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# Thermodynamics - PowerPoint PPT Presentation

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.

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### Thermodynamics

Spontaneity, Entropy & Free Energy

Chapter 16

On the A.P. Exam:

5 Multiple Choice Questions

Free Response—Every Year

• 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).

• 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.

• 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

• 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.)

• 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)

• 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.

• 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?

• 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

• 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

• 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.

• G

• Uses enthalpy, entropy and temperature to predict the spontaneity of a reaction

• If DG is negative, then a process is spontaneous.

• 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

• 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.

• 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?

• 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.)

• 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

• 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)

• 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

• 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

• 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

Reverse one equation (change sign of DG) and add them together.

• 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

• 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)

• 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)

• 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)

• 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

Calculate K : DG = -RTlnK