Transcript Chapter Seventeen - DePaul University

1
Chapter Seventeen
Thermodynamics: Spontaneity,
Entropy, and Free Energy
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Seventeen
2
Introduction
• Thermodynamics examines the relationship
between heat and work.
• Spontaneity is the notion of whether or not a
process can take place unassisted.
• Entropy is a mathematical concept describing the
distribution of energy within a system.
• Free energy is a thermodynamic function that
relates enthalpy and entropy to spontaneity, and
can also be related to equilibrium constants.
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Chapter Seventeen
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Why Study Thermodynamics?
• With a knowledge of thermodynamics and by making a few
calculations before embarking on a new venture, scientists
and engineers can save themselves a great deal of time,
money, and frustration.
– “To the manufacturing chemist thermodynamics gives information
concerning the stability of his substances, the yield which he may
hope to attain, the methods of avoiding undesirable substances, the
optimum range of temperature and pressure, the proper choice of
solvent.…” - from the introduction to Thermodynamics and the Free
Energy of Chemical Substances by G. N. Lewis and M. Randall
• Thermodynamics tells us what processes are possible.
– (Kinetics tells us whether the process is practical.)
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Spontaneous Change
• A spontaneous process is one that can occur in a system
left to itself; no action from outside the system is necessary
to bring it about.
• A nonspontaneous process is one that cannot take place in
a system left to itself.
• If a process is spontaneous, the reverse process is
nonspontaneous, and vice versa.
• Example: gasoline combines spontaneously with oxygen.
• However, “spontaneous” signifies nothing about how fast
a process occurs.
• A mixture of gasoline and oxygen may remain unreacted
for years, or may ignite instantly with a spark.
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Spontaneous Change (cont’d)
• Thermodynamics determines the equilibrium state of a
system.
• Thermodynamics is used to predict the proportions of
products and reactants at equilibrium.
• Kinetics determines the pathway by which equilibrium is
reached.
• A high activation energy can effectively block a reaction
that is thermodynamically favored.
• Example: combustion reactions are thermodynamically
favored, but (fortunately for life on Earth!) most such
reactions also have a high activation energy.
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Example 17.1
Indicate whether each of the following processes is
spontaneous or nonspontaneous. Comment on
cases where a clear determination cannot be made.
(a) The action of toilet bowl cleaner, HCl(aq), on “lime”
deposits, CaCO3(s).
(b) The boiling of water at normal atmospheric
pressure and 65 °C.
(c) The reaction of N2(g) and O2(g) to form NO(g) at
room temperature.
(d) The melting of an ice cube.
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Spontaneous Change (cont’d)
• Early chemists proposed that spontaneous chemical
reactions should occur in the direction of
decreasing energy.
• It is true that many exothermic processes are
spontaneous and that many endothermic reactions
are nonspontaneous.
• However, enthalpy change is not a sufficient
criterion for predicting spontaneous change …
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Spontaneous Change (cont’d)
Water falling (higher to
lower potential energy) is
a spontaneous process.
Conclusion: enthalpy alone is
not a sufficient criterion for
prediction of spontaneity.
H2 and O2 combine
spontaneously to form water
(exothermic) BUT …
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… liquid water vaporizes
spontaneously at room
temperature; an
endothermic process.
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Chapter Seventeen
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The Concept of Entropy
When the valve
is opened …
… the gases mix spontaneously.
• There is no significant enthalpy change.
• Intermolecular forces are negligible.
• So … why do the gases mix?
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The Concept of Entropy (cont’d)
• The other factor that drives reactions is a
thermodynamic quantity called entropy.
• Entropy is a mathematical concept that is difficult
to portray visually.
• The total energy of the system remains unchanged
in the mixing of the gases …
• … but the number of possibilities for the
distribution of that energy increases.
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Chapter Seventeen
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Formation of an Ideal Solution
Benzene and toluene have similar
intermolecular forces, so there is no
enthalpy change when they are mixed.
They mix completely because
entropy of the mixture is
higher than the entropies of
the two substances separated.
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Increase in Entropy in the
Vaporization of Water
Evaporation is
spontaneous because of
the increase in entropy.
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The Concept of Entropy
• The spreading of the energy among states, and increase of
entropy, often correspond to a greater physical disorder at
the microscopic level (however, entropy is not “disorder”).
• There are two driving forces behind spontaneous
processes: the tendency to achieve a lower energy state
(enthalpy change) and the tendency for energy to be
distributed among states (entropy).
• In many cases, however, the two factors work in
opposition. One may increase and the other decrease or
vice versa. In these cases, we must determine which factor
predominates.
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Assessing Entropy Change
• The difference in entropy (S) between two states is the
entropy change (DS).
• The greater the number of configurations of the
microscopic particles (atoms, ions, molecules)
among the energy levels in a particular state of a
system, the greater is the entropy of the system.
• Entropy generally increases when:
– Solids melt to form liquids.
– Solids or liquids vaporize to form gases.
– Solids or liquids dissolve in a solvent to form nonelectrolyte
solutions.
– A chemical reaction produces an increase in the number of
molecules of gases.
– A substance is heated.
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Chapter Seventeen
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Example 17.2
Predict whether each of the following leads to an
increase or decrease in the entropy of a system. If in
doubt, explain why.
(a) The synthesis of ammonia:
N2(g) + 3 H2(g)  2 NH3(g)
(b) Preparation of a sucrose solution:
C12H22O11(s) H2O(l)
C12H22O11(aq)
(c) Evaporation to dryness of a solution of urea,
CO(NH2)2, in water:
CO(NH2)2(aq)  CO(NH2)2(s)
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Entropy Change
• Sometimes it is necessary to obtain quantitative values of
entropy changes.
The expansion
DS = qrxn/T
can be reversed
• where qrxn is reversible heat, a state function.
by allowing the
sand to return,
one grain at a
time.
A reversible process can be
reversed by a very small
change, as in the expansion
of this gas. A reversible
process is never more than a
tiny step from equilibrium.
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Chapter Seventeen
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Entropy as a Function of Temperature
Entropy always
increases with
temperature …
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… and it increases
dramatically during
a phase change.
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Chapter Seventeen
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Standard Molar Entropies
• According to the Third Law of Thermodynamics, the
entropy of a pure, perfect crystal can be taken to be zero at
0 K.
• The standard molar entropy, S°, is the entropy of one
mole of a substance in its standard state.
• Since entropy increases with temperature, standard molar
entropies are positive—even for elements.
DS = Svp S°(products) – Svr S°(reactants)
Does the form of this equation
look familiar? (remember
calculating enthalpy change
from ΔHf°?)
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Chapter Seventeen
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Example 17.3
Use data from Appendix C to calculate the standard
entropy change at 25 °C for the Deacon process, a
high-temperature, catalyzed reaction used to convert
hydrogen chloride (a by-product from organic
chlorination reactions) into chlorine:
4 HCl(g) + O2(g)  2 Cl2(g) + 2 H2O(g)
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Chapter Seventeen
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The Second Law of Thermodynamics
• Entropy can be used as a sole criterion for spontaneous
change …
• … but the entropy change of both the system and its
surroundings must be considered.
• The Second Law of Thermodynamics establishes that all
spontaneous processes increase the entropy of the
universe (system and surroundings).
• If entropy increases in both the system and the
surroundings, the process is spontaneous.
– Is it possible for a spontaneous process to exhibit a decrease in
entropy? Yes, if the surroundings ____________________.
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Chapter Seventeen
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The Second Law of Thermodynamics
ΔStotal = ΔSuniverse = ΔSsystem + ΔSsurroundings
For a spontaneous process:
ΔSuniverse > 0
Since heat is exchanged
with the surroundings:
qsurr = –qp = ΔHsys
ΔSsurr
and:
Therefore:
ΔSuniv = ΔSsys
Multiply by –T:
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qsurr
–ΔHsys
= –––– = ––––––
T
T
ΔHsys
– ––––––
T
–TΔSuniv = ΔHsys – TΔSsys
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Chapter Seventeen
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Free Energy and Free Energy Change
• What is the significance of: –TΔSuniv = ΔHsys – TΔSsys ?
• The entropy change of the universe—our criterion for
spontaneity—has now been defined entirely in terms of the
system.
• The quantity –TΔSuniv is called the free energy change
(DG).
• For a process at constant temperature and pressure:
DGsys = DHsys – TDSsys
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Chapter Seventeen
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Free Energy and Free Energy Change
• If DG < 0 (negative), a process is spontaneous.
• If DG > 0 (positive), a process is nonspontaneous.
• If DG = 0, neither the forward nor the reverse process is
favored; there is no net change, and the process is at
equilibrium.
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Chapter Seventeen
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Case 3 illustrated
At high T, the size of
TΔS is large, and –TΔS
predominates.
ΔH is (+) and is moreor-less constant with T.
At low T, the size of
TΔS is small, and ΔH
(+) predominates.
Since ΔS is (+), the
slope TΔS is also (+).
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Chapter Seventeen
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Example 17.4
Predict which of the four cases in Table 17.1 you expect
to apply to the following reactions.
(a) C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
ΔH = –2540 kJ
(b) Cl2(g)  2 Cl(g)
Example 17.5
A Conceptual Example
Molecules exist from 0 K to a few thousand kelvins. At
elevated temperatures, they dissociate into atoms. Use
the relationship between enthalpy and entropy to
explain why this is to be expected.
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Chapter Seventeen
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Standard Free Energy Change, ΔG°
• The standard free energy change, DG°, of a reaction is the
free energy change when reactants and products are in their
standard states.
• The standard free energy of formation, DGf°, is the free
energy change for the formation of 1 mol of a substance in
its standard state from the elements in their standard states.
DG° = Svp DGf°(products) – Svr DGf°(reactants)
The form of this
equation should appear
very familiar by now!
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Chapter Seventeen
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Example 17.6
Calculate ΔG° at 298 K for the reaction
4 HCl(g) + O2(g)  2 Cl2(g) + 2 H2O(g)
ΔH° = –114.4 kJ
(a) using the Gibbs equation (17.8) and (b) from
standard free energies of formation.
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Free Energy Change and Equilibrium
• At equilibrium, DG = 0 (reaction is neither spontaneous
nor nonspontaneous).
• Therefore, at the equilibrium temperature, the free energy
change expression becomes:
DH = TDS
and
DS = DH/T
• Trouton’s rule states that the entropy change is about the
same when one mole of a substance is converted from
liquid to vapor (at the normal boiling point).
•
DS°vapn for many substances is about 87 J mol–1 K–1.
• This rule works best with nonpolar substances.
• It generally fails for liquids with a more ordered structure,
such as those with extensive hydrogen bonding.
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Illustrating Trouton’s Rule
The three substances have different
entropies and different boiling
points, but DS of vaporization is
about the same for all three.
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Example 17.7
At its normal boiling point, the enthalpy of vaporization
of pentadecane, CH3(CH2)13CH3, is 49.45 kJ/mol. What
should its approximate normal boiling point temperature
be?
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Chapter Seventeen
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Raoult’s Law Revisited
… higher entropy for
the vapor from the
solution than from the
pure solvent.
Entropy of a vapor increases
if the vapor expands into a
larger volume—lower vapor
pressure.
Entropy of vaporization
of the solvent is about
the same in each case,
which means …
A pure solvent has a
lower entropy than a
solution containing
the solvent.
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Chapter Seventeen
Relationship of DG° to the Equilibrium
Constant, Keq
•
DG = 0 is a criterion for equilibrium at any temperature.
•
DG° = 0 is a criterion for equilibrium at a single
temperature, that temperature at which the equilibrium
state has all reactants and products in their standard states.
32
• ΔG and DGo are related through the reaction quotient, Q:
DG = DG° + RT ln Q
• When DG = 0, then Q = Keq, and the equation becomes:
DG° = -RT ln Keq
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The Equilibrium Constant, Keq
• The concentrations and partial pressures we have used in
Keq are approximations.
• Activities (a) are the correct variables for Keq. But
activities are very difficult to determine. That is why we
use approximations.
• For pure solid and liquid phases: a = 1.
• For gases: Assume ideal gas behavior, and replace the
activity by the numerical value of the gas partial pressure
(in atm).
• For solutes in aqueous solution: Replace solute activity
by the numerical value of the solute molarity (M).
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Chapter Seventeen
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Example 17.8
Write the equilibrium constant expression, Keq, for the
oxidation of chloride ion by manganese dioxide in an acidic
solution:
MnO2(s) + 4 H+(aq) + 2 Cl–(aq) 
Mn2+(aq) + Cl2(g) + 2 H2O(l)
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Calculating Equilibrium Constants
• Rearranging Equation (17.12):
DG° = -RT ln Keq
DG°
ln Keq = - –––––
RT
• The units of DG° and R must be consistent; both must
use kJ or both must use J.
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Chapter Seventeen
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Example 17.9
Determine the value of Keq at 25 °C for the reaction
2 NO2(g)
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N2O4(g)
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Chapter Seventeen
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The Sign and Magnitude of DG°
Large, negative
DG°; equilibrium
lies far to right.
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Large, positive
DG°; equilibrium
lies far to left.
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Intermediate DG°;
equilibrium lies in
intermediate position.
Chapter Seventeen
38
Coupled Reactions
• A nonspontaneous reaction may be coupled with a
spontaneous reaction.
• The decomposition of copper(I) oxide is quite
nonspontaneous at room temperature:
Cu2O(s)
2 Cu(s) + ½ O2(g)
ΔG°298 = +149.9 kJ
• By coupling this decomposition with the formation of CO
from carbon,
C(graphite) + ½ O2(g)
CO(g)
ΔG°298 = –137.2 kJ
we can reduce the nonspontaneity of Cu2O and make the
overall reaction occur slightly above room temperature:
Cu2O(s) + C(graphite)
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2 Cu(s) + CO(g) ΔG°298 = +12.7 kJ
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Chapter Seventeen
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The Dependence of DG° and Keq on
Temperature
• To obtain equilibrium constants at different temperatures, it
will be assumed that DH° does not change much with
temperature.
• To obtain Keq at the desired temperature, the van’t Hoff
equation is used:
-DH°
ln Keq = ––––– + constant or
RT
K2
DH°
ln ––– = ––––
K1
R
[ ]
1
1
––– – –––
T1
T2
• The form used depends on whether we have a single value
of Keq available, or multiple values.
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Chapter Seventeen
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Example 17.10
Consider this reaction at 298 K:
CO(g) + H2O(g)
CO2(g) + H2(g)
ΔH°298 = – 41.2 kJ
Determine Keq for the reaction at 725 K.
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Chapter Seventeen
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Cumulative Example
Waste silver from photographic solutions or laboratory
operations can be recovered using an appropriate redox
reaction. A 100.0-mL sample of silver waste is 0.200 M in
Ag+(aq) and 0.0200 M in Fe3+(aq). Enough iron(II) sulfate
is added to make the solution 0.200 M in Fe2+(aq). When
equilibrium is established at 25 °C, how many moles of
solid silver will be present?
Ag+(aq) + Fe2+(aq)  Ag(s) + Fe3+(aq)
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Seventeen