(g)
c) Benzene vapor, C6H6(g), at a pressure of 1atm, condenses to a liquid at the normal boiling point of benzene, 80.1oC

7. What is the driving force for reactions?
Thought used to be enthalpy – most spontaneous reactions are exothermic
negative ∆H is “good” – more favorable
leads to lower potential energy, which is more stable

8. What is the driving force for reactions?
Most spontaneous reactions are exothermic.
BUT…….
We do have reactions that are endothermic that occur spontaneously
So another driving force could be ENTROPY (S)
Units = J/mol x K
** NOTE THE UNITS! The standard for this is in J, while ∆H is in kJ/mol

9. Entropy
Measure of degree of disorder, randomness, or chaos in a system
Described as the # of possible arrangements that are available to the system
Reactions will “tend towards disorder”

10. Spontaneous Reaction Scorecard
Enthalpy, H, negative is “good”
lowering of energy
Entropy, S, positive is “good”
increase in possibilities/randomness

11. Increasing Entropy Entropy tends to increase with increases in
Temperature
Volume
The number of independently moving molecules
State change, solubility

12. # Moving particles 1. Change in State Solubility
Entropy increases with the freedom of motion of molecules.
Therefore, S(g) > S(l) > S(s)
Solubility
Generally, when a solid is dissolved in a solvent, entropy increases

13. Example
1. Choose the sample of matter that has a greater entropy in each pair:
1mol NaCl(s) or 1mol HCl(g)
2mol HCl(g) or 1mol HCl(g)
1mol of HCl(g) or 1mol Ar(g)

14. Example
2. Predict whether S is positive or negative for each of the following processes:
a) H2O(l)  H2O(g)
b) Ag+(aq) + Cl-(aq)  AgCl(s)
c) 4Fe(s) + 3O2(g)  2Fe2O3(s)
d) N2(g) + O2(g)  2NO(g)

15. Second Law of Thermodynamics
The entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.

16. Main Entropy Calculation
Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:
S = S(products) —S(reactants)

17. Example Calculate the entropy of the following reaction:
2NiS(s) + 3O2(g) --> 2SO2(g) + 2NiO(s)
SO2 = 248 J/mol K NiO = 38 J/mol K
O2 = 205 J/mol K NiS = 53 J/mol K

18. Example Calculate the entropy of the following reaction:
2NiS(s) + 3O2(g) --> 2SO2(g) + 2NiO(s)
S =(2(248) + 2(38)) - (2(53) + 3(205))

19. Example (496+76) - (106+615) 572-721 S = -149 J / K
Calculate the entropy of the following reaction:
2NiS(s) + 3O2(g) --> 2SO2(g) + 2NiO(s)
S = (2(248) + 2(38)) - (2(53) + 3(205)
(496+76) - ( )
S = -149 J / K

20. Example Why? Gas molecules are decreasing
Calculate the entropy of the following reaction:
2NiS(s) + 3O2(g) --> 2SO2(g) + 2NiO(s)
S is decreasing “good or bad?”
Why? Gas molecules are decreasing

21. Example
Calculate the standard entropy change for the reaction below at 298K:
Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g)

22. 3rd Law of Thermodynamics
The entropy of a pure solid at Absolute Zero is equal to 0
ONLY AT ABSOLUTE ZERO!
Hf of an element/diatomic = 0, but the S of one is usually NOT – be careful

23. Gibbs Free Energy (G) ΔG = ΔH - T ΔS
Can be thought of as the amount of available energy to do useful work.
Mathematically combines enthalpy and entropy to predict whether a reaction at constant temperature and pressure will be spontaneous
ΔG = ΔH - T ΔS

24. Gibbs Free Energy If DG < 0 If DG = 0 If G > 0
the forward reaction is spontaneous.
If DG = 0
the system is at equilibrium.
If G > 0
the reaction is spontaneous in the reverse direction.

25. Calculations G = H - T S Problem with units: H = kJ/mole
S = J / mole x K
T = K
*****Convert J to kJ******

26. Example ΔG = ΔH - T ΔS ΔG = (180.7 kJ) – (298 K)( 𝟐𝟒.𝟕 𝟏𝟎𝟎𝟎 kJ/K)
1. Calculate the standard free energy change for the formation of NO(g) from N2(g) and O2(g) at 298K given that ΔHo=180.7kJ and ΔSo=24.7J/K. Is the reaction spontaneous under these conditions?
ΔG = ΔH - T ΔS
ΔG = (180.7 kJ) – (298 K)( 𝟐𝟒.𝟕 𝟏𝟎𝟎𝟎 kJ/K)
ΔG = kJ,
Not spontaneous

27. Example
2. A particular reaction has a ΔHo = 24.6kJ and ΔSo =132J/K at 298K. Calculate ΔGo. Is the reaction spontaneous under these conditions?

28. Standard Free Energy Changes
Analogous to standard enthalpies of formation are standard free energies of formation, G.
DG = SDG products)  SG (reactants)

29. Example
1. Use data from appendix C to calculate the free-energy change for the following reaction at 298K: P4(g) + 6Cl2(g)  4PCl3(g) . What is ΔGo for the reverse reaction?

30. Example
2. The ΔHo for the combustion of propane is -2220kJ : C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
a) Predict whether ΔGo for this reaction is more or less negative than ΔHo
b) Calculate ΔGo. Was your prediction correct?

31. Free Energy and TemperatureΔH ΔS ΔG
Reaction Characteristics - +
Spontaneous at all temperatures
Nonspontaneous at all temperatures
+ or -
Spontaneous at low T; nonspontaneous at high T
Spontaneous at high T; nonspontaneous at low T
ΔG = ΔH - T ΔS

32. Example
1. The Haber process is used for the production of ammonia. Assume ΔHo and ΔSo for this reaction do not change with temperature.
a) Predict the direction in which ΔGo for this reaction changes with increasing temperature
b) Calculate the ΔGo values for the reaction at 25oC and 500oC.