Spontaneity, Entropy and Free Energy

1
Chapter 16

• Spontaneity, Entropy and Free Energy
• (Chemical Thermodynamics)

2
THERMODYNAMICS EQUATIONS

• ?E q w
• w - P?V -?nRT
• q ?H
• ?Suniverse ?Ssystem ?Ssurroundings
• ?S ?H/T
• ?So ? n Soproducts - ? n Soreactants
• ?Ho ? n Hoproducts - ? n Horeactants
• ?Gfo ? n Gfoproducts - ? n Gforeactants
• ?G ?H - T ?S
• ?G ?Go RT ln Q
• Q ?(Pproducts)a / ? (Preactants)b
• R 8.314 joules/mol x K
• ?Go - RT ln Qequilibrium
• and Qequilibrium K

3
THERMODYNAMICS

• The study of energy changes in chemical and
  physical systems.
• Processes run in a continuum
• Spontaneous processes
• Equilibrium processes
• Nonspontaneous processes

4
The domains of kinetics and thermodynamics
5
Exothermic versus endothermic
Surroundings
Surroundings
EXOTHERMIC
ENDOTHERMIC
Energy
Energy
System ?H ? 0 ?E ? 0 (-)
System ?H ? 0 ?E ? 0 ()
6

• Given the following data
• S(s) 3/2 O2(g) ?? SO3(g) ?Ho
  -395.2 kJ
• 2SO2(g) O2(g) ?? 2SO3(g) ?Ho
  -198.2 kJ
• Calculate ?Ho for the reaction
• S(s) O2(g) ?? SO2(g)

7

• Given the following data
• N2(g) 2O2(g) ?? 2NO2(g) ?Ho 67.7 kJ
• N2(g) 2O2(g) ?? N2O4(g) ?Ho 9.7 kJ
• Calculate ?Ho for the dimerization of NO2
• 2NO2(g) ?? N2O4(g)

8

• Given the following data
• C2H2(g) 5/2 O2(g) ?? 2CO2(g) H2O(l)
  ?Ho -1300. kJ
• C(s) O2(g) ?? CO2(g) ?Ho -394 kJ
• H2(g) ½ O2(g) ?? H2O(l) ?Ho -286 kJ
• Calculate ?Ho for the reaction
• 2C(s) H2(g) ?? C2H2(g)

9

• Given the following data
• 2O3(g) ?? 3O2(g) ?Ho -427 kJ
• O2(g) ?? 2O(g) ?Ho 495 kJ
• NO(g) O3(g) ?? NO2(g) O2(g)
• ?Ho -199 kJ
• Calculate ?Ho for the reaction
• NO(g) O(g) ?? NO2(g)

10

• Keep in Mind the Following Key Concepts When
  Doing Enthalpy calculations
• When a reaction is reversed, the magnitude of ?H
  remains the same but its sign changes.
• When the balanced equation for a reaction is
  multiplied by an integer, the value of ?H for
  that reaction must be multiplied by the same
  integer.
• The change in enthalpy for a given reaction can
  be calculated from the enthalpies of formation of
  the reactants and products
• ?Horeaction ? np ?Hof(products) - ? nr
  ?Hof(reactants)
• Elements in their standard states are not
  included in the ?Hreaction calculations. That
  is, ?Hof for an element in its standard state is
  zero.

11
Spontaneous Processes

• ?H, Enthalpy ?? Heat
• Exothermic, ?H lt 0 (-)
• Endothermic, ?H gt 0 ()
• ?S, Entropy ?? Randomness
• Increased randomness, ?S gt 0 ()decrease
  randomness, ?S lt 0 (-)

12
? S, positive entropy changes

• Solids ?? Liquids ?? Gases
• and into solutions of each

13

• In each case tell which reaction, the forward or
  the reverse, is favored by entropy changes.
• a. 4NH3(g) 3O2(g) ?? 2N2(g) 6H2O(g)
• b. 4NO(g) 6H2O(g) ?? 4NH3(g) 5O2(g)
• c. C(s) CO2(g) ?? 2CO(g)
• d. 2O3(g) ?? 3O2(g) e. CH3OH(l) ??CH3OH(g)

14

• C6H4(OH)2(aq) ?? C6H4O2(aq) H2(g)
• ?Ho 177.4 kJ
• H2(g) O2(g) ?? H2O2 (aq)
• ?Ho -191.2 kJ
• H2(g) ½ O2(g) ?? H2O(g)
• ?Ho -241.8 kJ
• H2O(g) ?? H2O(l)
• ?Ho -43.8 kJ

15
Spontaneous Processes

• ?H, Enthalpy ? Heat
• Exothermic, ?H lt 0 (-)
• Endothermic, ?H gt 0 ()
• ?S, Entropy ? Randomness
• Increased randomness, ?S gt 0 ()decrease
  randomness, ?S lt 0 (-)

16
SECOND LAW OF THERMODYNAMICS

• Entropy (randomness) of the universe
• tends toward a maximum (increases).
• ? Suniverse ? Ssystem ? Ssurroundings
• ? Suniverse gt 0 ?? spontaneous
• ? Suniverse lt 0 ?? spontaneous in the
  opposite direction
• ? Suniverse 0 ?? equilibrium
• ?S Sfinal - Sinitial

17
Interplay of ?Ssys and ?Ssurr in Determining the
Sign of ?Suniv

• Signs of Entropy Changes
• ?Ssys ?Ssurr ?Suniv Process Spontaneous?
• Yes
• - - - No (reaction will occur in
  opposite direction)
• - ? Yes, if ?Ssys has a larger
  magnitude than ?Ssurr
• - ? Yes, if ?Ssurr has a large
  magnitude than ?Ssys

18
STANDARD ENTROPY VALUES, So

• Conditions solid or liquid - pure substance
• gas one atmosphere pressure solution - one
  mole/ liter, M
• temperature of 25oC or 298 K
• These follow Hesss Law as with Enthalpy, so use
  values for So from tables to calculate.
• ?So ? n Soproducts - ? n Soreactants

19

• Calculate ?So for each of the following
  reactions.
• a. H2(g) ½ O2(g) ?? H2O(g)
• b. 3O2(g) ?? 2O3(g)
• c. N2(g) O2(g) ?? 2NO(g)
• d. N2(g) 3H2(g) ?? 2NH3(g)
• e. HCl(g) ?? H(aq) Cl-(aq)

20

• For the reaction
• CS2(g) 3O2(g) ?? CO2(g) 2SO2(g)
• ?So is equal to 143 J/K. Use this value and
  data to calculate the value of So for CS2(g).
• For the reaction
• 2Al(s) 3Br2(l) ?? 2AlBr3(s)
• ?So is equal to 144 J/K. Use this value and
  data to calculate the value of So for solid
  aluminum bromide.

21
THIRD LAW OF THERMODYNAMICS

• Absolute Entropy, S, 0 at 0 Kelvins for
• a perfect crystal of a pure substance.
• Changes in entropy, ? S, are used to calculate
• in an equilibrium system,
• ? S ? H / T

22

• The boiling point of chloroform (CHCl3) is
  61.7oC. The enthalpy of vaporization is 31.4
  kJ/mol. Calculate the entropy of vaporization.
• For mercury, the enthalpy of vaporization is
  58.51 kJ/mol and the entropy of vaporization is
  92.92 J/K.mol. What is the normal boiling point
  of mercury?

23
?G, Gibbs Free Energy

• Is the maximum amount of energy available
• to do useful work on the surroundings.
• ?G ? H T ? S
• ?G lt 0 (-) ?? Spontaneous
• ?G gt 0 () ?? Spontaneous in the opposite
• direction
• ?G 0 ?? equilibrium

24
?G, Gibbs Free Energy

• Is the maximum amount of energy available
• to do useful work on the surroundings.
• ?G ? H T ? S
• ?G lt 0 (-) ?? Spontaneous
• ?G gt 0 () ?? Spontaneous in the opposite
• direction
• ?G 0 ?? equilibrium

25
Variation of free energy with extent of reaction
Spontaneous in Opposite direction
Equal
?G is spontaneous
26

• Given the values of ?H and ?S, which of the
  following changes will be spontaneous at constant
  T and P?
• a. ?H 25 kJ, ?S 5 J/K, T 300K
• b. ?H 25 kJ, ?S 100 J/K, T
  300K
• c. ?H -10 kJ, ?S 5 J/K, T
  298K
• d. ?H -10 kJ, ?S -40 J/K, T
  200K
• e. ?H -10 kJ, ?S -40 J/K, T
  300K
• f. ?H -10 kJ, ?S -40 J/K, T
  500K
• g. ?H 5 kJ, ?S -5 J/K, T 10K
• h. ?H 5 kJ, ?S -5 J/K, T 1000K

27

• ?G ?H - T ?S ?G 25 x 103 J (300 K)(5
  J/K)
• Note We must be consistent on units. Typically
  enthalpy values are in kJ and entropy in J/K. We
  must use the same energy units for both if we are
  using the equation
• ?G ?H - T ?S.
• ?G 23,500 J ? 24,000 K Not spontaneous
• ?G 25,000 J (300 K)(100 J/K) -5000 J
  Spontaneous
• Without calculating ?G, we know this reaction
  will be spontaneous at all temperatures. ?H is
  negative and ?S is positive (- T ?S lt 0), ?G will
  always be less than zero.
• ?G -10,000 J (200 K)(-40 J/K) -2000J

28

• ?G -10,000 J (300 K)(-40 J/K) 2000J Not
  spontaneous
• ?G -10,000 J (500 K)(-40 J/K) 10,000J
  Not spontaneous
• Since ?H is positive and ?S is negative (-T ?S is
  positive) the value of ?G will be greater than
  zero regardless of the temperature. Thus, this
  change is not spontaneous at any temperature.
• Not spontaneous, see (c).

29

• From data in Tables, calculate ?Ho, ?So, and ?Go
  for each of the following reactions at 25oC.
• a. CH4(g) 2O2(g) ?? CO2(g) 2H2O(g)
• b. 6CO2(g) 6H2O(l) ?? C6H12O6(s) 6O2(g)
• glucose
• c. P4O10(s) 6H2O(l) ?? 4H3PO4(s)
• d. HCl(g) NH3(g) ?? NH4Cl(s)

Hesss Law also work for ?G ?Go ?(n
?Gofproducts) - ? (n ?Gofreactants) Calculate
?Ho,?So, ?Go ?H - T ?S
30

• Standard free energies of formation in kJ/mole at
  25oC
• Al(s) 0.0 CCl4(g) -60.63
• Al2O3(s) -1582. C2H5OH(l) -174.9
• Br2(l) 0.0 C2H5OH(g) -168.6
• Br2(g) 3.142 CH3COOH(l) -390.0
• C(graphite) 0.0 CH3COOH(g) -374.0
• C(diamond) 2.900 Ca(s) 0.0
• CO(g) -137.15 CaO(s) -604.2
• CO2(g) -394.36 Ca(OH)2(s) -896.76
• CH4(g) -50.75 CaSO4(s) -1320.3
• C2H6(g) -32.9 Cl2(g) 0.0
• C2H4(g) 68.12 Cu(s) 0.0
• C2H2(g) 209.2 CuO(s) -130.
• C3H8(g) -23.49 F2(g) 0.0
• C6H6(l) 124.50 Fe(s) 0.0
• C6H6(g) 129.66 Fe2O3(s) -742.2
• CH3OH(l) -166.4 H2(g) 0.0
• CH3OH(g) -162.0 H2O(l) -237.18
• CS2(l) 65.27 H2O(g) -228.59

31

• HCl(g) -95.299 Na2O2(s) -451.
• HBr(g) -53.43 NaOH(s) -381.
• HI(g) 1.7 O2(g) 0.0
• I2(s) 0.0 Pb(s) 0.0
• K(s) 0.0 PbO2(s) -217.4
• KCl(s) -408.32 PbSO4(s) -811.2
• Mg(s) 0.0 S(s) 0.0
• MgCl2 -592.33 SO2(g) -300.19
• MgO(s) -569.57 SO3(g) -371.1
• Mg(OH)2(s) -833.75 H2S(g) -33.6
• N2(g) 0.0 H2SO4(l) -690.101
• NH3(g) -16.5 Si(s) 0.0
• N2O(g) 104.2 SiO2(s) -856.67
• NO(g) 86.57 SiH4(g) 56.9
• NO2(g) 51.30 SiF4(g) -1572.7
• Na(s) 0.0 SiCl4(l) -619.90
• NaF(s) -541.0 SiCl4(g) -617.01
• NaCl(s) -384.03 Zn(s) 0.0
• NaBr(s) -347 ZnO(s) -318.3

32
?Go, STANDARD FREE ENERGY CHANGES

• Will the reaction or phase change be spontaneous
• for standard state substances?
• ?Go ?Ho - T ?So
• First calculate ?Ho and ?So using Hesss Law for
  each and the tables of values. Then calculate
  ?Go as above

33
Various Possible Combinations of ?H and ?S for a
Process and the ResultingDependence of
spontaneity on Temperature

• Case Result
• ?S positive, ?H negative Spontaneous at all
  temperature
• ?S positive, ?H positive Spontaneous at high
  temperatures
• (where exothermicity is relatively unimportant)
• ?S negative, ?H negative Spontaneous at low
  temperatures
• (where exothermicity is dominant)
• ?S negative, ?H positive Process not spontaneous
  at any temperature (reverse process is
  spontaneous at all temperatures)

34

• For the reaction
• SF4(g) F2(g) ?? SF6(g)
• The value of ?Go is 374 kJ. Use this value and
  data from Tables to calculate the value of ?Gof
  for SF4(g).
• Two crystalline forms of white phosphorus are
  known. Both forms contain P4 molecules, but the
  molecules are packed together in different ways.
  The ? form is always obtained when the liquid
  freezes. However, below 76.9oC, the ? form
  spontaneously converts to the ? form
• P4(s, ?) ?? P4(s, ?)
• a. Predict the signs of ?H and ?S for this
  process
• b. Predict which form of phosphorus has the
  more ordered crystalline structure.

35

• Hydrogen cyanide is produced industrially by the
  following reactions
• 1000oC
• 2NH3(g) 3O2(g) 2CH4(g) ????
  Pt-Rh
• 2 HCN(g) 6H2O(g)
• Is the high temperature needed for thermodynamic,
  or kinetic, reasons?

36

• Hydrogen cyanide is produced industrially by the
  following reactions
• 2NH3(g) 3O2(g) 2CH4(g) ?? 2HCN(g)
  6H2O(g)
• ?Hfo -46 0 -75 135.1 -242
• (kJ/mole)
• So 193 205 186 202 189
• (J/K.mol)
• Is the high temperature needed for thermodynamic,
  or kinetic, reasons?
• Is ?Greaction , -, or 0
• How will we calculate ?Greaciton _at_100oC?
• Hesss Law? No it uses ?Gof values.
• ?Gorx ?Ho - T ?So
• ?Sorx ?(nSoproducts) - ?(n ?Hof reactants)
• Is ?Go , - at 1000oC (1273k)
• Is ?G , - at room temperature (300k)
• Why not use ?Gof ? (n ?Goproducts) - ? (n
  ?Gorx)
• Signs of ?H and ?S ? Rx is spontaneous at all
  temperatures, no thermodynamic reason for
  temperature to be high, therefore reason is to
  increase reaction rate.

1000oC
Pt-Rh
37

• 2 NH3(g) 3O2(g) 2CH4(g) ?? 2HCN(g)
  6H2O(g)
• ?Hof -46 0 -75 135.1 -242
• (kJ/mol)
• So 193 205 186 202 189
• ? Gorx ? Ho -T ? So
• ? Sorx ?(nSoproducts)- ? (nSoreactants)
• ? Hreaction ?(nSoproducts)- ? (nSoreactants)

38
?G 0FREE ENERGY AT EQUILIBRIUM

• ?G ?Go RT ln Q
• Where ?Go is found in APPENDIX R 8.317
  joules/ mol x k,
• T Kelvins,
• And Q Reaction Quotient.
• Q ?(Pproducts)a / ?(Preactants)b
• ? ?? multiplication of each pressure or
  concentration of product or reactant in the
  equation.
• a,b, ? coefficients of the equation become
  exponents

39
FREE ENERGY AND EQUILIBRIUM

• At ? G 0,
• ? G ? Go R T ln Q
• Where Q reaction quotient
• R 8.314 joules/ mol . K
• So ?Go -RT ln Qequilibrium
• And Qequiliibrium Kp (gases)
• Kc (solution)
• Where Standard States are used,
• (1 atmosphere, 1 M, 298 K)

40
Write the reaction quotient for each of the
following reactions. a. 2HF(g) ??? H2(g)
F2(g) b. 2NO(g) O2(g) ??? 2NO2(g) c. Ca(s)
Cl2(g) ??? CaCl2(s) d. SiF4(g) 2H2O(l)
??? SiO2(s) 4HF(g) e. 2HgO(s) ??? 2Hg(l)
O2(g) f. NH4Cl(s) ??? NH3(g) HCl(g)
41
K, EQUILIBRIUM CONSTANT

• At ? G 0,
• ? G ? Go RT ln Q
• So ? Go - RT ln Qequilibrium
• And Qequilibrium K
• Where K is the equilibrium constant.
• At K ltlt 1, mostly reactants present,
• K gtgt 1, mostly products present,
• K 1, equal products and reactants present
• Using ? Go values, the equilibrium constants may
  be calculated.

42

• The equilibrium constant for the following
  reaction at 45oC is 0.65
• N2O4(g) ??? 2NO2(g)
• What is the value of ?Go at 45oC for this
  reaction?

43

• The equilibrium constant at 440oC for the
  following reaction is 0.020.
• 2HI(g) ??? H2(g) I2(g)
• What would you expect to be the sign of the value
  of the change in standard free energy?

44

• One of the reactions that destroys ozone in the
  upper atmosphere is
• NO(g) O3(g) ??? NO2(g) O2(g)
• Using data from Tables, calculate ?Go and K (at
  298 K) for this reaction.
• Use Hess Law for ?Go values or ?Go ?H - T ?So,
  if ?Go values not available.

45

• Hydrogen sulfide can be removed from natural gas
  by the reaction.
• 2H2S(g) SO2(g) ??? 3S(s) 2H2O(g)
• Calculate ?Go and K (at 298 K) for this reaction.
  Would this reaction be favored at a high or low
  temperature?
• ?Go -RT LnQequil -RT LnK
• First calculate ?Go from tables than K.

46
LeCHATELIERS PRINCIPLE

• A system at equilibrium will shift to a new
  equilibrium to partially remove any external
  stress. Consider the effects of
• -- Concentration changes
• -- Pressure changes
• -- Temperature
• -- Catalysts

47
LeCHATELIERS PRINCIPLEConcentration changes

• K constant but look at Q
• ?G -, KgtQ causes a shift to increase Q.
• ?G , KltQ causes a shift to decrease Q.
• H2(g) Cl2(g) ??? 2HCl(g)
• So as HCl goes up, Q goes up, KltQ

48
LeCHATELIERS PRINCIPLETemperature changes

• At equilibrium,
• ? Go - RT ln K and ? Go ?Ho - T ?So
• so - RT ln K ? Ho - T ? So
• or ln K - ? Ho/RT ? So/R
• If T goes up, K goes up when ? Ho gt 0 i.e. ?H
• T goes up, K goes down when ? Ho lt 0. i.e.
  -?H
• At Two temperatures,
• What is effect of a large K value on ?Go? Ln K
  is and therefore reaction goes as written.
  What is effect of temperature on K? K large
  means more products.

49
SUMMARY

• Reaction spontaneity is the bottom line for
  chemical thermodynamics.
• The sign of ?G, - , , or 0 tells the
  difference between initial and final states.
• Maximum work, wmax ?G
• The maximum possible useful work obtainable from
  a process at constant temperature and pressure is
  equal to the change in free energy.

50
THERMODYNAMICS EQUATIONS

• ?E q w
• w - P?V -?nRT
• q ?H
• ?Suniverse ?Ssystem ?Ssurroundings
• ?S ?H/T
• ?So ? n Soproducts - ? n Soreactants
• ?Ho ? n Hoproducts - ? n Horeactants
• ?Gfo ? n Gfoproducts - ? n Gforeactants
• ?G ?H - T ?S
• ?G ?Go RT ln Q
• Q ?(Pproducts)a / ? (Preactants)b
• R 8.314 joules/mol x K
• ?Go - RT ln Qequilibrium
• and Qequilibrium K

51

• Predict direction in which system will shift to
  reach equilibrium.
• Predicting direction calculate Q from
• ?G ?Go RT ln(Q)

52

• At 25oC
• 4Fe(s) 3O2(g) ??? 2Fe2O3(g)
• Calculate ?Go two ways.

53

• Free Energy equilibrium
• ? G - R T ln k
• ? G ? Go R T ln Q
• at equilibium ? G 0, and Q 0
• ? G - R T ln k
• When substance undergo chemical reaction,
  reaction proceeds with minimum fill energy
  (equil) which corresponds to
• Gproduct Greactant
• ?G Gproduct Greactant 0

54

• ?Go 0
• Gp Gr when all components are in standard
  states 1 atm for gases.
• ?Go lt 0 Gop Gor is negative A ??? B
• therefore Gop lt Gor k B/A K
  lt 1
• According to LeChatelier Principle will
  adjust to right to reach equilibrium
• ?Go gt 0 Gop Gor is positive A ??? B
• Gop gt Gor k B/A
  K gt 1
• According to LeChatelier Principle reaction
  will shift to left

55
Energy/enthalpy changes

• ?H heat changes at constant pressure.
• ?E q w
• If w - P ? V
• and q ?H
• then ?E ?H P ? V

56
STATE FUNCTIONS

• ?E, ?H, and ?V are state functions.
• They are dependent only on initial and final
  states regardless of the path between states.
• Capital letters are used for state functions.
• q and w are not state functions.