1. Chemical Equilibrium

2. Reaction Arrow Conventions
A single arrow indicates all the reactant molecules are converted to product molecules at the end
A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products
 in these notes

3. Reaction Dynamics
When a reaction starts, the reactants are consumed and products are made
forward reaction = reactants  products
[reactant] decrease and [product] increase
as [reactant] decreases, the forward reaction rate decreases
Eventually, the products can react to re-form some of the reactants, the reverse reaction
reverse reaction = products  reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate increases
Processes that proceed in both the forward and reverse direction are said to be reversible
reactants Û products

4. Hypothetical Reaction 2 Red Û Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 90 and 100 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red

5. Hypothetical Reaction 2 Red Û Blue

6. Reaction Dynamics Once equilibrium is established,
the forward and reverse reactions
proceed at the same rate, so the
concentrations of all materials
stay constant
Eventually, the reaction proceeds
in the reverse direction as fast as
it proceeds in the forward direction.
At this time equilibrium is established.
Because the reactant concentration
decreases, the forward reaction slows.
As the products accumulate, the
reverse reaction speeds up.
As the forward reaction proceeds
it makes products and uses reactants
Initially, only the forward
reaction takes place
Rate
Rate Forward
Rate Reverse
Time

7. Dynamic Equilibrium
Rate of forward reaction = Rate of reverse reaction
At equilibrium, the concentrations of all the chemicals remain constant
because the chemicals are being consumed and made at the same rate

8. H2(g) + I2(g) Û 2 HI(g) At t0: reactants only  only
forward reaction takes place
[H2] = 8, [I2] = 8, [HI] = 0
At t16: both reactants & products in the mixture  forward reaction & reverse reaction
[H2] = 6, [I2] = 6, [HI] = 4

9. H2(g) + I2(g) Û 2 HI(g)
At t32: [products] > [reactants]  forward reaction rate slows, reverse reaction rate accelerates
[H2] = 4, [I2] = 4, [HI] = 8
At t48: [products] = [reactants] forward rxn rate = reverse rxn rate
Equilibrium is reached.

10. H2(g) + I2(g) Û 2 HI(g)
Because the reactant concentrations are decreasing, the forward reaction rate slows down
And because the product concentration is increasing, the reverse reaction rate speeds up
As the reaction proceeds, the [H2] and [I2] decrease, and the [HI] increases
Because the [HI] at equilibrium is larger than the [H2] or [I2], we say the position of equilibrium favors products
At equilibrium, the forward reaction rate is the same as the reverse reaction rate
Once equilibrium is established, the concentrations no longer change
Equilibrium established

11. Equilibrium  Equal
The rates of the forward and reverse reactions are equal at equilibrium
[reactants]  [products]
When “Equilibrium favors products”,
[products]>[reactants]
When “Equilibrium favors reactants”,
[reactants]>[products]

12. Equilibrium Constant Law of Mass Action aA + bB Û cC + dD
The relationship between [reactants] and [products]
Must know the balanced chemical equation
aA + bB Û cC + dD
Lowercase letters = coefficients
In determining, relationship, Productscoefficients/Reactantscoefficients
K = equilibrium constant, a unit-less number

13. Writing Equilibrium Constant Expressions
For the reaction
aA(aq) + bB(aq) Û cC(aq) + dD(aq) the equilibrium constant expression
So for the reaction
2 N2O5(g) Û 4 NO2(g) + O2(g)
the equilibrium constant
expression

14. Implications of Keq When Keq >> 1 When Keq << 1
[product molecules] > [reactant molecules]
equilibrium favors products
When Keq << 1
[reactant molecules] > [product molecules]
equilibrium favors reactants

15. Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following
2 SO2(g) + O2(g) Û 2 SO3(g) K = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) K = 3 x 10−17
favors products
favors reactants

16. Relationships between K and Chemical Equations
When the reaction is written backward, the equilibrium constant is inverted
For the reaction
aA + bB Û cC + dD
the equilibrium constant expression is
For the reaction
cC + dD Û aA + bB
the equilibrium constant expression is

17. Relationships between K and Chemical Equations
When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor
For the reaction
aA + bB Û cC
the equilibrium constant expression is
For the reaction
2aA + 2bB Û 2cC
the equilibrium constant expression is

18. Relationships between K and Chemical Equations
When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations
For the reactions
aA Û bB & (2) bB Û cC
the equilibrium constant expressions are
For the reaction
aA Û cC
the equilibrium constant expression is

19. Practice – When the reaction A(aq) Û 2 B(aq) reaches equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M. When the reaction 2 B(aq) Û Z(aq) reaches equilibrium [B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M. Calculate the equilibrium constant for each of these reactions and the equilibrium constant for the reaction 3 Z(aq) Û 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2 B Û A K3 = 1/Krxn 1 = 6.25 x 10−5
Z Û 2 B K4 = 1/Krxn 2 = 2
Z Û A K3K4 = 1.25 x 10−4
3 Z Û 3 A (K3K4)3 = 2.0 x 10−12

20. K for Reactions Involving Gases
In a mixture, [gas] α to its Pp (partial pressure)
Therefore, K can be expressed as the ratio of the Pp of the gases
For aA(g) + bB(g) Û cC(g) + dD(g) the equilibrium constant expressions are or

21. Kc and Kp In calculating Kp, the units of Pp = atm (always)
Therefore, values of Kp ≠ Kc , necessarily
difference in units
Kp = Kc when Dn = 0
The relationship between them is
Dn is the difference between the number of moles of reactants and moles of products

22. Deriving the Relationship between Kp and Kc

23. Deriving the Relationship Between Kp and Kc
for aA(g) + bB(g)  cC(g) + dD(g)
substituting

24. Calculate the value of Kp or Kc for each of the following at 27 °C:
2 SO2(g) + O2(g) Û 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) Kp = 3 x 10−17

25. Heterogeneous Equilibria
[Pure solids] and [pure liquids] do not change during the course of a reaction
Because it is not in solution, the amount changes, but the amount of it in solution doesn’t
Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression
For the reaction aA(s) + bB(aq) Û cC(l) + dD(aq) the equilibrium constant expression is

26. Heterogeneous Equilibria
The amount of C is different, but the amounts of CO and CO2 remain the same. Therefore the amount of C has no effect on the position of equilibrium.

27. Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following
HNO2(aq) + H2O(l) Û H3O+(aq) + NO2−(aq) K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) Û CaSO4(s) + 2 HNO3(aq) K = 1 x 104
Tro: Chemistry: A Molecular Approach, 2/e

28. Calculating Equilibrium Constants from Measured Equilibrium Concentrations
Most direct way of finding K:
measure the amounts of reactants and products in a mixture at equilibrium
A mixture at equilibrium may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of the initial amounts of reactants and products

29. Initial and Equilibrium Concentrations for H2(g) + I2(g) Û 2HI(g) @ 445 °C
Constant
[H2]
[I2]
[HI]
0.50
0.0
0.11
0.78
0.055
0.39
0.165
1.17
1.0
0.5
0.53
0.033
0.934
Tro: Chemistry: A Molecular Approach, 2/e

30. Calculating Equilibrium Concentrations
To use stoichiometry to determine equilibrium concentrations of reactants and products
You must know initial concentrations and one equilibrium concentration
Use the change in the concentration of the material that you know to determine the change in the other chemicals in the reaction

31. Calculating Equilibrium Concentrations: An Example
2 A(aq) + B(aq) Û 4 C(aq)
Suppose you have a reaction 2 A(aq) + B(aq) Û 4 C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M.
Product C increased by 0.50 mol/L
Because Product C increased by 0.50 mol/L, you must have used ½ that amount of Reactant A, according to the stoichiometry.
Because Product C increased by 0.50 mol/L, you must have used ¼ that amount of Reactant B, according to the stoichiometry.
-½(0.50)
-¼(0.50)
+0.50
0.75
0.88

32. Practice –The following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc.

33. Practice – A second experiment was done and the following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc. Compare them to the first experiment.

34. The Reaction Quotient
In a mixture containing both reactants and products not at equilibrium  how can we determine in which direction it will proceed?
compare the current concentration ratios to the equilibrium constant
The reaction quotient, Q is the concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients)
for the gas phase reaction aA + bB Û cC + dD
the reaction quotient is:

35. The Reaction Quotient: Predicting the Direction of Change
If Q > K, the reaction will proceed fastest in the reverse direction: [products]  and [reactants] 
If Q < K, the reaction will proceed fastest in the forward direction: [products] and [reactants] 
If Q = K, the reaction is at equilibrium there will be no Δ in [products] and [reactants]
If Q = 0, the reaction mixture contains only reactants, the reaction will go forward
If Q = ∞, a reaction mixture contains only products, the reaction will reverse

36. Q, K, and the Direction of Reaction

37. Practice – For the reaction CH4(g) + 2 H2S(g) Û CS2(g) + 4 H2(g) Kc = 3.59 at 900 °C. For each of the following measured concentrations determine whether the reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to get to equilibrium?

38. Practice – A sample of PCl5(g) is placed in a 0
Practice – A sample of PCl5(g) is placed in a L container and heated to 160 °C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc =
PCl5 Û PCl3 + Cl2
if in addition you calculate Kp from Kc you find that it is 2.26, slightly greater than 1 – showing that the equilibrium constant may be unreliable for predicting the position of equilibrium when it is close to 1

39. Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the coefficient of x
the x change is + for materials on the side the reaction is proceeding toward
the x change is  for materials on the side the reaction is proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the quadratic formula
may be able to simplify and approximate answer for very large or small equilibrium constants

40. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? (Hint: you will need to use the quadratic formula to solve for x)

41. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0 and the reaction must proceed forward

42. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?

43. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
0.500  = 0.498
[I2] = M
2( ) =
[I] = M
x =  

44. Approximations to Simplify the Math
When the K is very small, the position of equilibrium favors the reactants
For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial  ax)  [X]initial
we are approximating the equilibrium concentration of reactant to be the same as the initial concentration
assuming the reaction is proceeding forward

45. Checking the Approximation and Refining as Necessary
Check approximation by comparing the approximate value of x to the initial concentration
If the approximate value of x is less than 5% of the initial concentration, the approximation is valid

46. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? (use the simplifying assumption to solve for x)

47. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0 and the reaction must proceed forward

48. the approximation is valid!!
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!

49. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
0.500  = 0.498
[I2] = M
2( ) =
[I] = M
x =  

50. Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1
Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is M, what will be the equilibrium concentrations of [N2O4] and [NO2]?

51. Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1
Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is M, what will be the equilibrium concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0, Q = 0 and the reaction must proceed forward

52. the approximation is valid!!
Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is M, what will be the equilibrium concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!

53. Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1
Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is M, what will be the equilibrium concentrations of [N2O4] and [NO2]?
[N2O4]=  1.83 x 10−4
[N2O4] = M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4  

54. Disturbing and Restoring Equilibrium
Once a reaction is at equilibrium, the [reactants] and [products] remain the same
However, if conditions change, the concentrations of all the chemicals will change until equilibrium is restored
The new concentrations will be different, but the equilibrium constant will be the same
unless you change the temperature

55. Le Châtelier’s Principle
Le Châtelier's Principle helps us in predict the effect various changes in conditions have on the position of equilibrium
if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance
disturbances all involve making the system open

56. An Analogy: Population Changes
When an influx of population enters Country B
from somewhere outside Country A, it disturbs the
equilibrium established between Country A and Country B.
The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established; the new populations will have different numbers of people than the old ones.
When the populations of Country A and Country B
are in equilibrium, the emigration rates between the
two countries are equal so the populations stay constant.

57. Disturbing Equilibrium: Adding or Removing Reactants
After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium constant expression (not a solid or liquid)…
How will this affect the rate of the forward reaction? Initially,…
Adding a reactant increases the forward reaction rate
How will it affect the rate of the reverse reaction?
No effect on the reverse reaction rate
The reaction proceeds to the right until equilibrium is restored
How will it affect the value of K?
At the new equilibrium position, you will have
more of the products than before
less of the non-added reactants than before
Slightly less of the added reactant
At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

58. Disturbing Equilibrium: Adding or Removing Reactants
After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium constant expression
i.e., not a solid or liquid
How will this affect the rate of the forward reaction?
Initial decrease in forward reaction rate, no effect on the reverse reaction rate
How will it affect the rate of the reverse reaction?
Reverse reaction goes faster
Reaction proceeds to the left until equilibrium is restored
How will it affect the value of K?
At the new equilibrium position, you will have
less of the products than before
more of the non-removed reactants than before
Slightly more of the removed reactant
At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

59. The Effect of Concentration Changes on Equilibrium
Adding a reactant will
decrease the amounts of the other reactants
increase the amount of the products
Re-establish a new position of equilibrium is found
K stays the same
Removing a product will
increase the amounts of the other products
decrease the amounts of the reactants.
Drive the reaction to completion
Equilibrium shifts away from side with added chemicals or toward side with removed chemicals
Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium

60. The Effect of Concentration Changes on Equilibrium
When NO2 is added, some of it combines to make more N2O4

61. The Effect of Concentration Changes on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2

62. The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium
Adding a gaseous reactant increases its partial pressure  equilibrium shifts to the right
increase in partial pressure  increase in concentration
No increase the partial pressure of the other gases in the mixture
Adding an inert gas to the mixture has no effect on the position of equilibrium
does not affect the partial pressures of the gases in the reaction

63. Effect of Volume Change on Equilibrium
Decrease container volume  increase in all [gas]
Increase in gas Pp
It does not change the concentrations of solutions!!
As Pp increase, the Ptotal increases
According to Le Châtelier’s Principle
equilibrium should shift to remove pressure
system reduces the pressure by reducing the number of gas molecules in the container
When the volume decreases, the equilibrium shifts to the side with fewer gas molecules

64. Disturbing Equilibrium: Changing the Volume
After equilibrium is established, the container volume is decreased
How will it affect the concentration of solids, liquid, solutions, and gases?
How will this affect the total pressure of solids, liquid, and gases?
How will it affect the value of K?

65. Disturbing Equilibrium: Reducing the Volume
Decreasing the container volume increases the concentration of all gases, but not solids, liquid, or solutions
same number of moles, but different number of liters, resulting in a different molarity
Decreasing the container volume will increase the total pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the gases will increase – Dalton’s Law of Partial Pressures
Because the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules
shift toward the side with fewer gas molecules
K doesn’t change: at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that K is the same

66. The Effect of Volume Changes on Equilibrium
Because there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the side with fewer molecules to decrease the pressure
When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side

67. The Effect of Temperature Changes on Equilibrium Position
Exothermic reactions release energy and endothermic reactions absorb energy
To useLe Châtelier’s Principle to predict the effect of temperature changes, write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction

68. The Effect of Temperature Changes on Equilibrium for Exothermic Reactions
For an exothermic reaction, heat is a product
Increasing the temperature is like adding heat
According to Le Châtelier’s Principle, the equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will
decrease the concentrations of products
increase the concentrations of reactants
decrease the value of K
How will decreasing the temperature affect the system?
aA + bB  cC + dD + Heat

69. The Effect of Temperature Changes on Equilibrium for Endothermic Reactions
For an endothermic reaction, heat is a reactant
Increasing the temperature is like adding heat
According to Le Châtelier’s Principle
the equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will
decrease the concentrations of reactants
increase the concentrations of products
increase the value of K
How will decreasing the temperature affect the system?
Heat + aA + bB  cC + dD

70. The Effect of Temperature Changes on Equilibrium

71. Not Changing the Position of Equilibrium: The Effect of Catalysts
Catalysts provide an alternative, more efficient mechanism
Catalysts work for both forward and reverse reactions
Catalysts affect the rate of the forward and reverse reactions by the same factor
Therefore, catalysts do not affect the position of equilibrium

72. Practice – Le Châtelier’s Principle
2 SO2(g) + O2(g) Û 2 SO3(g)
DH° = −198 kJ at equilibrium
How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture

73. Practice – Le Châtelier’s Principle
2 SO2(g) + O2(g) Û 2 SO3(g)
DH° = −198 kJ at equilibrium
How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored?
adding more O2 to the container shift to SO3
condensing and removing SO3 shift to SO3
compressing the gases shift to SO3
cooling the container shift to SO3
doubling the volume of the container shift to SO2
warming the mixture shift to SO2
adding helium to the container no effect
adding a catalyst to the mixture no effect