Equilibrium
Chemical Equilibrium – Reversible Reactions
The conversion of reactants to products and the
conversion of products to reactants occur simultaneously.
Example:
Forward Reaction:
Reverse Reaction:
2 SO2(g) + O2(g)  2 SO3(g)
2 SO2(g) + O2(g)  2 SO3(g)
The 2 equations can be combined using a double arrow.
2 SO2(g) + O2(g) ↔ 2 SO3(g)
Chemical Equilibrium – Reversible Reactions
2SO2(g) + O2(g) ↔ 2SO3(g)
Molecules of SO2 and O2 react to give SO3. Molecules of SO3 decompose to
give SO2 and O2. At equilibrium, all 3 types of molecules are present in the
mixture.
Chemical Equilibrium – Equilibrium Constants
Equilibrium Constant (Keq)
Concentrations in MOLARITY
 Ratio of product concentrations to reactant
concentrations at equilibrium, with each
concentration raised to a power equal to the number
of moles of that substance in the balanced equation.
𝐾𝑒𝑞
𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
=
𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
Chemical Equilibrium – Equilibrium Constants
Position of equilibrium in terms of numerical values:
aA + bB ↔ cC + dD
𝑲𝒆𝒒
𝑪𝒄𝑫𝒅
=
𝑨𝒂𝑩𝒃
The value of Keq depends on temperature. If the temperature
changes the value of Keq also changes. Pure solids and liquids do
NOT factor into Keq
Chemical Equilibrium – Equilibrium Constants
Equilibrium constants show whether products or
reactants are favored at equilibrium.
aA + bB ↔ cC + dD
𝐾𝑒𝑞 =
𝐶𝑐𝐷𝑑
𝐴𝑎𝐵𝑏
Keq > 1, products favored at equilibrium
Keq < 1, reactants favored at equilibrium
Chemical Equilibrium – Equilibrium Constants
Practice: Write the equilibrium expression.
a. 4NH3(g) + 5O2(g) ↔ 6H2O(g) + 4NO(g)
6
4
H2O NO
Keq =
NH 4 O 5
3
b. 2SO3(g) ↔ 2SO2(g) + O2(g)
2
SO2 2 O2
Keq =
SO 2
3
Chemical Equilibrium – Equilibrium Constants
Dinitrogen tetroxide, a colorless gas, and nitrogen dioxide, a dark
brown gas, exist in equilibrium with each other.
N2O4(g) ↔ 2NO2(g)
A 3.0 liter sample of the gas mixture at 10oC at equilibrium contains
0.0045 mol N2O4 and 0.030 mol NO2. Write equilibrium expression
and calculate the equilibrium constant (Keq) for the reaction.
NO2 2
Keq =
N2O4
0.067
Chemical Equilibrium – Equilibrium Constants
Analysis of an equilibrium mixture of nitrogen, hydrogen and
ammonia contained in a 2.0 L flask at 300oC gives the following
results:
Hydrogen 0.15 mol; Nitrogen 0.25 mol; Ammonia 0.10 mol
N2(g) + 3 H2(g) ↔ 2 NH3(g)
Write the equilibrium expression and calculate the equilibrium
constant for the equation.
47
Chemical Equilibrium – Equilibrium Constants
Bromine monochloride decomposes to form chlorine and bromine.
2BrCl(g) ↔ Cl2(g) + Br2(g)
At a certain temperature, the equilibrium constant for the reaction
is 11.1, and the equilibrium mixture contains 4.00 mol Cl2. How
many moles of Br2 and BrCl are present in the equilibrium mixture?
Assume that initially only pure BrCl existed and that the container
has a volume of 1.0 L.
[Br2] = 4.0M
[BrCl] = 1.2M
Chemical Equilibrium – Le Chatelier’s Principle
If an external stress is applied to a system at equilibrium,
the system adjusts in such a way that the stress is partially
offset as the system reaches a new equilibrium position.
Stresses that upset the equilibrium of a chemical systems
include:
1. Changes in concentrations of reactants or products
2. Changes in temperature
3. Changes in pressure
Chemical Equilibrium – Le Chatelier’s Principle
1. Changes in Concentration of Reactants or Products The system will adjust to minimize the effects of the change.
Shifts to partially offset the change.
Example – The decomposition of carbonic acid in aqueous
solution to produce carbon dioxide and water
Equilibrium position
H2CO3(aq) ↔ CO2(aq) + H2O(l)
<1%
>99%
What would happen to the equilibrium if more CO2 is
added?
What would happen to the equilibrium if CO2 is removed?
Chemical Equilibrium – Le Chatelier’s Principle
2. Changes in Temperature –
 For exothermic reactions, increasing the
temperature causes Keq to decrease, which shifts
the reaction to the left
 For endothermic reactions, increasing the
temperature causes Keq to increases, which shifts
the reaction to the right
Chemical Equilibrium – Le Chatelier’s Principle
2. Changes in Temperature –
Example – The exothermic reaction that occurs when SO3 is
produced from the reaction of SO2 and O2.
2 SO2(g) + O2(g) ↔ 2 SO3(g) + heat
What would happen to the equilibrium if temperature
increases?
What would happen to the equilibrium if temperature
decreases?
Chemical Equilibrium – Le Chatelier’s Principle
3. Changes in Pressure – A change in pressure on a system
affects only gaseous equilibrium that have unequal number of
moles of gaseous reactants vs. products.
Example – The reaction that occurs when NH3 is produced
from the reaction of N2 and H2.
N2(g) + 3 H2(g) ↔ 2 NH3(g)
What would happen to the equilibrium if pressure increases?
What would happen to the equilibrium if pressure decreases?
Chemical Equilibrium – Le Chatelier’s Principle
3. Changes in Pressure –
Decrease in volume = Increase in pressure
P1V1 = P2V2
Chemical Equilibrium – Le Chatelier’s Principle
3. Changes in Pressure – A change in pressure on a system
affects only gaseous equilibrium that have unequal number of
moles of gaseous reactants vs. products.
Example – The reaction that occurs when HF is produced from
the reaction of F2 and H2.
F2(g) + H2(g) ↔ 2 HF(g)
Equal # of moles on both sides
What would happen to the equilibrium if pressure increases?
No Shift
What would happen to the equilibrium if pressure decreases?
No Shift
Chemical Equilibrium – Le Chatelier’s Principle
What effect do each of the following changes have on
the equilibrium position for this reversible reaction?
PCl5(g) + heat ↔ PCl3(g) + Cl2(g)
a. Addition of Cl2
Shift Left 
b. Increase in pressure
Shift Left 
c. Decrease in temperature
Shift Left 
d. Removal of PCl3 as its formed
Shift Right 
Chemical Equilibrium – Le Chatelier’s Principle
What effect do each of the following changes have on
the equilibrium position for this reversible reaction?
C(s) + H2O(g) + heat ↔ CO(g) + H2(g)
a. Decrease in temperature
Shift Left 
b. Decrease in pressure
Shift Right 
c. Removal of hydrogen
Shift Right 
d. Addition of water vapor
Shift Right 
Chemical Equilibrium –
Equilibrium Concentrations
When solving equilibrium problems
1. Write a balanced equation for the Reaction
2. List the Initial concentrations
3. Define the Change needed to reach equilibrium
4. Define the Equilibrium concentrations by applying the
change to the initial concentrations.
5. Write the equilibrium expression
6. Substitute the equilibrium concentrations into the equilibrium
expression.
7. Solve for the unknown
Chemical Equilibrium –
Equilibrium Concentrations
Carbon monoxide reacts with steam to produce carbon dioxide
and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate
the equilibrium concentrations of all species if 1.000 mol of each
reactant is mixed in a 1.000 L flask.
Reaction
Initial
1.000 M
1.000 M
0
0
Change
-x
-x
+x
+x
x
x
Equilibrium
CO(g) + H2O(g)
1.000 M - x 1.000 M - x
↔
CO2(g)
+
H2(g)
CO2 H2
1.000 M−x 1.000 M−x
1.000 M−x 2
Keq =
→ 5.10 =
=
x x
CO H2 O
x 2
Chemical Equilibrium –
Equilibrium Concentrations
Carbon monoxide reacts with steam to produce carbon dioxide
and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate
the equilibrium concentrations of all species if 1.000 mol of each
reactant is mixed in a 1.000 L flask.
Reaction
Initial
1.000 M
1.000 M
0
0
Change
-x
-x
+x
+x
x
x
Equilibrium
x = 0.307M
CO(g) + H2O(g)
↔
1.000 M - x 1.000 M - x
CO2(g)
+
H2(g)
[CO] = [H2O] = 0.693M
[CO2] = [H2] = 0.307M
Chemical Equilibrium –
Equilibrium Concentrations
Gaseous NOCl decomposes to form the gases NO and Cl2. At
35oC the equilibrium constant is 1.6 x 10-5. In an experiment in
which 1.0 mole of NOCl is placed in a 2.0 L flask, what are the
equilibrium concentrations?
Reaction
Initial
2NOCl
↔
2NO
+
Cl2
0.50 M
0
0
Change
- 2x
+ 2x
+x
Equilibrium
0.50 M - 2x
2x
x
2x
3
NO 2 Cl2
2x
4x
−5
Keq =
→ 1.6 x 10 =
=
NOCl 2
0.50 M −2x 2
0.50 M −2x 2
Chemical Equilibrium –
Equilibrium Concentrations
Gaseous NOCl decomposes to form the gases NO and Cl2. At
35oC the equilibrium constant is 1.6 x 10-5. In an experiment in
which 1.0 mole of NOCl is placed in a 2.0 L flask, what are the
equilibrium concentrations?
Reaction
Initial
2NOCl
↔
2NO
+
Cl2
0.50 M
0
0
Change
- 2x
+ 2x
+x
Equilibrium
0.50 M - 2x
2x
x
3
NO 2 Cl2
2x 2 x
4x
−5
Keq =
→
= 1.6 x 10 =
Quadratic?
NOCl 2
0.50 M −x 2
0.50 M −2x 2
Chemical Equilibrium –
Equilibrium Concentrations
Gaseous NOCl decomposes to form the gases NO and Cl2. At
35oC the equilibrium constant is 1.6 x 10-5. In an experiment in
which 1.0 mole of NOCl is placed in a 2.0 L flask, what are the
equilibrium concentrations?
because K is so small we know that the value of x is going to be TINY in
relation to2the initial concentration. So small, that if we subtract x
2 xa
3
If your
enough
(by
NOKeq isClsmall
2x
4x
−5
2
from
the
initial
concentration
and
round
to
the
Keq
=
→
= sig figs, the
magnitude
of 3 degrees),
you can = 1.6 x 10 correct
2
2
2
NOCl
0.50
M
−x
0.50
M
−2x
value has not changed.
ignore the x in the denominator
Chemical Equilibrium –
Equilibrium Concentrations
Gaseous NOCl decomposes to form the gases NO and Cl2. At
35oC the equilibrium constant is 1.6 x 10-5. In an experiment in
which 1.0 mole of NOCl is placed in a 2.0 L flask, what are the
equilibrium concentrations?
Reaction
Initial
2NOCl
0.50 M
0
0
Change
- 2x
+ 2x
+x
Equilibrium
0.50 M - 2x
2x
x
x = 0.010M
↔
2NO
+
Cl2
[NOCl] = 0.48M
[NO] = 0.020M
[Cl2] = 0.010M
Chemical Equilibrium –
Equilibrium Concentrations
Assume that the reaction for the formation of gaseous hydrogen
fluoride from hydrogen and fluorine has an equilibrium constant of
1.15 x 102 at a certain temperature. In a particular experiment,
3.000 moles of each component were added to a 1.500 L flask.
Calculate the equilibrium concentrations of all species.
Reaction
H2
+
F2
↔
2HF
Initial
2.00 M
2.00 M
2.00M
Change
Equilibrium
HF 2
-
X
2.00M – x
-
X
2.00 M – x
+ 2x
2.00 M + 2x
2.00 M +2x 2
2.00 M+2x 2
Keq =
→ 115 =
=
2.00 M−x 2.00 M−x
H2 F2
2.00 M−x 2
Chemical Equilibrium –
Equilibrium Concentrations
Assume that the reaction for the formation of gaseous hydrogen
fluoride from hydrogen and fluorine has an equilibrium constant of
1.15 x 102 at a certain temperature. In a particular experiment,
3.000 moles of each component were added to a 1.500 L flask.
Calculate the equilibrium concentrations of all species.
Reaction
H2
+
F2
↔
2HF
Initial
2.00 M
2.00 M
2.00M
Change
Equilibrium
x = 1.53M
-
X
2.00M – x
-
X
2.00 M – x
+ 2x
2.00 M + 2x
[H2] = [F2] = 0.47M
[HF] = 5.06M
Chemical Equilibrium –
Equilibrium Concentrations
At a certain temperature a 1.00 L flask initially contained 0.298 mol
PCl3(g) and 8.70 x 10-3 mol PCl5(g). After the system had reached
equilibrium, 2.00 x 10-3 mol Cl2(g) was found in the flask. Calculate
the equilibrium concentrations of all species and calculate the
equilibrium constant.
Reaction
Initial
PCl5
↔
PCl3
+
Cl2
0.00870 M
0.298 M
0
Change
- 0.00200
- x M
+ 0.00200
+x M
+x
Equilibrium
0.00670 M
0.300 M
0.00200 M
PCl3 Cl2
0.300 M 0.00200 M
Keq =
→
= 0.0896
PCl5
0.00670 M
Chemical Equilibrium – Equilibrium Constants
Equilibrium involving gases can also be described in
terms of pressure.
aA(g) + bB(g) ↔ cC(g) + dD(g)
𝑷𝑪
𝑲𝒑 =
𝑷𝑨
𝒄
𝒂
𝑷𝑫
𝑷𝑩
𝒅
𝒃
Kp represents an equilibrium constant in terms of partial
pressures for GASES ONLY.
Chemical Equilibrium –
Equilibrium Pressures
The reaction for the formation of nitrosyl chloride was studied at
25oC. The pressures at equilibrium were found to be:
PNOCl = 1.2 atm, PNO = 5.0 x 10-2 atm, PCl2 = 3.0 x 10-1 atm.
Write the equilibrium expression and calculate the value of Kp for
this reaction at 25oC.
2 NO(g) +
NO 2 Cl2
Kp =
→
NOCl
Cl2(g)
↔ 2 NOCl(g)
2
−2
−1
5.0×10 atm (3.0×10 atm)
1.2 atm
6.25x10-4
Chemical Equilibrium –
Equilibrium Pressures
Solid ammonium carbamate decomposes into ammonia and
carbon dioxide gas according to the equation below. At 25oC,
calculate Kp when 0.116 atm of NH3 & CO2 are present at
equilibrium.
NH4H2NCO2(s)
Kp = PNH
3
2
↔ 2 NH3(g) + CO2(g)
2
CO2 → 0.116 atm 0.116 atm
0.00156
Chemical Equilibrium –
Equilibrium Pressures
The equilibrium constant, Kp, for the synthesis of ammonia from
nitrogen and hydrogen is 4.31 x 10-4. In a certain experiment a
student starts with 0.862 atm of nitrogen and 0.373 atm of
hydrogen in a constant volume vessel at 375oC. Calculate the
partial pressure of all species when equilibrium is reached.
Reaction
Initial
N2(g)
+ 3H2(g)
↔
2NH3(g)
0.862 atm
0.373 atm
0
Change
-x
- 3x
+ 2x
Equilibrium
0.862 atm - x
0.373 atm – 3x
2x
Chemical Equilibrium –
Equilibrium Pressures
The equilibrium constant, Kp, for the synthesis of ammonia from
nitrogen and hydrogen is 4.31 x 10-4. In a certain experiment a
student starts with 0.862 atm of nitrogen and 0.373 atm of
hydrogen in a constant volume vessel at 375oC. Calculate the
partial pressure of all species when equilibrium is reached.
Kp =
NH3
2
N2 H2
3
−4
→ 4.31×10 =
x = 0.00220
2x
2
0.862 atm − x 0.373 atm − 3x
3
[H2] = 0.366 atm
[N2] = 0.860 atm
[NH3] = 0.00440 atm