1. Chapter Fourteen
Chemical Equilibrium

2. Chemical equilibrium is achieved when:
Chapter Fourteen/ Chemical Equilibrium
The Concept of Equilibrium
and the Equilibrium Constant
Few chemical reactions proceed in only one direction, most are reversible.
At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process begins to take place and reactant molecules are formed from product molecules.
Chemical equilibrium is achieved when:
The rates of the forward and reverse reactions are equal and
The concentrations of the reactants and products remain constant
Physical equilibrium
H2O (l)
H2O (g)
Chemical equilibrium
N2O4 (g)
2NO2 (g)

3. Start with NO2 Start with N2O4 Start with NO2 & N2O4
Chapter Fourteen/ Chemical Equilibrium
The Concept of Equilibrium
and the Equilibrium Constant
N2O4 (g) NO2 (g)
equilibrium
equilibrium
equilibrium
Start with NO2
Start with N2O4
Start with NO2 & N2O4

4. Where K is equilibrium constant.
Chapter Fourteen/ Chemical Equilibrium
The Concept of Equilibrium
and the Equilibrium Constant
Where a, b, c, & d: are the stoichiometric coefficients for A, B, C , & D.
Where K is equilibrium constant.
The equilibrium constant: for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, K, called the equilibrium constant.
Always
The concentration of solids and pure liquids and solvent are not included in the expression for the equilibrium constant.
K dos not have a unit
Law of Mass Action

5. Chapter Fourteen/ Chemical Equilibrium
The Concept of Equilibrium
and the Equilibrium Constant
The magnitude of the equilibrium constant tells us whether an equilibrium reaction favors the products or reactants.
If K is much greater than 1 (that is, K >> 1), the equilibrium will lie to the right and favors the products.
if the equilibrium constant is much smaller than 1 (that is, K << 1), the equilibrium will lie to the left and favor the reactants.
In this context, any number greater than 10 is considered to be much greater than 1, and any number less than 0.1 is much less than 1.

6. Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria
homogeneous equilibrium applies to reactions in which all reacting species are in the same phase.
Example
K can be given as
Note that the subscript in Kc indicates that the concentrations of the reacting species
are expressed in molarity or moles per liter.

7. Relationship between Kc and Kp: Kc ≠ Kp
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria
Relationship between Kc and Kp:
Kc ≠ Kp
∆n = moles of gaseous products – moles of gaseous reactants
Kc = Kp, when ∆n = 0

8. Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria
Example
Write expressions for Kc , and Kp if applicable, for the following reversible reactions at equilibrium?

9. Because there are no gases then there is no Kp
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria
Always do not write solvent in the expression of equilibrium constant. Normally water is solvent
Because there are no gases then there is no Kp

10. Writing Equilibrium Constant Expressions
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria

11. Because there are no gases then there is no Kp
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria
Always do not write solvent in the expression of equilibrium constant. Normally water is solvent
Because there are no gases then there is no Kp

12. The following equilibrium process has been studied at 230 oC:
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria
Example
The following equilibrium process has been studied at 230 oC:
In one experiment, the concentrations of the reacting species at equilibrium are found to be [NO] = M, [O2] = M, and [NO2] = 15.5 M. Calculate the equilibrium constant (Kc) of this reaction at this temperature.
Stopped here

13. The equilibrium constant Kp for the decomposition of
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria
Example
The equilibrium constant Kp for the decomposition of
Is found to be at 250 oC . If the equilibrium partial pressures of PCl5, and PCl3 are atm and atm, respectively, what is the equilibrium partial pressure of Cl2 at 250 oC.

14. ∆n = moles of products – moles of reactants
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Homogeneous Equilibria
Example
The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
∆n = moles of products – moles of reactants
CO (g) + Cl2 (g) COCl2 (g)
[COCl2]
[CO][Cl2]
Dn = 1 – 2 = -1
R =
Kc =
T = = 347 K =
0.14
0.012 x 0.054
= 220
Kp = 220 x ( x 347)-1 = 7.7

15. Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Heterogeneous Equilibria
heterogeneous equilibrium applies to reactions in which all reacting species are in different phases.
Example
All rules applied for homogeneous equilibria also applies for heterogeneous.
CaCO3 (s) CaO (s) + CO2 (g)

16. Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Example
Write the equilibrium constant expression Kc, and Kp if applicable, for each of the following heterogeneous systems:
Heterogeneous Equilibria

17. X Writing Equilibrium Constant Expressions Heterogeneous Equilibria
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Heterogeneous Equilibria X

18. AgCl is solid therefore it dose not go to equilibrium constant.
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
AgCl is solid therefore it dose not go to equilibrium constant.
There are no gases thus there are no Kp,.
Heterogeneous Equilibria

19. Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
P4 is solid and PCl3 is liquid therefore they do not go to constant of equilibrium.
Heterogeneous Equilibria

20. Consider the following heterogeneous equilibrium:
Chapter Fourteen/ Chemical Equilibrium
Writing Equilibrium Constant Expressions
Example
Consider the following heterogeneous equilibrium:
At 800 oC, the pressure of CO2 is atm. Calculate (a) Kp and (b) Kc for the reaction at this temperature.
CaCO3 and CaO are solid therefore they do not go to equilibrium of constant.
(b)
Heterogeneous Equilibria
R =
T = = 1073 K
∆n = moles of products – moles of reactants
Dn = 1 – 0 = 1

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