Atkins & de Paula:
Elements of Physical Chemistry: 5e
Chapter 7:
Chemical Equilibrium: The Principles
End of chapter 7 assignments
Discussion questions:
• 2
Exercises:
• 1, 2, 3, 4, 5, 6, 7, 10, 11, 15, 18
Use Excel if data needs to be graphed
Homework Assignment
• How many of you have already read
all of chapter 7 in the textbook?
• In the future, read the entire chapter
in the textbook before we begin
discussing it in class
Homework Assignment
• Connect to the publisher’s website and
access all “Living Graphs”
• http://bcs.whfreeman.com/elements4e/
• Change the parameters and observe the
effects on the graph
• Sarah: these “Living Graphs” are not
really living; this is just a hokey name!
Homework Assignments
• Read Chapter 7.
• Work through all of the “Illustration”
boxes and the “Example” boxes and
the “Self-test” boxes in Chapter 7.
• Work the assigned end-of-chapter
exercises by the due date
Principles of chemical equilibrium
Central Concepts:
• Thermodynamics can predict whether
a rxn has a tendency to form products,
but it says nothing about the rate
• At constant T and P, a rxn mixture
tends to adjust its composition until its
Gibbs energy is at a minimum
Gibbs Energy vs Progress of Rxn
• Fig 7.1 (158)
• (a) does not go
• (b) equilibrium with
amount of reactants
~amount of products
• (c) goes to completion
Example Rxns
• G6P(aq)

F6P(aq)
• N2(g) + 3 H2(g)  2 NH3(g)
• Reactions are of this form: aA + bB  cC + dD
• If n is small enough, then,
G = (F6P x n) – (G6P x n) --now divide by
n
• rG = G/n = F6P – G6P
The Rxn Gibbs Energy
• rG = G/n = F6P – G6P
• rG is the difference of the chemical
potentials of the products and reactants at the
composition of the rxn mixture
• We recognize that rG is the slope of the
graph of the (changing) G vs composition of
the system (Fig 7.1, p154)
Effect of composition on rG
• Fig 7.2 (154)
• The relationship of
G to composition of
the reactions
• rG changes as n
(the composition)
changes
Reaction Gibbs energy
• Consider this reaction:
aA + bB  cC + dD
• rG = (cC + dD) – (aA + bB)
μJtμJ+ RT ln aJ 
(derived in sec 6.6)
• Chemical potential (μ) changes as [J] changes
• The criterion for chemical equilibrium at
constant T,P is:
 rG = 0
(7.2)
Meaning of the value of rG
•
•
•
•
•
Fig 7.3 (155)
When is rG<0?
When is rG=0?
When is rG>0?
What is the significance of each?
Variation of rG with composition
For solutes in an ideal solution:
• aJ = [J]/c, the molar concentration of J
relative to the standard value c = 1 mol/dm3
For perfect gases:
• aJ = pJ/p, the partial pressure of J relative to
the standard pressure p = 1 bar
For pure solids and liquids, aJ = 1
p155f
Variation of rG with composition
• rG = (cC + dD) – (aA + bB)
(7.1c)




• rG = (cC + dD) – (aA + bB)
(7.4a)




• rG = {cGm(C)+dGm(D)} – {(aGm(A)+bGm(B)}
(7.4b)
• 7.4a and 7.4b are the same
Variation of rG with composition
rG = rG +
RT ln
Since
Q=
aCc aDd
( )
a
b
aA aB
aCc aDd
a
b
aA aB
Then
rG = rG + RT ln Q
Reactions at equilibrium
• Again, consider this reaction:
aA + bB  cC + dD
c d
Q=
aC a D
a b
aA aB
c
K=
d
aC a D
( )
a
b
aA aB
• Q, arbitrary position; K, equilibrium
• 0 = rG + RT ln K and
• rG = –RT ln K
equilibrium
Equilibrium constant
• With these equations….
0 = rG + RT ln K
rG = –RT ln K
(7.8)
• We can use values of rG from a
data table to predict the equilibrium
constant
• We can measure K of a reaction and
calculate rG
Relationship between rG and K
• Fig 7.4 (157)
• Remember,
• rG = –RT ln K
• So, ln K = –
(rG/RT)
• If rG<0, then K>1;
& products
predominate at
equilibrium
• And the rxn is thermodynamically feasible
At K > 1, rG < 0
At K = 1, rG = 0
At K < 1, rG > 0
Relationship between rG and K
• On the other hand…
• If rG>0, then K<1 and the
reactants predominate at
equilibrium…
• And the reaction is not thermodynamically feasible HOWEVER….
• Products will predominate over
reactants significantly if K1 (>103)
• But even with a K<1 you may have
products formed in some rxns
Relationship between rG and K
• For an endothermic rxn to have rG<0, its
rS>0; furthermore,
• Its temperature must be high enough for its
TrS to be greater than rH
• The switch from rG>0 to rG<0
corresponds to the switch from K<1 to K>1
• This switch takes place at a temperature at
which rH - TrS = 0, OR….
rH
T=
rS
Table 7.1 Thermodynamic criteria of spontaneity
G = H – T S
1 If the reaction is exothermic (rHo 0) and rSo  0
rGo  0 and K  1 at all temperatures
2 If the reaction is exothermic (rHo  0) and rSo  0
rGo  0 and K  1 provided that T  rHo/rSo
3 If the reaction is endothermic (rHo  0) and rSo  0
rGo  0 and K  1 provided that T  rHo/rSo
4 If the reaction is endothermic (rHo  0) and rSo  0
rGo  0 and K  1 at no temperature
2 If the reaction is exothermic (rH  0) and rS  0
o Thermodynamic criteria of spontaneity
Table7.1
0 and K  1 provided that T  rHo/rSo
rG 
3 If the reaction is endothermic (rHo  0) and rSo  0
 H – T S
 Go  0 and KG
 1=provided
that T   Ho/ So
r
r
r
4 If the reaction is endothermic (rHo  0) and rSo  0
rGo  0 and K  1 at no temperature
4. If H is positive and S is negative, G will always
be positive—regardless of the temperature.
These two statements are an attempt to say the
same thing.
G = H – TS
1. If H is negative and S is positive, then G will
always be negative regardless of temperature.
2. If H is negative and S is negative, then G will be
negative only when TS is smaller in magnitude than
H. This condition is met when T is small.
3. If both H and S are positive, then G will be
negative only when the TS term is larger than H.
This occurs only when T is large.
4. If H is positive and S is negative, G will always be
positive—regardless of the temperature.
G = H – TS
Factors Affecting the Sign of G
Gibbs Free Energy (G)
G = H – TS
All quantities in the above equation refer to the system
For a constant-temperature process:
The change in Gibbs free energy (G)
G = Hsys – TSsys
If G is negative (G<0),
there is a release of usable energy,
and the reaction is spontaneous!
If G is positive (G>0), the reaction is not spontaneous!
18.4
Gibbs Free Energy (G)
For a constant-temperature process:
G = Hsys – TSsys
G < 0
The reaction is spontaneous in the forward direction.
G > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
G = 0
The reaction is at equilibrium.
18.4
Gibbs Free Energy (G)
The standard free-energy of reaction (G0rxn )
is the free-energy change for a reaction
when it occurs under standard-state conditions.
aA + bB
cC + dD
0
Grxn
= [cG0f (C) + dG0f (D) ] – [aG0f (A) + bG0f (B) ]
0
Grxn
= S nG0f (products) – S mG0f (reactants)
7.10
p158
Who will explain this graph to the class?
Relationship between rG and K
• For an endothermic rxn to have rG<0, its
rS>0; furthermore,
• Its temperature must be high enough for its
TrS to be greater than rH
• The switch from rG>0 to rG<0
corresponds to the switch from K<1 to K>1
• This switch takes place at a temperature at
which rH - TrS = 0, OR….
rH
T=
rS
Reactions at equilibrium
• Fig 7.5 (162)
• An endothermic
rxn with K>1 must
have T high
enough so that the
result of subtracting TrS from
rH is negative
• Or rH–TrS
<0
• Set rH–
TrS=0 and
rG = rH – TrS
equilibrium
T=
rH
rS
Reactions at equilibrium
rG = rH – TrS
equilibrium
Table 7.2 Standard Gibbs energies of
formation at 298.15 K* (gases)
Substance
fGo/(kJ mol )
1
Gases
Ammonia, NH3
Carbon dioxide, CO2
Dinitrogen tetroxide, N2O4
Hydrogen iodide, HI
Nitrogen dioxide, NO2
16.45
394.36
97.89
1.70
51.31
Sulfur dioxide, SO2
300.19
Water, H2O
228.57
Table 7.2 Standard Gibbs energies of
formation at 298.15 K* (liquids & solids)
Substance
fGo/(kJ mol )
1
Liquids
Benzene, C6H6
124.3
Ethanol, CH3CH2OH
174.78
Water, H2O
237.13
Solids
Calcium carbonate, CaCO3
Iron(III) oxide, Fe2O3
1128.8
742.2
Silver bromide, AgBr
96.90
Silver chloride, AgCl
109.79
* Additional values are given in the Data section.
Standard Gibbs Energy of Formation
• Fig 7.6 (159)
• Analogous to
altitude above or
below sea level
• Units of kJ/mol
The equilibrium composition
• The magnitude of K is a qualitative indicator
• If K 1 (>103) then rG < –17 kJ/mol @
25ºC, the rxn has a strong tendency to form
products
• If K 1 (<10–3) then rG > +17 kJ/mol @
25ºC, the rxn will remain mostly unchanged
reactants
• If K  1 (10–3-103), then rG is between –17
to +17 kJ/mol @ 25ºC, and the rxn will have
significant concentrations of both reactants and
p.160
products
Calculating an
equilibrium concentration
• Example 7.1 (p165)
• Example 7.2 (p166)
Standard reaction Gibbs energy
• rG= 
Gm(products) –
Gm(reactants)



• rG = rH – TrS

7.6 Kc and Kp
aA + bB
Kc =
cC + dD
aA (g) + bB (g)
[C]c[D]d
Kp =
[A]a[B]b
In most cases
Kc  Kp
Kp = Kc(RT)n
When does Kp = Kc ?
cC (g) + dD (g)
c
d
a
pA
pB
pC pD
b
Derivation 7.1: Kc and Kp
Atkins uses
[ ]
vgas
cR
Tp

Substituting values for c, p, and R, we get
K = Kc 
[
K = Kc 
T
12.07K
]
vgas
What is this K?
What is vgas?
Work through Derivation 7.1, p.162
Coupled reactions
• Box 7.1 (164)
• Weights as analogy
to rxns
• A rxn with a large
rG can force
another rxn with a
smaller rG to run in
its nonspontan-eous
direction
• Enzymes couple
biochemical rxns
Coupled reactions
• Biological standard state (pH = 7)
• Typical symbols for standard state:
¤´°
¤
+
Read the last paragraph in Box 7.1 on p164
regarding ATP and the “high energy” bond
Equilibrium response to conditions
• What effect will a change in temperature,
in pressure, or the presence of a catalyst
have on the equilibrium position?
• Presence of a catalyst? None. Why?
• rG is unchanged, so K is not changed
• How about a change in temperature?
• Or a change in pressure?
Let’s see…
The effect of temperature
• Fig 7.7 (163)
• rG of a rxn that
results in fewer moles
of gas increases with
increasing T
• rG of a rxn with no
net change…
• rG of a rxn that
produces more moles
of gas decreases with
increasing T
Equilibrium response to conditions
• Le Chatelier’s principle suggests
When a system at equilibrium is
compressed, the composition of
a gas-phase equilibrium adjusts
so as to reduce the number of
molecules in the gas phase
p.172
The effect of pressure
• Fig 7.9 (174)
• A change in
pressure does not
change the value
of K, but it does
have other
consequences
(composition)
• As p0, xHI1
• What is [I2]?
1



2

K
4

p

   1 
  1
x HI  
 

K 
 2  p  

H2(g) + I2(s)

2 HI(g)
Key
Ideas
Key
Ideas
The End
…of this chapter…”
Spare parts to copy and paste
• μJtμJ+RT
 ln aJ
• Chemical potential (μ) changes as [J]
changes
Box 7.1 pp.172f
• O2 binding in
hemoglobin and
myoglobin…
• …In resting tissue
and in lung tissue
Chemical Potential
• Review pp.128-130, Partial molar properties
(e.g. partial molar volume)
• Read p.129, last two paragraphs
• Read handout, “Chemical Potential” by
Philip A. Candela
• Chemical potential ( ) is usually described
as the “partial molar Gibbs function” or
“partial molar Gibbs energy”
Chemical Potential
• The quantity G/n is so important that it is
given a special symbol () and its own
name (chemical potential)
• As the symbols G/n above indicate,
chemical potential is the Gibbs free energy
per mole of substance
• The chemical potential is an indication of
the potential of a substance to be
chemically active (p.130)
Excursus: Chemical Potential
• The standard chemical potential of a gas
(μJ), is identical to its standard molar
Gibbs energy (Gm) at 1 bar
• The greater the partial pressure of a gas,
the greater its chemical potential

Excursus: Chemical Potential
• Common expressions of chemical potential:
μJtμJ+RT ln aJ
μJtμJ+
RT ln
p
= 1 bar
μJtμJ+
RT ln p
p
p