2.
First Law of Thermodynamics
Chemical
Thermodynamics
3.
First Law of Thermodynamics
• You will recall from Chapter 5 that
energy cannot be created nor
destroyed.
Chemical
Thermodynamics
4.
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.
Chemical
Thermodynamics
5.
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. Chemical
Thermodynamics
7.
Spontaneous Processes
• Spontaneous processes
are those that can
proceed without any
outside intervention.
Chemical
Thermodynamics
8.
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
Chemical
Thermodynamics
9.
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
A process is either spontaneous, Chemical
non-spontaneous, or at equilibrium. Thermodynamics
10.
Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse
direction.
Chemical
Thermodynamics
11.
Melting a cube of water ice at −10°C
and atmospheric pressure is a
________ process.
1. Spontaneous
2. Nonspontaneous
Chemical
Thermodynamics
12.
Correct Answer:
Melting of an ice cube at
1. Spontaneous 1 atm pressure will only
be spontaneous above
2. Nonspontaneous the freezing point (0°C).
Chemical
Thermodynamics
14.
Spontaneous Processes
• Processes that are spontaneous at one
temperature may be nonspontaneous at other
temperatures.
Chemical
Thermodynamics
15.
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.
Chemical
Thermodynamics
16.
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
18.
No. Nonspontaneous processes can occur with
some external assistance.
Chemical
Thermodynamics
19.
SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes
Predict whether these processes are spontaneous as
described, spontaneous in the reverse direction, or in
equilibrium:
(a) When a piece of metal heated to 150°C is added
to water at 40°C, the water gets hotter.
(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.
20.
SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes
Predict whether these processes are spontaneous as
described, spontaneous in the reverse direction, or in
equilibrium:
(a) When a piece of metal heated to 150°C is added
to water at 40°C, the water gets hotter.
(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.
Analyze: We are asked to judge whether each process will proceed spontaneously in the direction indicated, in the reverse
direction, or in neither direction.
Plan: We need to think about whether each process is consistent with our experience about the natural direction of events or
whether it is the reverse process that we expect to occur.
Solve: (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is
transferred from the hotter object to the colder one. In this instance heat is transferred from the hot metal to the cooler water.
The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere
between the initial temperatures of the metal and the water. (b) Experience tells us that this process is not spontaneous—we
certainly have never seen hydrogen and oxygen gas bubbling up out of water! Rather, the reverse process—the reaction of
H2 and O2 to form H2O—is spontaneous once initiated by a spark or flame. (c) By definition, the normal boiling point is the
temperature at which a vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. Neither the
condensation of benzene vapor nor the reverse process is spontaneous. If the temperature were below 80.1°C,
condensation would be spontaneous.
21.
PRACTICE EXERCISE
Under 1 atm pressure CO2(s) sublimes at
–78°C. Is the transformation of CO2(s) to
CO2(g) a spontaneous process at –100ºC
and 1 atm pressure?
Chemical
Thermodynamics
22.
PRACTICE EXERCISE
Under 1 atm pressure CO2(s) sublimes at
–78°C. Is the transformation of CO2(s) to
CO2(g) a spontaneous process at –100ºC
and 1 atm pressure?
Answer: No, the reverse process is
spontaneous at this temperature.
Chemical
Thermodynamics
23.
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
24.
Irreversible Processes
Chemical
Thermodynamics
25.
Irreversible Processes
• Irreversible processes cannot be undone by
exactly reversing the change to the system.
Chemical
Thermodynamics
26.
Irreversible Processes
• Irreversible processes cannot be undone by
exactly reversing the change to the system.
• Spontaneous processes are irreversible.
Chemical
Thermodynamics
30.
Entropy
• Entropy (S) is a term coined by Rudolph
Clausius in the 19th century.
Chemical
Thermodynamics
31.
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
Chemical
Thermodynamics
33.
Entropy
• Entropy can be thought of as a measure
of the randomness of a system.
Chemical
Thermodynamics
34.
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.
Chemical
Thermodynamics
36.
Entropy
• Like total energy, E, and enthalpy, H,
entropy is a state function.
Chemical
Thermodynamics
37.
Entropy
• Like total energy, E, and enthalpy, H,
entropy is a state function.
• Therefore,
Chemical
Thermodynamics
38.
Entropy
• Like total energy, E, and enthalpy, H,
entropy is a state function.
• Therefore,
ΔS = Sfinal − Sinitial
Chemical
Thermodynamics
39.
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:
Chemical
Thermodynamics
40.
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
ΔS =
T
Chemical
Thermodynamics
42.
• ΔS depends not merely on q but on qrev.
There is only one reversible isothermal path
between two states regardless of the number of
possible paths.
Chemical
Thermodynamics
43.
SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change
The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point
of mercury is –38.9ºC, and its molar enthalpy
of fusion is ΔΗfusion = 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
44.
SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change
The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point
of mercury is –38.9ºC, and its molar enthalpy
of fusion is ΔΗfusion = 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
45.
SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change
The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point
of mercury is –38.9ºC, and its molar enthalpy
of fusion is ΔΗfusion = 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
46.
SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change
The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and
its molar enthalpy of fusion is ΔΗfusion = 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?
Solution
Analyze: We first recognize that freezing is an exothermic process; heat is transferred from the system to the
surroundings when a liquid freezes (q < 0). The enthalpy of fusion is ΔΗ for the melting process. Because freezing is
the reverse of melting, the enthalpy change that accompanies the freezing of
1 mol of Hg is –ΔΗfusion = 2.29 kJ/mol.
Plan: We can use –ΔΗfusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg:
We can use this value of q as qrev in Equation 19.2. We must first, however, convert the temperature to K:
Solve: We can now calculate the value of ΔSsys:
Check: The entropy change is negative because heat flows from the system making qrev negative.
Comment: The procedure we have used here can be used to calculate ΔS for other isothermal phase changes, such
as the vaporization of a liquid at its boiling point.
Chemical
Thermodynamics
47.
SAMPLE EXERCISE 19.2 continued
PRACTICE EXERCISE
The normal boiling point of ethanol,
C2H5OH, is 78.3°C (Figure 11.24), and
its molar enthalpy of vaporization is
38.56 kJ/mol. What is the change in
entropy in the system when 68.3 g of
C2H5OH(g) at 1 atm condenses to
liquid at the normal boiling point?
Chemical
Thermodynamics
48.
SAMPLE EXERCISE 19.2 continued
PRACTICE EXERCISE
The normal boiling point of ethanol,
C2H5OH, is 78.3°C (Figure 11.24), and
its molar enthalpy of vaporization is
38.56 kJ/mol. What is the change in
entropy in the system when 68.3 g of
C2H5OH(g) at 1 atm condenses to
liquid at the normal boiling point?
Answer: –163 J/K
Chemical
Thermodynamics
49.
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
50.
Second Law of Thermodynamics
Chemical
Thermodynamics
51.
Second Law of Thermodynamics
In other words:
For reversible processes:
ΔSuniv = ΔSsystem + ΔSsurroundings = 0
For irreversible processes:
ΔSuniv = ΔSsystem + ΔSsurroundings > 0
Chemical
Thermodynamics
52.
Second Law of Thermodynamics
These last truths mean that as a result of all
spontaneous processes
the entropy of the
universe increases.
Chemical
Thermodynamics
54.
The entropy of the universe for the process must
increase by a greater amount than the entropy of the
system decreases.
Chemical
Thermodynamics
55.
Entropy on the Molecular Scale
Chemical
Thermodynamics
56.
Entropy on the Molecular Scale
• Ludwig Boltzmann described the concept of
entropy on the molecular level.
Chemical
Thermodynamics
57.
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
58.
Entropy on the Molecular Scale
Chemical
Thermodynamics
59.
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 σ bonds.
Chemical
Thermodynamics
61.
A molecule can vibrate (atoms moving relative to
one another) and rotate (tumble); a single atom can
only rotate.
Chemical
Thermodynamics
62.
Entropy on the Molecular Scale
Chemical
Thermodynamics
63.
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
64.
Entropy on the Molecular Scale
Chemical
Thermodynamics
65.
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
microstates, W, associated with it.
Chemical
Thermodynamics
66.
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
microstates, W, associated with it.
• Entropy is
Chemical
Thermodynamics
67.
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
microstates, W, associated with it.
• Entropy is
S = k lnW
Chemical
Thermodynamics
68.
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.
Chemical
Thermodynamics
71.
Entropy on the Molecular Scale
Chemical
Thermodynamics
72.
Entropy on the Molecular Scale
• The change in entropy for a process,
then, is
ΔS = k lnWfinal − k lnWinitial
lnWfinal
ΔS = k ln
lnWinitial
Chemical
Thermodynamics
73.
Entropy on the Molecular Scale
• The change in entropy for a process,
then, is
ΔS = k lnWfinal − k lnWinitial
lnWfinal
ΔS = k ln
lnWinitial
• Entropy increases with the number of
microstates in the system. Chemical
Thermodynamics
74.
Entropy on the Molecular Scale
Chemical
Thermodynamics
75.
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
76.
Entropy and Physical States
Chemical
Thermodynamics
77.
Entropy and Physical States
• Entropy increases with
the freedom of motion
of molecules.
Chemical
Thermodynamics
78.
Entropy and Physical States
• Entropy increases with
the freedom of motion
of molecules.
• Therefore,
Chemical
Thermodynamics
79.
Entropy and Physical States
• Entropy increases with
the freedom of motion
of molecules.
• Therefore,
S(g) > S(l) > S(s)
Chemical
Thermodynamics
80.
Solutions
Generally, when
a solid is
dissolved in a
solvent, entropy
increases.
Chemical
Thermodynamics
82.
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
83.
The sign of entropy change, ΔS,
associated with the boiling of water is
_______.
1. Positive
2. Negative
3. Zero
Chemical
Thermodynamics
84.
Correct Answer:
1. Positive Vaporization of a liquid
to a gas leads to a large
2. Negative increase in volume and
hence entropy; ΔS
3. Zero must be positive.
Chemical
Thermodynamics
85.
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:
Chemical
Thermodynamics
86.
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 22.4 L as a gas at STP. 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.
87.
PRACTICE EXERCISE
Indicate whether each of the following
processes produces an increase or
decrease in the entropy of the system:
Chemical
Thermodynamics
88.
PRACTICE EXERCISE
Indicate whether each of the following
processes produces an increase or
decrease in the entropy of the system:
Answers: (a) increase, (b) decrease,
(c) decrease, (d) decrease Chemical
Thermodynamics
89.
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. Chemical
Thermodynamics
90.
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.
Chemical
Thermodynamics
91.
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.
Chemical
Thermodynamics
92.
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. (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.
Chemical
Thermodynamics
93.
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. (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.
Chemical
Thermodynamics
94.
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. (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
NO2(g) of at STP
Chemical
Thermodynamics
95.
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.
96.
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 H (g) at 100°C and 0.5 atm,
2
(b) 1 mol of H2O(l) at 25°C, (c) 1 mol of SO2(g) at STP,
(d) 2 mol NO (g) of at STP
97.
Which of the following would be
expected to have the highest entropy?
1. 1 mole N2(l)
2. 1 mole N2(g)
3. 1 mole Ar(g)
4. 1 mole Ar(s)
Chemical
Thermodynamics
98.
Correct Answer:
1. 1 mole N2(l) Given the same amounts, gases
2. 1 mole N2(g) have higher entropy than
liquids, which have higher
3. 1 mole Ar(g) entropy than solids. Of the
4. 1 mole Ar(s) gases, N2 has greater entropy
than Ar because N2 is a
molecule that can store energy
in ways not possible for Ar.
Chemical
Thermodynamics
99.
Third Law of Thermodynamics
The entropy of a pure crystalline
substance at absolute zero is 0.
Chemical
Thermodynamics
104.
Standard Entropies
• These are molar
entropy values of
substances in their
standard states.
Chemical
Thermodynamics
105.
Standard Entropies
• These are molar
entropy values of
substances in their
standard states.
• Standard entropies tend
to increase with
increasing molar mass.
Chemical
Thermodynamics
106.
Standard Entropies
Larger and more complex molecules have
greater entropies.
Chemical
Thermodynamics
107.
Which substance will
have the highest
entropy?
1. Butane
2. Isobutane
3. Pentane
4. Isopentane
5. Neopentane
Chemical
Thermodynamics
108.
Which substance will
have the highest
entropy?
1. Butane
2. Isobutane
3. Pentane
4. Isopentane
5. Neopentane
Chemical
Thermodynamics
110.
Predict whether entropy increases,
decreases, or stays constant for:
2 NO2(g) → 2 NO(g) + O2(g)
1. Increases
2. Decreases
3. Stays constant
The entropy increases because two molecules of gas
react to form three molecules of gas, leading to more
disorder.
Chemical
Thermodynamics
111.
Which process will have a
negative ΔS?
1. Dissolving a substance
2. Melting
3. Falling leaves
4. Oxidation of a metal
5. Decomposition of water
into H2 and O2 Dissolution of NaCl
Chemical
Thermodynamics
112.
Which process will have a
negative ΔS?
1. Dissolving a substance
2. Melting
3. Falling leaves
4. Oxidation of a metal
5. Decomposition of water
Dissolution of NaCl
into H2 and O2
Chemical
Thermodynamics
114.
Entropy Changes
Entropy changes for a reaction can be
estimated in a manner analogous to that by
which ΔH is estimated:
ΔS° = ΣnΔS°(products) - ΣmΔS°(reactants)
where n and m are the coefficients in the
balanced chemical equation. Chemical
Thermodynamics
115.
SAMPLE EXERCISE 19.5 Calculating ΔS from Tabulated Entropies
Calculate ΔS° for the
synthesis of ammonia
from N2(g) and H2(g) at
298 K:
Chemical
Thermodynamics
116.
SAMPLE EXERCISE 19.5 Calculating ΔS from Tabulated Entropies
Calculate ΔS° for the synthesis of ammonia from N2(g) and H2(g) at 298 K:
Solution
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
Substituting the appropriate S° values from Table 19.2 yields
Chemical
Thermodynamics
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.
117.
PRACTICE EXERCISE
Using the standard
entropies in Appendix C,
calculate the standard
entropy change, ΔS°, for
the following reaction at
298 K:
Chemical
Thermodynamics
118.
PRACTICE EXERCISE
Using the standard
entropies in Appendix C,
calculate the standard
entropy change, ΔS°, for
the following reaction at
298 K:
Answer: 180.39 J/K Chemical
Thermodynamics
119.
Calculate ΔS° for the equation below using the
standard entropy data given:
2 NO(g) + O2(g) → 2 NO2(g)
ΔS° values (J/mol-K): NO2(g) = 240, NO(g) = 211,
O2(g) = 205.
1. +176 J/mol 3. −147 J/mol
2. +147 J/mol 4. −76 J/mol
Chemical
Thermodynamics
121.
Entropy Changes in Surroundings
Chemical
Thermodynamics
122.
Entropy Changes in Surroundings
• Heat that flows into or out of the
system changes the entropy of the
surroundings.
Chemical
Thermodynamics
123.
Entropy Changes in Surroundings
• Heat that flows into or out of the
system changes the entropy of the
surroundings.
• For an isothermal process:
−qsys
ΔSsurr =
T
Chemical
Thermodynamics
124.
Entropy Changes in Surroundings
• Heat that flows into or out of the
system changes the entropy of the
surroundings.
• For an isothermal process:
−qsys
ΔSsurr =
T
• At constant pressure, qsys is simply
ΔH° for the system. Chemical
Thermodynamics
125.
−qsys
ΔSsurr =
T
• At constant pressure, qsys is simply
ΔH° for the system. Chemical
Thermodynamics
126.
always increase
−qsys
ΔSsurr =
T
• At constant pressure, qsys is simply
ΔH° for the system. Chemical
Thermodynamics
127.
Entropy Change in the Universe
Chemical
Thermodynamics
128.
Entropy Change in the Universe
• The universe is composed of the system
and the surroundings.
Chemical
Thermodynamics
129.
Entropy Change in the Universe
• The universe is composed of the system
and the surroundings.
• Therefore,
Chemical
Thermodynamics
130.
Entropy Change in the Universe
• The universe is composed of the system
and the surroundings.
• Therefore,
ΔSuniverse = ΔSsystem + ΔSsurroundings
Chemical
Thermodynamics
131.
Entropy Change in the Universe
• The universe is composed of the system
and the surroundings.
• Therefore,
ΔSuniverse = ΔSsystem + ΔSsurroundings
• For spontaneous processes
Chemical
Thermodynamics
132.
Entropy Change in the Universe
• The universe is composed of the system
and the surroundings.
• Therefore,
ΔSuniverse = ΔSsystem + ΔSsurroundings
• For spontaneous processes
ΔSuniverse > 0
Chemical
Thermodynamics
133.
Entropy Change in the Universe
Chemical
Thermodynamics
134.
Entropy Change in the Universe
• This becomes:
−ΔHsystem
ΔSuniverse = ΔSsystem +
T
Multiplying both sides by −T,
−TΔSuniverse = ΔHsystem − TΔSsystem
Chemical
Thermodynamics
136.
Gibbs Free Energy
• −TΔSuniverse is defined as the Gibbs free
energy, ΔG.
Chemical
Thermodynamics
137.
Gibbs Free Energy
• −TΔSuniverse is defined as the Gibbs free
energy, ΔG.
• When ΔSuniverse is positive, ΔG is negative.
Chemical
Thermodynamics
138.
Gibbs Free Energy
• −TΔSuniverse is defined as the Gibbs free
energy, ΔG.
• When ΔSuniverse is positive, ΔG is negative.
• Therefore, when
ΔG is negative,
a process is spontaneous.
Chemical
Thermodynamics
143.
Gibbs Free Energy
1. If ΔG is negative, the
forward reaction is
spontaneous.
Chemical
Thermodynamics
144.
Gibbs Free Energy
1. If ΔG is negative, the
forward reaction is
spontaneous.
2. If ΔG is 0, the system
is at equilibrium.
Chemical
Thermodynamics
145.
Gibbs Free Energy
1. If ΔG is negative, the
forward reaction is
spontaneous.
2. If ΔG is 0, the system
is at equilibrium.
3. If ΔG is positive, the
reaction is
spontaneous in the
reverse direction.
Chemical
Thermodynamics
146.
Standard Free Energy Changes
Analogous to standard enthalpies of
formation are standard free energies of
formation, ΔG°.
f
Chemical
Thermodynamics
147.
Standard Free Energy Changes
Analogous to standard enthalpies of
formation are standard free energies of
formation, ΔG°.
f
ΔG° = ΣnΔG°(products) − ΣmΔG°(reactants)
f f
where n and m are the stoichiometric
coefficients.
Chemical
Thermodynamics
148.
Calculate ΔG° for the equation below using
the Gibbs free energy data given:
2 SO2(g) + O2(g) 2 SO3(g)
ΔG° values (kJ): SO2(g) = −300.4, SO3(g) = −370.4
1. +70 kJ
2. +140 kJ
3. −140 kJ
4. −70 kJ
Chemical
Thermodynamics
151.
It indicates the process has taken place under standard
conditions.
Chemical
Thermodynamics
152.
It indicates the process has taken place under standard
conditions.
Chemical
Thermodynamics
153.
It indicates the process has taken place under standard
conditions.
298K = 25°C this is NOT STP (273K)
Chemical
Thermodynamics
154.
It indicates the process has taken place under standard
conditions.
298K = 25°C this is NOT STP (273K)
1 atm for gases
Chemical
Thermodynamics
155.
It indicates the process has taken place under standard
conditions.
298K = 25°C this is NOT STP (273K)
1 atm for gases
1M for solutions
Chemical
Thermodynamics
156.
Calculate ΔG° for the equation below using
the thermodynamic data given:
N2(g) + 3 H2(g) → 2 NH3(g)
ΔH° (NH3) = −46 kJ; ΔS° values (J/mol-K) are
NH3 = 192.5, N2 = 191.5, H2 = 130.6.
1. +33 kJ 3. −66 kJ
2. +66 kJ 4. −33 kJ
Chemical
Thermodynamics
158.
SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy
Change from Free Energies of Formation
(a) By using data from Appendix C,
calculate the standard free-energy change
for the following reaction at 298 K:
(b) What is ΔG° for the reverse of the
above reaction?
Chemical
Thermodynamics
159.
SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy Change
from Free Energies of Formation
(a) By using data from Appendix C, calculate the standard free-energy change for the following reaction at
298 K:
(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
19.13: 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 is zero for this reactant. P4(g), however, is not in its standard
state, so is not zero for this reactant. From the balanced equation and using Appendix C, we have:
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 PCl3. 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, Chemical
using the result from part (a): Thermodynamics
160.
PRACTICE EXERCISE
By using data from Appendix C,
calculate ΔG° at 298 K for the
combustion of methane:
Chemical
Thermodynamics
161.
PRACTICE EXERCISE
By using data from Appendix C,
calculate ΔG° at 298 K for the
combustion of methane:
Answer: –800.7 kJ
Chemical
Thermodynamics
162.
SAMPLE EXERCISE 19.7 Estimating and Calculating ΔG°
In Section 5.7 we used Hess’s law to
calculate ΔH° for the combustion of
propane gas at 298 K:
(a) Without using data from Appendix C,
predict whether ΔG° for this reaction is
more negative or less negative than ΔH°.
(b) Use data from Appendix C to calculate
the standard free-energy change for the
reaction at 298 K. Is your prediction from
part (a) correct?
163.
SAMPLE EXERCISE 19.7 Estimating and Calculating ΔG°
In Section 5.7 we used Hess’s law to calculate ΔH° for the combustion of propane gas at 298 K:
(a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH
°. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your
prediction from part (a) correct?
Solution
Analyze: In part (a) we must predict the value for ΔG° relative to that for ΔH° on the basis of the balanced equation for
the reaction. In part (b) we must calculate the value for ΔG° and compare with our qualitative prediction.
Plan: The free-energy change incorporates both the change in enthalpy and the change in entropy for the reaction
(Equation 19.11), so under standard conditions:
ΔG° = ΔH° – TΔS°
To determine whether ΔG° is more negative or less negative than ΔH° we need to determine the sign of the term TΔS°.
T is the absolute temperature, 298 K, so it is a positive number. We can predict the sign of ΔS° by looking at the
reaction.
Solve: (a) We see that the reactants consist of six molecules of gas, and the products consist of three molecules of gas
and four molecules of liquid. Thus, the number of molecules of gas has decreased significantly during the reaction. By
using the general rules we discussed in Section 19.3, we would expect a decrease in the number of gas molecules to
lead to a decrease in the entropy of the system—the products have fewer accessible microstates than the reactants.
We therefore expect ΔS° and TΔS° to be negative numbers. Because we are subtracting TΔS°, which is a negative
number, we would predict that ΔG° is less negative than ΔH°.
(b) Using Equation 19.13 and values from Appendix C, we can calculate the value of ΔG°:
Chemical
Notice that we have been careful to use the value of as in the calculation of ΔH values, the Thermodynamics
phases of the reactants and products are important. As we predicted, ΔG° is less negative than ΔH° because
of the decrease in entropy during the reaction.
164.
PRACTICE EXERCISE
Consider the combustion of propane to
form CO2(g) and H2O(g) at 298 K:
Would you expect ΔG° to be more
negative or less negative than ΔH°?
Chemical
Thermodynamics
165.
PRACTICE EXERCISE
Consider the combustion of propane to
form CO2(g) and H2O(g) at 298 K:
Would you expect ΔG° to be more
negative or less negative than ΔH°?
Answer: more negative
Chemical
Thermodynamics
167.
Free Energy Changes
At temperatures other than 25°C,
ΔG° = ΔH° − TΔS°
How does ΔG° change with temperature?
Chemical
Thermodynamics
168.
Free Energy and Temperature
Chemical
Thermodynamics
169.
Free Energy and Temperature
• There are two parts to the free energy
equation:
ΔH°— the enthalpy term
TΔS° — the entropy term
• The temperature dependence of free
energy, then comes from the entropy
term.
Chemical
Thermodynamics
170.
Upon heating, limestone (CaCO3) decomposes to
CaO and CO2. Speculate on the sign of ΔH and
ΔS for this process.
CaCO3(s) CaO(s) + CO2(g)
Limestone Cave
Carlsbad Caverns, New Mexico
Chemical
Thermodynamics
171.
Upon heating, limestone (CaCO3) decomposes to
CaO and CO2. Speculate on the sign of ΔH and
ΔS for this process.
CaCO3(s) CaO(s) + CO2(g)
Limestone Cave
Carlsbad Caverns, New Mexico
Chemical
Thermodynamics
172.
Free Energy and Temperature
Chemical
Thermodynamics
175.
ΔH < TΔS
ΔG(Negative/spontaneous)=
Chemical
Thermodynamics
176.
ΔH < TΔS
ΔG(Negative/spontaneous)=
ΔH (Positive/vaporization) - TΔS (positive/liquid to gas)
Chemical
Thermodynamics
177.
ΔH < TΔS
ΔG(Negative/spontaneous)=
ΔH (Positive/vaporization) - TΔS (positive/liquid to gas)
↑must be larger
Chemical
Thermodynamics
178.
Both the entropy, ΔS° and the enthalpy ΔH°
have positive values. Near absolute zero (0 K),
will the process be spontaneous or
nonspontaneous?
1. Spontaneous
2. Nonspontaneous
Chemical
Thermodynamics
179.
Correct Answer:
1. Spontaneous
2. Nonspontaneous
At low temperatures, the enthalpy
term dominates the entropy term,
leading to a positive ΔG and a
nonspontaneous process.
Chemical
Thermodynamics
180.
A reaction will be spontaneous at
lower temperatures when the sign
of ΔH and ΔS is
Chemical
Thermodynamics
181.
A reaction will be spontaneous at
lower temperatures when the sign
of ΔH and ΔS is
Chemical
Thermodynamics
182.
SAMPLE EXERCISE 19.8 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
this reaction changes with increasing
temperature.
(b) Calculate the values of ΔG° for the
reaction at 25°C and 500°C.
183.
SAMPLE EXERCISE 19.8 continued
Solution
Analyze: In part (a) we are asked to predict the direction in which ΔG° for the ammonia synthesis reaction changes as
temperature increases. In part (b) we need to determine ΔG° for the reaction at two different temperatures.
Plan: In part (a) we can make this prediction by determining the sign of ΔS° for the reaction and then using that
information to analyze Equation 19.20. In part (b) we need to calculate ΔH° and ΔS° for the reaction by using the data
in Appendix C. We can then use Equation 19.20 to calculate ΔG°.
Solve: (a) Equation 19.20 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –TΔS°. The
temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because
the number of molecules of gas is smaller in the products. Because ΔS° is negative, the term
–TΔS° is positive and grows larger with increasing temperature. As a result, ΔG° becomes less negative (or more
positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing
temperature.
(b) We calculated the value of ΔH° in Sample
Exercise 15.14, and the value of ΔS° was
determined in Sample Exercise 19.5: ΔH° = –92.38
kJ and ΔS° = –198.4 J/K. If we assume that these
values don’t change with temperature, we can
calculate ΔG° at any temperature by using
Equation 19.20. At T = 298 K we have:
Notice that we have been careful to convert –TΔS° into units of kJ so that it can be added to ΔH°, which has units of kJ.
Comment: Increasing the temperature from 298 K to 773 K changes ΔG° from –33.3 kJ to +61 kJ. Of course, the result at 773 K
depends on the assumption that ΔH° and ΔS° do not change with temperature. In fact, these values do change slightly with
temperature. Nevertheless, the result at 773 K should be a reasonable approximation. The positive increase in ΔG° with
increasing T agrees with our prediction in part (a) of this exercise. Our result indicates that a mixture of N2(g), H2(g), and NH3(g),
each present at a partial pressure of 1 atm, will react spontaneously at 298 K to form more NH3(g). In contrast, at 773 K the
positive value of ΔG° tells us that the reverse reaction is spontaneous. Thus, when the mixture of three gases, each at a partial
pressure of 1 atm, is heated to 773 K, some of the NH (g) spontaneously decomposes into N (g) and H (g).
184.
PRACTICE EXERCISE
(a) Using standard enthalpies of
formation and standard entropies in
Appendix C, calculate ΔH° and ΔS° at
298 K for the following reaction:
(b) Using the values obtained in part (a),
estimate ΔG° at 400 K.
185.
PRACTICE EXERCISE
(a) Using standard enthalpies of
formation and standard entropies in
Appendix C, calculate ΔH° and ΔS° at
298 K for the following reaction:
(b) Using the values obtained in part (a),
estimate ΔG° at 400 K.
Answer: (a) ΔH° = –196.6 kJ, ΔS° = –189.6 J/K;
(b) ΔG° = –120.8 kJ
186.
Chapter 19- Chemical
Thermodynamics
Section 19.7
Free Energy & Keq
187.
Free Energy and Equilibrium
Chemical
Thermodynamics
188.
Free Energy and Equilibrium
Under any conditions, standard or
nonstandard, the free energy change
can be found this way:
ΔG = ΔG° + RT lnQ
(Under standard conditions, all concentrations are 1 M,
so Q = 1 and lnQ = 0; the last term drops out.)
Chemical
Thermodynamics
189.
SAMPLE EXERCISE 19.9 Relating ΔG to a Phase Change at Equilibrium
The normal boiling point is the temperature
at which a pure liquid is in equilibrium with
its vapor at a pressure of 1 atm.
(a) Write the chemical equation that
defines the normal boiling point of liquid
carbon tetrachloride, CCl4(l).
(b) What is the value of ΔG° for the
equilibrium in part (a)?
(c) Use thermodynamic data in Appendix C
and Equation 19.20 to estimate the
normal boiling point of CCl4.
190.
SAMPLE EXERCISE 19.9 Relating ΔG to a Phase Change at Equilibrium
As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its
vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon
tetrachloride, CCl4(l). (b) What is the value of ΔG° for the equilibrium in part (a)? (c) Use thermodynamic data in
Appendix C and Equation 19.20 to estimate the normal boiling point of CCl4.
Solution
Analyze: (a) We must write a chemical equation that describes the physical equilibrium between liquid and gaseous
CCl4 at the normal boiling point. (b) We must determine the value of ΔG° for CCl4 in equilibrium with its vapor at the
normal boiling point. (c) We must estimate the normal boiling point of CCl4, based on available thermodynamic data.
Plan: (a) The chemical equation will merely show the change of state of CCl4 from liquid to solid. (b) We need to
analyze Equation 19.21 at equilibrium (ΔG = 0). (c) We can use Equation 19.20 to calculate T
when ΔG = 0.
Solve: (a) The normal boiling point of CCl4 is the temperature at which pure liquid CCl4 is in equilibrium with its vapor at
a pressure of 1 atm:
(b) At equilibrium ΔG = 0. In any normal boiling-point equilibrium both the liquid and the vapor are in their standard
states (Table 19.2). As a consequence, Q = 1, ln Q = 0, and ΔG = ΔG° for this process. Thus, we conclude that ΔG° = 0
for the equilibrium involved in the normal boiling point of any liquid. We would also find that ΔG° = 0 for the equilibria
relevant to normal melting points and normal sublimation points of solids.
(c) Combining Equation 19.20 with the result from part (b), we see that the equality at the normal boiling point, Tb, of
CCl4(l) or any other pure liquid is
Chemical
Thermodynamics
191.
SAMPLE EXERCISE 19.9 continued
Solving the equation for Tb, we obtain
Strictly speaking, we would need the values of ΔH° and ΔS° for the equilibrium between CCl4(l) and
CCl4(g) at the normal boiling point to do this calculation. However, we can estimate the boiling point by
using the values of ΔH° and ΔS° for CCl4 at 298 K, which we can obtain from the data in Appendix C
and Equations 5.31 and 19.8:
Notice that, as expected, the process is endothermic (ΔH > 0) and produces a gas in which energy
can be more spread out (ΔS > 0). We can now use these values to estimate Tb for CCl4(l):
Note also that we have used the conversion factor between J and kJ to make sure that the units of ΔH
° and ΔS° match.
Check: The experimental normal boiling point of CCl (l) is 76.5°C. The small deviation of our
4
estimate from the experimental value is due to the assumption that ΔH° and ΔS° do not change with
temperature.
Chemical
Thermodynamics
192.
PRACTICE EXERCISE
Use data in Appendix C to estimate the
normal boiling point, in K, for elemental
bromine, Br2(l). (The experimental
value is given in Table 11.3.)
Chemical
Thermodynamics
193.
PRACTICE EXERCISE
Use data in Appendix C to estimate the
normal boiling point, in K, for elemental
bromine, Br2(l). (The experimental
value is given in Table 11.3.)
Answer: 330 K Chemical
Thermodynamics
194.
SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change
Under Nonstandard Conditions
We will continue to explore the Haber
process for the synthesis of ammonia:
Calculate ΔG at 298 K for a reaction
mixture that consists of 1.0 atm N2,
3.0 atm H2, and 0.50 atm NH3.
Chemical
Thermodynamics
195.
SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change Under Nonstandard Conditions
We will continue to explore the Haber process for the synthesis of ammonia:
Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3.
Solution
Analyze: We are asked to calculate ΔG under nonstandard conditions.
Plan: We can use Equation 19.21 to calculate ΔG. Doing so requires that we calculate the value of the reaction
quotient Q for the specified partial pressures of the gases and evaluate ΔG°, using a table of standard free energies of
formation.
Solve: Solving for the reaction quotient gives:
In Sample Exercise 19.8 we calculated ΔG° = 33.3 kJ for this reaction. We will have to change the units of this quantity
in applying Equation 19.21, however. In order for the units in Equation 19.21 to work out, we will use kJ/mol as our
units for ΔG°, where “per mole” means “per mole of the reaction as written.” Thus,
ΔG° = –33.3kJ/mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3.
We can now use Equation 19.21 to calculate ΔG for these nonstandard conditions:
Comment: We see that ΔG becomes more negative, changing from –33.3 kJ/mol to –44.9 kJ/mol as the pressures of N2, H2,
and NH3 are changed from 1.0 atm each (standard conditions, ΔG° ) to 1.0 atm, 3.0 atm, and 0.50 atm, respectively. The larger
negative value for ΔG indicates a larger “driving force” to produce NH3.
We would have made the same prediction on the basis of Le Châtelier’s principle. • (Section 15.6) Relative to standard
conditions, we have increased the pressure of a reactant (H2) and decreased the pressure of the product (NH3). Le Châtelier’s
principle predicts that both of these changes should shift the reaction more to the product side, thereby forming more NH3.
196.
PRACTICE EXERCISE
Calculate ΔG at 298 K for the reaction
of nitrogen and hydrogen to form
ammonia if the reaction mixture
consists of 0.50 atm N2, 0.75 atm H2,
and 2.0 atm NH3.
Chemical
Thermodynamics
197.
PRACTICE EXERCISE
Calculate ΔG at 298 K for the reaction
of nitrogen and hydrogen to form
ammonia if the reaction mixture
consists of 0.50 atm N2, 0.75 atm H2,
and 2.0 atm NH3.
Answer: –26.0 kJ/mol
Chemical
Thermodynamics
198.
Decomposition of CaCO3 has a ΔH° =
178.3 kJ/mol and ΔS° = 159 J/mol • K.
At what temperature does this become
spontaneous?
1. 121°C
2. 395°C
3. 848°C
4. 1121°C
Chemical
Thermodynamics
199.
Correct Answer:
1. 121°C T = ΔH°/ΔS°
2. 395°C
3. 848°C T = 178.3 kJ/mol/0.159 kJ/mol • K
4. 1121°C
T = 1121 K
T (°C) = 1121 – 273 = 848
Chemical
Thermodynamics
200.
Free Energy and Equilibrium
Chemical
Thermodynamics
201.
Free Energy and Equilibrium
• At equilibrium, Q = K, and ΔG = 0.
Chemical
Thermodynamics
202.
Free Energy and Equilibrium
• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
Chemical
Thermodynamics
203.
Free Energy and Equilibrium
• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
0 = ΔG° + RT lnK
Chemical
Thermodynamics
204.
Free Energy and Equilibrium
• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
0 = ΔG° + RT lnK
• Rearranging, this becomes
Chemical
Thermodynamics
205.
Free Energy and Equilibrium
• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
0 = ΔG° + RT lnK
• Rearranging, this becomes
ΔG° = −RT lnK
Chemical
Thermodynamics
206.
Free Energy and Equilibrium
• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
0 = ΔG° + RT lnK
• Rearranging, this becomes
ΔG° = −RT lnK
or,
Chemical
Thermodynamics
207.
Free Energy and Equilibrium
• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
0 = ΔG° + RT lnK
• Rearranging, this becomes
ΔG° = −RT lnK
or,
K = e−ΔG°/RT
Chemical
Thermodynamics
208.
Suppose ΔG° is a large, positive value.
What then will be the value of the
equilibrium constant, K?
1. K=0
2. K=1
3. 0<K<1
4. K>1
Chemical
Thermodynamics
209.
Correct Answer:
ΔG° = −RTlnK
1. K=0
Thus, large positive values of ΔG°
2. K=1
lead to large negative values of lnK.
3. 0<K<1 The value of K itself, then, is very
4. K>1 small, approaching zero.
Chemical
Thermodynamics
210.
N2(g) + 3 H2(g) 2 NH3(g) ΔH° = -92.4 kJ/mol
If the temperature is decreased, what is the effect
on the equilibrium constant and ΔG?
1. K increases, ΔG increases
2. K decreases, ΔG decreases
3. K increases, ΔG decreases
4. K decreases, ΔG increases
5. Temperature does not affect
Josiah Willard Gibbs
equilibrium or ΔG (1839-1903)
Chemical
Thermodynamics
211.
N2(g) + 3 H2(g) 2 NH3(g) ΔH° = -92.4 kJ/mol
If the temperature is decreased, what is the effect
on the equilibrium constant and ΔG?
1. K increases, ΔG increases
2. K decreases, ΔG decreases
3. K increases, ΔG decreases
4. K decreases, ΔG increases
5. Temperature does not affect
Josiah Willard Gibbs
equilibrium or ΔG (1839-1903)
Chemical
Thermodynamics
212.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
Use standard free energies of
formation to calculate the equilibrium
constant, K, at 25°C for the reaction
involved in the Haber process:
The standard free-energy change for
this reaction was calculated in Sample
Exercise 19.8:
ΔG° = –33.3 kJ/mol = –33,300 J/mol.
213.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
Haber process:
The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.
Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
214.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
Haber process:
The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.
Solution
Analyze: We are asked to calculate K for a reaction, given ΔG°.
Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form
Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
215.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
Haber process:
The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.
Solution
Analyze: We are asked to calculate K for a reaction, given ΔG°.
Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form
In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG°
when using Equations 19.21, 19.22, or 19.23).
Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
216.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
Haber process:
The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.
Solution
Analyze: We are asked to calculate K for a reaction, given ΔG°.
Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form
In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG°
when using Equations 19.21, 19.22, or 19.23).
Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have
Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
217.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
Haber process:
The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.
Solution
Analyze: We are asked to calculate K for a reaction, given ΔG°.
Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form
In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG°
when using Equations 19.21, 19.22, or 19.23).
Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have
We insert this value into Equation 19.23 to obtain K:
Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
218.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
Haber process:
The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.
Solution
Analyze: We are asked to calculate K for a reaction, given ΔG°.
Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form
In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG°
when using Equations 19.21, 19.22, or 19.23).
Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have
We insert this value into Equation 19.23 to obtain K:
Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
Thermodynamics can tell us the direction and extent of a
reaction, but tells us nothing about the rate at which it occurs.
If a catalyst were found that would permit the reaction to proceed at a rapid rate at room temperature, high pressures would not be
needed to force the equilibrium toward NH3.
219.
PRACTICE EXERCISE
Use data from Appendix C to calculate
the standard free-energy change, ΔG°,
and the equilibrium constant, K, at 298
K for the reaction
Chemical
Thermodynamics
220.
PRACTICE EXERCISE
Use data from Appendix C to calculate
the standard free-energy change, ΔG°,
and the equilibrium constant, K, at 298
K for the reaction
Answer: ΔG° = –106.4 kJ/mol, K = 5 × 1018
Chemical
Thermodynamics
221.
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together
Consider the equilibria in which these salts dissolve in water to form aqueous
solutions of ions:
(a) Calculate the value of ΔG° at 298 K for each of the
preceding reactions.
(b) The two values from part (a) are very different. Is this
difference primarily due to the enthalpy term or the
entropy term of the standard free-energy change?
(c) Use the values of ΔG° to calculate the Ksp values for
the two salts at 298 K.
(d) Sodium chloride is considered a soluble salt, whereas
silver chloride is considered insoluble. Are these
descriptions consistent with the answers to part (c)?
(e) How will ΔG° for the solution process of these salts
change with increasing T? What effect should this
change have on the solubility of the salts?
222.
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together
Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to
form aqueous solutions of ions:
(a) Calculate the value of ΔG° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very
different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change?
(c) Use the values of ΔG° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a
soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part
(c)? (e) How will ΔG° for the solution process of these salts change with increasing T? What effect should this change
have on the solubility of the salts?
Solution (a) We will use Equation 19.13 along with values from Appendix C to calculate the
values for each equilibrium. (As we did in Section 13.1, we use the subscript “soln” to indicate that these are
thermodynamic quantities for the formation of a solution.) We find
Chemical
Thermodynamics
223.
SAMPLE INTEGRATIVE EXERCISE continued
(b) We can write as the sum of an enthalpy term, and an entropy term,
We can calculate the values of and by using Equations 5.31 and 19.8.
We can then calculate at T = 298 K. All these
calculations are now familiar to us. The results are summarized in the following table:
The entropy terms for the solution of the two salts are very similar. That seems sensible because each
solution process should lead to a similar increase in randomness as the salt dissolves, forming hydrated ions.
• (Section 13.1) In contrast, we see a very large difference in the enthalpy term for the solution of the two
salts. The difference in the values of is dominated by the difference in the values of
(c) The solubility product, Ksp, is the equilibrium constant for the solution process. • (Section 17.4) As
such, we can relate Ksp directly to by using Equation 19.23:
We can calculate the Ksp values in the same way we applied Equation 19.23 in Sample Exercise 19.13.
We use the values we obtained in part (a), remembering to convert them from to J/mol:
Chemical
Thermodynamics
The value calculated for the Ksp of AgCl is very close to that listed in Appendix D.
224.
SAMPLE INTEGRATIVE EXERCISE continued
(d) A soluble salt is one that dissolves appreciably in water. • (Section 4.2) The Ksp value for NaCl
is greater than 1, indicating that NaCl dissolves to a great extent. The Ksp value for AgCl is very small,
indicating that very little dissolves in water. Silver chloride should indeed be considered an insoluble
salt.
(e) As we expect, the solution process has a positive value of ΔS for both salts (see the table in part b).
As such, the entropy term of the free-energy change, is negative. If we assume that
and do not change much with temperature, then an increase in T will serve to make more
negative. Thus, the driving force for dissolution of the salts will increase with increasing T, and we therefore
expect the solubility of the salts to increase with increasing T. In Figure 13.17 we see that the solubility of
NaCl (and that of nearly any salt) increases with increasing temperature. • (Section 13.3)
Chemical
Thermodynamics
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