1. Ch. 13 Equilibrium

2. Chemical Equilibrium
The state where the concentrations of all reactants and products remain constant with time.
On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.
(BDVD)

3. Dynamic Equilibrium Reactions continue to take place.
A + B C + D ( forward)
C + D A + B (reverse)
Initially there is only A and B so only the forward reaction is possible
As C and D build up, the reverse reaction speeds up while the forward reaction slows down.
Eventually the rates are equal.

4. Forward Reaction
Reaction Rate
Equilibrium
Reverse reaction
Time

5. What is equal at Equilibrium?
Rates are equal.
Concentrations are not.
Rates are determined by concentrations and activation energy.
The concentrations do not change at equilibrium.

6. 13.2 Law of Mass Action jA + kB lC + mD j, k, l, m are coefficients
The law of mass action is represented by the equilibrium expression:
K = [C]l[D]m PRODUCTSpower [A]j[B]k REACTANTSpower
K is called the equilibrium constant.
is how we indicate a reversible reaction. [x] represents concentration.

7. Playing with K If we write the reaction in reverse. lC + mD jA + kB
Then the new equilibrium constant is
K’ = [A]j[B]k = 1/K [C]l[D]m

8. K’ = [A]nj[B]nk = ([A] j[B]k)n = Kn [C]nl[D]nm ([C]l[D]m)n
Playing with K
If we multiply the equation by a constant
njA + nkB nlC + nmD
Then the equilibrium constant is
K’ = [A]nj[B]nk = ([A] j[B]k)n = Kn [C]nl[D]nm ([C]l[D]m)n

9. K is CONSTANT At any temperature. Temperature affects rate.
The equilibrium concentrations don’t have to be the same only K.
Equilibrium position is a set of concentrations at equilibrium.
One value at each temperature, but there are an unlimited number of possibilities.
Usually written without units.

10. Equilibrium Expression
4NH3(g) + 7O2(g) NO2(g) + 6H2O(g)
What is the equilibrium expression for this reaction? #19a
N2(g) + O2(g) NO(g)
What is the equilibrium expression for this reaction? #19b
N2O4 (g) 2NO2(g)

11. Calculate K N2(g) + 3H2(g) 2NH3 (g)
In the above reaction, at a given temperature. K =
Calculate the value of K in the following reactions. #21ab
1/2N2(g) + 3/2H2(g) NH3(g)
2NH3(g) N2(g) + 3H2(g)

12. Calculate K N2(g) + 3H2(g) 2NH3(g) Initial At Equilibrium
[N2]0 =1.000 M [N2] = 0.921M
[H2]0 =1.000 M [H2] = 0.763M
[NH3]0 =0 M [NH3] = 0.157M

13. Calculate K N2(g) + 3H2(g) 2NH3(g) Initial At Equilibrium
[N2]0 = 0 M [N2] = M
[H2]0 = 0 M [H2] = M
[NH3]0 = M [NH3] = 0.203M
K is the same no matter what the amount of starting materials (at the same temperature).

14. 13.3 Equilibrium and Pressure
Some reactions are gaseous
PV = nRT
P = (n/V)RT
P = CRT
C is a concentration in moles/Liter
C = P/RT

15. Equilibrium and Pressure
 2SO2(g) + O2 (g) 2SO3 (g)
In term of partial pressures:
Kp = (PSO3) (PSO2)2 (PO2)
In terms of concentration:
Kc = [SO3] [SO2]2 [O2]

16. Practice N2(g) + 3H2(g) 2NH3(g)
In the reaction above, the following equilibrium pressures were observed.
PNH3 = 0.89 atm
PN2 = 0.62 atm
PH2 = atm
Calculate the value for the equilibrium constant Kp at this temperature.

17. K v. Kp For jA + kB lC + mD Kp = K(RT)n
n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.

18. Practice Kp = K(RT)n At 25°C, K = 2.01 x 104 for the reaction
CO(g) + Cl2(g) COCl2(g)
What is the value of Kp at this temperature?

19. Practice #32
For which reaction in #31 is Kp = K?

20. Homogeneous Equilibria
So far every example dealt with reactants and products where all were in the same phase.
We can use K in terms of either concentration or pressure.
Units depend on reaction.

21. 13.4 Heterogeneous Equilibria
Are equilibria that involve more than one phase.
If the reaction involves pure solids or pure liquids, the concentration of the solid or the liquid doesn’t change.
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.
As long as they are not used up we can leave them out of the equilibrium expression.

22. For Example H2(g) + I2(s) 2HI(g) K = [HI]2 [H2][I2] K = [HI]2 [H2]
But the concentration of I2 does not change, therefore:
K = [HI] [H2]

23. 13.5 Applying the Equilibrium Constant
Reactions with large K (>>1), essentially to completion. Large negative E.
Reactions with small K (<<1) consist mostly of reactants.
Time to reach equilibrium is related to rate and AE. It is not related to size of K.

24. The Reaction Quotient
Tells you the directing the reaction will go to reach equilibrium
Calculated the same as the equilibrium constant, but for a system not at equilibrium by using initial concentrations.
Q = [Products]coefficient [Reactants] coefficient
Compare value to equilibrium constant

25. What Q tells us If Q<K Not enough products Shift to right If Q>K
Too many products
Shift to left
If Q=K system is at equilibrium

26. Using the Reaction Quotient
For the reaction
2NOCl(g) NO(g) + Cl2(g)
K = 1.55 x 10-5 M at 35ºC
In an experiment mol NOCl, mol NO(g) and mol Cl2 are mixed in 2.0 L flask.
Which direction will the reaction proceed to reach equilibrium?

27. Using the Reaction Quotient
For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0x10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases. Ex. 13.7
[NH3]0 = 1.0 x 10-3 M; [N2]0 = 1.0 x 10-5 M; [H2]0 = 2.0 x 10-3 M
[NH3]0 = 1.0 x 10-4 M; [N2]0 = 5.0 M;
[H2]0 = 1.0 x 10-2 M

28. Problems Involving Pressure
For the reaction N2O4(g) NO2(g) KP = atm. At equilibrium, the pressure of N2O4 was found to be 2.71 atm? Calculate the equilibrium pressure of NO2(g) EX.13.8

29. Problems Involving Pressure
At a certain temperature a 1.00 L flask initially contained mol PCl3(g) and 8.70 x 10-3 mol PCl5(g). After the system reached equilibrium, 2.00 x 10-3 mol Cl2(g) was found in the flask. Calculate the equilibrium concentrations of all species and the value of K. EX.13.9
PCl5(g) PCl3(g) + Cl2(g)

30. No Equilibrium [x] Ex
Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the equilibrium constant is Calculate the equilibrium concentrations of all species if 1.00 mol of each component is mixed in a 1.00 L flask.

31. Intro to ICE 2 SO3(g) 2SO2 (g) + O2(g)
At a certain temperature mol SO3 is placed into a 3.0 L rigid container. The SO3 dissociates during the reaction. At equilibrium, 3.0 mol SO2 is present. What is the value of K for this reaction?
(#43)

32. Intro to ICE
2NH3 N2 + 3H2
At a certain temperature. 4.0 mol NH3 is introduced into a 2.0 L container. The NH3 partially dissociates during the reaction. At equilibrium, 2.0 mol NH3 remains. What is the value of K for this reaction?
(#44) WAssign

33. What if you’re not given equilibrium concentration?
H2 + F2 2HF
The Equilibrium constant for the above reaction is 115 at a certain temperature mol of each component was added to a L flask. Calculate the equilibrium concentrations of all species.
(ex.13.11)

34. Practical Example H2(g) + F2(g) 2HF(g) K = 1.15 x 102 at 25ºC
Calculate the equilibrium concentrations if a 3.00 L container initially contains 3.00 mol of H2 and 6.00 mol F2 .
[H2]0 = 3.00 mol/3.00 L = 1.00 M
[F2]0 = 6.00 mol/3.00 L = 2.00 M
[HF]0 = 0

35. Q= 0<K so more product will be formed.
Assumption since K is large reaction will go to completion.
Stoichiometry tells us H2 is LR, it will be smallest at equilibrium let it be x
Set up table of initial, change and equilibrium in concentrations.

36. H2(g) F2(g) 2HF(g) Initial 1.00 M 2.00 M 0 M Change
Equilibrium
For H2 and F2 the change must be -X
Using to stoichiometry HF must be +2X
Equilibrium = initial + change

37. H2(g) F2(g) 2HF(g) Initial 1.00 M 2.00 M 0 M Change -X -X +2X Equili 1.00 -X 2.00-X 2X
Therefore, ice chart looks like this.
Change in HF = twice change in H2

38. H2(g) F2(g) 2HF(g) Initial 1.00 M 2.00 M 0 M Change -X -X +2X Equili 1.00 -X 2.00-X 2X
Now plug these values into the equilibrium expression
K = 1.15 x 102 = (2X)2
(1.00-x)(2.00-x)
Solving this gives us a quadratic equation.
Quadratic Calculator

39. H2(g) F2(g) 2HF(g) Initial 1.00 M 2.00 M 0 M Change -X -X +2X Equili 1.00 -X 2.00-X 2X
Now plug these values into the equilibrium expression
K = 1.5 x 102 = (2X)2
(1.00-x)(2.00-x)
Solving this gives us a quadratic equation.
Quadratic gives us 2.14 mol/L and mol/L. Only is reasonable.

40. [H2] = 1.00 M M = 3.2 x 10-2M
[F2] = 2.00 M M = M
[HF] = 2(0.968 M) = M
If substituted into the equilibrium expression we get 1.13 x 102 which is very close to given K.

41. Practice H2O(g) + Cl2O(g) 2HClO(g) K = 0.090
In an experiment 1.0 g H2O(g) and 2.0 g Cl2O are mixed in a 1.00 L flask.
Calculate the equilibrium concentrations.
(#48)

42. 13.6 Solving Equilibrium Problems
Balance the equation.
Write the equilibrium expression.
List the initial concentrations.
Calculate Q and determine the shift to equilibrium.
Define equilibrium concentrations.
Substitute equilibrium concentrations into equilibrium expression and solve.
Check calculated concentrations by calculating K.

43. Problems with small K
K< .01

44. Process is the same
Set up table of initial, change, and equilibrium concentrations.
Choose X to be small.
For this case it will be a product.
For a small K the product concentration is small.

45. For example For the reaction 2NOCl 2NO +Cl2 K= 1.6 x 10-5
If 1.00 mol NOCl is put in a 2.0 L container, what are the equilibrium concentrations?
K = [NO]2[Cl2] = 1.6 x [NOCl]2
[NOCl]0= 0.50M, [NO]& [Cl2] =0

46. K = [NO]2[Cl2] = (2x)2(x) = 1.6 x 10-5 [NOCl]2 (0.50 -2x)2
2NOCl 2NO + Cl2
Initial
Change -2X X +X
Equil X X X
K = [NO]2[Cl2] = (2x)2(x) = 1.6 x [NOCl] ( x)2
Since K is so small, we we can make an approximation that x = 0.50
This makes the math much easier. X = 1.0 x 10-2

47. 5% Rule
Many of the systems we will deal with have very small equilibrium constants.
When this is the case, there will be very little shift to the right to reach equilibrium. Since x is so small, we will ignore it. However, the final value must be checked against the initial concentration. If the difference is less than 5%, then our assumption is valid.
In the previous problem X = 1.0 x 10-2
x = (1.0 x 10-2) = 0.48
Is 1.0 x 10-2 five percent or less than 0.50?
0.02/0.50 x 100 = 4%

48. Practice Problem For the reaction N2O4(g) 2NO2(g) K = 4.0 x 10-7
1.0 mol N2O4(g) is placed in a 10.0L vessel. Calculate the concentrations of all species at equilibrium.
(#52)

49. Practice Problem For the reaction COCl2(g) CO(g) + Cl2(g)
Kp = 6.8 x 10-9
If COCl2(g) at an initial pressure of 1.00 atm decomposes, calculate the equilibrium pressures of all species?
(#54)

50. Practice Problem At 25°C, Kp = 2.9 x 10-3 NH4OCONH2(s) 2NH3g) + CO2(g)
In an experiment, a certain amount of NH4OCONH2(s) is placed in an evacuated rigid container and allowed to come to equilibrium. Calculated the total pressure in the container at equilibrium.
(#55)

51. Practice Problem #56WA 2AsH3(g) 2As(s) + 3H2(g)
In an experiment pure AsH3 was placed in a rigid, sealed flask at a pressure of torr. After 48 hours the pressure was observed to be constant at torr.
Calculate the Equilibrium pressure of H2(g).
Calculate Kp for this reaction. (#56)

52. Le Chatelier’s Principle
If a change is applied to a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce the change.
3 Types of change: Concentration (adding or reducing reactant or product), Pressure, Temperature.

53. Change amounts of reactants and/or products
Adding product makes Q>K
Removing reactant makes Q>K
Adding reactant makes Q<K
Removing product makes Q<K
Determine the effect on Q, will tell you the direction of shift.
The system will shift away from the added component.

54. Change Pressure By changing volume.
When volume is reduced, the system will move in the direction that has the least moles of gas.
COCl2(g) CO(g) + Cl2(g)
When volume is increased, the system will move in the direction that has the greatest moles of gas.
Because partial pressures (and conc.) change a new equilibrium must be reached.
System tries to minimize the moles of gas.

55. Change in Pressure By adding an inert gas.
Partial pressures of reactants and product are not changed.
No effect on equilibrium position.

56. Change in Temperature
Affects the rates of both the forward and reverse reactions.
Doesn’t just change the equilibrium position, changes the equilibrium constant.
The direction of the shift depends on whether it is exo - or endothermic.

57. Exothermic H < 0 Releases heat. Think of heat as a product.
N2(g) + 3H2(g) 2NH3(g) + Heat
Raising temperature push toward reactants.
Shifts to left.

58. Endothermic H > 0 Absorbs heat. Think of heat as a reactant.
Heat + COCl2(g) CO(g) + Cl2(g)
Raising temperature push toward products.
Shifts to right.

63. Free Energy and Equilibrium
G tells us spontaneity at current conditions. When will it stop?
It will go to the lowest possible free energy which may be an equilibrium.
At equilibrium G = 0, Q = K
Gº = -RTlnK from [G = Gº + RTlnK]
Gº
= 0
< 0
> 0 K
= 1
> 1
< 1

64. Free Energy and Equilibrium
The overall reaction for the corrosion of iron by oxygen is
4Fe(s) + 3O2(g) 2Fe2O3(s)
Use the following data, calculate the equilibrium constant for this reaction at 25°C.
Substance H°f (kj/mol) S°(J/K•mol)
Fe2O3(s)
Fe(s)
O2(g) Ex
Gº = -RTlnK

65. Practice Calculate K for the following reaction (at 298 K)
H2(g) + Cl2(g) 2HCl
H°f (kj/mol) = -184 S°(J/K•mol) = 20.
If each gas is placed in a flask such that the pressure of each gas is 1 atm, in which direction will the system shift to reach equilibrium at 25°C?
Gº = -RTlnK