1. Chemical Equilibrium
Reactions Go
Both Ways

2. Equilibrium Plural is equilibria (for you Latin fans)
Reactions are reversible
they go forward and backward
There is always some product and reactant
How much of each is there?
Depends on the reaction

3. Equilibria is when….
The rate of the forward reaction and reverse reaction are the same
Sealed jar example

4. Add liquid to empty, dry jar
The rate of evaporation does not change,
but the rate of condensation continuously increases
Until the rate of evaporation = the rate of condensation
When first added, the air has no liquid vapor.
The vapor pressure of the liquid starts to populate the vapor
With molecules

5. Reactions Go Both Ways 2 NO2 (g)  N2O4 (g)  Brown Colorless
Colorless Brown

6. The forward reaction rate is the same as the reverse reaction
In an Equilibrium
The forward reaction rate is the same as
the reverse reaction

7. Chemical equilibrium
Different types of arrows are used in chemical equations associated with equilibria.
Single arrow
Assumes that the reaction proceeds to completion as written.
Two single-headed arrows
Used to indicate a system in equilibrium.
Two single-headed arrows of different sizes.
May be used to indicate when one side of an equilibrium system is favored.

8. Chemical equilibrium
Homogeneous equilibria - Equilibria that involve only a single phase.
Examples.
All species in the gas phase
H2 (g) + I2 (g) HI (g)
All species are in solution.
HC2H2O2 (aq) H+ (aq) + C2H3O2- (aq)

9. Chemical equilibrium A + B C Basic steps for reaching equilibrium:
For the general reaction:
A + B C
We can view the reaction as occurring in three steps.
Initial mixing
Kinetic region
Equilibrium region

10. Chemical equilibrium A + B C Initial mixing.
When A and B are first brought together, there is no C present.
The reaction proceeds as
A + B C
This is just at the very start of the reaction. Things change as soon as some C is produced.

11. Chemical equilibrium A + B C Kinetic region.
As soon as some C has been produced, the reverse reaction is possible.
A + B C
Overall, we still see an increase in the net concentration of C.
As we approach equilibrium, the rate of the forward reaction becomes slower.

12. Chemical equilibrium A + B C Equilibrium region.
A point is finally reached where the forward and reverse reactions occur at the same rate.
A + B C
There is no net change in the concentration of any of the species.

13. Chemical equilibrium Concentration Time C B A Equilibrium Region
Kinetic

14. Equilibrium A point is ultimately reached where the
rates of the forward
and reverse changes
are the same.
At this point,
equilibrium
is reached.
Reactants
Rate of Formation
Products
Time

15. Equilibrium
An equilibrium exists when no further change in concentration occurs. Note that the concentrations of products and reactants do not have to be equal!
Products
Kinetic Equilibrium
Region Region
Concentration
Reactants
Time

16. The Equilibrium Constant
Also note that the equilibrium concentrations of the reactants and products are the same regardless of the whether or not you start with only the reactants or only the products.
No products
No reactants

17. Equilibrium Constants
The equilibrium constant is related to the rate laws of the forward and reverse reactions.
Because the forward and reverse reactions are equal in rate, the rate laws must be equal to each other.
kf [N2O4] = kr[NO2]2
Isolating the concentrations and rate constants together…
[NO2]2/[N2O4] = kf/kr = a constant (Keq), or the equilibrium constant.

18. Equilibrium Constant
The position of the equilibrium can be described mathematically
We use the balanced chemical equation to develop a mathematical expression
Use molarity to describe concentration
Remember, that’s moles per liter of solution

19. Law of Mass Action pA + qB  rC + tD K = [C]r • [D]t [A]p • [B]q
[ ] = molarity
K = equilibrium constant
Products “over”
reactants

20. Law of Mass Action 4 NH3(g) + 7O2 (g)  4NO2 (g) + 6H2O (g)
K = [C]r • [D]t = [NO2]4 • [H2O]6 [A]p • [B]q [NH3]4 • [O2]7

21. Equilibrium Constant More products means a bigger K
Y Reactant  X Product (g)
Keq = [products]x [reactants]y
More products means a bigger K
Reaction lies to the right
More reactants means smaller K
Reaction lies to the left

22. N2(g) + 3H2(g)  2NH3(g)
K = [NH3]2
[N2] • [H2]3
K = 6.02 x 10-2

23. N2(g) + 3H2(g)  2NH3(g) K = [NH3]2 [N2] • [H2]3 K = 6.02 x 10-2

24. N2(g) + 3H2(g)  2NH3(g) K = [NH3]2 [N2] • [H2]3 K = 6.02 x 10-2

25. Equilibria It doesn’t matter where you start
The “ratio” of products to reactants is the same
Because the equilibria is established by the rate of the forward reaction vs the rate of the reverse reaction

26. Equilibrium Constant, Keq
The equilibrium constant, Keq, is the ratio of the concentrations of the products compared to the concentrations of the reactants.
Keq > 1 indicate that forming products is favored. Keq < 1 indicate that reactants don’t react easily.

27. Equilibrium Constant, Keq
3H2(g)+N2(g) NH3(g)
The equilibrium constant, Keq, for the reaction above is determined by:
Keq = [ NH3 ] 2
[ H2 ] 3 [ N2 ]
At 472C, Keq = 0.105

28. Equilibrium Constant, Keq
2NH3(g) H2(g)+N2(g)
Reversing a reaction will result in the new equilibrium constant, Keq new, that is equal to 1 / Keq old
Keq new = [ H2 ] 3 [ N2 ] = 1 / 0.105
[ NH3 ] or 9.52

29. Equilibrium Constant, Keq
NH3(g) H2(g)+0.5N2(g)
Changing the number of moles of reactants and products will exponentially change the equilibrium constant: Keq new= (Keq old)^n
Keq new = [ H2 ]1.5 [ N2 ]0.5 = (9.52)^0.5
[ NH3 ] or 3.09

30. Partial pressure equilibrium constants
At constant temperature, the pressure of a gas is proportional to its molarity.
Remember, for an ideal gas: PV = nRT
and molarity is: M = mole / liter or n/V
so: P = MR T
where R is the gas law constant
T is the temperature, K.

31. Partial pressure equilibrium constants
For equilibria that involves gases, partial pressures can be used instead of concentrations.
aA (g) + bB (g) eE (g) + fF (g)
Kp =
Kp is used when the partial pressures are expressed in units of atmospheres.
pEe pFf
pAa pBb

32. Partial pressure equilibrium constants
In general, Kp ≠ Kc, instead
Kp = Kc (RT)Dng
Dng is the number of moles of gaseous products minus the number of moles of gaseous reactants.
Dng = (e + f) - (a + b)

33. Partial pressure equilibrium constants
For the following equilibrium, Kc = 1.10 x 107 at 700. oC. What is the Kp?
2H2 (g) + S2 (g) H2S (g)
Kp = Kc (RT)Dng
T = = 973 K
R =
Dng = ( 2 ) - ( 2 + 1) = -1
atm L
mol K

34. Partial pressure equilibrium constants
Kp = Kc (RT)Dng
= 1.10 x ( ) (973 K)
= x105 -1 ] [
atm L
mol K

35. Equilibrium constants and expressions
Heterogeneous equilibria
Equilibria that involve more than one phase.
CaCO3 (s) CaO (s) + CO2 (g)
Equilibrium expressions for these types of systems do not include the concentrations of the pure solids (or liquids).
Kc = [CO2]

36. Equilibrium constants and expressions
Heterogeneous equilibria - We don’t include the pure solids and liquids because their concentrations do not vary. These values end up being included in the K value.
CO2
CaO &
CaCO3
As long as the temperature is constant and some solid is present, the amount of solid present has no effect on the equilibrium.

37. Writing an equilibrium expression
Write a balanced equation for the equilibrium.
Put the products in the numerator and the reactants in the denominator.
Omit pure solids and liquids from the expression
Omit solvents if your solutes are dilute (<0.1M).
The exponent of each concentration should be the same as the coefficient for the species in the equation.

38. Writing an equilibrium expression
Example.
What would be the equilibrium expression for the following?
(NH4)2CO3(s) NH3(g) + CO2(g) + H2O(g)
Kc = [NH3]2 [CO2] [H2O]
(NH4)2CO3 is a pure solid so is not included Kc
We use [NH3]2 because the coefficient for NH3(g) in the equation is 2.

39. Equilibrium constant alphabet soup
The equilibrium constant expressed in concentration units Kc
The equilibrium constant expressed in pressure units Kp

40. Equilibrium constant alphabet soup
For the limited dissociation of insoluble solids.
Ksp
Ksp is called the
solubility product constant.

41. Equilibrium constant alphabet soupKa
For the dissociation of weak acids Kb
For the dissociation of weak bases
For the dissociation of water into H+ and OH- Kw

42. Equilibrium and rate of reaction
Chemical reactions tend to go to equilibrium providing that reaction takes place at a significant rate.
There is no relationship between the magnitude of the equilibrium constant and the rate of a reaction.
Example: 2H2 (g) + O2 (g) H2O (g)
Kc = 2.9 x = [H2O]2
[H2]2 [O2]
However, the reaction will take years to reach equilibrium at room temperature.

43. Determining equilibrium constants
Equilibrium constants can be found by experiment.
If you know the initial concentrations of all of the reactants, you only need to measure the concentration of a single species at equilibrium to determine the Kc value.
Lets consider the following equilibrium:
H2 (g) + I2 (g) HI (g)

44. Determining equilibrium constants
H2 (g) + I2 (g) HI (g)
Assume that we started with the following initial concentrations at 425.4oC.
H2 (g) M
I2 (g) M
HI (g) M
At equilibrium, we determine that the concentration of iodine is M

45. Determining equilibrium constants
The equilibrium expression for our system is:
Kc =
Based on the chemical equation, we know the equilibrium concentrations of each species.
I2 (g) = the measured amount
= M
That means that mol I2 reacts ( M M) to produce HI in 1.00 L of solution.
[HI]2
[H2] [I2]

46. Determining equilibrium constants
I2 (g) = M
HI (g) = M
= M
H2 (g) = M M
= M
Kc = = = 54
2 mol HI
1 mol I2
[HI]2
[H2] [I2]
( )2
( )( )

47. Equilibrium calculations
We can predict the direction of a reaction by calculating the reaction quotient.
Reaction quotient, Q
For the reaction: aA + bB eE + fF
Q has the same form as Kc with one important difference. Q can be for any set of concentrations, not just at equilibrium.
Q =
[E]e [F]f
[A]a [B]b

48. Reaction quotient
Any set of concentrations can be given and a Q calculated. By comparing Q to the Kc value, we can predict the direction for the reaction.
Q < Kc Net forward reaction will occur.
Q = Kc No change, at equilibrium.
Q > Kc Net reverse reaction will occur.

49. Reaction quotient example
For an earlier example
H2 (g) + I2 (g) HI (g)
we determined the Kc to be 54 at oC.
If we had a mixture that contained the following, predict the direction of the reaction.
[H2] = 4.25 x 10-3 M
[I2] = 3.97 x 10-1 M
[HI] = 9.83 x 10-2 M

50. Reaction quotient example= =
Since Q is < Kc, the system is not in equilibrium and will proceed in the forward direction.
[ HI ]2
[ H2 ] [ I2 ]
(9.83 x 10-2)2
(4.25 x 10-3)(3.97 x 10-1)

51. Calculating equilibrium concentrations
If the stoichiometry and Kc for a reaction is known, calculating the equilibrium concentrations of all species is possible.
Commonly, the initial concentrations are known.
One of the concentrations is expressed as the variable x.
All others are then expressed in terms of x.

52. Equilibrium calculation example
A sample of COCl2 is allowed to decompose. The value of Kc for the equilibrium
COCl2 (g) CO (g) + Cl2 (g)
is 2.2 x at 100 oC.
If the initial concentration of COCl2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?

53. Equilibrium calculation example
COCl2 (g) CO (g) Cl2 (g)
Initial conc., M
Change in conc. - X + X + X
due to reaction
Equilibrium
Concentration, M ( X) X X
Kc = =
[CO ] [Cl2 ]
[ COCl2 ] X2
( X)

54. Equilibrium calculation example
Kc = 2.2 x =
Rearrangement gives
X x X x = 0
This is a quadratic equation. Fortunately, there
is a straightforward equation for their solution

55. Quadratic equations An equation of the form a X2 + b X + c = 0
Can be solved by using the following
x =
Only the positive root is meaningful in equilibrium problems.
-b b2 - 4ac 2a

56. Equilibrium calculation example
X x X x = 0
a b c
-b b2 - 4ac 2a
X =
- 2.2 x [(2.2 x 10-10)2 - (4)(1)( x 10-11)]1/2 2
X = x 10-6 M

57. Equilibrium calculation example
Now that we know X, we can solve for the concentration of all of the species.
COCl2 = X = M
CO = X = 9.1 x 10-6 M
Cl2 = X = 9.1 x 10-6 M
In this case, the change in the concentration of is COCl2 negligible.

58. Summary of method of calculating equilibrium concentrations
Write an equation for the equilibrium.
Write an equilibrium constant expression.
Express all unknown concentrations in terms of a single variable such as x.
Substitute the equilibrium concentrations in terms of the single variable in the equilibrium constant expression.
Solve for x.
Use the value of x to calculate equilibrium concentrations.

59. You Try!
A 40.0g sample of solid (NH4)2CO3 is placed in a closed evacuated 3L flask and heated to 400oC. It decomposes to produce NH3, H2O, and CO2 :
(NH4)2CO3 ↔ 2 NH3 + H2O + CO2
The equilibrium constant, Kp , is 0.295
Write the Kp expression.
Calculate Kc
Calculate the partial pressure of NH3
Calculate the total pressure in the flask at equilibrium.
Calculate the number of grams of solid at equilibrium.
What is the minimum amount of solid needed to establish equilibrium?

60. Answers Kp = PNH32 PH2O PCO2 3.17x10-8 1.04 atm 2.08 atm 37.3g
More than 2.72g

61. Predicting Shifts: LeChatelier
Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress.
You can put stress on a system by adding or removing something from one side of a reaction.
Co(H2O) Cl CoCl H2O
How can you cause the color to change from pink to blue?

62. Predicting shifts in equilibria
Equilibrium concentrations are based on:
The specific equilibrium
The starting concentrations
Other factors such as:
Temperature
Pressure
Reaction specific conditions
Altering conditions will stress a system, resulting in an equilibrium shift.

63. Le Chatelier’s principle
Co(H2O) Cl CoCl H2O
Note: ∆H = + value
Add HCl (adds Cl1-)
Add heat
Add acetone
Add Ag1+
(removes Cl1- )
(removes H2O)
Add ice
What other stresses could be placed on this system to change the color back and forth between pink and blue?

64. Le Chatelier’s principle
Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress.
You can put stress on a system by adding or removing something from one side of a reaction.
N2(g) + 3H2 (g) NH3 (g)
What effect will there be if you added more
ammonia? How about more nitrogen?

65. Changes in concentration
Changes in concentration do not change the value of the equilibrium constant at constant temperature.
When a material is added to a system in equilibrium, the equilibrium will shift away from that side of the equation.
When a material is removed from a system in equilibrium, the equilibrium will shift towards that sid of the equation.

66. Le Châtelier’s Principle
Change in Reactant or Product Concentrations
Consider the Haber process
If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier).
That is, the system must consume the H2 and produce products until a new equilibrium is established.
Therefore, [H2] and [N2] will decrease and [NH3] increases.

67. Le Châtelier’s Principle Change in Reactant or Product Concentrations

68. Le Châtelier’s Principle
Change in Reactant or Product Concentrations
Adding a reactant or product shifts the equilibrium away from the increase.
Removing a reactant or product shifts the equilibrium towards the decrease.
To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier).
We illustrate the concept with the industrial preparation of ammonia

69. Le Châtelier’s Principle
Change in Reactant or Product Concentrations
The catalyst bed is kept at C under high pressure.
N2 and H2 are pumped into a chamber.
The pre-heated gases are passed through a heating coil to the catalyst bed.
In the refrigeration unit, ammonia liquefies but not N2 or H2.
The product gas stream (containing N2, H2 and NH3) is passed over a cooler to a refrigeration unit.

70. Le Châtelier’s Principle
Change in Reactant or Product Concentrations
The unreacted nitrogen and hydrogen are recycled with the new N2 and H2 feed gas.
The equilibrium amount of ammonia is optimized because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added.
Effects of Volume and Pressure
As volume is decreased pressure increases.
Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase by producing fewer moles of gas.

71. Changes in pressure
Changing the pressure does not change the value of the equilibrium constant at constant temperature.
Solids and liquids are not effected by pressure changes.
Changing pressure by introducing an inert gas will not shift an equilibrium.
Pressure changes only effect gases that are a portion of an equilibrium.

72. Changes in pressure
In general, increasing the pressure by decreasing volume shifts equilibria towards the side that has the smaller number of moles of gas.
H2 (g) + I2 (g) HI (g)
N2O2 (g) NO2 (g)
Unaffected by pressure
Increased pressure, shift to left

73. Le Châtelier’s Principle
Effects of Volume and Pressure
That is, the system shifts to remove gases and decrease pressure.
An increase in pressure favors the direction that has fewer moles of gas.
In a reaction with the same number of product and reactant moles of gas, pressure has no effect.
Consider

74. Le Châtelier’s Principle Effects of Volume and Pressure
An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4.
The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased.
The system moves to reduce the number moles of gas (i.e. the forward reaction is favored).
A new equilibrium is established in which the mixture is lighter because colorless N2O4 is favored.

75. Le Châtelier’s Principle
Effect of Temperature Changes
The equilibrium constant is temperature dependent.
For an endothermic reaction, H > 0 and heat can be considered as a reactant.
For an exothermic reaction, H < 0 and heat can be considered as a product.
Adding heat (i.e. heating the vessel) favors away from the increase:
if H > 0, adding heat favors the forward reaction,
if H < 0, adding heat favors the reverse reaction.

76. Le Châtelier’s Principle
Effect of Temperature Changes
Removing heat (i.e. cooling the vessel), favors towards the decrease:
if H > 0, cooling favors the reverse reaction,
if H < 0, cooling favors the forward reaction.
for which DH > 0.
Co(H2O)62+ is pale pink and CoCl42- is blue.

77. Le Châtelier’s Principle Effect of Temperature Changes

78. Le Châtelier’s Principle Effect of Temperature Changes
If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue.
Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl42-.
If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink.
Since H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H2O)62+. Co

79. A catalyst does not effect the composition of the equilibrium mixture.
Le Châtelier’s Principle
The Effect of Catalysts
A catalyst lowers the activation energy barrier for the reaction.
Therefore, a catalyst will decrease the time taken to reach equilibrium.
A catalyst does not effect the composition of the equilibrium mixture.

80. You try! H2(g) + S(s) → H2S(g) ∆Hrxn = -20.17kJ/mol
An amount of solid S and an amount of gaseous H2 are placed in an evacuated container at 25oC. At equilibrium, some solid S remains in the container. Predict and explain the following:
Effect on partial pressure of H2S when S is added.
Effect on partial pressure of H2 when H2S is added
Effect on mass of S when volume is increased.
Effect on S when the temperature increased.
Effect of adding a catalyst to initial concentrations.

81. Answers Solids have no effect on equilibrium P H2 will increase.
S remains the same, equal numbers of moles of gases on both sides means no change in equilibrium.
S decreases. Equilibrium is not effected by solids, but solid amounts can be changed by equilibrium. If more gases are produced, solid must go down.
Catalysts make them reach equilibrium faster, but do not change equilibrium positions.