Spontaneity, Entropy and Free energy

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Chapter 16

• Spontaneity, Entropy and Free energy

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16.1 Spontaneous Process and Entropy

• A process that will occur without outside
  intervention.
• Thermodynamics cant determine how fast the
  process is (may be fast or slow).
• Kinetics tell us that the rate of reaction
  depends on activation energy temperature,
  concentration and catalysts i.e, it depends on
  the pathway of process.
• Thermodynamics compares initial and final states
  and does not require knowledge of the pathway.
• Kinetics describes pathway between reactants and
  products.
• We need both thermodynamics and kinetics to
  describe a reaction completely.

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Spontaneity

• process that will occur without outside
• intervention.
• Rolling a ball down a hill (gravity) steel
  rusting (?) wood burning (exothermic process)
  transfer of heat from hot to cold (exothermic)
  freezing of water (exothermic) Melting of ice
  (endothermic!!) etc.
• What common characteristic derives all those
  processes to be spontaneous?
• It is an increase in a property called Entropy,
  S.
• Thus all spontaneous processes occur as a result
  of an increase of the S of the universe.

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What is Entropy?

• Entropy, S, is measure of randomness, or
  disorder in the system. For example molecular
  randomness.
• Naturally, things move from order to disorder!
  (from lower S to higher S.)

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Entropy

• S defined in terms of probability.
• Substances take the most likely arrangement (that
  have more probabilities).
• The most likely arrangement is the most random.
• Can we calculate the number of arrangements for a
  system?

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• 2 possible arrangements
• 50 chance of finding the left empty

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• 4 possible arrangements
• 25 chance of finding the left empty (1 chance
  only)
• 50 chance of them being evenly dispersed

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• 6 possible arrangements
• 8 chance of finding the left empty
• 50 chance of them being evenly dispersed

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Entropy and Gases

• Gas placed in one bulb of a container will
  spontaneously expand to fill the entire vessel
  (vessel of two bulbs) evenly.
• Probabilities of finding equal number of
  molecules in each bulb are huge.
• Finding molecules in one bulb is highly
  improbable thus the process does not occur
  spontaneously.

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Entropy and state of matter

• Gases completely fill their chamber because there
  are many more ways to do that than to leave half
  empty.
• Ssolid ltSliquid ltltSgas
• In solids, molecules are very close and thus they
  have relatively few positions available to them
• There are many more ways for the molecules to be
  arranged as a liquid than a solid.
• Gases have a huge number of positions possible.

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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 i.e., the highest positional
• entropy
• Probability depends on the number of
• configurations in space ( positional
• microstates)

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16.2 Entropy and the second law of thermodynamics

• Solutions form because there are many more
• possible arrangements of dissolved pieces than
  if
• they stay separate there is an increase in
  entropy
• Generally, in any spontaneous process, there is
  
• always an increase in the entropy of the
  universe
• This is the second law of thermodynamics
• DSuniv DSsys DSsurr
• If DSuniv is positive the process is
  spontaneous.
• If DSuniv is negative the process is spontaneous
  in
• the opposite direction.
• ?Suniv gt 0 for any spontaneous process
• First law The energy of a universe is constant

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16.3 The effect of temperature on spontaneity

• For exothermic processes DSsurr is positive
• For endothermic processes DSsurr is negative
• Consider this process (endothermic) H2O(l)
  H2O(g)
• DSsys is positive (change from L to G)
• DSsurr is negative (random motion of atoms in the
  surrounding decreases)
• Which one will control the process DSsys or
  DSsurr?
• DSuniv
• DSuniv depends on temperature.

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Temperature and Spontaneity

• Entropy changes in the surroundings are
  determined by the heat flow.
• An exothermic process is a driving force for
  spontaneity because by giving up heat the entropy
  of the surroundings increases.
• The significance of exothermicity as the driving
  force depends on the temperature at which the
  process occurs
• The size of DSsurr depends on temperature at
  which the heat is transferred
• DSsurr -DH/T
• for a reaction that takes place at constant
  temperature and pressure.
• DH change in enthalpy

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Entropy of DSys and DSsurr in determining the
sign of DSuniv
DSsurr
DSuniv
Spontaneous?
DSsys
-
-
-
No



Yes

-
?
Yes at High temp.
DSsurr -DH/T
Yes at Low temp.
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16.4 Free Energy

• GH-TS G Gibbs free energy Thus the change in
  G is given by
• DGDH-TDS at constant temperature
• All quantities refer to the system.
• When no subscript the quantity refers to the
  system
• Divide by -T
• -DG/T -DH/T DS
• -DG/T DSsurr DS
• -DG/T DSuniv at constant T and P
• If DG is negative at constant T and P, the
  process is spontaneous.
• The process is spontaneous in the direction DG
  decreases
• - DG means DSuniv

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Check

• For the reaction H2O(s) H2O(l)
• DSº 22.1 J/K mol DHº 6030 J/mol
• Calculate DG at 10ºC and -10ºC
• Look at the equation DGDH-TDS
• We can also say DGoDHo-TDSo
• The symbol (o) indicates that all substances are
  in their standard states
• At -10ºC DGo 220 At 0ºC DGo 0
• At 10ºC DGo -220
• Spontaneity can be predicted from the sign of DH
  and DS.

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DGDH-TDS

-
At all Temperatures
Spont. at high temperatures, ?H is unimportant


Spont. at low temperatures, ?H is important
-
-
Not spont at any temperature, Reverse is
spontaneous

-
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16.5 Entropy changes in chemical reactionsThe
Third Law of Thermodynamics

• In a chemical reaction, when the number of
• gaseous molecules increases, the positional
• disorder will increase and consequently ?S
  would
• be Positive and Visa Versa
• Consider the following examples
• N2(g) 3H2(g) 2NH3(g)
• 4NH3(g) 5O2 (g) 4NO(g)
  6H2O(g)
• CaCO3(s) CaO
  (s) CO2(g)

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• The changes in enthalpy determines the
  exothermicity and endothermicity at constant P
• The changes in entropy determines the
• spontaneity at constant P and T
• Are there values given for S?
• The entropy of a pure crystal at 0 K is 0.
• Molecular motion is almost zero and only one
  arrangement is possible.
• Thus, the entropy of a perfect crystal is zero
  this statement is called the Third Law of
  Thermodynamics

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• At T gt0, some changes in order will occur,
  consequently, S gt0
• This value gives us a starting point.
• Standard Entropies Sº ( at 298 K and 1 atm) of
  substances are listed.
• DSº can then be determined for products and
  reactants
• ?S? ?np?S?(products) ? ?nr?S?(reactants)
• Entropy is a state function of the system (It is
  not pathway dependent)

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• Entropy is an extensive property. It depends on
  the amount of substance present
• Number of moles of reactants and products must be
  taken into account
• For calculations go to examples 16.7 and 16.8
  in the book
• What is the expected ?S (0, 0r ve or
• ve) for the following reaction?
• Al2O3(s) 3H2(g) 2Al(s)
  3H2O(g)
• More complex molecules possess higher Sº

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16.6 Free Energy and Chemical Reactions

• DGº standard free energy change.
• Free energy change that will occur if reactants
  in their standard state turn to products in their
  standard state.
• N2(g) 3H2(g) 2NH3(g) DGº
  -33.3 kJ
• DGº cant be measured directly, but can be
  calculated from other measurements.
• DGºDHº-TDSº
• Use Hesss Law with known reactions.

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Free Energy in Reactions

• There are tables of DGºf .
• The standard free energy of formation for any
  element in its standard state is 0.
• Remember- Spontaneity tells us nothing about rate.

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Free Energy in Reactions

• Why DGº is useful?
• To compare relative tendencies of reactions to
  occur.
• The more ve the value of DGº, the further a
  reaction will go to the right to reach equilibrium

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• How do we calculate DGº?
• 1. Use the equation
• DGºDHº-TDSº
• This equation is applicable for reactions taking
  place at constant temperature
• Use Hesss Law with known reactions.
• Exercise 16.9 p. 802/803

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2. Claculation of DGº using Hesss law

• Free energy is a state function just like
  enethalpy, i.e., it depends upon the pathway of
  the reaction
• It can be found same as DH using Hesss law
• Exercise 16.10 p.804

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3. Calculation of DGº using the standard free
energies of formation DGºf

• DGºf of a substance is
• the change in free energy that accompanies the
  formation of 1 mole of that substance from its
  constituent elements with all reactants and
  products in their standard states
• DGº for a specific reaction is obtained from the
  equation
• ?G? ?np?Gf?(products) ??nr?Gf?(reactants)

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• There are tables of DGºf .
• The standard free energy of formation for any
  element in its standard state is 0.
• Remember also that the number of moles of each
  reactant and product must be used in calculating
  DGº for a reaction
• Exercise 16.11 / p. 805

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Exercise

• Is the following reaction spontaneous under the
  standard conditions?
• C2H4(g) H2O(l) C2H5OH(l)
• Find DGº for the reaction using DGºf values from
  the table
• If DGº is Ve, the process is spontaneous and the
  formation of ethanol is favorable
• Of course to judge whether the reaction is
  feasible or not, we have to study its kinetics to
  know whether the reaction is fast or slow.
• If it is slow adding a catalyst may be considered
  
• Also, remember that DGº depends on temp.
• DGoDHo-TDSo

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16.7 The dependence of Free energy on Pressure

• At a large volume the gas has many more positions
  available for its molecules than at low volumes
• S large V gt S small V
• S low P gt S high P
• Since S depends on P then G will depend on P. It
  can be shown that
• G Go RTln(P)
• (Go free energy at 1 atm)

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• By proper derivation we got
• DG DGº RTln(Q)
• where Q is the reaction quotient (P of the
  products /P of the reactants).
• CO(g) 2H2(g) CH3OH(l)
• Would the reaction be spontaneous at 25ºC with
  the H2 pressure of 5.0 atm and the CO pressure of
  3.0 atm? (Ex 16.13 p. 808)
• DGºf CH3OH(l) -166 kJ
• DGºf CO(g) -137 kJ DGºf H2(g) 0 kJ

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16.8 Free energy and equilibrium

• According to thermodynamics the equilibrium
  occurs at the lowest value of free energy
  available
• At equilibrium DG 0, Q K
• DG DGº RTln(Q)
• Thus, DGº -RTlnK
• Solve example 16.15 p 813

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DGº K

• DGº -RTlnK

• 0 1
• lt0 gt0
• gt0 lt0

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Temperature dependence of K

• DGº -RTlnK DHº - TDSº
• ln(K) - DHº/RT DSº/R
• Plot lnK VS 1/T
• A straight line gives a
• Slope DHº/R
• Intercept DSº/R

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Example
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The temperature dependence of K
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16.9 Free energy and Work

• Free energy is that energy free to do work.
• The maximum amount of work possible at a given
  temperature and pressure.
• Never really achieved because some of the free
  energy is changed to heat during a change, so it
  cant be used to do work.

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Reversible v. 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)

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Reversible v. 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).