1. CHEMICAL EQUILIBRIUM

2. Chemical Equilibrium Reversible Reactions:
A chemical reaction in which the products
can react to re-form the reactants
Chemical Equilibrium:
When the rate of the forward reaction
equals the rate of the reverse reaction
and the concentration of products and
reactants remains unchanged
2HgO(s)  2Hg(l) + O2(g)
Arrows going both directions (  ) indicates equilibrium in a chemical equation

3. 2NO2(g)  2NO(g) + O2(g) Remember this from Chapter 12?
Why was it so important to measure reaction rate at the start of the reaction
(method of initial rates?)

4. 2NO2(g)  2NO(g) + O2(g)

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

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

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

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

9. N2(g) + 3H2  2NH3
The following equilibrium concentrations were observed for the above process at 127oC:
[NH3] = 3.1 x 10-2 mol/L
[N2] = 8.5 x 10-1 mol/L
[H2] = 3.1 x 10-3 mol/L
Calculate the value of K at 127oC for this reaction.
B. Calculate the value of the equilibrium constant at 127oC for the reaction:
2NH3(g)  N2(g) + 3H2(g)
Calculate the value of the equilibrium constant at 127oC for the reaction given by the equation:
1/2N2(g) + 3/2H2(g)  NH3(g)

10. [NH3]2 A. (3.1 x 10-2)2 K = = [N2] [H2]3 (8.5 x 10-1)(3.1 x 10-3)3
Solution
[NH3]2 A.
(3.1 x 10-2)2
K = =
[N2] [H2]3
(8.5 x 10-1)(3.1 x 10-3)3
= 3.8 x 104
What are the units for K?

11. [N2][H2]3 K’ = [NH3]2 [N2][H2]3 K’ = 1 1 = = [NH3]2 K 3.8 x 104
Solution
B. This reaction is written in the reverse order from the equation given in part A.
[N2][H2]3
K’ =
[NH3]2
Which is the reciprocal of the expression used in part A.
[N2][H2]3
K’ = 1 1 = =
[NH3]2 K
3.8 x 104
= 2.6 x 10-5

12. ( ) [NH3] K’’ = [N2]1/2[H2]3/2 1/2 [NH3] [NH3]2 = [N2]1/2[H2]3/2
Solution
C. We use the law of mass action:
[NH3]
K’’ =
[N2]1/2[H2]3/2
If we compare this expression to that obtained in part A.
we see that since:
1/2 ( )
[NH3]
[NH3]2 =
[N2]1/2[H2]3/2
[N2][H2]3
Thus: K’’ = K1/2 =
(3.8 x 104)1/2 = 1.9 x 102

13. Conclusions about Equilibrium Expressions
The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse
2NO2(g)  2NO(g) + O2(g)
2NO(g) + O2(g)  2NO2(g)

14. Conclusions about Equilibrium Expressions
When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power.
2NO2(g)  2NO(g) + O2(g)
NO2(g)  NO(g) + ½O2(g)

15. Equilibrium Expressions Involving Pressure
For the gas phase reaction:
3H2(g) + N2(g)  2NH3(g)
Δn = the sum of the coefficients of the gaseous products minus
the sum of the coefficients of the gaseous reactants.

16. Heterogeneous Equilibria
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present
Write the equilibrium expression for the reaction:
PCl5(s)  PCl3(l) + Cl2(g)
Pure
solid
Pure
liquid

17. 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

18. 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

19. Try me! For the synthesis of ammonia at 500oC , the
equilibrium constant is 6.0 x Predict
the direction in which the system will shift
to reach equilibrium in each of the following
cases.
[NH3]0 = 1.0 x 10-3M; [N2]0 = 1.0 x 10-5M; [H2]0 = 2.0 x 10-3M
[NH3]0 = 2.0 x 10-4M; [N2]0 = 1.5 x 10-5M; [H2]0 = 3.54 x 10-1M
[NH3]0 = 1.0 x 10-4M; [N2]0 = 5.0M; [H2]0 = 1.0 x 10-2M

20. First we calculate the value of Q
Q = 1.3 x 107
Since K = 6.0 x 10-2, Q is MUCH greater
than K. To attain equilibrium, the concentrations
of the products must be decreased and
the concentrations of the reactants increased.
So…. The system will shift to the LEFT:
N2 + 3H NH3

21. N2 + 3H2 2NH3 B. First we calculate the value of Q Q = 6.01 x 10-2
In this case Q = K, so the system is at
equilibrium. NO shift will occur.
C. Calculate Q
Q = 2.0 x 10-3
Here Q is less than K, so the system will shift
to the right to attain equilibrium by increasing
the concentration of the product and decreasing
the reactant concentrations
N2 + 3H NH3

22. 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” – the most important problem solving technique in chemistry! Lol (ok…maybe sometimes).

23. 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:

24. 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

25. 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

26. 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
Equilibrium:
8-4=4
6-4=2 4

27. A Weak Base Equilibrium Problem
What are the equilibrium concentrations of NH3 and NH4+ in a 0.5M solution of ammonia? (Kb x 10-5)
Balanced Chemical Rxn and…ICE it!
NH3 + H2O  NH4+ + OH- I C E
0.50
- x +x +x x x x

28. A Weak Base Equilibrium Problem
Set up the law of mass action
NH3 + H2O  NH4+ + OH- E x x x

29. A Weak Base Equilibrium Problem
Solve for x, which is also [OH-]
NH3 + H2O  NH4+ + OH- E x x x
[OH-] = 3.0 x 10-3 M

30. Gaseous NOCl decomposes to form the
gases NO and Cl2. At 35oC the equilibrium
constant is 1.6 x In an experiment
in which 1.0 mol NOCl is placed in a 2.0 L
flask, what are the equilibrium concentrations?
2NOCl(g) NO(g) + Cl2(g)

31. Start the following problems: Work cooperatively!
Practice time!
Start the following problems: Work cooperatively!
Pages :
#42,44,46,52,68

32. A Weak Base Equilibrium Problem
What are the equilibrium concentrations of NH3 and NH4+ in a 0.5M solution of ammonia? (Kb x 10-5)
Balanced Chemical Rxn and…ICE it!
NH3 + H2O  NH4+ + OH- I C E
0.50
- x +x +x x x x

33. A Weak Base Equilibrium Problem
Set up the law of mass action
NH3 + H2O  NH4+ + OH- E x x x

34. A Weak Base Equilibrium Problem
Solve for x, which is also [OH-]
NH3 + H2O  NH4+ + OH- E x x x
[OH-] = 3.0 x 10-3 M

35. Start the following problems:
Practice time!
Start the following problems:
Pages :
#42,44,46,52,68

36. Assume that the reaction foe 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, mol of each component was
added to L flask.
Calculate the equilibrium concentrations of all
species.

37. THE HABER PROCESS

39. LeChatelier’s Principle
WHO
AM I?
Fritz Haber
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.
Translated: The system undergoes a temporary shift in order to restore equilibrium.

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

41. 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)
The system temporarily shifts to the _______ to restore equilibrium.
left

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

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

44. LeChatelier Example #5
The preparation of liquid phosphorus trichloride by the below reaction. The volume is decreased.
P4 (s) Cl2 (g)  4 PCl3 (l)
The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation.
right

45. LeChatelier Example #6
The preparation of gaseous phosphorus pentachloride by the below reaction. The volume is decreased.
PCl3 (g) + Cl2 (g)  PCl5 (g)
The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation.
right

46. LeChatelier Example #7
The preparation of phosphorus trichloride with ammonia by the below reaction. The volume is decreased.
PCl3 (g) + 3 NH3 (g)  P(NH2)3 (g) + 3 HCl (g)
Both sides have four gaseous molecules. A change in volume will have no effect on the equilibrium position. There is NO SHIFT in this case.

47. LeChatelier Example #8 N2O4 (g)  2 NO2 (g) ΔHo = 58 kJ Change Shift
Right
Left
None
Addition of N2O4 (g)
Addition of NO2 (g)
Removal of N2O4 (g)
Removal of NO2 (g)
Addition of He (g)
Decrease container volume
Increase container volume
Increase temperature
Decrease temperature