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### Thermodynamics

Spontaneity, Entropy & Free Energy

Chapter 16

On the A.P. Exam:

5 Multiple Choice Questions

Free Response—Every Year

Standard State Conditions

- Indicated by a degree symbol
- Gases at 1 atm
- Liquids are pure.
- Solids are pure.
- Solutions are at 1 M.
- Temperature is 25o C (298 K).

First Law of Thermodynamics

- Conservation of Energy—Energy can be neither created nor destroyed.
- The amount of energy in the universe is constant.
- Energy can change forms, but the amount does not change.

Example

- Chemical (potential) energy is stored in the bonds of methane gas.
- Combustion releases the energy stored in the bonds as heat.
- Energy exchange between system & environment = enthalpy

Spontaneity

- Spontaneous processes need no outside cause; they happen naturally.
- Spontaneous = Fast
- (Thermodynamics is not concerned with how a reaction happens—just whether it will happen.)

What Causes Spontaneity??

- Exothermicity?? (Tendency to reach lowest potential energy)
- Partial explanation, but does not always hold true
- Probability—how likely are arrangements of atoms (randomness or entropy)

Entropy

- The amount of disorder or randomness in a system
- Higher entropy is favored because random arrangements are more likely.
- Ex.: Gas molecules are more likely to be evenly dispersed than clumped up in one place.

Which has higher entropy?

- 1 mole of gas or 1 mole of liquid?
- Pure water or a 0.5 M salt solution?
- Gas at 25o C or gas at 100o C?
- 1 mole of gas or 5 moles of gas?

Entropy Increases:

- In state changes--solid to liquid, liquid to gas or solid to gas
- When solids dissolve
- When temperature increases
- When volume of a gas increases (or pressure decreases)
- When a reaction occurs that produces more moles of gases

Second Law of Thermodynamics

- In any spontaneous process, there is an increase in entropy of the universe.
- Think of the universe as being inclusive of a system and its surroundings—combined entropy must increase (Cell)

DSsurr

- Sign depends on direction of heat flow—heat added to surroundings increases entropy & heat absorbed from surroundings decreases entropy

DSsurr

- Magnitude—depends on temperature—greater impact of exothermic system if temperature is low
- Analogy—giving $50 to a millionaire or to you—to whom does it mean more?

DSsurr

- Equal to –DH/T
- negative—because the surroundings are opposite from the system
- DH –change in heat in the system (from system’s perspective)
- T—temperature in Kelvin

What’s the Point???

- Dssurris larger at lower temperatures because the denominator of the fraction is smaller .
- The effect of heat flow for a process is more significant when temperature is lower.

Free Energy

- G
- Uses enthalpy, entropy and temperature to predict the spontaneity of a reaction
- If DG is negative, then a process is spontaneous.

Calculating DG

- DG = DH – TDS
- DGis free energy
- DH is change in enthalpy
- DSis change in entropy
- All terms are from the point of view of the system.

DG = DH – TDS

- Mathematical relationships:
- If DH is negative and DS is postive—always spontaneous
- If DH is positive and DS is negative—never spontaneous
- If both DH and DS have the same sign—temperature dependent

Example

- At what temperature is the following process spontaneous?

Br2(l) Br2(g)

DH = 31.0 kJ/mol & DS = 93.0 J/K mol

- Calculate temperature when DG is zero. (DG = DH – TDS)

DG and Boiling Point

- At the exact boiling point, DG is zero.
- Set known values of DH, DS and/or T equal to zero and solve for unknowns.

Example

- The enthalpy of vaporization of mercury is 58.51 kJ /mol and the entropy of vaporization is 92.92 J/K mol. What is mercury’s normal boiling point?

Third Law of Thermodynamics

- The entropy of a perfect crystal at 0 K is zero.
- Standard entropy values have been determined—Appendix 4

DS

- To determine entropy change for a reaction, subtract the entropy values
- DS = SnpSoproducts – SnrSoreactants
- n = number of moles (State functions depend on amount of substance present.)

Example

- Calculate DSo at 25o C for the reaction:

2NiS(s) + 3O2 (g) 2SO2 (g) + 2NiO (s)

- So in J/K mol:
- SO2 248 J/K mol
- NiO 38 J/K mol
- O2 205 J/K mol
- NiS 53 J/K mol

Example

- Calculate DSo for the reduction of aluminum oxide by hydrogen gas:

Al2O3(s) + 3H2(g) 2 Al(s) + 3 H2O(g)

Note: equal number of gas mol

Ans: 179 /K (large & positive because of water’s asymmetry)

Free Energy

- Cannot be directly measured
- Can be calculated using enthalpy and entropy changes (DG = DH – TDS)
- Can be calculated from already determined standard free energy values:

DG = SnpGoproducts – SnrGoreactants

Example:

- Calculate DH, DS, and DG using the following information for the reaction 2SO2 (g) + O2 (g) 2SO3(g).
- SO2: DH = -297 kJ/mol; So = 248 J/K mol
- SO3: DH = -396 kJ/mol; So = 257 J/K mol
- O2: DH = 0 kJ/mol; So = 205 J/K mol

Example:

- Using the following data, calculate DG for the reaction Cdiamond (s) C graphite (s)
- Cdiamond (s) + O2 (g) CO2 (g) DG = -397 kJ
- Cgraphite (s) + O2 (g) CO2 (g) DG = -394 kJ

Example

- Given the following free energies of formation, calculate the DG for the reaction

2CH3OH (g) + 3O2(g) 2 CO2(g) + 4 H2O (g)

CH3OH (g): DG = -163 kJ/mol

O2 (g): DG = 0 kJ/mol

CO2 (g): DG = -394 kJ/mol

H2O (g): DG = -229 kJ/mol

Free Energy & Pressure

- Entropy depends on pressure because changes in pressure affect volume.
- G = Go + RTln(P)
- G = free energy at new pressure
- Go = free energy at 1 atm
- R = gas constant (8.3145 J/K*mol)
- T = Kelvin temperature
- P = new pressure

DG at Nonstandard Conditions

- From the previous equation, a relationship between DG and Q or K can be derived.
- (See p. 807)
- DG = DGo + RT ln (Q)

Example

- CO (g) + 2H2 (g) CH3OH (l)
- Calculate DG for this process where CO is at 5.0 atm and H2 is at 3.0 atm.
- Calculate DGo using standard values for reactants & products.
- Then use DG = DGo + RT ln (Q)

DG and Equilibrium

- At equilibrium, a system has used all of the free energy available.
- DG equals zero because forward & reverse reactions have the same free energy.
- DG = DGo + RT ln (Q)
- 0 = DGo + RT ln (K)
- DGo = - RT ln (K)

Example

- For the synthesis of ammonia, DGo = -33.33 kJ/mol of N2 consumed at 25o C.
- Predict how each system below will shift in order to reach equilibrium:
- NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2 atm
- NH3 = 1.00 atm; N2 = 1.00 atm; H2 = 1.00 atm

NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2atm

- DG = DGo + RT ln (Q)

NH3 = 1.00 atm; N2 = 1.00 atm; H2 = 1.00 atm

- DG = DGo + RT ln (Q)

Example

- 4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)
- Use the data to calculate K for this reaction at 25o C.
- Fe2O3: DH = -826 kJ/mol; So = 90 J/K mol
- Fe: DH = 0 kJ/mol; So = 27 J/K mol
- O2: DH = 0 kJ/mol; So = 205 J/K mol

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