1. Chemical Equilibrium
Portions adapted from

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
A(g) B(g)

3. The Equilibrium Condition
Dynamic Chemical Equilibrium
Reversible Reactions
Equilibrium is a DYNAMIC situation!!
Cars crossing a bridge analogy or workers going on lunch break as we will use in future.

4. The Equilibrium Constant (Keq or Kc)
aA + bB  cC + dD
At a given temperature, Keq for a reaction is always the same.
Keq has no units. (K, Kc, or Keq may be used)
The numerical value of K indicates the extent to which the reactants are converted to products.
K>>1 Signifies the reaction is “product favored”
K=1
K<<1 Signifies the reaction is “reactant favored”
[] = concentration at equilibrium in Molarity (mol/L)
K is UNITLESS!!
K>>1: When equilibrium is achieved, most reactant has been converted to product
K=1
K<<1: When equilibrium is acheved, very little reactant has been converted to product

5. Write the equilibrium expression for the following reactions:
4NH3(g) + 7O2(g)  4NO2(g) + 6H20(g) C3H8(g) + O2(g)  CO2(g) + H2O(g) PCl5(g)  PCl3(g) + Cl2(g)
Be sure equation is balanced!!

6. a. Calculate the value of Keq at 127°C for this reaction.
The following equilibrium concentrations were observed for the Haber process at 127°C :
[NH3] = 3.1 x 10-2 mol/L
[N2] = 8.5 x 10-1 mol/L
[H2] = 3.1 x 10-3 mol/L
a. Calculate the value of Keq at 127°C for this reaction.
N2(g) + 3 H2(g) ↔ 2 NH3(g)
b. Calculate the value of the equilibrium constant at 127°C for the reaction
2 NH3(g) ↔ N2(g) + 3 H2(g)
Answer to a: Keq = 3.8 x 104
Answer to b: K’eq= 2.6 x 10-5

7. Manipulating K Expressions
If all of the coefficients in a balanced equation are multiplied by a number, each term in the K expression must be raised to that power. Knew = (Koriginal)n
The K for a reverse reaction is the reciprocal of the K for the forward reaction.
For a reaction which is the sum of two or more reactions, the overall K is the product of the expressions for K for the individual steps: Koverall = K1 x K2 x K3 …
For a reaction at a given temperature, there are many equilibrium positions (set of equilibrium concentrations)but only one value for K.

8. Calculate the value of the equilibrium constant at 127°C for the reaction given by the equation
½ N2(g) + 3/2 H2(g)  NH3(g)
A: 1.9 x 102 or square root of (3.8x104)

9. At 25°C, the following reactions have the equilibrium constants shown:
2 CO(g) + O2(g) ↔ 2 CO2(g) Kc = 3.3 x 1091
2 H2(g) + O2(g) ↔ 2 H2O(g) Kc = 9.1 x 1080
Use this data to calculate Kc for the reaction
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
Square root of 3.3 x 1091 multiplied by
Plus the square root and reciprocal of 9.1 x 1080
= 1.9x105

10. Experiment 1 Experiment 2
The following results were collected for two experiments involving the reaction at 600°C between gaseous sulfur dioxide and oxygen to form gaseous sulfur trioxide:
Experiment 1 Experiment 2
Show that the equilibrium constant is the same in both cases.
K1=4.36 ; K2=4.32

11. Equilibrium Expressions Involving Gases & Pressure (Kp)
PV=nRT For the gas phase reaction: 3H2(g) + N2(g)  2NH3(g) PNH3, PN2, PH2 are equilibrium partial pressures

12. Example The reaction for the formation of nitrosyl chloride
2 NO(g) + Cl2(g) ↔ 2 NOCl(g)
Was studied at 25°C. 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
Calculate the value of Kp for this reaction at 25°C.
1.9 x 103

13. Relationship between Keq and Kp
NOT INTERCHANGEABLE!!
Kp = Keq(RT)Δn
Δn = moles of gas product – moles of gas reactants
Sum of the products coefficients minus the sum of the reactants coefficients.
What if there is no change in the number of moles of gas?

14. Using the value of Kp obtained in previous example problem, calculate the value of K at 25°C for the reaction:
2 NO(g) + Cl2(g) ↔ 2 NOCl(g)
4.6 x 104

15. 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)
Homogeneous all reactants and products are in the same phase.
Heterogeneous reactants and products are in different phases.
K=[Cl2]
Kp=Pcl2

16. Write the expressions for K and Kp for the following processes:
a. CaCO3(s) ↔ CaO(s) + CO2(g)
b. 2 CO(g) ↔ CO2(g) + C(s)
c. 2 Hg(l) + Cl2(g) ↔ Hg2Cl2(s)
d. NH3(g) + HCl(g) ↔ NH4Cl(s)
e. Ag2CrO4(s) ↔ 2 Ag+(aq) + CrO42-(aq) – solubility of a salt
f. 2 NaHCO3(s) ↔ Na2CO3(s) + H2O(g) + CO2(g)
g. CuSO4∙5H2O(s) ↔ CuSO4(s) + 5 H2O(g)

17. Calculate Kc for the following reaction:
CaCO3(s) ↔ CaO(s) + CO2(g) Kp = 2.1 x 10-4 at 1000 K

18. Individual Practice Part 1
Page 614
#17, 18, 19, 21, 25, 27, 29
Unit 6 Review Packet

19. Applications of the Equilibrium Constant (Reaction Quotient, Q)
aA + bB ↔ cC + dD
Qc = [C]c[D]d / [A]a[B]b
Qp = PCc PDd / PAa PBb
Q < K
 Q > K
 Q = K

20. Example
For the synthesis of ammonia at 500°C, 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:
a. [NH3]0 = 1.0 x 10-3 M; [N2]0 = 1.0 x 10-5 M; [H2]0 = 2.0 x 10-3 M
A: shifts left   
b. [NH3]0 = 2.00 x 10-4 M; [N2]0 = 1.50 x 10-5 M; [H2]0 = 3.54 x 10-1 M
A: no shift 
c. [NH3]0 = 1.0 x 10-4 M; [N2]0 = 5.0 M; [H2]0 = 1.0 x 10-2 M
A: shifts right

21. Equilibrium Calculations (ICE)
Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp = At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2(g).
Answer: 0.600atm

22. Equilibrium Calculations (ICE)
At a certain temperature in a 1.00-L flask initially contained 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. Gaseous PCl5 decomposes according to the following reaction
PCl5(g) ↔ PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of all species and the value of K.
A: [Cl2] = 2.00 x 10-3 M
[PCl3] = M
[PCl5] = 6.70 x 10-3 M
K = 8.96 x 10-2

23. Equilibrium Calculations (ICE)
Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is Calculate the equilibrium concentrations of all the species if mol of each component is mixed in a L flask.
A: [CO] = [H2O] = M
[CO2] = [H2] = M

24. Equilibrium Calculations (ICE)
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, mol of each component was added to a L flask. Calculate the equilibrium concentrations of all species.
A: [H2] = [F2] = M
[HF] = M

25. Equilibrium Calculations (ICE)
What are the steps to solving equilibrium problems?

26. Equilibrium Calculations
Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00 x 102. Suppose HI at x 10-1 atm, H2 at x 10-2 atm, and I2 at x 10-3 atm are mixed in a L flask. Calculate the equilibrium pressures of all species.
A: PHI = 4.29 x 10-1 atm
PH2 = 4.55 x 10-2 atm
PI2 = 4.05 x 10-2 atm

27. Le Chatelier’s Principle POGIL
Partner (Work TOGETHER!)
Complete in your notes!
Search for/identify “Key Ideas”
Share out responses and Key Ideas
_________minutes

28. Stressing a system
If a stress is applied to a system, the system will shift in the direction that relieves the stress
Change in Concentration Change in Volume
Change in Pressure Change in Temperature

29. Le Châtelier’s Principle:
1. The effect of a change in concentration
Example:
Predict the direction of the shift of the equilibrium position in response to each of the following changes in conditions:
As4O6(s) + 6 C(s) ↔ As4(g) + 6 CO(g)
a. Addition of carbon monoxide.
b. Addition or removal of carbon or tetraarsenic hexoxide.
c. Removal of gaseous arsenic (As4).

30. Changes in Concentration
Option A:
If you increase the reactant, the reaction shifts to the right to produce more product.
Option B:
If you increase the product, the reaction shifts to the left to produce more reactants.

31. Le Châtelier’s Principle:
2. The Effect of a Change in Pressure
Three ways to change the pressure of a reaction system involving gaseous components: 1. 2. 3.

32. Change in Volume/Pressure
Look at equation and pay attention to only gaseous substances (g)
If you increase the pressure (ie. decrease the volume) the reaction will shift to the side with fewest gas particles

33. Pressure Example
Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced.
a. P4(s) + 6 Cl2(g) ↔ 4 PCl3(l) 3
b. PCl3(g) + Cl2(g) ↔ PCl5(g)
c. PCl3(g) + NH3(g) ↔ P(NH2)3(g) + 3 HCl(g)

34. Le Châtelier’s Principle:
3. The Effect of a Change in Temperature
Remember, the value of K changes with temperature! Changes in concentration and pressure do not alter the equilibrium constant, K.
N2(g) + 3 H2(g) ↔ 2 NH3(g) + 92 kJ
556 kJ + CaCO (s) ↔ CaO(s) + CO2(g)

35. Changes in Temperature
Determine if reaction is exothermic or endothermic.
-Exothermic—heat is a product
-Endothermic—heat is a reactant
Exothermic
-Add heat, left shift
-Remove heat, right shift
Endothermic
-Add heat, right shift
-Remove heat, left shift

36. Temperature Example
For each of the following reactions, predict how the value of K changes as the temperature is increased. a. N2(g) + O2(g) ↔ 2 NO(g) ΔH° = 181 kJ b. 2 SO2(g) + O2(g) ↔ 2 SO3(g) ΔH° = –198 kJ

37. Le Châtelier’s Principle:
4. The effect of adding a catalyst
Example:
The reaction
N2O4(g) ↔ 2 NO2(g)
Is endothermic, with ΔH° = kJ. How will the amount of NO2 at equilibrium be affected by:
a. adding N2O4
b. increasing the volume of the container
c. raising the temperature
d. adding a catalyst to the system
e. Which of these changes alters the value of Kc?

38. Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)
Practice Catalyst
Iron (III) oxide reacts with carbon monoxide in a blast furnace, reducing it to iron metal:
Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)
Use Le Châtelier’s Principle to predict the direction of net reaction when an equilibrium mixture is disturbed by:
a. removing CO2
b. removing CO
c. adding CO
d. adding Fe2O3
e. decreasing the volume of the container
f. Which of these changes alters the value of Kc?

39. Let’s Practice Concentration: N2 (g) + 3H2 (g) 2NH3 (g)
Q: What type of shift will occur when the following changes are made to the system?
Removing hydrogen from the system
Adding ammonia to the system
Adding hydrogen to the system

40. Let’s Practice…Again Pressure/Volume: 2SO2 (g) + O2 (g) 2SO3 (g)
Q: What type of shift will occur when the following changes are made to the system?
How many gas particles are on the reactant and product side of the reaction?
You increase the volume?
You increase the pressure?

41. Let’s Practice…Last One
Temperature:
C2H2 (g) H2O (g) CH3CHO (g) ΔH = -151 kJ
Q: Would you raise or lower the temperature to obtain the following results?
An increase in the amount of CH3CHO (g)
A decrease in the amount of C2H2 (g)
An increase in the amount of H2O (g)

42. Individual Practice Page 615 Unit 6 Review Packet Equilibrium
33, 35, 43, 45, 47, 50, 53, 55, 57, 59, 61
Unit 6 Review Packet Equilibrium