Chapter 14 Chemical Equilibrium
Watch animation…chemical reaction
reaching equilibrium
equilibrium being dynamic
Dynamic equilibrium
Physical processes:
solid-liquid
vapor-liquid….
This chemical engineer is
testing a process for the
formation of new liquid fuels
from coal and petroleum.
Such methods may transform
the generation of energy
worldwide.
Chemical reactions:
reactantsproducts
Dynamic equilibria are responsive
to changes in the conditions.
Assignment for Chapter 14
14.23,14.41.14.51.14.55
Figure 14.1 In the synthesis of ammonia, the molar
concentrations of N2, H2, and NH3 change with time until there
is no further net change and the concentrations settle into
values corresponding to a mixture in which all three
substances are present.
N2(g)+3H2(g)2NH3(g)
N2(g)+3H2(g)2NH3(g)
No chemical reaction is ‘complete’,
No chemical reaction is ‘impossible’.
Figure 14.2 When we plot the rates of the forward and reverse
reactions for the formation of ammonia on one graph, we can see that
as the forward rate decreases, the reverse rate increases, until they
are equal. At this point, the reaction is at equilibrium and the rates
remain constant.
ReactantsProducts
Figure 14.3 In an experiment showing that equilibrium is dynamic, a
reaction mixture in which N2 (pairs of blue spheres), D2 (pairs of
yellow spheres), and ND3 have reached equilibrium is mixed with one
with the same concentrations of N2, H2 (pairs of gray spheres), and
NH3. After some time, the concentrations of nitrogen, hydrogen, and
ammonia are found to be the same, but the D atoms are distributed
among the hydrogen and ammonia molecules.
N2(g)+3D2(g)2ND3(g)
N2(g)+2HD (g)+D2(g)2NHD2(g)
N2(g)+H2(g)+2HD(g)2NH2Dg)
N2(g)+3H2(g)2NH3(g)
Chemical reaction is dynamic, forward
and reverse reactions taking place all the time.
Try Yourself to Define the
Equilibrium Constant?
aA+bB+cC+…pP+qQ+rR+…
Kc=[P]+[Q]+[R]+…-[A]-[B]-[C]-…?
Kc=p[P]+q[Q]+r[R]+…-a[A]-b[B]-c[C]-…?
Kc=[P][Q][R]…/[A][B][C]…?
Kc=pqr…[P][Q][R]…/abc…[A][B][C]…?
Equilibrium Constant
[Molar concentrations of products] Stoichiometric coefficients
Kc  [Molar concentrations of reactants] Stoichiometric coefficients
aA+bB+cC+…pP+qQ+rR+…
Kc 
p
q
r
a
b
c
[ P ] [ Q ] [ R ] ...
[ A] [ B ] [ C ] ...
Cato Guldberg & Peter Waage (1864)
‘Justification’ of the Definition of
Equilibrium Constant
aA+bB+cC+…pP+qQ+rR+…
A+A+…B+B+…+C+C+…P+P+..+Q+Q+…+R+R+…
Kc 
[ P ][ P ]...[Q ][Q ]...[ R ][ R ]...
[ A][ A]...[ B ][ B ]...[C ]...
[ P ] p [Q ]q [ R ]r ...
 [ A]a [ B]b [C ]c ...
[Molar concentrations of products] Stoichiometric coefficients
Kc  [Molar concentrations of reactants] Stoichiometric coefficients
Example
2SO2 ( g )  O2 ( g )  2SO3 ( g )
[ SO3 ]2
K c  [ SO ]2 [O
2
2]
Is the same for all experiment no matter what
the initial compositions are.
4NH3 (g)  5O2 (g)  4NO(g)  6H 2O(g)
K c  [NH
4
6
[NO] [H 2 O]
4
5
]
[O
]
3
2
Classroom Exercise
Write the equilibrium constant for the following reactions:
2
(a) 3ClO (aq)
(b) 2O3 (g)
2ClO (aq)  Cl (aq)
3
3O2 (g)
(a) K c 
[ClO3- ]2 [Cl- ]
[ClO-2 ]3
[O2 ]3
(b) Kc  [O ]2
3
-
Using smallest possible stoichiometric coefficients to write equilibrium
constants
H 2 (g)  I 2 (g)  2HI(g)
K c1 
[HI]2
[H 2 ][I 2 ]
 54
• Multiply a reaction:
2H 2 (g)  2I 2 (g)  4HI(g)
Kc2 
[HI]4
2
 54  2.9  10
2
[H 2 ] [I 2 ]
2
• Reverse a reaction:
2HI  H 2 (g)  I 2 (g)
K c3 
[H 2 ][I 2 ]
2
[HI]

1
54
 1.8  10
2
3
Classroom Exercise
• Write down the equilibrium constant of the
following reaction:
H2+D22HD
Suppose the equilibrium constant of above reaction at 500 K
is 3.6, what is the equilibrium constant of its reverse reaction?
Kc=[HD]2/[H2][D2]
Reverse reaction: Kc-1=0.28
Multi-Step Reactions
[PCl3 ]2
2P(g)  3Cl2 (g)  2PCl3 (g) K c  [P]2 [Cl
'
3
2]
5]
PCl3 (g)  Cl2 (g)  PCl5 (g) K "c  [PCl[PCl
3 ][Cl 2 ]
[PCl5 ]2
2P(g)  5Cl2 (g)  2PCl5 (g) K c  [P]2 [Cl
5
2]
The third reaction is the sum of the first two :
2P(g)  3Cl2 (g)  2PCl3 (g)
 PCl3 (g)  Cl2 (g)  PCl5 (g)
 PCl3 (g)  Cl2 (g)  PCl5 (g)
________________________
2P(g)  5Cl2 (g)  2PCl5 (g)
[PCl5 ]2
K c  [P]2 [Cl
5
2]
[PCl3 ]2
 [P]2 [Cl
 K c'  K c"  K c"
2
[PCl5 ]
[PCl5 ]


[PCl3 ][Cl 2 ]
[PCl3 ][Cl 2 ]
]3
Equilibrium constant depends on reaction conditions
Reaction Rate and Equilibrium
2A  C  D
K c  [C[ A][]D2 ]
Suppose elementary reactions:
AA CD
Rate  k[A]
CDAA
Rate  k ' [C][D]
2
At equilibrium: k[A]  k [C][D]
2
Kc 
k
k'
'

(equilibrium constant equals the
ratio of the forward to the reverse reaction rates)
Figure 14.4 The equilibrium constant for a reaction is equal to the
ratio of the rate constants for the forward and reverse elementary
reactions that continue in a state of dynamic equilibrium. (a) A
relatively large forward rate constant means that the forward rate can
match the reverse rate even though only a small amount of reactants
is present. (b) Conversely, if the reverse rate constant is relatively
large, then the forward and reverse rates are equal when only small
amounts of products are present.
Multi-Step Reactions
R 1,a  R1,b  ...  P1,a  P1,b  ...
k1
k1'
k2
 R 2,a  R 2,b  ... 
P

P

...
' 
2,a
2,b
k
2

...
______________________________________
R a  R b  ...  Pa  Pb  ...
Kc 
k1
k1'
 k '  ...
k2
2
Homogeneous Equilibria:
All products and reactants are of the
same phase.
Heterogeneous Equilibria:
Reacting systems with more than
one phase.
2

Ca(OH) 2 (s)  Ca (aq)  2OH (aq)
2
 2
K c  [Ca ][OH ]
Ignored!
Molar concentration of pure solid and liquid is a
constant, independent of the amount present . It is
ignored in the calculation of equilibrium constant.
Another way of understanding: the concentration of pure solid/liquid
Is always 100%1. Since 1a =1, it does not affect Kc.
Classroom Exercise
Ag 2S(s)
+
2-
2Ag (aq)+S (aq)
Kc  ?
Ignored!
 2
2
K c  [Ag ] [S ]
Gaseous Equilibria
CaCO3 (s)  CaO(s)  CO2 (g)
K c  [CO 2 ] 
n CO2
V

PCO2
RT
Thus we may introduce
K p  PX for gaseous reactant/p roduct X.
Relationsh ip between K c and K p .
aA(g)  bB(g)  cC(g)  dD(g)
aA(g)  bB(g)  cC(g)  dD(g)
Kp 
PCc PDd
PAa PBb

[[ C ] RT ]c [[ D ] RT ]d
[[ A ] RT ]a [[ B ] RT ]b

[ C ]c [ D ]d
[ A ]a [ B ]b
Kp 
Kc 
PCc PDd
PAa PBb
[ C ]c [ D ]d
[ A ]a [ B ]b
 ( RT )cd a b
 K c  ( RT ) n
n  change in the number of the gas phase molcules.
N2(g)+3H2(g)2NH3(g)
T=500 K, equilibrium concentrations:
[NH3]=0.796 mol/L, [N2]=0.305 mol/L,
[H2]=0.324 mol/L.
Kc 
[ NH 3 ]2
[ N 2 ][ H 2 ]3

0.7962
0.3050.3243
 61.1.
K p  K c ( RT ) n  61.1 (0.08206  500)213
 1.027  105
Classroom Exercise
2S(s)+3O2(g)2SO3(g)
T=300K, equilibrium concentrations: [O2]=0.25 mol/L,
[SO3]=0.3 mol/L. calculate the equilibrium constant Kc
and Kp.
Kc 
[ SO3 ]2
3
[ O2 ]

0.32
0.253
n
 5.76.
K p  K c ( RT )  5.76  (0.08206  300)
 1.42 10
1
2 3
Gaseous Equilibria
• Equilibrium constants for gaseous
reactions can be written by using
either molar concentrations or
partial pressures.
Summary of Equilibrium Constants
The equilibrium constant is the ratio of the concentrations or partial pressures of the
products to those of the reactants, each concentration raised to a power equal to its
stoichiometric coefficient in the balanced equation.
Reaction 1
K1
+ Reaction 2
K2
+ Reaction 3
K3
+
_______________________
Overall reaction = Reaction 1+Reaciton 2+Reaction 3+...
K overall  K1 K 2 K 3
k1
R 1,a  R1,b  ... 
 P1,a  P1,b  ...
k'
1
n
K p  ( RT ) K c ,
n  change in the number
of the gas phase molcules.
k2
 R 2,a  R 2,b  ... 
 P2,a  P2,b  ...
k'
2

...
______________________________________
R a  R b  ...  Pa  Pb  ...
Kc 
k1
k1'
 kk2'  ...
2
Using Equilibrium Constants
A. The Extent of Reaction
B. The Direction of Reaction
C. Equilibrium Tables
The Extent of Reaction
• K>1000, product dominant;
• 0.001<K<1000, neither reactants nor
products dominate equilibrium;
• K<0.001,reactants dominant.
Figure 14.6 The size of the equilibrium constant indicates
whether the reactants (blue squares) or the products (yellow
squares) are favored. Note that reactants are favored when Kc
is small (left), products are favored when Kc is large (right),
and reactants and products are in almost equal abundance
when Kc is close to 1 (middle). Here, for simplicity, we compare
reactions with Kc  102 and Kc  102.
Calculating equilibrium concentration
• H2(g)+Cl2(g)2HCl(g)
17
17
[H 2 ]  1.0  10 mol/L; [Cl2 ]  1.0  10 mol/L;
K c  4.0  1031at300K. [HCl]  ?
Solution:
Kc 
16
[HCl]2
[H 2 ][Cl2 ]
 [HCl]  K c [H 2 ][Cl2 ]  0.28
10 times higher than [Cl 2 ] and [H 2 ]!
Classroom Exercise
• Suppose that the equilibrium molar concentrations of H2 and Cl2
at 300 K are both 1.0×10-16 mol/L. What is the equilibrium molar
concentration of HCl, given Kc= 4.0×1031?
• H2(g)+Cl2(g)2HCl(g)
[H 2 ]  [Cl2 ]  1.0  1016 mol/L; K c  4.0  1031 at 300K. [HCl]  ?
Solution:
Kc 
[HCl]2
[H 2 ][Cl2 ]
 [HCl]  K c [H 2 ][Cl2 ]  4.0  1031  1.0  1032  0.63,
almost 1016 times higher than [Cl2 ] and [H 2 ]!
Using K to determine a partial pressure
PCl5 (g)  PCl3 (g)  Cl2 (g) K p  25 at 298K.
PPCl5  0.0021 atm, PCl2  0.48 atm, PPCl3  ?
Solution :
Kp 
PPCl3 PCl2
PPCl5
 PPCl3  K p 
PPCl5
PCl2
 0.11
Classroom Exercise:
Using K to determine a partial pressure
2NOCl(g)  2NO(g)  Cl2 (g) K p  0.018 at 500 K.
PNO  0.11 atm, PCl2  0.84 atm, PNOCl  ?
Solution :
Kp 
2
PNO
PCl2
P 2 NOCl
 PNOCl 
2
PNO
PCl2
Kp

0.112 atm 2 0.84 atm
0.018 atm
 0.75 atm
The Direction of Reaction
Reaction quotient:
molar concentrations of products at any stage
molar concentrations of reactants at any stage
partial pressures of products at any stage

partial pressures of reactants at any stage
Qc 
Qp
Qc (equilibrium)  K c Q p (equilibrium)  K p
Example
H 2 (g)  I 2 (g)  2HI(g)
At certain stage: PH 2  0.1 atm, PI 2  0.2 atm, PHI  0.4 atm,
then Q p 
2
PHI
PH 2 PI 2
 0.8
A reaction has a tendency to form products if Q  K and to form reactants if Q  K .
Figure 14.7 The relative sizes of the reaction quotient Q and the
equilibrium constant K indicate the direction in which a reaction
mixture tends to change. The arrows point from reactants to products
when Q, K (left) or from products to reactants when Q  K (right).
There is no tendency to change once the reaction quotient has
become equal to the equilibrium constant.
Quiz
PCl5 (g)  PCl3 (g)  Cl2 (g) K p  25 at 298K.
At certain stage, it is measured that
PPCl5  0.0021 atm, PCl2  0.48 atm, PPCl3  0.08 atm
Will the reaction be moving to formation of more product or not?
Solution :
Qp 
PPCl3 PCl2
PPCl5

0.080.48
0.0021
 18.3  K p
The reaction will be moving to formation of more product.
Equilibrium Tables
A table that shows the initial concentrations,
the changes needed to reach equilibrium, and the
final equilibrium compositions.
Problem: 0.15 mol PCl5 is placed in reaction vessel of
volume 500 mL and allowed to reach equilibrium with its
decomposition products at 250 o C and K c  1.8. What is the
composition of the equilibrium mixture?
3 ][Cl 2 ]
PCl5 (g)  PCl3 (g)  Cl 2 (g) K c  [PCl[PCl
5]
The initial concentration [PCl5 ]  0.150 mol/0.5L  0.30 mol/L
Draw an equilirium table:
Species
PCl
Step 1: Initial concentration
0.30
Step 2: Change in the molar concentration
-x
Step 3: equilibrium concentration
K
c

5
0.30 - x
2
x x
 x  1.8 x  0.54  0  x  0.262(positive root).
0.30 x
PCl
3
Cl
2
0
0
x
x
x
x
Classroom Exercise
A mixture of 0.002 mol/L hydrogen gas and 0.002 mol/L iodine
is allolwed to form hydroegn iodide at 773 o C by the reaction
H 2 (g)+I 2 (g)  2HI(g). At equilibrium, the concentration of HI is
0.002 mol/L. Calculate the equilibrium constant at this temperature.
The initial concentration [H2 ]  0.002 mol/L, [I2 ]  0.002 mol/L
Draw an equilirium table:
Species
H
Step 1: Initial concentration
2
0.002
Step 2: Change in the molar concentration
Step 3: equilibrium concentration
x  0.001
2
(2 x )2
0.002
K 

 4.0
c
(0.002 x )(0.002 x )
2
0.001
-x
I
2
0.002
x
0.002 - x 0.002  x
HI
0
 2x
0.002
Le Chatelier’s Principle
When a stress is applied to a system in
dynamic equilibrium, the equilibrium
tends to adjust to minimize the effect of
the stress.
Henry Louis LE CHATELIER
(1850-1936)
Use of Le Chatelier Principle (I)
• 
• Adding a reactant or removing a product 
reaction tends to form products.
• Adding a product or removing a reactant 
more reactant tends to form.
Use of Le Chatelier Principle (II)
• 
• Compression of a reaction mixture  the
reaction that reduces the number of gas-phase
molecules.
• Increasing by introducing an inert gas  no
effect on the equilibrium!
Commercial ammonia synthesis vessel
Introducing inert gas (yellow) has no
effect on the equilibrium composition.
More generally, introducing anything that does not react with any product
and reactant of a reaction would not change the equilibrium composition
of the reaction. (A catalyst can increase reaction rate but not equilibrium
constant.)
Cheating equilibrium (1)
Cheating equilibrium (2)
Use of Le Chatelier Principle (III)
• 
• Raising the temperature of an exothermic
reaction  reaction tends to form more
reactants.
• Raising the temperature of an endothermic
reaction  reaction tends to form more
products.
(a) Endothermic reaction
(b) Exothermic reaction
Radar image of Venus:
high partial pressure of carbon dioxide
Catalysts
• A catalyst can increase the rate of a chemical
reaction.
• A catalyst has no effect on the equilibrium
composition of a reaction mixture.
Haber’s Achievements
Fritz Haber (1868-1934，
Nobel Prize for Chemistry, 1916)
N2(g)+3H2(g)2NH3(g)
Assignment for Chapter 14
14.23,14.41.14.51.14.55