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Chapter 20
Thermodynamics:
Entropy, Free Energy and the
Direction of Chemical Reactions
20-1
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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
20-2
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First law of thermodynamics
 Law
of conservation of energy: Energy can
be neither created nor destroyed.
 ∆E= q + w (q=heat, w=work, E=internal
energy)
 E univ= E sys + E surr
 Heat gained by system is lost by
surroundings and vice-versa.
 Total energy of Universe is constant ∆E
sys + ∆E surr=0=∆E univ
20-3
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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, this does not tell us anything about the direction of
change in the universe.
20-4
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Spontaneous
 It
20-5
occurs without outside intervention
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Figure 20.1
A spontaneous endothermic chemical reaction.
water
Ba(OH)2.8H2O(s) + 2NH4NO3(s)
Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)
DH0rxn = +62.3 kJ
20-6
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Enthalpy
 H=E+PV,
 E=internal
energy,
 P=pressure,
 V=volume
20-7
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The Concept of Entropy (S)
Entropy refers to the state of order.
A change in order is a change in the number of ways of
arranging the particles, and it is a key factor in determining the
direction of a spontaneous process.
more order
solid
more order
crystal + liquid
more order
crystal + crystal
20-8
less order
liquid
gas
less order
ions in solution
less order
gases + ions in solution
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1877 Ludwig Boltzman
S = k ln W
where S is entropy, W is the number of ways of arranging the
components of a system, and k is a constant (the Boltzman constant),
R/NA (R = universal gas constant, NA = Avogadro’s number.
•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.
DSuniverse = DSsystem + DSsurroundings > 0
This is the second law of thermodynamics.
20-9
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Entropy:





S, A measure of molecular randomness or
disorder.
Thermodynamic function that describes number
of arrangements that are available to a system
existing in a given state.
Probability of occurrence of a particular
arrangement(state) depends on the number of
ways(microstates) in which it can be arranged.
S=k ln W, k= Boltzmann constant=R/Av
number=1.38 x 10-23J/K.
Un its of S are J/K
20-10
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Entropy
 More
the microstates available, higher is
the entropy. Ex: solid →liquid→gas
 Entropy is a state function depends only
on present state and not its path.
 ∆ Ssys= Sfinal – Sinitial , ∆ Ssys > 0, as
entropy increases,
 ∆ Ssys < 0, as entropy decreases.
20-11
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Entropy
 If
there are n number of molecules,
relative probability of finding all molecules
in one place is: (1/2)n
 ∆ Ssys= 5.76 J/K for 1 mol
 ∆ Ssys=qrev/T , T=temp, q=heat
absorbed, rev=reversible process
20-12
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Figure 20.2
Spontaneous expansion of a gas
stopcock
closed
1 atm
evacuated
stopcock
opened
0.5 atm
Spontaneous expansion of gas
20-13
0.5 atm
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Figure 20.3
Expansion of a gas and the increase in number of microstates.
20-14
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Entropy and Second Law of
Thermodynamics:





Second Law of Thermo: In any spontaneous
process there is always an increase in the
entropy of the universe. OR The entropy of the
universe is increasing.
∆Suniv= ∆Ssys + ∆Ssurr
If ∆Suniv= +ve, process is spontaneous in
direction written.
∆Suniv= -ve, process is spontaneous in the
opposite direction.
∆Suniv= 0, process has no tendency to
occur, system is at equilibrium
20-15
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Third Law of Thermodynamics
 the
entropy of a perfect crystal at 0 K is
zero
20-16
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Figure 20.4
Random motion in a crystal
The third law of
thermodynamics.
A perfect crystal has
zero entropy at a
temperature of
absolute zero.
Ssystem = 0 at 0 K
20-17
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Predicting Relative S0 Values of a System
1. Temperature changes
S0 increases as the temperature rises.
2. Physical states and phase changes
S0 increases as a more ordered phase changes to a less ordered
phase.
3. Dissolution of a solid or liquid
S0 of a dissolved solid or liquid is usually greater than the S0 of
the pure solute. However, the extent depends upon 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.
20-18
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Figure 20.5 The increase in entropy from solid to liquid to gas.
20-19
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Figure 20.6
The entropy change accompanying the dissolution of a salt.
pure solid
MIX
pure liquid
solution
20-20
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Figure 20.7
The small increase in entropy when ethanol dissolves in water.
Ethanol
20-21
Water
Solution of
ethanol
and water
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Figure 20.8
The large decrease in entropy when a gas dissolves in a liquid.
O2 gas
O2 gas in H2O
20-22
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Figure 20.9
Entropy and vibrational motion.
NO
NO2
20-23
N 2 O4
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Answer the following:

1. Will solid or gaseous CO2 have higher
positional entropy?
 2. Will N2 gas at 1 atm or 1.0 x 10-2 atm have
higher positional entropy?
 3. When solid sugar is dissolved in water to form
a solution, what will be the sign of entropy
change?
 4. When iodine vapors are condensed to form
crystals, what will be the sign of entropy
change?
20-24
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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) 1mol of SO2(g) or 1mol of SO3(g)
(b) 1mol of CO2(s) or 1mol of CO2(g)
(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)
(d) 1mol of KBr(s) or 1mol of KBr(aq)
(e) Seawater in midwinter at 20C or in midsummer at 230C
(f) 1mol of CF4(g) or 1mol of CCl4(g)
PLAN: In general less ordered systems have higher entropy than ordered
systems and entropy increases with an increase in temperature.
SOLUTION:
(a) 1mol of SO3(g) - more atoms
(d) 1mol of KBr(aq) - solution > solid
(b) 1mol of CO2(g) - gas > solid
(e) 230C - higher temperature
(c) 3mol of O2(g) - larger #mols
(f) CCl4 - larger mass
20-25
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Entropy Changes in the System
DS0rxn - the entropy change that occurs when all reactants
and products are in their standard states.
DS0rxn = DS0products - DS0reactants
The change in entropy of the surroundings is directly
related to an opposite change in the heat of the system
and inversely related to the temperature at which the
heat is transferred.
DSsurroundings = -
20-26
DHsystem
T
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Sample Problem 20.2:
PROBLEM:
Calculating the Standard Entropy of Reaction, DS0rxn
Calculate DS0rxn for the combustion of 1mol of propane at 250C.
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
PLAN: Use summation equations. It is obvious that entropy is being lost
because the reaction goes from 6 mols of gas to 3 mols of gas.
SOLUTION:
Find standard entropy values in the Appendix or other table.
DS = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)] - [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)]
DS = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) +
(5 mol)(205.0J/mol*K)]
DS = - 374 J/K
20-27
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Entropy changes in the
surrounding:





Exothermic change: Heat is lost by the system
and gained by the surroundings.
q sys< 0 , q surr >0, ∆ Ssurr > 0
Endothermic Change: Heat gained by the
system is lost by the surroundings.
q sys> 0 , q surr <0, ∆ Ssurr < 0
The change is entropy of the surroundings is α
opposite charge in the heat of a system 1/α
temperature at which heat is transferred.
20-28
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Entropy changes in the surrounding
∆Ssurr = -q sys/T
∆Ssurr = -∆H sys/T
Sign of ∆Ssurr depends on the direction of heat
flow. When an exothermic process occurs
∆Ssurr is + ve. Opposite is true for endothermic
process. Nature tends to seek the lowest
possible energy.
 Magnitude of ∆Ssurr depends on the
temperature. Transfer of heat at low temperature
produces a greater percent change in the
randomness of the surroundings.



20-29
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Sample Problem 20.3:
PROBLEM:
Determining Reaction Spontaneity
At 298K, the formation of ammonia has a negative DS0sys;
N2(g) + 3H2(g)
2NH3(g)
DS0sys = -197 J/K
Calculate DS0rxn, and state whether the reaction occurs
spontaneously at this temperature.
PLAN: DS0universe must be > 0 in order for this reaction to be spontaneous, so
DS0surroundings must be > 197 J/K. To find DS 0surr, first find DHsys; DHsys =
DHrxn which can be calculated using DH0f values from tables. DS0universe
= DS0surr + DS0sys.
SOLUTION:
DH0rx = [(2 mol)(DH0fNH3)] - [(1 mol)(DH0fN2) + (3 mol)(DH0fH2)]
DH0rx = -91.8 kJ
DS0surr = -DH0sys/T = -(-91.8x103J/298K) = 308 J/K
DS0universe = DS0surr + DS0sys = 308 J/K + (-197 J/K) = 111 J/K
DS0universe > 0 so the reaction is spontaneous.
20-30
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Problem
 P.3.
Sb2 S3(s) + 3Fe(s)______ 2Sb (s) +
3FeS (s) ∆H=-125kJ

Sb4 O6(s) + 6C(s)______ 4Sb (s) + 6
CO (s) ∆H= 778 kJ

Calculate ∆Ssurr for both of these
reactions at 25 C and 1 atm
20-31
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Figure B20.1
Lavoisier studying human respiration as a form of combustion.
Figure B20.2
A whole-body calorimeter.
20-32
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Figure 20.10
Components of DS0universe for spontaneous reactions
exothermic
endothermic
system becomes more disordered
exothermic
system becomes more disordered
∆Ssurr must be smaller
system becomes more ordered
∆Ssurr must be larger
20-33
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Gibbs Free Energy (G)
DG, the change in the free energy of a system, is a
measure of the spontaneity of the process and of the
useful energy available from it.
DG0system = DH0system - TDS0system
DG < 0 for a spontaneous process
DG > 0 for a nonspontaneous process
DG = 0 for a process at equilibrium
DG0rxn =  mDG0products -  nDG0reactants
20-34
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Free Energy






G= H-TS, H= enthalpy, T=temp in K, S= entropy.
Also called as Gibbs Free Energy.
Free energy change (∆G) is a measure of the
spontaneity of a process and of the useful
energy available from it.
∆ G= ∆H-T∆S(constant temp)
∆Suniv = -∆G/T
Process is spontaneous in the direction in which
the free energy decreases. (constant T and P)
20-35
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Free Energy
 ∆S
univ =0, at equilibrium
 ∆S univ > 0, for a spontaneous process.
 ∆S univ < 0, for a nonspontaneous
process.
 ∆G=0, at equilibrium
 ∆G > 0, for a nonspontaneous process.
 ∆G< 0, for a spontaneous process.
20-36
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Standard free energy of
formation
∆Gf0 is defined as a change in free energy that
accompanies the formation of 1 mole of that
substance from its constituent elements with all
reactants and products in their standard states.
 Standard free energy of formation of an element
in its standard state is zero.
 An equation coefficient x ∆Gf0 by the number.
 Reversing a reaction changes ∆Gf0

20-37
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Problem
 P.4.
For the equation 2CH3OH (g) + 3
O2(g) _____ 2 CO 2(g) + 4 H2O (g)
 Calculate ∆G0
-163
0
-394
-229
∆Gf0
kJ/mol
20-38
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Sample Problem 20.6:
PROBLEM:
Determining the Effect of Temperature on DG0
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 298K, DG0 = -141.6kJ; DH0 = -198.4kJ; and DS0 = -187.9J/K
(a) Use the data to decide if this reaction is spontaneous at 25 0C, and predict
how DG0 will change with increasing T.
(b) Assuming DH0 and DS0 are constant with increasing T, is the reaction
spontaneous at 900.0C?
PLAN:
The sign of DG 0 tells us whether the reaction is spontaneous
and the signs of DH0 and DS0 will be indicative of the T effect.
Use the Gibbs free energy equation for part (b).
SOLUTION: (a) The reaction is spontaneous at 250C because DG0 is (-).
Since DH0 is (-) but DS0 is also (-), DG0 will become less
spontaneous as the temperature increases.
20-39
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Sample Problem 20.6:
Determining the Effect of Temperature on DG0
continued
(b) DG0rxn = DH0rxn - T DS0rxn
DG0rxn = -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/103J)
DG0rxn = 22.0 kJ; the reaction will be nonspontaneous at 900.0C
20-40
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Problems
P.6. Use ∆Gf0 values to calculate free energy change at 25
C for the following:
2NO (g) + O2(g) → 2NO2 (g)
P.7. For the equation 2CH3OH (g) + 3 O2(g) _____ 2 CO
2(g) + 4 H2O (g)
Calculate ∆G0
-163
0
394
-229
∆Gf0 kJ/mol
P.8. Is the reaction C2H4 (g) + H2O (l) _______ C2H5OH
(l) spontaneous under standard conditions?
20-41
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DG and the Work a System Can Do
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 maximum work that must be
done to the system as the process takes place: DG = workmax
An example
20-42
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∆G and work a system can do:
∆G < 0, for a spontaneous process., ∆G=-wmax,
maximum useful work obtainable from the system.
 ∆G > 0, for a nonspontaneous process., minimum work
that must be done to the system to make the process
take place.
 In any real process work is performed irreversibly and
some of the free energy is lost to the surroundings as
heat.
 You can never obtain maximum possible work.
 You need more than minimum energy to start the
process for a nonspontaneous type.
 At equilibrium a reaction can no longer do any work.

20-43
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Effect of Temperature on
Reaction spontaneity
1.Reaction is spontaneous at all temperatures:
∆H < 0, ∆S >0, -T ∆S= -ve, ∆G =-ve
2. Reaction is nonspontaneous at all temperatures:
∆H > 0, ∆S <0, -T ∆S= +ve,∆G =+ve
3. Reaction is spontaneous at higher temperatures:
∆H > 0, ∆S >0, -T ∆S favors spontaneity at high temp , ∆G
=-ve and reaction is spontaneous.
4. Reaction is spontaneous at lower temperatures:
∆H < 0, ∆S <0, ∆H favors spontaneity at low temp ,
Reaction occurs if -T ∆S< ∆H
20-44
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Table 20.1 Reaction Spontaneity and the Signs of DH0, DS0, and DG0
DH0
DS0
-TDS0
DG0
-
+
-
-
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
20-45
Description
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Figure 20.11
The effect of temperature on reaction spontaneity.
Reaction between Cu2O and C.
Relatively const ∆H and steadily increasing T∆S
20-46
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Free Energy, Equilibrium and Reaction Direction
•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)
DG = RT ln Q/K = RT lnQ - RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1
and ln Q = 0 so
DG0 = - RT lnK
20-47
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Table 20.2 The Relationship Between DG0 and K at 250C
DG0(kJ)
100
3x10-18
50
2x10-9
10
2x10-2
1
7x10-1
0
1
1.5
-10
5x101
-50
6x108
-100
3x1017
-200
1x1035
Essentially no forward reaction;
reverse reaction goes to completion
Forward and reverse reactions
proceed to same extent
Forward reaction goes to
completion; essentially no reverse
reaction
REVERSE REACTION
9x10-36
Significance
FORWARD REACTION
200
-1
20-48
K
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Free Energy and Equilibrium:
 DG
= DG + RT ln(Q)
 Q = reaction quotient from the law of mass
action.
 Equilibrium point occurs at the lowest
value of free energy available to the
reaction system.
 DG = RT ln(K)
20-49
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Sample Problem 20.7:
Calculating DG at Nonstandard Conditions
PROBLEM: The oxidation of SO2, which we considered in Sample Problem 20.6
2SO2(g) + O2(g)
2SO3(g)
is too slow at 298K 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 298K and at 973K. (DG0298 = -141.6kJ/mol of reaction as
written using DH0 and DS0 values at 973K. DG0973 = -12.12kJ/mol of reaction
as written.)
(b) In experiments to determine the effect of temperature on reaction
spontaneity, two sealed containers are filled with 0.500atm of SO2,
0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. 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 slide
20-50
.
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Sample Problem 20.7:
Calculating DG at Nonstandard Conditions
continued (2 of 3)
SOLUTION: (a) Calculating K at the two temperatures:
DG0
(DG / RT)
= -RTlnK so K  e
0
(-141.6kJ/mol)(103J/kJ)
At 298, the exponent is -DG0/RT = -
= 57.2
(8.314J/mol*K)(298K)
K  e(DG
0
/ RT)
= e57.2 = 7x1024
(-12.12kJ/mol)(103J/kJ)
At 973, the exponent is -DG0/RT
= 1.50
(8.314J/mol*K)(973K)
K  e(DG
20-51
0
/ RT)
= e1.50 = 4.5
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Calculating DG at Nonstandard Conditions
Sample Problem 20.7:
continued (3 of 3)
pSO32
(b) The value of Q =
(pSO2)2(pO2)
(0.100)2
=
= 4.00
(0.500)2(0.0100)
Since Q is < K at both temperatures the reaction will shift right; for 298K
there will be a dramatic shift while at 973K the shift will be slight.
(c) The nonstandard DG is calculated using DG = DG0 + RTlnQ
DG298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(ln4.00)
DG298 = -138.2kJ/mol
DG973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(ln4.00)
DG298 = -0.9kJ/mol
20-52
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Figure 20.12
The relation between free energy and the extent of reaction.
DG0 < 0
K >1
20-53
DG0 > 0
K <1