1. Chapter 13 Equilibrium
Dr. Walker
DE Chemistry

2. Chemical Equilibrium
The state where the concentrations of all reactants and products remain constant with time
This occurs when the rate of the forward reaction equals the rate of the reverse reaction
All reactions carried out in a closed vessel will reach equilibrium

3. Reversible Reactions
A chemical reaction in which the products can react to re-form the reactants
Usually represented by a double sided arrow
Equilibrium can be to the left or the right depending on conditions
2 HgO(s)  2Hg(l) + O2(g)

4. Dynamic Equilibrium Reactions continue to take place
Reactant molecules continue to be converted to product (forward reaction)
Product continues to be converted to reactant (reverse reaction)
Forward and reverse reactions take place at the same rate at equilibrium

5. How Equilibrium Occurs
Beginning of reaction
Only reactant molecules exist, so only reactant molecules may collide
Middle
As product concentration increases, collisions may take place that lead to the reverse reaction
At equilibrium
Rates of forward and reverse reactions are identical

6. 2NO2(g)  2NO(g) + O2(g) Last chapter, we studied
reaction rates by looking
at changes in
concentration over time
Kinetics are irrelevant at
equilibrium because the
concentration is now
constant!
At this point, the concentrations are no longer changing
This reaction is now at equilbirium

7. The Haber Process N2 (g) + 3 H2 (g)  2 NH3 (g)
Hydrogen is consumed at 3x the rate of nitrogen
Ammonia is formed at 2x the rate at which nitrogen is consumed

8. Law of Mass Action jA + kB  lC + mD For the reaction:
Where K is the equilibrium constant, and is unitless

9. Product Favored Equilibrium
Large values for K signify the reaction is “product favored” (“equilibrium to right”)
When equilibrium is achieved, most reactant has been converted to product

10. Reactant Favored Equilibrium
Small values for K signify the reaction is “reactant favored” (“equilibrium to the left”)
When equilibrium is achieved, very little reactant has been converted to product

11. Writing an Equilibrium Expression
Write the equilibrium expression for the reaction:
2NO2(g)  2NO(g) + O2(g)
K = ???

12. Example
For the reaction N2 + 3 H2  2 NH3, the concentrations at equilibrium are as follows:
[N2] = M
[H2] = M
[NH3] = M
Calculate the equilibrium constant.

13. Example K = K = K = 6.03 x 10-2 [0.157]2 [0.763]3[0.921]
For the reaction N2 + 3 H2  2 NH3, the concentrations at equilibrium are as follows:
[N2] = M
[H2] = M
[NH3] = M
Calculate the equilibrium constant.
[NH3]2
[H2]3[N2]
K =
[0.157]2
[0.763]3[0.921]
K =
K = x 10-2

14. Equilibrium Expressions Involving Pressure
For the gas phase reaction:
3H2(g) + N2(g)  2NH3(g)
KP is equilibrium constant in terms of partial pressure
Dn = sum of gaseous product coefficients
– sum of gaseous reactant coefficients

15. Examples Using Pressure

16. Examples Using Pressure
[NOCl]2 [1.2]2 [NO]2[Cl2] [0.05]2[0.3]
Kp = =
= 1.9 x 103

17. Examples Using Pressure

18. Examples Using Pressure
1.9 x 103 = K [( Latm/mol K) (298 K)]-1
K = 4.7 x 104
From coefficients
2 – (2-1) = -1

19. Heterogeneous Equilibria
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present
Pure solids/liquids NOT included in equation
No concentration involved
Write the equilibrium expression for the reaction:
PCl5(s)  PCl3(l) + Cl2(g)
Pure
solid
Pure
liquid

20. jA + kB  lC + mD The Reaction Quotient
For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action.
jA + kB  lC + mD

21. Significance of the Reaction Quotient
If Q = K, the system is at equilibrium
If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved
If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

22. Writing Heterogeneous Equilibrium Expressions

23. Writing Heterogeneous Equilibrium Expressions
K = [Cl2] KP = PCl2 Other materials are pure
liquids or solids
K = [H2O]5 KP = (PH2O)5
All other materials are pure solids.
Coefficient stays on the equation

24. Determining Equilibrium Concentrations
This is a technique that will be used not just here, but in chapters as well.
As a result, it is critical that you can set up what we will go over without confusion….

25. Determining Equilibrium Concentrations
In typical chemical reactions
Reactants are broken down and their concentration decreases
Products are formed and their concentration increases
Higher [products] = higher K, higher [reactants] = lower K
Concentrations of all reagents change until equilibrium (Q = K is reached)

26. Solving for Equilibrium Concentration
Consider this reaction at some temperature:
H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0
Assume you start with 8 molecules of H2O and 6 molecules of CO. How many molecules of H2O, CO, H2, and CO2 are present at equilibrium?
Here, we learn about “ICE” – an important problem solving technique that we will use over the next few chapters
-Determines concentration of unknown reagents given an equilibrium constant

27. Solving for Equilibrium Concentration
H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0
Step #1: We write the law of mass action for the reaction:

28. Solving for Equilibrium Concentration
Step #2: We “ICE” the problem, beginning with the Initial concentrations
H2O(g) + CO(g)  H2(g) + CO2(g)
Initial:
Change:
Equilibrium: 8 6 -x -x +x +x
8-x
6-x x x

29. Solving for Equilibrium Concentration
Step #3: We plug equilibrium concentrations into our equilibrium expression, and solve for x
H2O(g) + CO(g)  H2(g) + CO2(g)
Equilibrium:
8-x
6-x x
x = 4

30. Solving for Equilibrium Concentration
Step #4: Substitute x into our equilibrium concentrations to find the actual concentrations
H2O(g) + CO(g)  H2(g) + CO2(g)
Equilibrium:
8-x
6-x x
x = 4
Equilibrium:
8-4=4
6-4=2 4

31. ICE Example
What will we do to solve this?

32. ICE Example What will we do to solve this? Set up an ICE table
equilibrium is 6.61 x 10-4 M

33. ICE Example What will we do to solve this? Set up an ICE table
equilibrium is 6.61 x 10-4 M
This means [I-] also is 6.61 x 10-4 M at equilibrium
Solve for x, which is [I3-], solve for K

34. ICE Example What will we do to solve this? Set up an ICE table
equilibrium is 6.61 x 10-4 M
This means [I-] also is 6.61 x 10-4 M at equilibrium
Solve for x, which is [I3-], solve for K

35. Le Chatelier’s Principle

36. LeChatelier’s Principle
When a system at
equilibrium is placed under
stress, the system will
undergo a change in such
a way as to relieve that
stress and restore a
state of equilibrium.
Henry Le Chatelier

37. Le Chatelier Translated:
When you take something away from a system at equilibrium, the system shifts in such a way as to replace some what you’ve taken away.
When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

38. LeChatelier Example #1
A closed container of ice and water is at equilibrium. Then, the temperature is raised.
Ice + Energy  Water
right
The system temporarily shifts to the _______ to restore equilibrium.

39. LeChatelier Example #2
A closed container of N2O4 and NO2 is at equilibrium. NO2 is added to the container.
N2O4 (g) + Energy  2 NO2 (g)
left
The system temporarily shifts to the _______ to restore equilibrium.

40. LeChatelier Example #3
A closed container of water and its vapor is at equilibrium. Vapor is removed from the system.
water + Energy  vapor
right
The system temporarily shifts to the _______ to restore equilibrium.

41. LeChatelier Example #4
A closed container of N2O4 and NO2 is at equilibrium. The pressure is increased.
N2O4 (g) + Energy  2 NO2 (g)
left
The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation.