1. Chemical Equilibrium 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠⇋𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
If the absorption of O2 by hemoglobin was not reversible we would suffocate, O2 absorption illustrates the principle of chemical equilibrium
Reactions go forwards and backwards

2. Dynamic Equilibrium
as the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate
dynamic equilibrium is the condition where the rates of the forward and reverse reactions are equal
once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant

3. Dynamic Equilibrium: H2 + I2  2HI
This reaction occurs in one elementary step
The rate law is connected to the molecularity
𝐻 2 + 𝐼 2 →2HI 𝑟𝑎𝑡𝑒 𝑓 = 𝑘 𝑓 𝐻 2 𝐼 2
But as with all reactions that go forward, the reaction can also go backward
2HI→𝐻 2 + 𝐼 𝑟𝑎𝑡𝑒 𝑏 = 𝑘 𝑏 𝐻𝐼 2
We write these two equations as one like so
𝐻 2 + 𝐼 2 ⇌2HI
At equilibrium the rate to go forward = rate to go backward kf kb

4. Reaction Dynamics

5. Reaction Dynamics } }
Rate we use up H2
Rate we make H2

6. Reaction Dynamics

7. Reaction Dynamics
𝑘 𝑓 𝑘 𝑏 = 𝐻𝐼 𝑒𝑞 𝐻 2 𝑒𝑞 𝐼 2 𝑒𝑞 = 𝐾 𝑐

8. Enthalpy DH and Equilibrium Constant Kc
𝐾 𝑐 = 𝑘 𝑓 𝑘 𝑏 = 𝐴 𝑓 𝐴 𝑏 ∙ 𝑒 − 𝐸 𝑎 𝑅𝑇 𝑒 − (𝐸 𝑎 −Δ𝐻) 𝑅𝑇
𝑘 𝑓 𝑘 𝑏 = 𝐻𝐼 𝑒𝑞 𝐻 2 𝑒𝑞 𝐼 2 𝑒𝑞 = 𝐾 𝑐
𝑖𝑓 𝐴 𝑓 = 𝐴 𝑏
𝐾 𝑐 = 𝑘 𝑓 𝑘 𝑏 = 𝑒 − Δ𝐻 𝑅𝑇 Ea
𝑖𝑓 𝐴 𝑓 ≠ 𝐴 𝑏
𝐾 𝑐 = 𝑘 𝑓 𝑘 𝑏 = 𝑒 − Δ𝐻 𝑅𝑇 + Δ𝑆 𝑅
DH = -|DH|
Here DH < 0 and let’s assume
for now that Af ≈ Ab
𝐾 𝑐 ≈ 𝑒 |Δ𝐻| 𝑅𝑇
“The speed of a chemical reaction is determined by kinetics, but the extent of the reaction is determined by the relative stability of reactants and product – this is the dominion of thermodynamics”
As T gets big Kc  1
As T gets small Kc gets big

9. Equilibrium: Multistep Reaction
Overall

10. Equilibrium: Multistep Reaction
Overall

11. Law of Mass Action
The overall equilbrium constant is the same as you would get if you treated the overall reaction as occuring in one elementary reaction

12. Equilibrium Constant for a Reaction
aA(aq) + bB(aq) cC(aq) + dD(aq)
2 N2O NO2 + O2

13. Interpreting Keq Keq >> 1
more product molecules present than reactant molecules
the position of equilibrium favors products
Keq << 1
more reactant molecules present than product molecules
the position of equilibrium favors reactants

14. Keq >> 1
Lots of HBr but very little H2 or Br2

15. Keq << 1
Lots of N2 and O2 but very little NO

16. Relationships between K and Chemical Equations
when the reaction is written backwards, the equilibrium constant is inverted
for aA + bB cC + dD
for cC + dD aA + bB

17. Relationships between K and Chemical Equations
when the reaction is written backwards, the equilibrium constant is inverted
for aA + bB cC + dD
for cC + dD aA + bB

18. 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 aA + bB cC
the equilibrium constant expression is:
for 2aA + 2bB cC
the equilibrium constant expression is:

19. 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 (1) aA bB
(2) bB cC
for aA cC

20. Equilibrium Constants for Reactions Involving Gases
the concentration of a gas in a mixture is proportional to its partial pressure
therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases
for aA(g) + bB(g) cC(g) + dD(g) or

21. Deriving the Relationship between Kp and Kc

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

23. Kc and Kp in calculating Kp, the partial pressures are always in atm
the values of Kp and Kc are not necessarily the same
because of the difference in units
Kp = Kc when Δn = 0
the relationship between them is:
Δn is the difference between the number of moles of reactants and moles of products

24. Heterogeneous Equilibria
Consider
The reaction occurs at the surface where the solid and water meet
Instead of the concentration of NaCl(s) the rate of dissolving is related to the surface area ANaCl
K is independent of the amount of solid!

25. Heterogeneous Equilibria
Consider
The Equilibrium constant would be
But what is [H2O]?
What is [H+]?
The [H2O] is effectively a constant for changes in Equilibrium and an be folded into the equilibrium constant so the rate of the backward reaction doesn’t change

26. Heterogeneous Equilibria
pure liquids are materials whose concentration doesn’t change (much) during the course of a reaction and is essentially constant compared to other reagents
For solids, the forward and backward reactions depend on the surface area making the equilibrium constant independent of the amount of solid
for the reaction aA(s) + bB(aq) cC(l) + dD(aq) the equilibrium constant expression is:

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

28. Initial and Equilibrium Concentrations for H2(g) + I2(g) 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

29. Calculating Equilibrium Concentrations
Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know ..
Initial concentrations
One equilibrium concentration
Example: suppose you have a reaction 2 A(aq) + B(aq) 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. What are [A] and [B] at equilibrium?
[A]
[B]
[C]
Initial molarity
1.00
Change in concentration
Equilibrium molarity
0.50
-½(0.50)
-¼(0.50)
+0.50
0.75
0.88

30. Find the value of Kc for the reaction 2 CH4(g) C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4] = M and the equilibrium [C2H2] = M
[CH4]
[C2H2]
[H2]
initial
0.115
0.000
change
equilibrium
0.035
use the known change to determine the change in the other materials
add the change to the initial concentration to get the equilibrium concentration in each column
use the equilibrium concentrations to calculate Kc + +
-2(0.035)
+0.035
+3(0.035)
0.045
0.105

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

32. The Reaction Quotient: Predicting the Direction of Change
if Q > K, the reaction will proceed fastest in the reverse direction
the [products] will decrease and [reactants] will increase
if Q < K, the reaction will proceed fastest in the forward direction
the [products] will increase and [reactants] will decrease
if Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction
if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction

33. Q, K, and the Direction of Reaction

34. If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Q
For the reaction below, which direction will it proceed if PI2 = atm, PCl2 = atm & PICl = atm?
Given:
Find:
for I2(g) + Cl2(g) ICl(g), Kp = 81.9
direction reaction will proceed
Concept Plan:
Relationships:
If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Q
PI2, PCl2, PICl
Solution:
I2(g) + Cl2(g) ICl(g) Kp = 81.9
since Q (10.8) < K (81.9), the reaction will proceed to the right

35. Finding Equilibrium Concentrations: Given the Equilibrium Constant and Initial Concentrations or Pressures
first decide which direction the reaction will proceed
compare Q to K
define the changes of all materials in terms of x
use the coefficient from the chemical equation for 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
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

36. since Qp(1) < Kp(81.9), the reaction is proceeding forward
For the reaction I2(g) + Cl2(g) °C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations
Construct an ICE table for the reaction
determine the direction the reaction is proceeding
[I2]
[Cl2]
[ICl]
Initial
0.100
Change
Equilibrium
since Qp(1) < Kp(81.9), the reaction is proceeding forward

37. For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81. 9
For the reaction I2(g) + Cl2(g) °C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations
represent the change in the partial pressures in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression and solve for x
[I2]
[Cl2]
[ICl]
initial
0.100
change
equilibrium -x -x
+2x
0.100-x
0.100-x x

38. For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81. 9
For the reaction I2(g) + Cl2(g) °C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations
[I2]
[Cl2]
[ICl]
initial
0.100
change
equilibrium
substitute into the equilibrium constant expression and solve for x -x -x
+2x
0.100-x
0.100-x x

39. For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25°C, Kp = 81. 9
For the reaction I2(g) + Cl2(g) °C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations
[I2]
[Cl2]
[ICl]
initial
0.100
change
equilibrium
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K
2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures

40. For the reaction I2(g) 2 I(g) Kc = 3. 76 x 10-5 at 1000 K If 1
For the reaction I2(g) I(g) 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]? (Hint: you will need to use the quadratic formula to solve for x)

41. For the reaction I2(g) 2 I(g) the value of Kc = 3. 76 x 10-5 at 1000 K
For the reaction I2(g) 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]?
since [I]initial = 0, Q = 0 and the reaction must proceed forward
[I2]
[I]
initial
0.500
change -x
+2x
equilibrium x 2x

42. For the reaction I2(g) 2 I(g) the value of Kc = 3. 76 x 10-5 at 1000 K
For the reaction I2(g) 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]?
[I2]
[I]
initial
0.500
change -x
+2x
equilibrium x 2x

43. For the reaction I2(g) 2 I(g) the value of Kc = 3. 76 x 10-5 at 1000 K
For the reaction I2(g) 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]?
[I2]
[I]
initial
0.500
change -x
+2x
equilibrium
0.498
= 0.498
[I2] = M
2( ) =
[I] = M
x = 

44. Approximations to Simplify the Math
when the equilibrium constant 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

45. Checking the Approximation Refining as Necessary
we can check our approximation afterwards 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. For the reaction I2(g) 2 I(g) the value of Kc = 3. 76 x 10-5 at 1000 K
For the reaction I2(g) 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]?
Since [I]initial = 0, Q = 0 and the reaction must proceed forward
Kc is very small so equilibrium stays to the left and x is small
[I2]
[I]
initial
0.500
change -x
+2x
equilibrium x 2x

47. Is the approximation valid?
For the reaction I2(g) 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]?
[I2]
[I]
initial
0.500
change -x
+2x
equilibrium x 2x
Is the approximation valid?
Yes

48. Disturbing and Re-establishing Equilibrium
once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same
however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established
the new concentrations will be different, but the equilibrium constant will be the same
unless you change the temperature

49. Le Châtelier’s Principle
Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium
it says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance

50. The Effect of Concentration Changes on Equilibrium

51. Disturbing Equilibrium: Adding Reactants
Adding a reactant increases the rate of the forward reaction
[Reactant] decreases and [Product] increases
Equilibrium shifts toward the product side (Equilibrium shifts right)
you can use this to drive a reaction to completion!
Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium

52. Disturbing Equilibrium: Adding Products
Adding a product increases the rate of the backward reaction
[Reactant] increases and [Product] decreases
Equilibrium shifts toward the reactant side (Equilibrium shifts to the left)

53. Disturbing Equilibrium: Removing Reactants
Removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction
so the reaction is going faster in reverse
The reaction proceeds to the left until equilibrium is re-established
at the new equilibrium position, you will have less of the products than before, more of the non-removed reactants than before, and more of the removed reactant
but not as much of the removed reactant as you had before the removal
at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

54. Disturbing Equilibrium: Removing Products
Removing a product initially decreases the rate of the backward reaction, but has no initial effect on the rate of the forward reaction
so the reaction is going faster in the forward direction
The reaction proceeds to the right until equilibrium is re-established
At the new equilibrium position, you will have less of the reactants than before, more of the non-removed products than before, and more of the removed product
but not as much of the removed product as you had before the removal
at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

55. Effect of Volume Change on Gas- Phase Equilibria
Decreasing the size of the container increases the concentration of all the gases in the container and their partial pressures
Partial pressures increase, then the total pressure in the container will increase
according to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure
the way the system reduces the pressure is to reduce the number of gas molecules in the container
when the volume decreases, the equilibrium shifts to the side with fewer gas molecules

56. Effect of Volume Change on Gas- Phase Equilibria
Consider
Decrease V
P increase equilibrium shift to reduce moles of gas
Increase V
P decreases equilibrium shifts to increase the moles of gas

57. Disturbing Equilibrium: Changing the Volume
How will decreasing the container volume affect the total pressure of solids, liquid, and gases?
How will it affect the concentration of solids, liquid, solutions, and gases?
What will this cause?
How will it affect the value of K?

58. Disturbing Equilibrium: Reducing the Volume
for solids, liquids, or solutions, changing the size of the container has no effect on the concentration, therefore no effect on the position of equilibrium
decreasing the container volume increases the concentration of all gases
since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules
at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same

59. Volume Changes: Equilibrium
Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products.
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.

60. Temperature and Equilibrium
exothermic reactions release energy and endothermic reactions absorb energy
if we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes
Temperature changes K

61. Temperature and Equilibrium

62. Temperature Changes: Exothermic Reactions
for an exothermic reaction, heat, dq, 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 and increase the concentrations of reactants (shift left)
adding heat to an exothermic reaction will decrease the value of K

63. Temperature Changes: 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 and increase the concentrations of products (shifts right)
adding heat to an endothermic reaction will increase the value of K

64. Temperature and Equilibrium

65. Not Changing the Position of Equilibrium:Catalysts
catalysts provide an alternative, more efficient mechanism
works for both forward and reverse reactions
affects the rate of the forward and reverse reactions by the same factor
therefore catalysts do not affect the position of equilibrium

66. Practice: Le Châtelier’s Principle
The reaction 2 SO2(g) + O2(g) SO3(g) with ΔH° = -198 kJ at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established?
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
right
right
right
right
left
left
No change
no change