Chapter 20
Thermodynamics
Entropy, Free Energy and the Direction
of Chemical Reactions
20-1
Thermodynamics
Entropy, Free Energy, and the Direction of Chemical
Reactions
20.1 The Second Law of Thermodynamics: Predicting
spontaneous change
20.2 Calculating the change in entropy of a reaction
20.3 Entropy, free energy and work
20.4 Free energy, equilibrium and reaction direction (we will
consider later in the semester)
20-2
The Larger Question
A
B
What factors must be identified in order to predict
whether the reaction will proceed spontaneously
in the direction written (as opposed to in the opposite
direction)?
Enthalpy and entropy changes are combined to yield a
new term, the Gibbs free energy. The sign of the latter
allows predictions of reaction spontaneity.
These are thermodynamic, not kinetic, considerations!
20-3
Limitations of the First Law of Thermodynamics
DE = q + w
Euniverse = Esystem + Esurroundings
DEsystem = -DEsurroundings
DEsystem + DEsurroundings = 0 = DEuniverse
The total energy-mass of the universe is
constant.
However, these energy changes do not explain the
direction of change in the universe!
20-4
A spontaneous endothermic chemical reaction
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water
Ba(OH)2.8H2O(s) + 2NH4NO3(s)
Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)
DHorxn = + 62.3 kJ
Figure 20.1
20-5
The sign of the enthalpy change is insufficient to predict
reaction spontaneity!
The Missing Factor: Entropy (S)
Entropy refers to the state of order/disorder of a system.
A change in order is a change in the number of ways of
arranging the particles. It is a key factor in determining the
direction of a spontaneous process.
more order
solid
more order
crystal + liquid
more order
crystal + crystal
20-6
less order
liquid
gas
less order
ions in solution
less order
gases + ions in solution
The number of ways to arrange a deck of playing cards
Figure 20.2
20-7
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Spontaneous expansion of a gas
This process occurs
spontaneously
without a change in
the total internal
energy of the
system!
stopcock
closed
evacuated
1 atm
This is an entropydriven process!
stopcock
opened
Figure 20.3
20-8
0.5 atm
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0.5 atm
S = k ln W
Ludwig Boltzmann, 1877
S = entropy, W = the number of ways of arranging the components of
a system having equivalent energy, and k = the Boltzmann constant =
R/NA (R = universal gas constant, NA = Avogadro’s number) = 1.38 x
10-23 J/K.
A system with relatively few equivalent ways to arrange its
components (smaller W) has relatively less disorder and low entropy.
A system with many equivalent ways to arrange its components
(larger W) has relatively more disorder and high entropy.
The Second Law of Thermodynamics
DSuniv = DSsys + DSsurr > 0
Entropy is a state function.
20-9
All processes occur spontaneously
in the direction that increases
the entropy of the universe
(system + surroundings).
20-10
Random motion in a crystal
(increased arrangements, greater S)
The Third Law of
Thermodynamics
A perfect crystal has
zero entropy at a
temperature of
absolute zero.
Ssystem = 0 at 0 K
Figure 20.4
20-11
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The Third Law leads to absolute values for the entropies of substances!
Standard Molar Entropies (So)
Standard states: 1 atm for gases; 1 M for solutions; pure substance
in its most stable form for solids and liquids.
units: J/mol/K at 25 oC
This value equals the entropy increase in a substance
upon raising its temperature from 0 K to the specified temperature.
20-12
Predicting relative So values of a system
1. Temperature changes
So increases as temperature rises.
2. Physical states and phase changes
So increases as a more ordered phase changes to a less
ordered phase.
3. Dissolution of a solid or liquid
So of a dissolved solid or liquid is usually greater than So of the
pure solute. However, the extent depends on the nature of the
solute and solvent.
4. Dissolution of a gas
A gas becomes more ordered when it dissolves in a liquid or solid.
5. Atomic size or molecular complexity
In similar substances, increases in mass relate directly to entropy.
In allotropic substances, increases in complexity (e.g., bond
flexibility) relate directly to entropy.
20-13
The increase in entropy from solid to liquid to gas
Large changes in S occur
at phase transitions!
Figure 20.5
20-14
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The entropy change accompanying the dissolution of a salt
pure solid
MIX
pure liquid
solution
salt becomes more disordered (dissociation): S increases
water becomes more ordered (ion-dipole interactions): S decreases
Figure 20.6
20-15
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The small increase in entropy when ethanol dissolves in water
ethanol
water
Freedom of movement remains essentially unchanged;
S increase is due solely to random mixing effects.
Figure 20.7
20-16
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solution of
ethanol and
water
The large decrease in entropy when a gas dissolves in a liquid
O2 gas
O2 dissolved
Figure 20.8
20-17
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Entropy and vibrational motion
more motions, greater S
NO
NO2
N 2O 4
Internal molecular motions influence entropy!
Figure 20.9
20-18
Sample Problem 20.1
Predicting relative entropy values
PROBLEM: Choose the member with the higher entropy in each of the following
pairs, and justify your choice. Assume constant temperature,
except in part (e).
(a) 1 mol of SO2(g) or 1 mol of SO3(g)
(b) 1 mol of CO2(s) or 1 mol of CO2(g)
(c) 3 mol of oxygen gas (O2) or 2 mol of ozone gas (O3)
(d) 1 mol of KBr(s) or 1 mol of KBr(aq)
(e) seawater in mid-winter at 2 oC or in mid-summer at 23 oC
(f) 1 mol of CF4(g) or 1 mol of CCl4(g)
PLAN: In general less ordered systems have higher entropy than ordered
systems and entropy increases with increasing temperature.
SOLUTION:
(a) 1 mol of SO3(g) - more atoms
(d) 1 mol of KBr(aq) - solution > solid
(b) 1 mol of CO2(g) - gas > solid
(e) 23 oC - higher temperature
(c) 3 mol of O2(g) - larger # mols
(f) CCl4 - larger mass
20-19
Standard Entropies of Reaction, DSorxn
By analogy to calculating standard heats of reaction: DHorxn
DHorxn = S mDHof (products) - S nDHof
(reactants)
From Chapter 6
DSorxn = S mSo (products) - S nSo (reactants)
As before, m and n are the appropriate coefficients
in the balanced chemical equation.
20-20
Sample Problem 20.2
PROBLEM:
Calculating the standard entropy of reaction, DSorxn
Calculate DSorxn for the combustion of 1 mol of propane at 25 oC.
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
PLAN: Use summation equations. Entropy is expected to decrease
because the reaction goes from 6 moles of gas to 3 moles of gas.
SOLUTION:
Find standard entropy values in an appropriate data table.
DSorxn = [(3 mol)(So CO2) + (4 mol)(So H2O)] - [(1 mol)(So C3H8) + (5 mol)(So O2)]
DSorxn = [(3 mol)(213.7 J/mol.K) + (4 mol)(69.9 J/mol.K)] - [(1 mol)(269.9
J/mol.K) + (5 mol)(205.0 J/mol.K)]
DSorxn = - 374 J/K
20-21
We can’t forget entropy changes in the surroundings!
For spontaneous reactions in which a decrease in the entropy of
the system occurs: must be outweighed by a concomitant increase
in the entropy of the surroundings (DSuniv > 0)
For an exothermic process: qsys < 0, qsurr > 0, DSsurr > 0
For an endothermic process: qsys > 0, qsurr < 0, DSsurr < 0
The initial temperature of the surroundings affects the
magnitude of DSsurr.
DSsurr a -qsys
20-22
and
DSsurr a 1/T
Combining these ideas......
DSsurr = -qsys/T
For a process of constant pressure, where qp = DH:
DSsurr = -DHsys/T
Implication: DSsurr can be calculated by measuring
DHsys and the temperature at which the change occurs.
20-23
Sample Problem 20.3
PROBLEM:
Determining reaction spontaneity
At 298 K, the formation of ammonia has a negative DSosys.
N2(g) + 3H2(g)
2NH3(g)
DSosys = -197 J/K
Calculate DSouniv and state whether the reaction
occurs spontaneously at this temperature.
PLAN: DSouniv must be > 0 in order for the reaction to be spontaneous, so
DSosurr must be > +197 J/K. To find DS osurr, first find DHsys. DHosys =
DHorxn which can be calculated using DHof values from data tables.
Then apply DSouniv = DSosurr + DSosys.
SOLUTION:
DHorxn = [(2 mol)(DHof NH3)] - [(1 mol)(DHof N2) + (3 mol)(DHof H2)]
DHorxn = -91.8 kJ
DSosurr = -DHosys/T = -(-91.8 x 103J/298 K) = +308 J/K (>+197 J/K)
DSouniv = DSosurr + DSosys = 308 J/K + (-197 J/K) = +111 J/K
DSouniv > 0: the reaction is spontaneous!
20-24
A whole-body calorimeter
Living systems do not violate the Second Law
DSuniv = DSsys + DSsurr > 0
Figure B20.2
20-25
What happens when equilibrium is reached?
DSouniv = DSosurr + DSosys = 0
Examples: phase changes (fusion, vaporization)
When a system reaches equilibrium, neither the forward
nor the reverse reaction is spontaneous; neither proceeds further
because there is no driving force.
20-26
Components of DSouniv for spontaneous reactions
exothermic
endothermic
system becomes more disordered
exothermic
system becomes more disordered
system becomes more ordered
Figure 20.10
20-27
But measuring DSsys and DSsurr is inconvenient
to determine reaction spontaneity
More convenient to use parameters that apply only to the system!
The Gibbs free energy (G) accomplishes this goal, where
G = H - TS or DG = DH - TDS
G combines the system’s enthalpy and entropy; thus
G is a state function.
Where does the equation come from?
20-28
Deriving the Gibbs free energy equation
DSuniv = DSsys + DSsurr
DSsurr = -DHsys/T
DSuniv = DSsys - DHsys/T
-TDSuniv = DHsys - TDSsys
DGsys = DHsys - TDSsys
DGsys = -TDSuniv
DSuniv > 0 or DGsys < 0 for a spontaneous process
DSuniv < 0 or DGsys > 0 for a nonspontaneous process
DSuniv = 0 or DGsys = 0 for a process at equilibrium
20-29
Some Key Concepts
If a process is nonspontaneous in one direction (DG > 0), then it is
spontaneous in the opposite direction (DG < 0).
The magnitude and sign of DG are unrelated to the rate
(speed) of the reaction.
20-30
Standard Free Energy Changes
A reference state; similar to DHo and So
All components are in their standards states.
DGosys = DHosys - TDSosys
20-31
Sample Problem 20.4
PROBLEM:
Calculating DGo from enthalpy and entropy values
Potassium chlorate, one of the common oxidizing agents in
explosives, fireworks and match heads, undergoes a solid-state
redox reaction when heated. In this reaction, the oxidation
number of Cl in the reactant is higher in one of the products and
lower in the other (a disproportionation reaction).
+5
+7
-1
D
4KClO3(s)
3KClO4(s) + KCl(s)
Use DHof and So values to calculate DGosys (DGorxn) at 25 oC for this reaction.
PLAN:
Obtain appropriate thermodynamic data from a data table; insert
them into the Gibbs free energy equation and solve.
SOLUTION:
DHorxn = S mDHof (roducts) - S nDHof (reactants)
DHorxn = [(3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol)] [(4 mol)(-397.7 kJ/mol)]
DHorxn = -144 kJ
20-32
Sample Problem 20.4
(continued)
DSorxn = S mSoproducts - S nSoreactants
DSorxn = [(3 mol)(151 J/mol.K) + (1 mol)(82.6 J/mol.K)] [(4 mol)(143.1 J/mol.K)]
DSorxn = -36.8 J/K
DGorxn = DHorxn - TDSorxn
DGorxn = -144 kJ - (298 K)(-36.8 J/K)(kJ/103J)
DG0rxn = -133 kJ
The reaction is spontaneous.
20-33
Another way to calculate DGorxn
From standard free energies of formation, DGof
DGof = the free energy change that occurs when 1 mol of compound
is made from its elements, with all components in their standard states.
Thus,
DGorxn = S mDGof (products) - S nDGof (reactants)
DGof values have properties similar to DHof values.
20-34
Sample Problem 20.5
PROBLEM:
Calculating DGorxn from DGof values
Use DGof values to calculate DGorxn for the following reaction:
4KClO3(s)
PLAN:
D
3KClO4(s) + KCl(s)
Use the DGorxn summation equation.
SOLUTION:
DGorxn = S mDGof (products) - S nDGof (reactants)
DGorxn = [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] [(4 mol)(-296.3 kJ/mol)]
DGorxn = -134 kJ
20-35
DG and Work
For a spontaneous process, DG is the maximum work obtainable from
the system as the process takes place: DG = workmax
For a nonspontaneous process, DG is the minimum work that must be
done to the system to make the process take place.
An example of a
gas doing work
20-36
More Key Concepts
A reversible process: one that can be changed in either direction by an
infinitesimal change in a variable. The maximum work from a spontaneous
process is obtained only if the work is carried out reversibly.
In most real processes, work is performed irreversibly; thus,
maximum work is not obtained; some free energy is lost to
the surroundings as heat and is thus unavailable to do work.
A reaction at equilibrium cannot do work.
20-37
Effect of Temperature on Reaction Spontaneity
DGosys = DHosys - TDSosys
The sign of DGosys is T-independent when DHosys and
DSosys have opposite signs.
The sign of DGosys is T-dependent when DHosys and
DSosys have the same signs.
20-38
Table 20.1
Reaction Spontaneity and the Signs of
DHo, DSo and DGo
DHo
DSo
-TDSo
DGo
-
+
-
-
spontaneous at all T
+
-
+
+
nonspontaneous at all T
+
+
-
+ or -
spontaneous at higher T;
nonspontaneous at lower T
-
-
+
+ or -
spontaneous at lower T;
nonspontaneous at higher T
20-39
description
Sample Problem 20.6
PROBLEM:
Determining the effect of temperature on DGo
An important reaction in the production of sulfuric acid is the
oxidation of SO2(g) to SO3(g):
2SO2(g) + O2(g)
2SO3(g)
At 298 K, DGo = -141.6 kJ, DHo = -198.4 kJ, and DSo = -187.9 J/K.
(a) Use these data to decide if this reaction is spontaneous at 25 oC, and
predict how DGo will change with increasing T.
(b) Assuming DHo and DSo are constant with increasing T, is the reaction
spontaneous at 900. oC?
PLAN:
The sign of DG o tells us whether the reaction is spontaneous
and the signs of DHo and DSo will be indicative of the T effect.
Use the Gibbs free energy equation for part (b).
SOLUTION: (a) The reaction is spontaneous at 25 oC because DGo is (-).
Since DHo is (-) and DSo is (-), DGo will become less negative and
the reaction less spontaneous as temperature increases.
20-40
Sample Problem 20.6
(continued)
(b) DGorxn = DHorxn - TDSorxn
DGorxn = -198.4 kJ - [(1173 K)(-187.9 J/mol.K)(kJ/103J)]
DGorxn = +22.0 kJ; the reaction will be nonspontaneous at 900.oC
20-41
Determining the Temperature at which a
Reaction Becomes Spontaneous
(only when DH and DS have the same sign)
DGo = DHo - TDSo = 0
(solve for this condition)
DHo = TDSo
T = DHo/DSo
20-42
The effect of temperature on reaction spontaneity
Cu2O(s) + C(s)
2Cu(s) + CO(g)
DHo = +58.1 kJ
DSo = +165 J/K
DHo: relatively
insensitive to T
DSo: increases
with increasing T
Figure 20.11
20-43
The cycling of metabolic free energy through ATP
Figure B20.4
20-44
The coupling of a nonspontaneous reaction to
the hydrolysis of ATP
D-glucose
Figure B20.3
20-45
+ ATP
D-glucose
6P + ADP
(catalyzed by the enzyme, hexokinase)
Why is ATP a high-energy molecule?
The DGo of ATP hydrolysis is -31 kJ/mol;
considerably more (-) under in vivo conditions!
Figure B20.5
20-46
End of Assigned Material
20-47
Free energy, equilibrium and reaction direction
DG = RT ln Q/K = RT ln Q - RT ln K
If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (DG < 0)
If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (DG > 0)
If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (DG = 0)
Under standard conditions (1 M concentrations, 1 atm for gases), Q =
1 and ln Q = 0. Therefore:
DGo = - RT ln K
20-48
Table 20.2
The relationship between DGo and K at 25 oC
DGo (kJ)
100
3 x 10-18
50
2 x 10-9
10
2 x 10-2
1
7 x 10-1
0
1
Essentially no forward reaction; reverse
reaction goes to completion
Forward and reverse reactions
proceed to the same extent
1.5
-10
5 x 101
-50
6 x 108
-100
3 x 1017
-200
1 x 1035
Forward reaction goes to
completion; essentially no reverse
reaction
REVERSE REACTION
9 x 10-36
significance
FORWARD REACTION
200
-1
20-49
K
Sample Problem 20.7
PROBLEM:
Calculating DG at non-standard conditions
The oxidation of SO2,
2SO2(g) + O2(g)
2SO3(g)
is too slow at 298 K to be useful in the manufacture of sulfuric acid. To
overcome this low rate, the process is conducted at an elevated temperature.
(a) Calculate K at 298 K and at 973 K. (DGo298 = -141.6 kJ/mol for the
reaction as written using DHo and DSo values. At 973 K, DGo973 =
12.12 kJ/mol for the reaction as written.)
(b) In experiments to determine the effect of temperature on reaction
spontaneity, two sealed containers are filled with 0.500 atm of SO2,
0.0100 atm of O2, and 0.100 atm of SO3 and kept at 25 oC and at
700. oC. In which direction, if any, will the reaction proceed to reach
equilibrium at each temperature?
(c) Calculate DG for the system in part (b) at each temperature.
PLAN: Use the equations and conditions found on previous slides.
20-50
Sample Problem 20.7
(continued)
SOLUTION: (a) Calculating K at the two temperatures:
DGo
(DG / RT)
= -RTln K so K  e
At 298 K, the exponent is -DGo/RT
K  e(DG
0
/ RT)
20-51
0
/ RT)
(-141.6 kJ/mol)(103 J/kJ)
=-
(8.314 J/mol.K)(298 K)
= 57.2
= e57.2 = 7 x 1024
At 973 K, the exponent is -DGo/RT = -
K  e(DG
0
= e1.50 = 4.5
(-12.12 kJ/mol)(103 J/kJ)
(8.314 J/mol.K)(973 K)
= 1.50
Sample Problem 20.7
(continued)
(b) The value of Q =
pSO32
(pSO2)2(pO2)
=
(0.100)2
(0.500)2(0.0100)
= 4.00
Since Q is < K at both temperatures the reaction will shift right; for 298 K
there will be a dramatic shift while at 973 K the shift will be slight.
(c) The non-standard DG is calculated using DG = DGo + RTlnQ
DG298 = -141.6 kJ/mol + (8.314 J/mol.K)(kJ/103J)(298 K)(ln 4.00)
DG298 = -138.2 kJ/mol
DG973 = -12.12 kJ/mol + (8.314 J/mol.K)(kJ/103J)(973 K)(ln 4.00)
DG298 = -0.9 kJ/mol
20-52
The relation between free energy and the extent of reaction
DGo < 0
K >1
DGo > 0
K <1
Figure 20.12
20-53
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