Thermodynamics

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# Thermodynamics - PowerPoint PPT Presentation

Thermodynamics. Spontaneity, Entropy &amp; 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.

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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
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