Chem 2 - Chemical Equilibrium VI: Heterogeneous Equilibria

1. Chemical Equilibrium (Pt. 6)
Heterogeneous Equilibria
By Shawn P. Shields, Ph.D.
This work is licensed by Dr. Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0
International License.

2. Heterogeneous Equilibria
A heterogeneous equilibrium system is
one in which more than one phase is
present.
For instance, and reaction involving both
gases and solids would be a
heterogeneous system.

3. Heterogeneous Equilibria
Calculation of equilibrium concentrations or
partial pressures for heterogeneous equilibria is
very similar to the process for homogeneous
equilibria (already discussed).
The essential difference is that the
activity for solids and pure liquids in the
equilibrium constant expression is always
equal to “1”.

4. General Example for Heterogeneous
Equilibria
For reactions involving more than one phase
(solids, liquids, and gases),
𝐚𝐀(𝒂𝒒) + 𝐛𝐁(𝒂𝒒) ⇌ 𝐜𝐂(𝒂𝒒) + 𝐝𝐃(𝒔)
The relationship between the value of the
equilibrium constant K and the concentrations
of reactants and products is
𝐊 =
𝐂 𝐜
𝟏 𝐝
𝐀 𝐚 𝐁 𝐛
“1” to any
power is still
just “1” 

5. Example: Decomposition of CaCO3(s)
CaCO3 (s) in a sealed
container at 25C
Let time pass…
CaCO3(s)
CaCO3 (s) CO2 (g) + CaO (s)

6. Example: Decomposition of CaCO3(s)
Let time pass…
CaCO3(s)
CaCO3 (s) in a sealed
container at 25C
CaCO3 (s), CaO(s), and CO2 (g)
in a sealed container at 25C
Now, CaCO3(s)
and CaO(s)
CaCO3 (s) CO2 (g) + CaO (s)

7. Example: Decomposition of CaCO3(s) and
the Equilibrium Constant Expression
The activity for solids and pure
liquids is “1”
𝐊 =
𝑷 𝑪𝑶 𝟐
(𝟏)
(𝟏)
CaCO3 (s) CO2 (g) + CaO (s)

8. What Does This Mean?
It doesn’t matter how much CaCO3(s) we
start with, the partial pressure of CO2(g)
in the closed container is always the same
(at a given T).
𝐊 =
𝑷 𝑪𝑶 𝟐
(𝟏)
(𝟏)
CaCO3 (s) CO2 (g) + CaO (s)

9. Use the ICE Tables to organize given
information and variables for
heterogeneous equilibrium calculations.
ICE Tables and Heterogeneous Equilibria
I
C
E
CaCO3 (s) CO2 (g) + CaO (s)

10. Allow CaCO3(s) to decompose in a closed
container. At a certain temperature, the
equilibrium constant is 1.04. Determine
the partial pressure of CO2 once the
system reaches equilibrium.
Example Calculation
CaCO3 (s) CO2 (g) + CaO (s)

11. Allow CaCO3(s) to decompose in a closed container. At a
certain temperature, the equilibrium constant is 1.04.
Determine the partial pressure of CO2 once the system
reaches equilibrium.
ICE Tables and Heterogeneous Equilibria
I
C
E
CaCO3 (s) CO2 (g) + CaO (s)
0

12. Notice: We don’t specify “how much” of each
solid is decomposed or formed.
ICE Tables and Heterogeneous Equilibria
I
C
E
CaCO3 (s) CO2 (g) + CaO (s)
0
+xused some made some

13. ICE Tables and Heterogeneous Equilibria
CaCO3 (s) CO2 (g) + CaO (s)
I
C
E
0
+xused some
𝐊 =
𝑷 𝑪𝑶 𝟐
(𝟏)
(𝟏)
𝟏. 𝟎𝟒 =
𝒙 (𝟏)
(𝟏)
𝐱 = 𝟏. 𝟎𝟒
x
made some
𝑷 𝑪𝑶 𝟐
= 𝟏. 𝟎𝟒

14. A system has to have at least
“some” of everything at equilibrium.
As long as we have a little of each solid,
the system has the same equilibrium
partial pressure for CO2(g).
Central Concept
CaCO3 (s) CO2 (g) + CaO (s)

15. If a reaction has to have at least “some”
of everything at equilibrium, would it
matter if we added more CaCO3?
(Assume no effect on volume.) Would the
partial pressure for CO2(g) change?
NO!
Central Concept- Adding Solids (or Pure
Liquids)
CaCO3 (s) CO2 (g) + CaO (s)
𝐊 =
𝑷 𝑪𝑶 𝟐
(𝟏)
(𝟏)

16. Next up,
The Reaction Quotient Q for
Nonequilibrium Systems
(Pt 7)