thermodynamics
Download
Skip this Video
Download Presentation
Thermodynamics

Loading in 2 Seconds...

play fullscreen
1 / 44

Thermodynamics - PowerPoint PPT Presentation


  • 101 Views
  • Uploaded on

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.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Thermodynamics' - dale-mejia


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
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
slide22
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
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
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
ad