Download Chapter 19 Chemical Thermodynamics

Lecture Presentation
Chapter 19
Chemical Thermodynamics
Dr. Subhash C. Goel
South GA State College
Douglas, GA
© 2012 Pearson Education, Inc.
Review
Units of Energy
• The SI unit of energy is the joule (J):
kg m2
1 J = 1 
s2
• An older, non-SI unit is still in
widespread use: the calorie (cal):
1 cal = 4.184 J
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Heat
• Energy can also
be transferred as
heat.
• Heat flows from
warmer objects to
cooler objects.
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First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the
universe is a constant; if the system loses
energy, it must be gained by the
surroundings, and vice versa.
• Energy can, however, be converted from
one form to another or transferred from a
system to the surroundings or vice versa.
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Internal Energy
The internal energy of a system is the
sum of all kinetic and potential energies of
all components of the system; we call it E.
By definition, the change in internal
energy, E, is the final energy of the
system minus the initial energy of the
system:
E = Efinal − Einitial
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E, q, w, and Their Signs
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Work
We can measure the work done by the gas if
the reaction is done in a vessel that has been
fitted with a piston:
w = −PV
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Enthalpy
• If a process takes place at constant
pressure (as the majority of processes we
study do) and the only work done is this
pressure–volume work, we can account
for heat flow during the process by
measuring the enthalpy of the system.
• Enthalpy is the internal energy plus the
product of pressure and volume:
H = E + PV
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Enthalpy
• When the system changes at constant
pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
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Enthalpy
• Since E = q + w and w = −PV, we
can substitute these into the enthalpy
expression:
H = E + PV
H = (q + w) − w
H = q
• So, at constant pressure, the change in
enthalpy is the heat gained or lost.
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Enthalpy of Reaction
The change in
enthalpy, H, is the
enthalpy of the
products minus the
enthalpy of the
reactants:
H = Hproducts − Hreactants
H = enthaly of
reaction or heat of
reaction
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Hess’s Law
Hess’s law states that
“[i]f a reaction is
carried out in a series
of steps, H for the
overall reaction will be
equal to the sum of
the enthalpy changes
for the individual
steps.”
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Enthalpies of Formation
An enthalpy of formation, Hf, is defined
as the enthalpy change for the reaction
in which a compound is made from its
constituent elements in their elemental
forms.
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Calculation of H
We can use Hess’s law in this way:
H = nHf,products – mHf°,reactants
where n and m are the stoichiometric
coefficients.
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Spontaneous Processes
• Spontaneous processes
are those that can
proceed without any
outside intervention.
• The gas in vessel B will
spontaneously effuse into
vessel A, but once the gas
is in both vessels, it will
not spontaneously return
to vessel B.
Chemical
Thermodynamics
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Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse direction.
Chemical
Thermodynamics
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Spontaneous Processes
• Processes that are spontaneous at one temperature may
be nonspontaneous at other temperatures.
• Above 0 C, it is spontaneous for ice to melt.
• Below 0 C, the reverse process is spontaneous.
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Exercise 1 Identifying Spontaneous Processes
Predict whether each process is spontaneous as described,
spontaneous in the reverse direction, or in equilibrium:
(a) Water at 40 °C gets hotter when a piece of metal heated to 150
°C is added.
(b) Water at room temperature decomposes into H2(g) and O2(g).
(c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to
liquid benzene at the normal boiling point of benzene, 80.1
°C.
Chemical
Thermodynamics
Reversible Processes
In a reversible process the system changes
in such a way that the system and
surroundings can be put back in their original
states by exactly reversing the process.
Chemical
Thermodynamics
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Irreversible Processes
• Irreversible processes cannot be undone by
exactly reversing the change to the system.
• Spontaneous processes are irreversible.
Chemical
Thermodynamics
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Entropy
• Entropy (S) is a term coined by Rudolph
Clausius in the nineteenth century.
• Clausius was convinced of the
significance of the ratio of heat
delivered and the temperature at which
it is delivered, q .
T
Chemical
Thermodynamics
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Entropy
• Entropy can be described as a
measure of how spread out or
dispersed the energy is among the
different possible ways that system can
contain energy.
• The greater is the dispersal, greater is
the entropy.
• It is related to the various modes of
motion in molecules.
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Entropy
• Like total energy, E, and enthalpy, H,
entropy is a state function.
• Therefore,
S = Sfinal  Sinitial
Chemical
Thermodynamics
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Entropy
For a process occurring at constant
temperature (an isothermal process), the
change in entropy is equal to the heat that
would be transferred if the process were
reversible divided by the temperature:
qrev
S =
T
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Exercise 2 Calculating ΔS for a Phase Change
Elemental mercury is a silver liquid at room temperature. Its
normal freezing point is –38.9 °C, and its molar enthalpy of
fusion is ΔHfusion = 2.29 kJ/mol. What is the entropy change of the
system when 50.0 g of Hg(l) freezes at the normal freezing point?
Chemical
Thermodynamics
Second Law of Thermodynamics
The second law of thermodynamics
states that the entropy of the universe
increases for spontaneous processes,
and the entropy of the universe does
not change for reversible processes.
Chemical
Thermodynamics
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Second Law of Thermodynamics
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
Chemical
Thermodynamics
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Second Law of Thermodynamics
These last truths mean that as a result
of all spontaneous processes, the
entropy of the universe increases.
Chemical
Thermodynamics
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Entropy on the Molecular Scale
• Ludwig Boltzmann described the concept
of entropy on the molecular level.
• Temperature is a measure of the average
kinetic energy of the molecules in a
sample.
Chemical
Thermodynamics
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Entropy on the Molecular Scale
• Molecules exhibit several types of motion:
– Translational: Movement of the entire molecule from
one place to another.
– Vibrational: Periodic motion of atoms within a molecule.
– Rotational: Rotation of the molecule about an axis or
rotation about  bonds.
Chemical
Thermodynamics
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Entropy on the Molecular Scale
• Boltzmann envisioned the motions of a sample of
molecules at a particular instant in time.
– This would be akin to taking a snapshot of all the
molecules.
• He referred to this sampling as a microstate of the
thermodynamic system.
Chemical
Thermodynamics
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Entropy on the Molecular Scale
• Each thermodynamic state has a specific number
of microstates, W, associated with it.
• Entropy is
S = k ln W
where k is the Boltzmann constant, 1.38  1023 J/K.
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Entropy on the Molecular Scale
• The change in entropy for a process,
then, is
S = k ln Wfinal  k ln Winitial
Wfinal
S = k ln
Winitial
• Entropy increases with the number of
microstates in the system.
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Entropy on the Molecular Scale
• The number of microstates and,
therefore, the entropy, tends to
increase with increases in
– Temperature
– Volume
– The number of independently moving
molecules.
Chemical
Thermodynamics
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Entropy and Physical States
• Entropy increases with the freedom of motion of
molecules.
• Therefore,
S(g) > S(l) > S(s)
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Solutions
Generally, when a
solid is dissolved
in a solvent,
entropy
increases.
Chemical
Thermodynamics
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Exercise 3 Predicting the Sign of ΔS
Predict whether is positive or negative for each process, assuming
each occurs at constant ΔS temperature:
(a) H2O(l)  H2O(g)
(b) Ag+(aq) + Cl–(aq)  AgCl(s)
(c) 4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
(d) N2(g) + O2(g)  2 NO(g)
Chemical
Thermodynamics
Exercise 4 Predicting Relative Entropies
In each pair, choose the system that has greater entropy and explain
your choice:
(a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C,
(b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C,
(c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.
Chemical
Thermodynamics
Entropy Changes
• In general, entropy increases when
– Gases are formed from liquids and solids;
– Liquids or solutions are formed from solids;
– The number of gas molecules increases;
– The number of moles increases.
Chemical
Thermodynamics
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Third Law of Thermodynamics
The entropy of a
pure crystalline
substance at
absolute zero is 0.
Chemical
Thermodynamics
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Standard Entropies
• These are molar entropy
values of substances in
their standard states.
• Standard entropies tend
to increase with
increasing molar mass.
Chemical
Thermodynamics
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Standard Entropies
Larger and more complex molecules have
greater entropies.
Chemical
Thermodynamics
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Entropy Changes
Entropy changes for a reaction can be
estimated in a manner analogous to that by
which H is estimated:
S = nS(products) — mS(reactants)
where n and m are the coefficients in the
balanced chemical equation.
Chemical
Thermodynamics
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Exercise 5 Calculating ΔS° from Tabulated Entropies
Calculate the change in the standard entropy of the system,
ΔS° , for the synthesis of ammonia from N2(g)
and H2(g) at 298 K:
N2(g) + 3 H2(g)  2 NH3(g)
Chemical
Thermodynamics
Entropy Changes in Surroundings
• Heat that flows into or out of the
system changes the entropy of the
surroundings.
• For an isothermal process:
qsys
Ssurr =
T
• At constant pressure, qsys is simply
H for the system.
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Chemical
Thermodynamics
Entropy Change in the Universe
• The universe is composed of the
system and the surroundings.
• Therefore,
Suniverse = Ssystem + Ssurroundings
• For spontaneous processes
Suniverse > 0
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Entropy Change in the Universe
• Since
Ssurroundings =
qsystem
T
and
qsystem = Hsystem
This becomes:
Suniverse = Ssystem +
Hsystem
T
Multiplying both sides by T, we get
TSuniverse = Hsystem  TSsystem
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Chemical
Thermodynamics
Gibbs Free Energy
• TSuniverse is defined as
the Gibbs free energy,
G.
• When Suniverse is
positive, G is negative.
• Therefore, when G is
negative, a process is
spontaneous.
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Gibbs Free Energy
1. If G is negative, the
forward reaction is
spontaneous.
2. If G is 0, the system
is at equilibrium.
3. If G is positive, the
reaction is
spontaneous in the
reverse direction.
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Exercise 6 Calculating Free–Energy Change from ΔH°, T, and
ΔS°
Calculate the standard free–energy change for the formation of
NO(g) from N2(g) and O2(g) at 298 K:
N2(g) + O2(g)  2 NO(g)
given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction
spontaneous under these conditions?
Chemical
Thermodynamics
Standard Free Energy Changes
Analogous to standard enthalpies of
formation are standard free energies of
formation, G:
f
G = nGf (products)  mGf (reactants)
where n and m are the stoichiometric
coefficients.
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Exercise 7
Calculating Standard Free–Energy Change
from Free Energies of Formation
(a) Use data from Appendix C to calculate the standard free–
energy change for the reaction P4(g) + 6 Cl2 (g) 
4 PCl3 (g) run at 298 K.
(b) What is ΔG° for the reverse of this reaction?
Chemical
Thermodynamics
Free Energy Changes
At temperatures other than 25 C,
G = H  TS
How does G change with
temperature?
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Free Energy and Temperature
• There are two parts to the free
energy equation:
H— the enthalpy term
– TS — the entropy term
• The temperature dependence of
free energy then comes from the
entropy term.
Chemical
Thermodynamics
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Free Energy and Temperature
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Exercise 8
Determining the Effect of Temperature on
Spontaneity
The Haber process for the production of ammonia involves the
equilibrium
Assume that ΔH° and ΔS° for this reaction do not change
with temperature. (a) Predict the direction in which ΔG° for
the reaction changes with increasing temperature. (b) Calculate
ΔG° at 25 °C and 500 °C.
Chemical
Thermodynamics
Free Energy and Equilibrium
Under any conditions, standard or
nonstandard, the free energy change
can be found this way:
G = G + RT ln Q
(Under standard conditions, all concentrations are 1
M, so Q = 1 and ln Q = 0; the last term drops out.)
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Exercise 9
Calculating the Free–Energy Change under
Nonstandard Conditions
Calculate at 298 K for a mixture of 1.0 atm N2, 3.0 atm
H2, and 0.50 atm NH3 being used in the Haber process:
Chemical
Thermodynamics
Free Energy and Equilibrium
• At equilibrium, Q = K, and G = 0.
• The equation becomes
0 = G + RT ln K
• Rearranging, this becomes
G = RT ln K
or
K = e G/RT
Chemical
Thermodynamics
© 2012 Pearson Education, Inc.
Exercise 10 Calculating an Equilibrium Constant from ΔG°
The standard free–energy change for the Haber process at was
obtained in Exercise 8 for the Haber reaction:
Use this value of ΔG° to calculate the equilibrium constant for
the process at 25 °C.
Chemical
Thermodynamics