Thermodynamics: Spontaneity, Entropy and Free Energy

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ThermodynamicsSpontaneity, Entropy and Free
Energy
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Spontaneity

• A spontaneous process is one that occurs
  without outside intervention. Examples include
• - a ball rolling downhill
• - ice melting at temperatures above 0oC
• - gases expanding to fill their container
• - iron rusts in the presence of air and water
• - two gases mixing

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Spontaneity

• Spontaneous processes can release energy (a
  ball rolling downhill), require energy (ice
  melting at temperatures above 0oC), or involve no
  energy change at all (two gases mixing) .

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Spontaneity

• There are three factors that combine to predict
  spontaneity. They are
• 1. Energy Change
• 2. Temperature
• 3. Entropy Change

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Entropy

• A measure of randomness or disorder

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Entropy

• Entropy, S, is a measure of randomness or
  disorder. The natural tendency of things is to
  tend toward greater disorder. This is because
  there are many ways (or positions) that lead to
  disorder, but very few that lead to an ordered
  state.

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Entropy

• The driving force for a spontaneous process is an
  increase in the entropy of the universe.

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Entropy
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?So and Phase Changes

Gases have more entropy than liquids or solids.
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?So and Mixtures

Mixtures have more entropy than pure substances.
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Entropy Values of Common Substances
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The 2nd Law of Thermodynamics

• In any spontaneous process there is always
• an increase in the entropy of the universe.

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The 2nd Law of Thermodynamics

• Water spontaneously freezes at a temperature
  below 0oC. Therefore, the process increases the
  entropy of the universe.
• The water molecules become much more ordered as
  they freeze, and experience a decrease in
  entropy. The process also releases heat, and
  this heat warms gaseous molecules in air, and
  increases the entropy of the surroundings.

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The 2nd Law of Thermodynamics

• Since the process is spontaneous below 0oC,
  ?Ssurr, which is positive, must be greater in
  magnitude than ?S of the water molecules.

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? S and Spontaneity
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Spontaneity

• Entropy, temperature and heat flow all play a
  role in spontaneity. A thermodynamic quantity,
  the Gibbs Free Energy (G), combines these factors
  to predict the spontaneity of a process.
• ?G ?H - T?S

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Spontaneity

• ?G ?H - T?S
• If a process releases heat (?H is negative) and
  has an increase in entropy (?S is positive), it
  will always be spontaneous.
• The value of ?G for spontaneous processes is
  negative.

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Spontaneity
?G ?H - T?S
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Spontaneity and ?G

• If ?G is negative, the process is spontaneous
  (and the reverse process is non-spontaneous).
• If ?G is positive, the process is
  non-spontaneous, and the reverse process is
  spontaneous.
• If ?G 0, the system is at equilibrium.

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?G

• Although ?G can be used to predict in which
  direction a reaction will proceed, it does not
  predict the rate of the reaction.
• For example, the conversion of diamond to
  graphite has a ?Go -3 kJ, so diamonds should
  spontaneously change to graphite at standard
  conditions. However, kinetics shows that the
  reaction is extremely slow.

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The Significance of ?G

• ?G represents the driving force for the
  reaction to proceed to equilibrium.

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The Significance of ?G

• If negative, the value of ?G in KJ is the
  maximum possible useful work that can be obtained
  from a process or reaction at constant
  temperature and pressure.
• If positive, the value of ?G in KJ is the
  minimum work that must be done to make the
  non-spontaneous process or reaction proceed.

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Predicting the sign of ?So

• For many chemical reactions or physical
  changes, it is relatively easy to predict if the
  entropy of the system is increasing or
  decreasing.
• If a substance goes from a more ordered phase
  (solid) to a less ordered phase (liquid or gas),
  its entropy increases.

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Predicting the sign of ?So

• For chemical reactions, it is sometimes
  possible to compare the randomness of products
  versus reactants.
• 2 KClO3(s) ? 2 KCl(s) 3 O2(g)
• The production of a gaseous product from a
  solid reactant will have a positive value of ?So.

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Calculating Entropy Changes

• Since entropy is a measure of randomness, it is
  possible to calculate absolute entropy values.
  This is in contrast to enthalpy values, where we
  can only calculate changes in enthalpy.
• A perfect crystal at absolute zero has an
  entropy value (S) 0. All other substances have
  positive values of entropy due to some degree of
  disorder.

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Calculating Entropy Changes

• Fortunately, the entropy values of most common
  elements and compounds have been tabulated. Most
  thermodynamic tables, including the appendix in
  the textbook, include standard entropy values,
  So.

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Entropy Values of Common Substances
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Calculating Entropy Changes

• For any chemical reaction,
• ? Soreaction Smolprod Soproducts- Smolreact
  Soreactants
• The units of entropy are joules/K-mol.

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Calculation of ?Go

• ?Go, the standard free energy change, can be
  calculated in several ways.
• ?Go ?Ho - T ?So
• It can be calculated directly, using the
  standard enthalpy change and entropy change for
  the process.

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Calculation of ?Go

• ?Go ?Ho - T ?So
• ?Ho is usually calculated by using standard
  enthalpies of formation, ?Hfo.
• ?Horxn Smolprod ?Hoproducts- Smolreact
  ?Horeactants

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Calculation of ?Go

• ?Go ?Ho - T ?So
• Once ?Ho and ?So have been calculated, the
  value of ?Go can be calculated, using the
  temperature in Kelvins.

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Calculation of ?Go

• ?Go can also be calculated by combining the
  free energy changes of related reactions. This
  is the same method used in Hess Law to calculate
  enthalpy changes. If the sum of the reactions
  gives the reaction of interest, the sum of the
  ?Go values gives ?Go for the reaction.

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Calculation of ?Go

• Lastly, ?Go can be calculated using standard
  free energies of formation, ?Gfo. Some tables of
  thermodynamic data, including the appendix of
  your textbook, include values of ?Gfo.
• ?Grxno Smolprod ?Gfo prod - Smolreact ?Gfo
  react

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Calculation of ?Go

• When calculating ?Go from standard free
  energies of formation, keep in mind that ?Gfo for
  any element in its standard state is zero. As
  with enthalpies of formation, the formation
  reaction is the reaction of elements in their
  standard states to make compounds (or
  allotropes).

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Calculation of ?Go
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Calculation of ?Go
Note the values of zero for nitrogen, hydrogen
and graphite.
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Spontaneity Problem

• Consider the reaction 
• CaCO3(s) ?CaO(s)    CO2(g)  at 25oC. 
• Calculate ?Go using the tables in the appendix
  of your textbook.  Is the process spontaneous at
  this temperature?  Is it spontaneous at all
  temperatures?  If not, at what temperature does
  it become spontaneous?

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Spontaneity Problem

• Consider the reaction 
• CaCO3(s) ?CaO(s)    CO2(g)  at 25oC. 
• Calculate ?Go using the tables in the appendix
  of your textbook.  Is the process spontaneous at
  this temperature? 
• Calculation of ?Grxno will indicate spontaneity
  at 25oC. It can be calculated using
• ?Gfo values or from ?Hfo and ?So values.

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Calculation of ?Go

• CaCO3(s) ?CaO(s)    CO2(g)
• ?Grxno Snprod ?Gfo prod - Snreact ?Gfo react

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Calculation of ?Go

• CaCO3(s) ?CaO(s)    CO2(g)
• ?Grxno (1 mol) (-604.0 kJ/mol) (1
  mol)(-394.4 kJ/mol) 1 mol(-1128.8 kJ/mol)

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Calculation of ?Go

• CaCO3(s) ?CaO(s)    CO2(g)
• ?Grxno (1 mol) (-604.0 kJ/mol) (1
  mol)(-394.4 kJ/mol) 1 mol(-1128.8 kJ/mol)
  130.4 kJ

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Spontaneity Problem

• Consider the reaction 
• CaCO3(s) ?CaO(s)    CO2(g)  at 25oC. 
• Calculate ?Go using the tables in the appendix
  of your textbook.  Is the process spontaneous at
  this temperature? 
• Since ?Grxno 130.4 kJ, the reaction is not
  spontaneous at 25oC.

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Spontaneity Problem

• Consider the reaction 
• CaCO3(s) ?CaO(s)    CO2(g)  at 25oC. 
•   Is it spontaneous at all temperatures?  If
  not, at what temperature does it become
  spontaneous?

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Spontaneity Problem

• Consider the reaction 
• CaCO3(s) ?CaO(s)    CO2(g)  at 25oC. 
•   Is it spontaneous at all temperatures?  If
  not, at what temperature does it become
  spontaneous?
• At 25oC, ?Grxno is positive, and the reaction
  is not spontaneous in the forward direction.

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Spontaneity Problem

• Consider the reaction 
• CaCO3(s) ?CaO(s)    CO2(g)  at 25oC. 
•   Is it spontaneous at all temperatures?  If
  not, at what temperature does it become
  spontaneous?
• Inspection of the reaction shows that it
  involves an increase in entropy due to production
  of a gas from a solid.

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Spontaneity Problem

• Consider the reaction 
• CaCO3(s) ?CaO(s)    CO2(g)  at 25oC. 
•   Is it spontaneous at all temperatures?  If
  not, at what temperature does it become
  spontaneous?
• We can calculate the entropy change and the
  enthalpy change, and then determine the
  temperature at which spontaneity will occur.

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CaCO3(s) ?CaO(s)    CO2(g)

• Since ?Go ?Ho - T?So, and there is an
  increase in entropy, the reaction will become
  spontaneous at higher temperatures.
• To calculate ?So, use the thermodynamic tables
  in the appendix.

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CaCO3(s) ?CaO(s)    CO2(g)
?Srxno 1mol(213.6J/K-mol)1mol(39.7J/K-mol) -
1mol(92.9J/K-mol) 160.4 J/K
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CaCO3(s) ?CaO(s)    CO2(g)

• ?Go ?Ho - T?So
• Since we know the value of ?Go (130.4 kJ) and
  ?So (160.4 J/K), we can calculate the value of
  ?Ho at 25oC.
• 130.4 kJ ?Ho (298K) (160.4 J/K)
• ?Ho 130.4 kJ (298K) (.1604 kJ/K)
• ?Ho 178.2 kJ

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CaCO3(s) ?CaO(s)    CO2(g)

• ?Go ?Ho - T?So
• If we assume that the values of ?Ho and ?So
  dont change much with temperature, we can
  estimate the temperature at which the reaction
  will become spontaneous.

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CaCO3(s) ?CaO(s)    CO2(g)

• ?Go ?Ho - T?So
• ?Go is positive at lower temperatures, and will
  be negative at higher temperatures. Set ?Go
  equal to zero, and solve for temperature.
• 0 ?Ho - T?So
• T ?Ho
• ?So

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CaCO3(s) ?CaO(s)    CO2(g)

• ?Go ?Ho - T?So
• 0 ?Ho - T?So
• T ?Ho
• ?So
• T (178.2 kJ)/(160.4 J/K)(10-3kJ/J)
• 1111K or 838oC
• The reaction will be spontaneous in the forward
  direction at temperatures above 838oC.

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?G for Non-Standard Conditions

• The thermodynamic tables are for standard
  conditions. This includes having all reactants
  and products present initially at a temperature
  of 25oC. All gases are at a pressure of 1 atm,
  and all solutions are 1 M.

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?G for Non-Standard Conditions

• For non-standard temperature, concentrations or
  gas pressures
• ?G ?Go RTlnQ
• Where R 8.314 J/K-mol
• T is temperature in Kelvins
• Q is the reaction quotient

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?G for Non-Standard Conditions

• For non-standard temperature, concentrations or
  gas pressures
• ?G ?Go RTlnQ
• For Q, gas pressures are in atmospheres, and
  concentrations of solutions are in molarity, M.

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?Go and Equilibrium

• A large negative value of ?Go indicates that
  the forward reaction or process is spontaneous.
  That is, there is a large driving force for the
  forward reaction. This also means that the
  equilibrium constant for the reaction will be
  large.

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?Go and Equilibrium

• A large positive value of ?Go indicates that
  the reverse reaction or process is spontaneous.
  That is, there is a large driving force for the
  reverse reaction. This also means that the
  equilibrium constant for the reaction will be
  small.
• When a reaction or process is at equilibrium,
  ?Go zero.

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?Go and Equilibrium
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?Go and Equilibrium

• ?G ?Go RT lnQ
• At equilibrium, ?G is equal to zero, and
• Q K.
• 0 ?Go RT lnK
• ?Go - RT lnK

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?Go and Equilibrium

• Calculate, ?Go and K at 25oC for
• C (s, diamond) ? C (s, graphite)

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?Go and Equilibrium

• Calculate, ?Go and K at 25oC for
• C (s, diamond) ? C (s, graphite)
• ?Go (1 mol) ?Gof (graphite) - (1 mol) ?Gof
  (diamond)
• 0 -(1 mol)(2.900 kJ/mol)
• -2.900 kJ
• The reaction is spontaneous at 25oC.

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?Go and Equilibrium

• Calculate, ?Go and K at 25oC for
• C (s, diamond) ? C (s, graphite)
• ?Go -2.900 kJ
• ?Go -2.900 kJ -RT ln K
• -2.900 kJ -(8.314J/mol-K) (298.2K)ln K
• ln K 1.170
• K e1.170 3.22

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C (s, diamond) ? C (s, graphite)

• The negative value of ?Go and the equilibrium
  constant gt1 suggest that diamonds can
  spontaneously react to form graphite. Although
  the reaction is thermodynamically favored, the
  rate constant is extremely small due to a huge
  activation energy. The disruption of the bonding
  in the diamond to form planar sp2 hybridized
  carbon atoms is kinetically unfavorable.