1
EQUILIBRIUM
- condition where opposing processes occur at the same time
- processes may be physical changes or chemical changes
Example: Ice slurry
- Temperature of slurry is
0- Melting
C.
of ice occurs at same rate as freezing of water.
- At 0 C, the conversion of liquid to solid to liquid is an
equilibrium process.
- All phase transitions at the transition temperature are equilibrium
processes.
Example: Saturated solution
- Dissolving of crystal occurs at same rate as crystallizing
of crystal.
- This is also an equilibrium process.
Example: Ionization of Acetic Acid
- Recall acetic acid (HC2H3O2) is a weak acid.
- It does not fully disassociate.
HC2H3O2 (aq)  H+ (aq) + C2H3O2- (aq)
- Balance exists between associated and dissociated forms of weak acid.
Example: Ozonolysis
Ozone = O3
2 O3 (g)  3 O2 (g)
3 O2 (g)  2 O3 (g)
2 O3 (g)  3 O2 (g)
- Decomposition and formation of ozone is happening at the same time.
- double arrow  indicates equilibrium
- A balance exists between amount ozone and diatomic oxygen.
Equilibrium expression
2
- Rate law cannot be determined by stoichiometry.
- Equ. expressions are determined by stoichiometry.
- For a general equilibrium a A + b B  c C + d D
 C c  D d
Kc 
 A a  B b
Note: products over reactants.
- Kc is constant; however, it will change with temperature.
- Equilibrium constants are dimensionless
- Technically, all the concentrations are divided by the standard concentration,
c = 1 M
c
d
 cC   c D 
 c   c 
Kc    a  b
 cA   cB 
 c    c  
- In practice, we do not include the standard concentration in our calculations.
Example: 2 O3 (g)  3 O2 (g)
O 

O 
3
Kc
2
2
3
Example: 3 H2 (g) + N2 (g)  2 NH3 (g)
 NH 

H  N 
2
Kc
3
3
2
2
Example: 2 N2O5 (g)  4 NO2 (g) + O2 (g)
 NO  O 

N O 
4
Kc
2
2
2
2
5
Example: Three experiments are done to find Kc at 460 C for the following
reaction, H2 (g) + I2 (g)  2 HI (g).
Calculate Kc for each trial and calculate the average.
3
1
2
3
[H2] (M)
6.47  10-3
3.84  10-3
1.43  10-3
[I2] (M)
5.94  10-4
1.52  10-3
1.43  10-3
[HI] (M)
0.0137
0.0169
0.0100
2
 0.0137 
 HI 
 48.8
 H 2  I2   6.47 103  5.94 104 
2
K1c 
2
3
Kc 
Kc 
 0.0169 
2
 3.84 10 1.52 10 
3
 0.0100 
3
2
1.4310 1.4310 
Kc 
3
 48.9
3
 48.9
48.8  48.9  48.9
 48.9
3
Note that one can have different amounts of each component at equilibrium; but,
the ratio between products and reactants (as in Kc) will always be the same.
Example: Kc for ozonolysis is 4.38  1028. If the molar concentration of oxygen
in the sea level atmosphere is 0.0089 M, calculate the molar concentration of
ozone.
O 
 Kc  2 2
2 O3 (g)  3 O2 (g)
 O3 
3
O   0.0089 3  1.6 1035
2
  O3   2 
28
3
Kc
O3  
4.38 10
1.6 1035  4.0 1018 M
For gases, equilibrium constants can be expressed in terms of pressure rather than
concentration.
General gas-phase equilibrium:
aA+bB cC+dD
4
p  p 

p  p 
c
Kp
C
d
D
a
A
b
B
Be Careful!! Kp is usually not equal to Kc.
Also: pressure must be in atmospheres
- standard pressure, p, is 1 atm.
Direction of chemical equation and Kc.
Kc = 4.38  1028
Reconsider 2 O3 (g)  3 O2 (g)
O 

O 
3
Kc
Note: Formation of oxygen is heavily favored.
2
2
3
What if we wrote equilibrium as 3 O2 (g)  2 O3 (g)?
O 
1
Kc  3 3 
 2.28 1029
28
O2  4.38 10
2
Note: Kc still indicates that oxygen is heavily favored.
Example: 2 N2O5 (g)  4 NO2 (g) + O2 (g)
 NO  O 

N O 
4
Kc
2
2
2
2
5
4 NO2 (g) + O2 (g)  2 N2O5 (g)
N O 
K 
 NO  O 
2
2
2
Key point:
K c 
5
4
c
2
1
Kc
Multiple Equilibria
- When multiple equilibria are occurring at the same time, we can add the
reactions together and multiply their equilibrium constants for the overall
reaction.
5
CH3OH (aq) + NaNH2 (aq)  NaOCH3 (aq) + NH3 (aq)
K c1 
NaOCH3 NH3 
CH3OHNaNH2 
C2H5COOH (aq) + NaOCH3  C2H5COOCH3 + NaOH (aq)
K c 2  
C H COOCH  NaOH 
C H COOH  NaOCH 
2
5
2
5
3
3
CH3OH (aq) + NaNH2 (aq) + C2H5COOH (aq) + NaOCH3
 NaOCH3 (aq) + NH3 (aq) + C2H5COOCH3 + NaOH (aq)
CH3OH (aq) + NaNH2 (aq) + C2H5COOH (aq)  NH3 (aq) + C2H5COOCH3 + NaOH (aq)
K c1  K c 2  K c
 NaOCH  NH   C H COOCH  NaOH 
CH OH  NaNH  C H COOH  NaOCH 
 NH C H COOCH  NaOH 

CH OH  NaNH C H COOH 

3
3
3
2
3
2
5
3
2
5
2
5
3
3
3
2
2
5
Heterogeneous Equilibria
- Equilibria expressions depend on concentration.
- Consider:
Ni (s) + 4 CO (g)  Ni(CO)4 (g)
- What is the concentration of the solid?
- Since moles of solid Ni per volume is constant, concentration is constant.
- The constant concentration is incorporated into the equilibrium constant.
Kc 
 Ni(CO) 
 CO 4
4
Kp 
( p Ni ( CO ) 4 )
p 
4
CO
- Heterogeneous equilibrium is equilibrium between different phases.
- Solid components or liquid solvent components are usually ignored in
writing equilibrium expression.
Reaction Quotient
Before equilibrium is reached, state of reaction can be evaluated with the reaction
quotient.
For a general reaction going toward equilibrium
aA+bB cC+dD
6
 C c  D  d
Q
 A  a  B b
**What is the difference between Kc and Q?**
Kc is only at equilibrium.
At a given temperature, Kc is a constant.
Q is at any point in reaction.
At a given temperature, Q can have any value.
If Q > Kc, amount of “products” is too much, reaction will shift to “reactant” side.
If Q < Kc, amount of “reactants” is too much, reaction will shift to “product” side.
If Q = Kc, system is at equilibrium.
Example: At 373 K, the equilibrium constant for the reaction,
COCl2 (g)  CO (g) + Cl2 (g), has the value, Kc = 2.19  10-10. Are the
following mixtures of COCl2, CO and Cl2 at equilibrium?
a) [COCl2] = 5.00  10-2 M
[CO] = 3.31  10-2 M
[Cl2] = 3.31  10-6 M
CO  Cl 2  3.31102  3.31106

Q

5.00 102
COCl2 
 2.19 106
- 2.19  10-6 > 2.19  10-10
- Reaction will shift toward COCl2.
b) [COCl2] = 3.50  10-3 M
[CO] = 1.11  10-5 M
[Cl2] = 3.25  10-6 M
CO  Cl2  1.11105  3.25 106

Q

3.50 103
COCl2 
 1.03108
- 1.03  10-8 > 2.19  10-10
- Reaction will shift toward COCl2.
c) [COCl2] = 1.45 M
[CO] = 1.56  10-6 M
[Cl2] = 1.56  10-6 M
Q
COCl2   1.56 106 1.56 106
1.45
COCl2 
 1.69 1012
- 1.69  10-12 < 2.19  10-10
- Reaction will shift toward CO and Cl2.
CALCULATING COMPONENT CONCENTRATIONS WITH
EQUILIBRIUM CONSTANTS
- Sometimes partial information about concentration is known.
- This information along with equilibrium expressions can be used to find missing
concentrations.
7
Example: At 490 C, the decomposition of HI vapor, 2 HI (g)  H2 (g) + I2 (g),
has an equilibrium constant of 0.0217. If 1.4 mol of HI is placed in a
2.0 L container, how many moles of H2, I2 and HI will be in the
container at equilibrium?
Strategy
1) Write equilibrium expression.
2) Create table of initial concentrations of components in equilibrium expression.
3) Indicate a change of a component by variable, x.
4) Changes in other variables are dictated by stoichiometry.
5) Add together initial concentration with changes to yield equilibrium
concentrations.
6) Substitute equilibrium concentrations into equilibrium expression.
7) Use algebra to solve for variable x.
8) Use solution of x to find equilibrium concentrations.
2 HI (g)  H2 (g) + I2 (g)
Continue with problem
Kc 
Initial
Change
Equil.
[H2]
0
x
x
 H I   0.0217
2
 HI
2
2
[I2]
0
x
x
[HI]
1.4 mol/2.0 L = 0.70 M
– 2x
0.70 – 2x
xx
 0.70  2 x 2
x
0147
.

0.70  2 x
0147
.  0.70  2 x  x
0103
.
 0.295x  x
x  0.0796 M
0103
.
 1295
. x
K c  0.0217 
molH = 0.0796 M  2.0 L = 0.16 mol
2
molI = 0.0796 M  2.0 L = 0.16 mol
2
molHI = 0.70 M – 2  0.0796 M
= 0.54 M  2.0 L = 1.1 mol
Example: For the reaction CO (g) + H2O (g)  CO2 (g) + H2 (g) Kc = 4.06 at
500 C. If [CO] = 0.100 M and [H2O] = 0.100 M are put in a reaction
vessel, what are the equilibrium concentrations of the products and
reactants?
Kc 
CO  H   4.06
 CO H O
2
2
2
8
Initial
Change
Equil.
[CO]
0.100 M
–x
0.100 – x
[H2O]
0.100 M
–x
0.100 – x
[CO2]
0
+x
x
[H2]
0
+x
x
xx
x2
Kc 

2  4.06
 0100
.
 x   0100
.
 x  0100
.
 x
x2
x
4.06 
 2.01
2 
 0100
 0100
.
 x
.
 x
 x = 0.201 – 2.01 x  3.01 x = 0.201
 x = 0.0668
[CO2] = 0.0668 M
[H2] = 0.0668 M
[CO] = [H2O] = (0.100 – 0.0668) M = 0.033 M
Example: At a certain temperature, Kc = 4.50 for the reaction N2O4 (g)  2 NO2
(g). If the concentration of N2O4 is 0.150 M in a 2.00 L container at this
temperature, what will be the equilibrium concentration of both gases?
O
O
N
O
N
O
O
N
O
N
O
O
 NO   4.50

N O 
2
Kc
2
2
Initial
Change
Equil.
Kc 
[NO2]
0
+ 2x
2x
 2 x 2
 0150
.
 x

4
[N2O4]
0.150 M
–x
0.150 – x
4x2
 4.50
 0150
.
 x
 4.50 (0.150 – x) = 4x2  4x2 + 4.50 x – 0.675 = 0
2
To solve for x, we must use the quadratic equation. ax  bx  c  0
b  b 2  4ac 4.50  4.502  4  4  0.675

2a
24
4.50  31.05 4.50  5.57


8
8
x
9
x = 0.134 M or –1.26 M
**Concentration can’t be negative**
Thus x = 0.134 M.
[NO2] = 2  0.134 M = 0.268 M
[N2O4] = 0.150 M – 0.134 M = 0.016 M
Example: A mixture consisting of 0.500 M N2 and 0.800 M H2 in a reaction
vessel reacts and reaches equilibrium. At equilibrium, the
concentration of ammonia is 0.150 M. Calculate the equilibrium
constant for the reaction N2 (g) + 3 H2 (g)  2 NH3 (g).
 NH 

 N  H 
2
Kc
3
3
2
Initial
Change
Change
Equil.
[N2]
0.500
–x
– 0.075
0.500 – 0.075 = 0.425
2
[H2]
0.800
– 3x
– 0.225
0.800 – 0.225 = 0.575
[NH3]
0
+ 2x
+ 0.150
0.150
2
 0150
. 
Kc 
 0.278
 0.425 0.575 3
**Often simplifying assumptions may be valid.**
Example: A mixture of 0.0482 M of N2 and 0.0933 M O2 is placed in a reaction
vessel. Calculate the equilibrium composition of the system at a
temperature at which Kc = 2.0  10-37 for the equilibrium
2 N2 (g) + O2 (g)  2 N2O (g).
10
 N O

 N  O 
2
Kc
2
2
2
Initial
Change
Equil.
[N2]
0.0482
- 2x
0.0482 – 2x
2
[O2]
0.0933
-x
0.0933 – x
[N2O]
0
+ 2x
2x
 2 x 2
Kc 
 0.0482  2 x 2  0.0933  x
This is a cubic equation! Yikes!!
Look at equilibrium constant
- Kc << 1 means that amount of N2O formed is very small.
- If N2O is very small, then x is very small.
- Therefore 0.0482 – 2x  0.0482
0.0933 – x  0.0933
Thus, the equilibrium expression becomes
 2x 
Kc 
 2.0 1037
2
 0.0482   0.0933
2
 4x2 = 2.0  10-37 (0.0482)2(0.0933) = 4.34  10-41
 x2 = 1.08  10-41  x = 3.29  10-21
[N2O] = 2x = 6.58  10-21 M
[N2] = 0.0482 M
[O2] = 0.0933 M
Note that our assumption was a good assumption.
* Always check assumption. *
If “x” is less than 5% of the quantity assumed to be large, the “x” is small
assumption was good.
Example: The partial pressures of carbon monoxide and hydrogen in a reaction
vessel at 225 C are 0.37 atm and 0.72 atm respectively. The
equilibrium constant for the reaction CO (g) + 2 H2 (g)  CH3OH (g) is
Kp = 0.0049. What are the partial pressures of the gases in the
reaction mixture at equilibrium?
11
Kp 
p CH 3OH
p CO p 2H 2
Recall that to use Kp, our pressure must be in atmospheres.
Initial
Change
Equil.
p(CO)
0.37 atm
-x
0.37 – x
Kp 
p(H2)
0.72 atm
- 2x
0.72 – 2x
p(CH3OH)
0
+x
x
x
 0.37  x 0.72  2 x 2
This is also a cubic equation also.
Maybe assuming x is small will help.
0.0049 
x
 0.37  0.72 
2
 x  9.4 104 atm
Check assumption
0.00094
100%  0.25%
0.37
2  0.00094 
100%  0.26%
0.72
Check assumption
0.37
-0.00094
Assumption was good.
0.37
p(CO) = 0.37 atm
p(H2) = 0.72 atm
p(CH3OH) = 9.4  10-4 atm
12
Example: For the equilibrium PCl5 (g)  PCl3 (g) + Cl2 (g), the equilibrium
constant at some temperature is 0.61. Calculate the equilibrium
concentrations of the chemical species when the reaction vessel starts
with only 3.0 M of PCl5.
Kc 
 PCl Cl   0.61
 PCl 
3
2
5
Initial
Change
Equil.
[PCl5]
3.0
–x
3.0 – x
[PCl3]
0
+x
x
[Cl2]
0
+x
x
x2
Kc 
 0.61
 3.0  x
 0.61 (3.0 – x) = x2  x2 + 0.61 x – 1.83 = 0
To solve for x, we must use quadratic equation.
2
b  b 2  4ac 0.61  0.61  4 1   1.83
x

 0.305  1.39
2a
2
x = 1.08 M or –1.695 M
**Concentration can’t be negative**
Thus x = 1.08 M.
[PCl3] = 1.1 M
[Cl2] = 1.1 M
[PCl5] = 3.0 M – 1.1 M = 1.9 M
Example: For the equilibrium FeO (s) + CO (g)  Fe (s) + CO2 (g), Kp = 0.403
at 1000 C. If the pressure of CO is 0.337 atm and no CO2 is present
initially, what are the equilibrium pressures?
Kp 
Initial
Change
Equil.
p CO2
p CO
p(CO)
p(CO2)
0.337 atm
0.00 atm
–x
+x
0.337 – x
x
x
0.403 
0.337  x
0.136  0.403x  x
pCO2 = 0.0968 atm
Relationship between Kc and Kp
0.136  1.403x
x  0.0968
pCO = 0.337 – 0.097 = 0.240 atm
13
A 
Recall for molarity
nA
V
For ideal gases, pV = nRT
n
p
p

or  A   A
V RT
RT
 p A   A  RT
Recall for general gas-phase equilibria:
aA+bB cC+dD
p  p 

p  p 
c
Kp
C
a
A
d
D
b
B
 C RTc  D RTd

 A RTa  B RT b
c
d
C  RT   D  RT 

Kp 
a
b
a
b
 A  RT   B  RT 
c
d
K p  K c  RT
c
d
cd

  C  D  
  RT 


a
b 
a b
A
B







  RT 
c  d  a  b 
 K c  RT
n
where n = coefficients of gaseous products – coefficients of gaseous reactants
Example: For the reaction 2 NO (g) + Cl2 (g)  2 NOCl (g) at 700 K,
Kp = 0.26, find Kc.
K p  Kc  RT
n
 Kc 
Kp
 RT n
n = 2 – 1 – 2 = – 1
Kc 
Kp
 RT 
1
 K p  RT   0.26   0.08206  700   15
Recall that equilibrium constants are dimenionless. Both Kc and Kp have no units.
LE CHÂTELIER’S PRINCIPLE
14
If a system in equilibrium is disturbed, the system will adjust to reestablish
equilibrium.
Ways to disturb equilibrium
1. Add or subtract chemical component
2. Increase or decrease temperature
3. Increase or decrease pressure (adjust volume)
Consider 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
What happens when H2 is added?
Balance has been upset: too much H2 on left.
To restore balance, process must reduce excess H2.
Therefore
N2 is produced.
H2O is produced.
NO is consumed.
H2 is consumed, but not all that was added.
What happens when H2 is removed?
Balance has been upset: not enough H2 on left.
To restore balance, process must produce H2 on the left. (Must replace missing H2.)
Therefore
N2 is consumed.
H2O is consumed.
NO is produced.
H2 is produced, but not all that was removed.
Sometimes the reaction quotient must be used to determine the direction of the
shift in the system.
Example: For the equilibrium C(s) + 2 H2 (g)  CH4 (g), Kp = 0.262 at 1000 C.
The initial pressure of H2 is 1.22 atm and the initial pressure of CH4 is
1.00 atm.
a) Use the reaction quotient to determine in which direction the equilibrium
must adjust.
p CH
K p  2 4  0.262
pH2
Q
p CH 4
p
2
H2

100
.
2  0.672
122
. 
Q > Kp  CH4 must decrease and H2 must increase
i. e., reaction will shift to the left.
b) Calculate the equilibrium pressures.
15
H2
1.22
+ 2x
1.22 + 2x
Initial
Change
Equil.
Kp 
CH4
1.00
–x
1.00 – x
1.00  x
1.22  2x 
2
 0.262
100
.  x  0.262122
.  2 x
100
.  x  0.262149
.  4.88x  4x2 
2
0.390  128
. x  105
. x2  100
. x
105
. x2  2.28x  0.610  0
2
b  b 2  4ac 2.28  2.28  4 1.05   0.610 
x

2a
2 1.05
x = 0.24 atm
pH = 1.22 + 2  0.24 = 1.70 atm
2
pCH = 1.00 – 0.24 = 0.76 atm
4
Effects of temperature on equilibrium
If reaction is exothermic, increasing temperature decreases products.
Exothermic reaction
Reactants  Products + heat
- increasing heat (T) shifts reaction to left.
CaC2 (s) + 2 H2O (l)  Ca(OH)2 (s) + C2H2 (g)
Hrxn = -127 kJ/mol
If reaction is endothermic, increasing temperature increases products.
Endothermic reaction
Reactants + heat  Products
- increasing heat (T) shifts reaction to right.
Ca(OH)2 (aq)  CaO (s) + H2O (l)
Hrxn = + 82 kJ/mol
Note: Temperature changes equilibrium constant.
Effects of pressure on equilibrium
Increasing pressure (decreasing volume) changes equilibrium to decrease total
number of moles of gas.
16
Consider N2O4 (g)  2 NO2 (g)
- Two moles of gas take more room than one mole of gas.
- Decreasing volume raises all concentrations.
- Reaction quotient adjusts by changing the total number of moles of gas.
- Increasing pressure for the above system shifts equilibrium to the left.
- External pressure does not change value of equilibrium constant.
Consider 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
Increase of pressure, decreases “reactant” side; because, concentrations must be
reduced.
Effect of catalysis on equilibrium
Catalyst has no effect on equilibrium.
Catalyst only affects speed of reaching equilibrium.