# Thermodynamics: Spontaneity, Entropy and Free Energy - PowerPoint PPT Presentation

Title:

## Thermodynamics: Spontaneity, Entropy and Free Energy

Description:

### Thermodynamics: Spontaneity, Entropy and Free Energy The Significance of G If negative, the value of G in KJ is the maximum possible useful work that can be ... – PowerPoint PPT presentation

Number of Views:204
Avg rating:3.0/5.0
Slides: 62
Provided by: Nancy390
Category:
Tags:
Transcript and Presenter's Notes

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
• All substances have zero disorder, and an
entropy of zero, at 0 K. As a result,
calorimetry can be used to measure absolute
values of entropy, rather than changes in
entropy.
• Entropy values are tabulated, and can be looked
up in thermodynamic tables of data.

14
Entropy Values of Common Substances
15
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.

16
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.

17
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.

18
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.

19
Entropy Values of Common Substances
20
Entropy Values
• For comparable structures, the entropy increases
with increasing mass

21
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.

22
Calculating Entropy Changes
• For any chemical reaction,
• ? Soreaction Smolprod Soproducts- Smolreact
Soreactants
• The units of entropy are joules/K-mol.

23
The 2nd Law of Thermodynamics
• In any spontaneous process there is always
• an increase in the entropy of the universe.

24
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.

25
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.

26
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.

27
? S and Spontaneity
28
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

29
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.

30
Spontaneity
?G ?H - T?S
31
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.

32
?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.

33
The Significance of ?G
• ?G represents the driving force for the
reaction to proceed to equilibrium.

34
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.

35
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.

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
Coupled Biological Processes
• Many biochemical reactions have positive values
of ?Go, and would not occur spontaneously. These
processes, such as the synthesis of ATP
(adenosine triphosphate), are driven by favorable
reactions such as the conversion of glucose to
carbon dioxide and water.
• ATP is essentially a way for the body to store
free energy until it is needed for cellular
processes.

61
Coupled Biological Processes