Transcript Chapter 19 Chemical Thermodynamics

Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 19
Chemical
Thermodynamics
John D. Bookstaver
St. Charles Community College
Cottleville, MO
Chemical
Thermodynamics
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First Law of Thermodynamics
• You will recall from Chapter 5 that
energy cannot be created nor
destroyed.
• Therefore, the total energy of the
universe is a constant.
• 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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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.
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Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse
direction.
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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.
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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.
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Irreversible Processes
• Irreversible processes cannot be undone by
exactly reversing the change to the system.
• Spontaneous processes are irreversible.
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Entropy
• Entropy (S) is a term coined by Rudolph
Clausius in the 19th century.
• Clausius was convinced of the
significance of the ratio of heat
delivered and the temperature at which
it is delivered, q .
T
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Entropy
• Entropy can be thought of as a measure
of the randomness of a system.
• It is related to the various modes of
motion in molecules.
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Entropy
• Like total energy, E, and enthalpy, H,
entropy is a state function.
• Therefore,
S = Sfinal  Sinitial
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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
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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.
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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
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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.
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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.
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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 on about an axis or
rotation about  bonds.
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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.
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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 lnW
where k is the Boltzmann constant, 1.38  1023 J/K.
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Entropy on the Molecular Scale
• Entropy increases with the number of
microstates in the system
• The number of microstates and, therefore,
the entropy tends to increase with increases
in
– Temperature.
– Volume.
– The number of independently moving molecules.
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Entropy and Physical States
• Entropy increases with
the freedom of motion
of molecules.
• Therefore,
S(g) > S(l) > S(s)
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Solutions
Generally, when
a solid is
dissolved in a
solvent, entropy
increases.
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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.
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Third Law of Thermodynamics
The entropy of a pure crystalline
substance at absolute zero is 0.
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Sample Exercise 19.3 Predicting the Sign of ΔS
Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at
constant temperature:
Solution
Analyze: We are given four equations and asked to predict the sign of ΔS for each chemical reaction.
Plan: The sign of ΔS will be positive if there is an increase in temperature, an increase in the volume in
which the molecules move, or an increase in the number of gas particles in the reaction. The question states
that the temperature is constant. Thus, we need to evaluate each equation with the other two factors in mind.
Solve:
(a) The evaporation of a liquid is accompanied by a large increase in volume. One mole of water (18 g)
occupies about 18 mL as a liquid and if it could exist as a gas at STP it would occupy 22.4 L. Because the
molecules are distributed throughout a much larger volume in the gaseous state than in the liquid state, an
increase in motional freedom accompanies vaporization. Therefore, ΔS is positive.
(b) In this process the ions, which are free to move throughout the volume of the solution, form a solid in
which they are confined to a smaller volume and restricted to more highly constrained positions. Thus, ΔS is
negative.
(c) The particles of a solid are confined to specific locations and have fewer ways to move (fewer
microstates) than do the molecules of a gas. Because O2 gas is converted into part of the solid product
Fe2O3, ΔS is negative.
(d) The number of moles of gases is the same on both sides of the equation, and so the entropy change will
be small. The sign of ΔS is impossible to predict based on our discussions thus far, but we can predict that
ΔS will be close to zero.
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Sample Exercise 19.3 Predicting the Sign of ΔS
Practice Exercise
Indicate whether each of the following processes produces an increase or decrease in the entropy of the
system:
Answer: (a) increase, (b) decrease, (c) decrease, (d) decrease
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Sample Exercise 19.4 Predicting Which Sample of Matter Has the Higher Entropy
Choose the sample of matter that has greater entropy in each pair, 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.
Solution
Analyze: We need to select the system in each pair that has the greater entropy.
Plan: To do this, we examine the state of the system and the complexity of the molecules it contains.
Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b)
The sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol.
Thus, the 2-mol sample has twice the number of microstates and twice the entropy when they are at the
same pressure. (c) The HCl sample has the higher entropy because the HCl molecule is capable of storing
energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms cannot.
Practice Exercise
Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or
1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(s) at 0 °C or 1 mol of H2O(l) at 25 °C,
(c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.
Answers: (a) 1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(l) at 25 °C, (c) 1 mol of SO2(g) at
STP, (d) 2 mol of NO2(g) at STP
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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.
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Standard Entropies
Larger and more complex molecules have
greater entropies.
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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.
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Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies
Calculate ΔSº for the synthesis of ammonia from N2(g) and H2(g) at 298 K:
Solution
N2(g) + 3 H2(g) → 2 NH3(g)
Analyze: We are asked to calculate the entropy change for the synthesis of NH3(g) from its constituent
elements.
Plan: We can make this calculation using Equation 19.8 and the standard molar entropy values for the
reactants and the products that are given in Table 19.2 and in Appendix C.
Solve: Using Equation 19.8, we have
ΔS° = 2S°(NH3) - [S°(N2) + 3S°(H2)]
Substituting the appropriate S° values from
Table 19.2 yields
ΔS° = (2 mol)(192.5 J/mol-K) - [(1 mol)(191.5 J/mol-K)
+ (3 mol)(130.6 J/mol-K)] = -198.3 J/K
Check: The value for ΔS° is negative, in agreement with our qualitative prediction based on the decrease in
the number of molecules of gas during the reaction.
Practice Exercise
Using the standard entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following
reaction at 298 K:
Al2O3(s) + 3 H2(g) → 2 Al(s) + 3 H2O(g)
Answers: 180.39 J/K
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Entropy Changes in Surroundings
• Heat that flows into or out of the
system changes the entropy of the
surroundings.
• For an isothermal process:
Ssurr =
qsys
T
• At constant pressure, qsys is simply
H for the system.
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Entropy Change in the Universe
• The universe is composed of the system
and the surroundings.
• Therefore,
Suniverse = Ssystem + Ssurroundings
• For spontaneous processes
Suniverse > 0
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Entropy Change in the Universe
• Since Ssurroundings =
qsystem
and qsystem = Hsystem
T
This becomes:
Hsystem
Suniverse = Ssystem +
T
Multiplying both sides by T, we get
TSuniverse = Hsystem  TSsystem
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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.
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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.
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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.
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Free Energy Changes
At temperatures other than 25°C,
G° = H  TS
How does G change with temperature?
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Sample Exercise 19.6 Calculating Free-Energy Change from ΔH°, T, 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 circumstances?
Solution
Analyze: We are asked to calculate ΔG° for the indicated reaction (given ΔH°, ΔS° and T) and to
predict whether the reaction is spontaneous under standard conditions at 298 K.
Plan: To calculate ΔG°, we use Equation 19.12, ΔG° = ΔH° – T ΔS°. To determine whether the
reaction is spontaneous under standard conditions, we look at the sign of ΔG°.
Solve:
Because ΔG° is positive, the reaction is not spontaneous under standard conditions at 298 K.
Comment: Notice that we had to convert the units of the T ΔS° term to kJ so that they could be added to
the ΔH° term, whose units are kJ.
Practice Exercise
A particular reaction has ΔH° = 24.6 kJ and ΔS° = 132 J/K at 298 K. Calculate ΔG°.
Is the reaction spontaneous under these conditions?
Answers: ΔG° = –14.7 kJ; the reaction is spontaneous.
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Sample Exercise 19.7 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 following reaction at
298 K:
P4(g) + 6 Cl2(g) → 4 PCl3(g)
(b) What is ΔG° for the reverse of the above reaction?
Solution
Analyze: We are asked to calculate the free-energy change for the indicated reaction and then to determine
the free-energy change of its reverse.
Plan: To accomplish our task, we look up the free-energy values for the products and reactants and use
Equation 19.14: We multiply the molar quantities by the coefficients in the balanced equation, and subtract
the total for the reactants from that for the products.
Solve:
(a) Cl2(g) is in its standard state, so ΔG°f is zero for this reactant. P4(g), however, is not in its standard
state, so ΔG°f is not zero for this reactant. From the balanced equation and using Appendix C, we have:
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Sample Exercise 19.7 Calculating Standard Free-Energy Change from
Free Energies of Formation
Solution (continued)
The fact that ΔG° is negative tells us that a mixture of P4(g), Cl2(g), and PCl3(g) at 25 °C, each present at
a partial pressure of 1 atm, would react spontaneously in the forward direction to form more PCl 3.
Remember, however, that the value of ΔG° tells us nothing about the rate at which the reaction occurs.
(b) Remember that ΔG = G (products) – G (reactants). If we reverse the reaction, we reverse the roles of the
reactants and products. Thus, reversing the reaction changes the sign of ΔG, just as reversing the reaction
changes the sign of ΔH. (Section 5.4) Hence, using the result from part (a):
4 PCl3(g) → P4(g) + 6 Cl2(g) ΔG° = +1102.8 kJ
Practice Exercise
By using data from Appendix C, calculate ΔG° at 298 K for the combustion of methane:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g).
Answer: –800.7 kJ
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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.
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Free Energy and Temperature
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Free Energy and Equilibrium
Under any conditions, standard or
nonstandard, the free energy change
can be found this way:
G = G + RT lnQ
(Under standard conditions, all concentrations are 1 M,
so Q = 1 and lnQ = 0; the last term drops out.)
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Free Energy and Equilibrium
• At equilibrium, Q = K, and G = 0.
• The equation becomes
0 = G + RT lnK
• Rearranging, this becomes
G = RT lnK
or,
-G
K = e RT
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