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Thermodynamics: Spontaneity, Entropy and Free Energy

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Thermodynamics: Spontaneity, Entropy and Free Energy Go and Equilibrium G = Go + RT lnQ At equilibrium, G is equal to zero, and Q = K. 0 = Go + RT lnK ... – PowerPoint PPT presentation

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Title: Thermodynamics: Spontaneity, Entropy and Free Energy


1
ThermodynamicsSpontaneity, Entropy and Free
Energy
2
Thermodynamics
  • Thermodynamics studies how changes in energy,
    entropy and temperature affect the spontaneity of
    a process or chemical reaction.
  • Using thermodynamics we can predict the
    direction a reaction will go, and also the
    driving force of a reaction or system to go to
    equilibrium.

3
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

4
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) .
  • Spontaneity is independent of the speed or rate
    of a reaction. A spontaneous process may proceed
    very slowly.

5
Spontaneity
  • There are three factors that combine to predict
    spontaneity. They are
  • 1. Energy Change
  • 2. Temperature
  • 3. Entropy Change

6
Entropy
  • A measure of randomness or disorder

7
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.

8
Entropy
  • The entropy of a system is defined by the
    Boltzmann equation
  • S k ln W
  • k is the Boltzmann constant, and W is the
    number of energetically equivalent ways to
    arrange the components of the system.

9
Entropy
  • Gases will spontaneously and uniformly mix
    because the mixed state has more possible
    arrangements (a larger value of W and higher
    entropy) than the unmixed state.

10
Entropy
  • The driving force for a spontaneous process is an
    increase in the entropy of the universe.

11
?So and Phase Changes

Gases have more entropy than liquids or solids.
12
?So and Mixtures

Mixtures have more entropy than pure substances.
13
Entropy Values of Common Substances
14
The 2nd Law of Thermodynamics
  • In any spontaneous process there is always
  • an increase in the entropy of the universe.

15
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.

16
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.

17
Entropy
  • Entropy can be viewed as the dispersal or
    randomization of energy. The freezing of water
    (an exothermic process) releases heat to the
    surroundings, and thus increases the entropy of
    the surroundings. The process is spontaneous at
    or below 0oC because the increase in entropy of
    the surroundings is greater than the decrease in
    entropy of the water as it freezes.

18
? S and Spontaneity
19
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

20
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.

21
Spontaneity
?G ?H - T?S
22
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.

23
?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.

24
The Significance of ?G
  • ?G represents the driving force for the
    reaction to proceed to equilibrium.

25
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.
  • In practice, some energy is always lost, so the
    actual work produced will be less than the
    calculated value.

26
The Significance of ?G
  • 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.
  • In practice, some additional work is required
    to make the non-spontaneous process or reaction
    proceed.

27
Reversibility
  • A reversible reaction is a reaction that
    achieves the theoretical limit with respect to
    free energy. That is, there is no loss of energy
    (usually as heat) to the surroundings.
  • All real reactions are irreversible, and do
    not achieve he theoretical limit of available
    free energy.

28
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.

29
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.

30
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.

31
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.

32
Entropy Values of Common Substances
33
Entropy Values
  • For comparable structures, the entropy increases
    with increasing mass

34
Entropy Values
  • For molecules with similar masses, the more
    complex molecule has greater entropy. The
    molecule with more bonds has additional ways to
    absorb energy, and thus greater entropy.

35
Calculating Entropy Changes
  • For any chemical reaction,
  • ? Soreaction Smolprod Soproducts- Smolreact
    Soreactants
  • The units of entropy are joules/K-mol.

36
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.

37
Calculation of ?Go
  • ?Go ?Ho - T ?So
  • ?Ho is usually calculated by using standard
    enthalpies of formation, ?Hfo.
  • ?Horxn Snprod ?Hoproducts- Snreact
    ?Horeactants

38
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.

39
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.

40
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.
  • ?Gorxn Smolprod ?Gfo prod - Smolreact ?Gfo
    react

41
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).

42
Calculation of ?Go
43
Calculation of ?Go
Note the values of zero for nitrogen, hydrogen
and graphite.
44
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?

45
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.

46
Calculation of ?Go
  • CaCO3(s) ?CaO(s)    CO2(g)
  • ?Grxno Snprod ?Gfo prod - Snreact ?Gfo react

47
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)

48
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

49
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.

50
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?

51
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.

52
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.

53
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.

54
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.

55
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
56
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

57
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.

58
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

59
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.

60
?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.

61
?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

62
?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.

63
?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.

64
?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.

65
?Go and Equilibrium
66
?Go and Equilibrium
  • ?G ?Go RT lnQ
  • At equilibrium, ?G is equal to zero, and
  • Q K.
  • 0 ?Go RT lnK
  • ?Go - RT lnK

67
?Go and Equilibrium
  • Calculate, ?Go and K at 25oC for
  • C (s, diamond) ? C (s, graphite)

68
?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.

69
?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

70
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.
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