Entropy

1
Entropy Gibbs Free Energy

• Chapter 19

2
The heat tax

• No matter what the process, heat always lost to
  surroundings
• More energy required to recharge batteries than
  energy available for work

3
Spontaneity

• Spontaneous process does NOT require outside
  energy
• Such as a car rusting
• Non-spontaneous process requires external energy
• Imagine trying to un-rust the jalopy above
• Not all spontaneous processes exothermic
• Ice melting above melting point

4
Entropy

• Disorder/randomness in a system
• Entropy increases in universe (2nd Law)
• Were all going to dissolution

5
Entropy the book definition

• Entropy (S) a thermodynamic function that
  increases with number of energetically equivalent
  ways to arrange components of a system to achieve
  a particular state
• HUH?!

6
Towards greater entropy

• Systems process in direction that has largest
  energetically equivalent ways to arrange its
  parts
• Which configuration has greater entropy i.e.,
  largest energetically equivalent ways to arrange
  itself?

7
Ludwig Boltzmann

• Distribution of energy over different energy
  states used as a way to calculate entropy
• S k ? log W
• k R/N 1.38 x 10-23 J/K Boltzmann constant
  (R 8.314 J/mol?K N Avogadros 6.022 x
  1023 atoms/mole)
• W of energetically equivalent ways to arrange
  components of system (unitless)

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Standard molar entropy, S

• For standard state elements (metals, gases, etc.)
  molar enthalpy of formation, Hf 0 J/mol
• For entropy, its different
• 3rd law of thermodynamics
• The entropy of a perfect crystal at absolute zero
  (0K) is zero.
• Since only one way to arrange (W 1)
• ? S k ? log W k ? log 1 0

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More on entropy

• Gas entropies are much larger than those for
  liquids
• Liquid entropies are larger than solids
• For ex
• I2(s) vs. Br2(l) vs. Cl2(g) standard entropies
  116.1, 152.2, 223.1 J/mol?K, respectively
• Solid C vs. gaseous C 5.6 vs. 158.1 J/mol?K

11
Even more on entropy!

• Generally, larger molecules have larger entropy
  than smaller molecules
• Due to greater molar mass
• Molecules with more complex structures have
  larger entropies than simpler molecules
• Since there are more ways for molecule to twist,
  rotate, vibrate in space
• For ex
• CH4, C2H6, C3H8 186.3, 229.2, 270.3 J/mol?K
• Ar, CO2, C3H8 154.9, 213.7, 270.3 J/mol?K

12
?Srxn

• Calculating standard changes in entropy
• Similar to calculating ?Hrxn
• Dont forget molar coefficients!
• ?Srxn Sproducts - Sreactants
• All reactants products in standard states
• J/(mol?K)

13
Appendix L, pages A-27-32

• Thermodynamic values ?H, ?S, ?G

14
Problem

• Compute ?Srxn for the following equation
• 4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g)
• Given
• 192.8 (J/mol?K) ammonia
• 205.2 oxygen gas
• 210.8 nitrogen monoxide
• 188.8 water

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Entropy changes and spontaneity

• According to the 2nd Law of Thermodynamics
• A spontaneous process is one that results in an
  increase of entropy in the universe.
• ?Suniv ?Ssystem ?Ssurrounding
• For a spontaneous process
• ?Suniv gt 0
• For a system at equilibrium
• ?Suniv 0
• For a nonspontaneous process
• ?Suniv lt 0

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Two questions

• Work on these individually and turn them in
• Does the entropy of surrounding increase or
  decrease in an exothermic process?
• Does the entropy of surrounding increase or
  decrease in an endothermic process?

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Temperature dependence of ?Ssurr

• ?Suniv ?Ssys ?Ssurr
• Therefore, for water freezing
• ?Ssys is negative (exothermic)
• ?Ssurr is positive (endothermic)
• ?S is large at low temperatures (closer to
  freezing point) has very little E, so big
  difference made
• ?S is small at high temps (further away from fp)
  already has more E due to higher temp ? little
  difference
• IOW, magnitude of ?S depends on temp (at above
  0C)

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Quantifying entropy changes in surrounding

• -?Hsys ? ?Ssurr
• ?Ssurr ? 1/T
• ?Ssurr ? -?Hsys/T
• ? exothermic processes have tendency to be more
  spontaneous at low temps than high
• They increase entropy of surrounding more if T is
  small
• As temp increases (greater E in surrounding), ?H
  decreases, producing a smaller ?Ssurr

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Problem

• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• Calculate entropy change in surrounding at 25C
  given ?Hrxn -2044 kJ
• Determine ?Ssys, given
• H2O(g) 188.84 J/mol?K
• CO2(g) 213.74 J/mol?K
• C3H8(g) 270.30 J/mol?K
• O2(g) 205.07 J/mol?K
• What is ?Suniv?

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Spontaneous or not?
Type
Hsys
Spontaneous?
S sys
1 Exothermic Less order Spontaneous
lt0 gt0 Sunivgt0
2 Exothermic More order Depends on H S
lt0 lt0 More favorable at lower temps
3 Endothermic Less order Depends on H S
gt0 gt0 More favorable at higher temps
4 Endothermic More order Nonspontaneous
gt0 lt0 Suniverlt0
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Examples of each type

• Type I
• Exothermic (?Hsyslt0) and less order (?Ssysgt0)
• Combustion reactions
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• Type IV
• Endothermic (?Hsysgt0) and more order (?Ssyslt0)
• N2(g) 2H2(g) ? N2H4(l)

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Examples of each type (cont.)

• Type II
• Exothermic (?Hsyslt0) and more order (?Ssyslt0)
• Temperature dependent
• More favorable at lower temperatures
• N2(g) 3H2(g) ? 2NH3(g)

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Examples of each type (cont.)

• Type III
• Endothermic (?Hsysgt0) and less order (?Ssysgt0)
• Temperature dependent
• More favorable at higher temperatures
• NH4Cl(s) ? NH3(g) HCl(g)

26
Gibbs Free Energy

• Measures spontaneity of a process with evaluation
  of only the system
• No longer needs evaluation of surrounding
• Free energy available energy to do work
• Sum of energies available from dispersal of
  energy and matter

27
G

• ? G (?Glt0) spontaneous process (?Sgt0)
• ? G (?Ggt0) non-spontaneous process (?Slt0)

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The effect of ?H, ?S, T on ?G

• Type I
• Exothermic (-?H)
• Entropy gt 0 (?S)
• -?G
• Spontaneous at all temps
• 2N2O(g) ? 2N2(g) O2(g)

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The effect of ?H, ?S, T on ?G

• Type IV
• Endothermic (?H)
• Entropy lt 0 (-?S)
• ?G
• Nonspontaneous at all temps
• 3O2(g) ? 2O3(g)

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The effect of ?H, ?S, T on ?G

• Type II
• Exothermic (-?H)
• Entropy lt 0 (-?S)
• ?G is temp-dependent
• H2O(l) ? H2O(s)
• -?H
• Water freezing is temp-dependent
• ?Spontaneous _at_ low temps
• Non-spontaneous _at_ high temps

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The effect of ?H, ?S, T on ?G

• Type III
• Endothermic (?H)
• Spontaneous (?S)
• ?G is temp-dependent
• H2O(l) ? H2O(g)
• ?H
• Water boiling is temp-dependent
• ?Spontaneous _at_ high temps
• Nonspontaneous _at_ low temps

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Calculating ?Grxn

• Calculate ?Grxn (at 25C) for
• SO2(g) ½ O2(g) ? SO3(g)
• SO2 ?Hf -296.8 kJ/mol ?S 248.2 J/(mol?K)
• O2 ?Hf ?, ?S 205.2J/(mol?K)
• SO3 ?Hf -395.7 kJ/mol, ?S 256.8 J/(mol?K)
• Is it spontaneous under standard conditions?

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Calculating ?Grxn from free energies of formation

• Free E of formation of pure elements in standard
  states 0 kJ/mol
• ?Grxn ?Gproducts - ?Greactants
• CH4(g) 8O2(g) ? CO2(g) 2H2O(g)
• -50.5 ? -394.4 -228.6
• Calculate ?Grxn

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Determine ?Grxn for 3C(s) 4H2(g) ? C3H8(g)

• Given
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• ?Grxn -2074 kJ
• C(s) O2(g) ? CO2(g)
• ?Grxn -394.4 kJ
• 2H2(g) O2(g) ? 2H2O(g)
• ?Grxn -457.1 kJ

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Non-standard conditions

• ?Grxn is at standard conditions
• 25 C, 273 K, 1 atm, 1M, etc.
• ?Grxn non-standard conditions
• Explains why water evaporates off floor even
  though ?Grxn 8.59 kJ/mol ? is a non-spont
  process.
• Partial pressure of water well below 1 atm ?
  Non-standard condition!

39
Free E change under non-standard conditions

• ?Grxn ?Grxn RT(lnQ)
• Q rxn quotient, R 8.314 J/(mol?K)

40
Problem

• Compute ?Grxn under the following conditions for
  
• 2NO(g) O2(g) ? 2NO2(g)
• PNO 0.100 atm PO2 0.100 atm PNO2 2.00 atm
  
• ?Grxn -71.2 kJ
• At 25C

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Relating ?Grxn to K

• ?Grxn ?Grxn RT(lnQ)
• At equilibrium, Q K ?Grxn 0
• 0 ?Grxn RT(lnK)
• ?Grxn -RT(lnK)
• When Klt1 (reactant-favored)
• lnK - ?Grxn
• ? spontaneous in reverse direction
• When Kgt1 (product-favored)
• lnK ?Grxn -
• ? spontaneous in fwd direction
• When K 1, lnK 0 ?Grxn 0
• ? Rxn _at_ equilibrium under standard conditions

43
Temp dependence of K

• How does one obtain the equation on the left?
• Useful for obtaining ?Hrxn ?Srxn
• Does the equation look familiar?
• How would we plot this?