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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.

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Thermodynamics

Thermodynamics

Spontaneity, Entropy & Free Energy

Chapter 16

On the A.P. Exam:

5 Multiple Choice Questions

Free Response—Every Year


Standard state conditions
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
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
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
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
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
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
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
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
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)


D s surr
DSsurr

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


D s surr1
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?


D s surr2
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
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
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 d g
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.


D g d h t d s
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


Example1
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)


D g and boiling point
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.


Example2
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
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.)


Example3
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


Example4
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 energy1
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


Example5
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


Example6
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


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


Example7
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
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


D g at nonstandard conditions
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)


Example8
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)


D g and equilibrium
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)


Example9
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


Nh 3 1 00 atm n 2 1 47 atm h 2 1 00 x 10 2 atm
NH3 = 1.00 atm; N2 = 1.47 atm; H2 = 1.00 x 10-2atm

  • DG = DGo + RT ln (Q)


Nh 3 1 00 atm n 2 1 00 atm h 2 1 00 atm
NH3 = 1.00 atm; N2 = 1.00 atm; H2 = 1.00 atm

  • DG = DGo + RT ln (Q)


Example10
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





Calculate k d g rtlnk
Calculate K : DG = -RTlnK


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