Spontaneity, Entropy and Free energy

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Chapter 16

• Spontaneity, Entropy and Free energy

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Contents

• Spontaneous Process and Entropy
• Entropy and the second law of thermodynamics
• The effect of temperature on spontaneity
• Free energy
• Entropy changes in chemical reactions The Third
  Law of Thermodynamics
• Free energy and chemical reactions
• The dependence of free energy on pressure
• Free energy and equilibrium
• Free energy and work

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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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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
• Entropy also describes the number of possible
  positions of a molecule
• 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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Example

• Which has higher positional S?
• Solid CO2 or gaseous CO2?
• N2 gas at 1 atm or N2 gas at 0.01 atm?
• Predict the sign of DS
• Solid sugar is added to water
• because larger volume
• Iodine vapor condenses on cold surface to form
  crystals
• - because g?s, less volume

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Positional Entropy

• Entropy also describes the number of
• possible positions of a molecule
• 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
• If the ?Ssys and ?Ssurr have different signs, the
  temperature determines the ?Suniv

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Determining ?Ssurr

• Sign of ?Ssurr
• depends on direction of heat flow
• ?Ssurr for exothermic reactions
• ?Ssurr - for endothermic reactions
• Magnitude of ?Ssurr
• Depends on temperature
• Heat flow ?H at constant P
• Very small at high T, increases as T decreases

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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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Example

• A process has a DH of 22 kJ and a DS of -13 J/K.
  At which temperatures is the process spontaneous?
• if there is no subscript, DS DSsys
• DSuniv gt 0 to be spontaneous
• DSsys DSsurr gt 0

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Example

• For methanol, the enthalpy of vaporization is
  71.8 kJ/mol and the entropy of vaporization is
  213 J/K. What is the normal boiling point of
  methanol?
• DSsys 213 J/K and DH 71.8 kJ/molK 71,800
  J/molK
• at the boiling point, the vaporization begins to
  be spontaneous
• DSuniv 0 to be at bp ? DSsys DSsurr 0

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16.4 Free Energy, G

• G Gibbs free energy
• G helps the determination of the T dependence
  on spontaneity
• G H - TS definition of G
• ?G ? H - T ? S for constant T
• All quantities refer to the system.
• When no subscript the quantity refers to the
  system

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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
• So what sign would the ?G of reaction with a
  ?Suniv have?
• Negative

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Example

• Is the following reaction spontaneous at -10C
  in
• the forward direction?
• H2O(s) ? H2O(l)
• where ?H6.03x103 J/mol and ?S22.1 J/mol.K

?Go ? Ho - T ? So
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• Another way to check for spontaneity of a
  reaction?
• Check to see if the ?Suniv is positive
• How can we solve for ?Ssurr?
• use ?H and T
• At -10C, is it spontaneous?

NO!- the reverse is spontaneous
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Example 2

• At what T is this reaction
• spontaneous at 1 atm
• Of pressure?
• Br2(l) ? Br2(g)
• ?Ho31.0 kJ/mol,
• ?So93.0 J/molK

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DGDH-TDS
-

At all Temperatures


Spontaneous at high temperatures
-
Spontaneous at low temperatures
-

-
Not spontaneous at any temperature, Reverse is
spontaneous
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16.5 Entropy changes in chemical reactionsThe
Third Law of Thermodynamics

• So far we dealt with physical changes
• 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?reaction ?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
• 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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Example

• Find the ?S at 25C for
• 2NiS(s) 3O2(g) ? 2SO2(g) 2NiO(s)
• 53 205 248 38

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16.6 Free Energy and Chemical Reactions

• DGº is needed when dealing with 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

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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.
• 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?There are three Ways to
Calculate ?G

• Use equation ?G ? H - T ? S
• Use Hesss Law
• Rearrange equations to get the given equation
• Add ?G values for the equations
• Use ?Gf standard free energy of formation

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• 1. Use the equation
• DGºDHº-TDSº
• This equation is applicable for reactions taking
  place at constant temperature

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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

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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

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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 temperature
  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).

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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 RT ln(Q)
• Thus, DGº -RT lnK

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DGº K

• DGº -RT lnK

• 0 1
• lt0 gt0
• gt0 lt0

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Temperature dependence of K

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

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Example
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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)