Spontaneity, Entropy,

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Spontaneity, Entropy, Free Energy

• Chapter 16

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1st Law of Thermodynamics

• The first law of thermodynamics is a statement of
  the law of conservation of energy energy can
  neither be created nor destroyed. The energy of
  the universe is constant, but the various forms
  of energy can be interchanged in physical and
  chemical processes.

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Spontaneous Processes and Entropy

• Thermodynamics lets us predict whether a process
  will occur but gives no information about the
  amount of time required for the process.
• A spontaneous process is one that occurs without
  outside intervention. This is also considered
  thermodynamically favored.

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Entropy

• The driving force for a spontaneous process is an
  increase in the entropy of the universe.
• Entropy, S, can be viewed as a measure of
  randomness, or disorder.
• Nature spontaneously proceeds toward the states
  that have the highest probabilities of existing.

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The expansion of an ideal gas into an evacuated
bulb.
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Positional Entropy

• A gas expands into a vacuum because the expanded
  state has the highest positional probability of
  states available to the system.
• Therefore,
• Ssolid lt Sliquid ltlt Sgas

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Positional Entropy

• Which of the following has higher positional
  entropy?
• a) Solid CO2 or gaseous CO2?
• b) N2 gas at 1 atm or N2 gas at 1.0 x 10-2 atm?

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Entropy

• What is the sign of the entropy change for the
  following?
• a) Solid sugar is added to water to form a
  solution?
• ?S is positive
• b) Iodine vapor condenses on a cold surface to
  form crystals?
• ?S is negative

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The Second Law of Thermodynamics

• . . . in any spontaneous(thermodynamically
  favored) process there is always an increase in
  the entropy of the universe.
• ?Suniv gt 0
• for a spontaneous process.

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?SUniverse

• ?Suniverse is positive -- reaction is
  spontaneous.
• ?Suniverse is negative -- reaction is spontaneous
  in the reverse direction.
• ?Suniverse 0 -- reaction is at equilibrium.

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?S? ?npS?(products) ? ?nrS?(reactants)

• ?H? ?npH?(products) ? ?nrH?(reactants)

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?Soreaction

• Calculate ?S? at 25 oC for the reaction
• 2NiS(s) 3O2(g) ---gt 2SO2(g) 2NiO(s)

SO2 (248 J/Kmol) NiO (38 J/Kmol) O2 (205
J/Kmol) NiS (53 J/Kmol)
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• ?S? ?npS?(products) ? ?nrS?(reactants)
• ?S? (2 mol SO2)(248 J/Kmol) (2 mol NiO)(38
  J/Kmol) - (2 mol NiS)(53 J/Kmol) (3 mol
  O2)(205 J/Kmol)
• ?S? 496 J/K 76 J/K - 106 J/K - 615 J/K
• ?S? -149 J/K gaseous molecules decreases!

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Effect of ?H and ?S on Spontaneity
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?G -- Free Energy

• Two tendencies exist in nature
• tendency toward higher entropy -- ?S
• tendency toward lower energy -- ?H
• If the two processes oppose each other (e.g.
  melting ice cube), then the direction is decided
  by the Free Energy, ?G, and depends upon the
  temperature.

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Free Energy

• ?G ?H ? T?S (from the standpoint of the
  system)
• A process (at constant T, P) is spontaneous in
  the direction in which free energy decreases
• ??Gsys means ?Suniv
• Entropy changes in the surroundings are primarily
  determined by the heat flow. An exothermic
  process in the system increases the entropy of
  the surroundings.

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Free Energy ?G

• ?G ?H ? T?S
• ?G negative -- spontaneous
• ?G positive -- spontaneous in opposite
  direction
• ?G 0 -- at equilibrium

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?G, ?H, ?S

• Spontaneous reactions are indicated by the
  following signs
• ?G negative
• ?H negative
• ?S positive

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Free Energy Change and Chemical Reactions

• ?G? standard free energy change that occurs if
  reactants in their standard state are converted
  to products in their standard state.
• ?G? ?np?Gf?(products) ? ?nr?Gf?(reactants)

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Temperature Dependence

• ?Ho ?So are not temperature dependent.
• ?Go is temperature dependent.
• ?G ?H ? T?S

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?G? Calculations

• Calculate ?H?, ?S?, ?G? for the reaction
• 2 SO2(g) O2(g) ----gt 2 SO3(g)
• ?H? ?np?Hf?(products) ? ?nr?Hf?(reactants)
• ?H? (2 mol SO3)(-396 kJ/mol)-(2 mol
  SO2)(-297 kJ/mol) (0 kJ/mol)
• ?H? - 792 kJ 594 kJ
• ?H? -198 kJ

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?G? CalculationsContinued

• ?S? ?npS?(products) ? ?nrS?(reactants)
• ?S? (2 mol SO3)(257 J/Kmol)-(2 mol SO2)(248
  J/Kmol) (1 mol O2)(205 J/Kmol)
• ?S? 514 J/K - 496 J/K - 205 J/K
• ?S? -187 J/K

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?G? CalculationsContinued

• ?Go ?Ho ? T?So
• ?Go - 198 kJ - (298 K)(-187 J/K)(1kJ/1000J)
• ?Go - 198 kJ 55.7 kJ
• ?Go - 142 kJ
• The reaction is spontaneous at 25 oC and 1 atm.

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The Third Law of Thermodynamics

• . . . The third law of thermodynamics the
  entropy of a perfect crystal at 0K is zero.
• not a lot of perfect crystals out there so,
  entropy values are RARELY ever zeroevenelements
• So what? This means the absolute entropy of a
  substance can then be determined at any temp.
  higher
• than 0 K. (Handy to know if you ever need to
  defend why G H for elements 0. . . . BUT S
  does
• not!).

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Hesss Law ?Go

• Cdiamond(s) O2(g) ---gt CO2(g) ?Go -397 kJ
• Cgraphite(s) O2(g) ---gt CO2(g) ?Go -394 kJ
• Calculate ?Go for the reaction
• Cdiamond(s) ---gt Cgraphite(s)
• Cdiamond(s) O2(g) ---gt CO2(g) ?Go -397 kJ
• CO2(g) ---gt Cgraphite(s) O2(g) ?Go 394 kJ
• Cdiamond(s) ---gt Cgraphite(s) ?Go -3 kJ
• Diamond is kinetically stable, but
  thermodynamically unstable.

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So

• So increases with
• solid ---gt liquid ---gt gas
• greater complexity of molecules (have a greater
  number of rotations and vibrations)
• greater temperature (if volume increases)
• lower pressure (if volume increases)

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?Go Temperature

• ?Go depends upon temperature. If a reaction must
  be carried out at temperatures higher than 25 oC,
  then ?Go must be recalculated from the ?Ho ?So
  values for the reaction.

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Free Energy Pressure

• The equilibrium position represents the lowest
  free energy value available to a particular
  system (reaction).
• ?G is pressure dependent
• ?S is pressure dependent
• ?H is not pressure dependent

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Free Energy and Pressure

• ?G ?G? RT ln(Q)
• Q reaction quotient from the law of mass action.

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Free Energy Calculations

• . CO(g) 2H2(g) ---gt CH3OH(l)
• Calculate ?Go for this reaction where CO(g) is
  5.0 atm and H2(g) is 3.0 atm are converted to
  liquid methanol.
• ?G? ?np?Gf?(products) ? ?nr?Gf?(reactants)
• ?G? (1 mol CH3OH)(- 166 kJ/mol)-(1 mol
  CO)(-137 kJ/mol) (0 kJ)
• ?G? - 166 kJ 137 kJ
• ?G? - 2.9 x 104 J

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Free Energy CalculationsContinued

2.2 x 10-2
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Free Energy CalculationsContinued

• ?G ?G? RT ln(Q)
• ?G (-2.9 x 104 J/mol rxn) (8.3145
  J/Kmol)(298 K) ln(2.2 x 10-2)
• ?G (- 2.9 x 104 J/mol rxn) - (9.4 x 103 J/mol
  rxn)
• ?G - 38 kJ/ mol rxn
• Note ?G is significantly more negative than
  ?G?, implying that the reaction is more
  spontaneous at reactant pressures greater than 1
  atm. Why?

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A system can achieve the lowest possible free
energy by going to equilibrium, not by going to
completion.
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As A is changed into B, the pressure and free
energy of A decreases, while the pressure and
free energy of B increases until they become
equal at equilibrium.
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Graph a) represents equilibrium starting from
only reactants, while Graph b) starts from
products only. Graph c) represents the graph
for the total system.
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Free Energy and Equilibrium

• ?G? ?RT ln(K)
• K equilibrium constant
• This is so because ?G 0 and Q K at
  equilibrium.

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?Go K

• ?Go 0 K 1
• ?G? lt 0 K gt 1 (favored)
• ?G? gt 0 K lt 1 (not favored)

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Equilibrium Calculations

• 4Fe(s) 3O2(g) lt---gt 2Fe2O3(s)
• Calculate K for this reaction at 25 oC.
• ?Go - 1.490 x 106 J
• ?Ho - 1.652 x 106 J
• ?So -543 J/K
• ?G? ?RT ln(K)
• K e - ?G?/RT
• K e 601 or 10261
• K is very large because ?G? is very negative.

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Temperature Dependence of K

• y mx b
• (?H? and S? ? independent of temperature over a
  small temperature range)
• If the temperature increases, K decreases for
  exothermic reactions, but increases for
  endothermic reactions.

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Free Energy Work

• The maximum possible useful work obtainable from
  a process at constant temperature and pressure is
  equal to the change in free energy
• wmax ?G

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Reversible vs. Irreversible Processes

• Reversible The universe is exactly the same as
  it was before the cyclic process.
• Irreversible The universe is different after
  the cyclic process.
• All real processes are irreversible -- (some work
  is changed to heat). ? w lt ?G
• Work is changed to heat in the surroundings and
  the entropy of the universe increases.

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Laws of Thermodynamics

• First Law You cant win, you can only break
  even.
• Second Law You cant break even.