1. IB Chemistry SL/HL
Equilibrium

2. 7.1 Dynamic Equilibrium
7.1.1 Outline the characteristics of chemical and physical systems in a state of equilibrium.

3. Escalator Analogy
Imagine you are on an escalator moving up and you start to walk down the steps.
If you do this at exactly the same speed (rate) as the escalator you will not have any net movement either up or down.
This means you are in a dynamic equilibrium: both you and the escalator are moving but there’s no change in your relative positions.

4. Completion?
Some reactions go to completion at room temperature as the products are energetically more stable and the activation energy is low, ex. NaOH + HCl  NaCl + H2O
Other reactions do not occur because the activation energy is too high and / or reactants are more stable than products.
Some reactions have a similar stability in their reactants and products so the reaction is reversible.

5. Reversible Reaction
A reversible reaction can occur in either direction.
Ex.NH3(g) + HCl(g)  NH4Cl (s)
When the gases mix ammonium chloride is seen as a white smoke of solid.
When ammonium chloride is heated, ammonia and chlorine gases form.

6. Chemical Equilibrium
If a system like this is in a closed vessel so no gases can escape and temperature is constant CHEMICAL EQUILIBRIUM is reached.
Chemical equilibrium is when the forward and reverse reactions occur at the same RATE.

7. Dynamic Equilibrium
It is called DYNAMIC equilibrium because the reactants continue to form products and the products continue to form reactants.
Ex. A + B  C + D
Initially when A and B are mixed they form C and D but the reverse reaction cannot occur as no C and D are present.

8. Dynamic Equilibrium A + B  C + D
As A and B react their concentration decreases and the concentration of C and D increases.
As soon as C and D start to form the reverse reaction can also occur and A and B reform.

9. Dynamic Equilibrium
Eventually the rates of the forward and reverse reactions are the same and the concentration of all substances are constant and equilibrium is established.
As well as concentration remaining constant, color, pH and density are also constant at equilibrium even though there is constant interconversion between the reactants and products.

10. Concentration vs. time

11. Rate vs. time

12. Concentration and Rate
The concentrations of the species at equilibrium reflect how easily they react.
If two species react easily to reach equilibrium concentration then their concentration at eqm will be low.
If two species do not react so well together it will take a higher concentration to reach eqm.

13. Example
1 mol of liquid dinitrogen tetroxide (N2O4) is placed into an evacuated sealed 1 dm3 flask at 80°C.
At first the colorless N2O4 will vaporise and then decompose into brown NO2.
The rate of this will fall once NO2 forms as some will dimerise to produce N2O4 again.

14. Example Eventually the 2 rates will become equal and eqm is reached.
This is seen by the fact that the brown color of the gas does not change any more and the pressure in the flask remains constant.
Eqm will occur when about 60% of the N2O4 reacts to form NO2.

15. NO2 N2O4 Equilibrium

16. Example Initial concn 1 mol dm-3 0 mol dm-3
N2O4(g)  2NO2(g)
Initial concn 1 mol dm-3 0 mol dm-3
Eqm concn mol dm mol dm-3
If 2 mol NO2 were cooled from 200°C to 80°C then the brown color would fade to a constant value and the same eqm would be reached.

17. Feature of Eqm state
Explanation 1
Equilibrium is dynamic
The reaction has not stopped but both backward and forward reactions are still occurring at the same rate 2
Equilibrium is achieved in a closed system
A closed system prevents exchange of matter with the surroundings, so the equilibrium is achieved where both reactants and products can react and recombine with each other. 3
The concentrations of products and reactants remain constant at equilibrium
They are being produced and destroyed at an equal rate. 4
At equilibrium there is no change in macroscopic properties
This refers to observable properties such as color and density. These do not change as they depend on concentrations of the components of the mixture. 5
Equilibrium can be reached from either direction
The same equilibrium mixture will result under the same conditions, no matter whether the reaction is started with all reactants, all products or a mixture of both.

18. Learning Check True or false: At Equilibrium
There are both reactants and products present.
The forward and reverse reactions occur at the same rate.
The concentrations of reactants and products are equal
The concentrations of reactants and products remain constant.

19. 7.2 The position of equilibrium
7.2.1 Deduce the equilibrium constant expression (Kc) from the equation for a homogeneous reaction. Consider gases, liquids and aqueous solutions.
7.2.2 Deduce the extent of a reaction from the magnitude of the equilibrium constant.

20. Position of Equilibrium
On that escalator you could be near the top, near the bottom or in the middle and be at equilibrium in any position as long as you are moving at the same rate.
In chemical equilibria we talk about the position by saying the forward reaction is favored or the equilibrium lies to the right if there are mostly products present.
If mainly reactants are present then the reverse reaction is favored and the equilibrium lies to the left.

21. Equilibrium Constant For the reaction A + B  C + D
If we assume the rate is 1st order in all the substances the rate expressions are
Forward rate = kf[A][B]
Reverse rate = kr[C][D]
At eqm these rates are equal so
kf[A][B] = kr[C][D]
And kf/kr = [C][D] / [A][B]

22. Equilibrium Constant kf[A][B] = kr[C][D] And kf/kr = [C][D] / [A][B]
As kf and kr are constants their ratio must also be constant.
This ratio is called the equilibrium constant, Kc
So the equilibrium constant is derived from the rate constants.

23. Equilibrium Constant
The more common way to find the equilibrium constant is by dividing the concentrations at equilibrium of the products by the concentration of the reactants at equilibrium.
Each concentration is raised to the power of its coefficient from the chemical equation
Kc = [products]x / [reactants]y
Ex. a A + bB  pP + qQ
Kc = [P]p[Q]q / [A]a[B]b

24. Learning Check Write the equilibrium expression for the eqm:
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

25. Learning Check
The units of Kc depend upon the eqm expression assuming all concentrations have the units mol dm-3.
You just need to work out how many cancel out between the top and bottom of the equilibrium expression.
Don’t bother reporting units for K.

26. Equilibrium Constant
If Kc is greater than 1 then the concentrations of the products are greater than the reactants and the eqm lies on the right hand side.
If Kc is very large then the reaction is considered as going to completion.
If Kc less than 1 then the concentration of the reactants is higher than the products and eqm lies on the left hand side.
If Kc is very small then the reaction may be considered not to occur.

27. Eqm Expression
Solids and pure liquids are assumed to have fixed density so have constant concentration and are left out of the eqm expression.
This is particularly important for H2O.
Exceptions to this are when pure liquids are reacting NOT in an aqueous solution.
Water should also be included in the expression if it’s a gas.

28. Learning Check Write the equilibrium expressions for the following:
Co(H2O)62+(aq) + 4Cl-(aq)  CoCl42-(aq) + 6H2O(l)
CH3COOH(l) + C2H5OH(l)  CH3COOC2H5(l) + H2O(l)
H2(g) + CO2(g)  H2O(g) + CO(g)

29. Kc
If an equilibrium is written the other way round (reverse reaction is forward) then the value of Kc is the reciprocal:
H2(g) + I2(g)  2HI(g) Kc= 2
2HI (g)  H2(g) + I2(g) Kc = 0.5 or ½

30. Combining Equilibria
You can also combine 2 related equilibria and the Kc is the product of the individual Kc’s.
H2O(g) + C(s)  H2(g) + CO(g)
Kc = [H2][CO] / [H2O] = 4.48x10-4 mol dm-3
H2O(g) + CO(g)  H2(g) + CO2(g)
Kc =[H2][CO2] / [H2O][CO] = 1.05
1+2. 2H2O(g) + C(s)  2H2(g) + CO2(g)
Kc = [H2]2[CO2] / [H2O]2 = 4.70x10-4 mol dm-3

31. 7.2 The position of equilibrium
7.2.3 Apply Le Chatelier’s principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of the equilibrium constant. Students are not required to state Le Chatelier’s principle.
7.2.4 State and explain the effect of a catalyst on an equilibrium reaction.

32. Le Chatelier’s Principle
“If a change is made to the conditions of a chemical equilibrium, then the position of equilibrium will readjust so as to minimize the change made.”
This means if temperature, pressure or concentration are changed equilibrium is disturbed and the concentrations of the species will change until the rates are equal and a new equilibrium is reached.

33. Le Chatelier’s Principle.
This means that increasing the concentration of a substance will result in a change that will decrease that substance’s concentration again.
The same is true for pressure and temperature.
The system reacts to try to return the system to the way it was.

34. Le Chatelier’s Principle

35. Find Out Why!
In pairs use the textbook to find an explanation for each of these effects on equilibrium:
Temperature
Pressure
Concentration
Catalyst

36. Concentration Consider this reaction PCl5(s)  PCl3(l) + Cl2(g)
∆H = +88 kJ mol-1
If [Cl] increases then the rate of the reverse reaction will increase but the forward will be unaffected so reaction rates are no longer equal.
To regain equilibrium the amount of PCl5 must increase and the amount of PCl3 must decrease.

37. Concentration PCl5(s)  PCl3(l) + Cl2(g)
Then the position of the eqm will shift to the left and the amount of Cl2 will decrease below its new higher level but will still be above the original level.
A decrease in Cl2 would have the opposite effect.
Kc remains unaffected.

38. Pressure PCl5(s)  PCl3(l) + Cl2(g)
If total pressure is increases the rate of the reverse reaction will increase as it is the only one with a gas.
Again the equilibrium will shift to the left to reduce the amount of chlorine and hence reducing the pressure to somewhere between the original and the new higher value.
A decrease in pressure would have the opposite effect.
Again Kc remains unchanged.

39. Temperature PCl5(s)  PCl3(l) + Cl2(g)
If temperature is increased then the rates of both forward and reverse reactions will be increased but NOT by the same amount.
The higher Ea the greater the effect of temperature on rate.
An increase in temperature will speed up the reaction in the endothermic direction (has greater Ea) than the exothermic direction.

40. Temperature PCl5(s)  PCl3(l) + Cl2(g)
In this example the forward reaction is endothermic so an increase in temperature will increase the Kc of the forward reaction more than that of the reverse reaction.
Therefore Kc increases and the reaction shifts to the right, making more Cl2 and PCl3 until the rates again become equal.

41. Temperature
The endothermic change absorbs heat and causes the temperature of the system to fall to below its new higher value.
A decrease in temperature will have the opposite effect and the value of Kc will decrease.

42. 7.2 The position of equilibrium
7.2.5 Apply the concepts of kinetics and equilibrium to industrial processes.
Suitable examples include the Haber and Contact processes.

43. Industrial Processes Many industrial processes involve equilibria.
The aim in industry is produce your chosen substance as efficiently as possible - that is, quickly but with the least input of energy.
This requires studying the kinetics and equilibrium of the reaction.

44. Haber Process
This process combines hydrogen and nitrogen to form ammonia.
A mixture of N2(g) and H2(g) in a 1:3 ratio by volume is compressed and passed over a heated, finely divided, iron catalyst where the equilibrium is established.
N2(g) + 3H2(g)  2NH3(g)
∆H = -92 kJ mol-1

45. Haber Process
From the equation it can be seen that high pressure will favor formation of the product as 4 mol of gas react to form 2 mol of gas.
It is expensive to create a high pressure environment so there has to be a compromise between pressure and price!

46. Haber Process
The forward reaction is exothermic so a low temperature would favor products.
However low temperatures result in a low rate so it would take a long time to make the NH3.
A moderate (compromise) temperature is found to produce the most NH3 per hour.
A catalyst is also used to increase the rate of reaction – this does not change the amount of product formed (yield).

47. Haber Process
Typical conditions for the Haber process is a pressure of 200 atm (20 MPa) and temperatures around 450°C (723K).
The reaction is not left long enough for equilibrium to be reached as the rate would decrease at eqm.

48. Haber Process
At any time only about 20% of the reactants are converted to ammonia.
The leftover reactants are not wasted though.
The reaction system is cooled and NH3 condenses to a liquid as it forms hydrogen bonds so can be separated and the N2 and H2 gases are recycled.
Separating the product in this way also favors the forward reaction to keep occurring.

49. Haber Process
About 80% of ammonia (NH3) is used to make fertilizers which are important for growing crops especially for food.
Other uses of ammonia and its compounds are explosives, plastics and refrigerants.
About 120 million tonnes of ammonia are produced worldwide in a year so it is to significant to the economy.

50. The Contact Process
This is the production of sulfuric acid via a series of reactions. It has the highest production of any chemical in the world with 150 tonnes made annually.
Sulfuric acid is used to make fertilizers, detergents, dyes, explosives, drugs, plastics plus other products.

51. The Contact Process Step 1. The combustion of sulfur:
S(s) + O2(g)  SO2(g)
Step 2. The oxidation of SO2:
2SO2(g) + O2(g)  2SO3(g) H= -196kJ mol-1
Step 3. Combination of SO3 with water:
SO3(g) + H2O(l)  2H2SO4(aq)

52. The Contact Process
The 3rd step is actually shown below. The SO3 is added to a flowing solution of concentrated sulfuric acid and then water is added. This avoids the violent direct reaction between H2SO4 and H2O:
SO3(g) + H2SO4(l)  H2S2O7(l) + H2O(l) 2H2SO4(aq)
A vanadium (V) oxide (V2O5) is used as a catalyst in this reaction.

53. Contact Process
The Contact process is the production of sulfuric acid by the oxidation of sulfur.
Initially sulfur is burnt in the air to make sulfur dioxide:
S(s) + O2(g)  SO2(g)
Sulfur dioxide is mixed with air and passed over a vanadium (V) oxide catalyst to produce sulfur trioxide:
2SO2(g) + O2(g)  2SO3(g) H= -196kJ mol-1

54. Contact Process
As with the Haber process a high pressure would favor the formation of the product.
In this case a high conversion rate occurs at only 2 atm so it is not as expensive as the Haber process.
The chosen compromise temperature is K (450°C) and the use of a catalyst (V2O5) help to increase the forward reaction rate.

55. 17.2 The equilibrium law
Solve homogeneous equilibrium problems using the expression for Kc. The use of quadratic equations will not be assessed.

56. Equilibrium Law
If the equilibrium concentrations are known, their values can be substituted into the equilibrium expression to find the equilibrium constant, Kc.
More commonly you only know the initial concentrations and need to know the equilibrium concentrations.
The equilibrium expression and the balanced chemical equation can help you to calculate the equilibrium concentrations.

57. Equilibrium Law
Homogeneous equilibrium: all substances are in the same phase
Heterogeneous equilibrium: substances are in 2 or more phases – we won’t be solving these.
The concentrations in the equilibrium expression will only equal Kc when at equilibrium.
If not at equilibrium this value is called the reaction quotient, Qc.

58. Qc Qc = [products]x / [reactants]y
The value of Qc indicates which way the reaction will go.
If Qc > Kc then some products must be converted into reactants to get to Kc - so the reverse reaction is favored.
If Qc<Kc then the system must shift to the right (forward reaction) until enough products form that Kc = Qc.

59. Example Problem
0.20 mol of PCl3(g) and 0.10 mol of Cl2(g) were placed into a 1 dm3 flask at 350°C. The reaction, which produced PCl5(g), was allowed to come to equilibrium and it was found the flask contained 0.12 mol PCl3. What is Kc for this reaction?

60. Example Solution
1. Write the balanced chemical equation: PCl3(g) + Cl2(g) PCl5(g)
2. Make a table of initial and eqm concentrations with the change. (The black values were given).
PCl3(g)
+ Cl2(g) 
 PCl5(g)
Initial (mol dm-3)
0.20
0.10
0.00
Change
(mol dm-3)
-0.08
+0.08
Eqm (mol dm-3)
0.12
0.02
0.08

61. Example Solution 3. Write the eqm expression:
Kc = [PCl5] / [PCl3][Cl2]
4. Plug in the numbers:
Kc = 0.08 / (0.12)(0.02) = 33
PCl3(g)
+ Cl2(g) 
 PCl5(g)
Initial (mol dm-3)
0.20
0.10
0.00
Change
(mol dm-3)
-0.08
+0.08
Eqm (mol dm-3)
0.12
0.02
0.08

62. Example Solution
Note that the volume in this example is 1 dm3 so concentrations were found easily – if volume is not 1 dm3 use c = n/V to find concentration.
There was a 1:1:1 ratio in this reaction, if it’s not that simple take that into account when calculating changes.

63. Learning Check
NO is oxidized to form NO2 in smog formation mol of NO reacts with 0.60 mol O2 in a 2 dm3 container at 500°C, the equilibrium mixture was found to contain 0.20 mol of NO2.
Calculate the equilibrium constant for the reaction at this temperature.

64. Calculating Equilibrium Concentrations from Kc
Q: The reaction CO(g) + 2H2(g)  CH3OH(g) has Kc = at 350K. If the concentrations at equilibrium are: CO = mol dm-3 and H2 = mol dm-3, what is the equilibrium concentration of CH3OH? A: Write equilibrium expression and substitute in known values.

65. Calculating Equilibrium Concentrations from Kc
Q: Kc for the reaction SO3(g) + NO(g)  NO2(g) + SO2(g) was found to be 6.78 at a specified temperature. If the initial concentration of both reactants was 0.03 mol dm-3, what would be the equilibrium concentration of each component? A: 1. Set up a table of changes

66. Calculating Equilibrium Concentrations from Kc
SO3(g) +
NO(g) 
 NO2(g)
+ SO2(g)
Initial (mol dm-3)
0.03
0.00
Change (mol dm-3) -x +x
Equilibrium (mol dm-3)
0.03-x x
The 1:1 ratio helps keep this simple.
2. Next write the eqm expression and substitute in values and terms.

67. Calculating Equilibrium Concentrations from Kc
Kc = [NO2][SO2] / [SO3][NO]
= x2 / (0.03 – x)2 = 6.78
Take the square root of both sides:
x / ( x) = √(6.78) = 2.60
x( ) = 2.60 – 0.03 = 0.078
x = / 3.60 =
3. Use table and value of x to find eqm concentrations.

68. Calculating Equilibrium Concentrations from Kc
x = = 0.022
[SO3] = 0.03 – = mol dm-3
[NO] = 0.03 – = mol dm-3
[NO2] = x = mol dm-3
[SO2] = x = mol dm-3

69. Calculating equilibrium concentrations when Kc is very small
In some reactions the forward reaction hardly occurs and the value of Kc is very small (less than 10-3).
This means that the change in reactant concentrations is near zero and [reactant]initial ~ [reactant]eqm.

70. Calculating equilibrium concentrations when Kc is very small
The thermal decomposition of water has a small Kc . At 1000°C, Kc = 7.3x10-18.
2H2O(g)  2H2(g) + O2(g)
A reaction is set up at this temperature with an intital H2O concentration of 0.10 mol dm-3. Calculate the H2 concentration at equilibrium.

71. Calculating equilibrium concentrations when Kc is very small
If change in [H2] is 2x then change in [O2] is x.
Change in [H2O] = -2x (due to 2:2:1)
2H2O(g) 
 2H2(g)
+ O2(g)
Initial (mol dm-3)
0.10
0.00
Change (mol dm-3)
-2x
+2x +x
Eqm ( mol dm-3)
0.10-2x 2x x
~ 0.10

72. Calculating equilibrium concentrations when Kc is very small
[H2O]initial ~ [H2O]eqm as Kc is so small.
Kc = [H2]2[O2] / [H2O]2
= (2x)2x / (0.10)2 = 7.3x10-18
This can now be solved for x
4x3 = (7.3x10-18)(0.010)
= 7.3x10-20
x = x10-7
[H2]eqm = 2x = 5.3x10-7 mol dm-3.

73. 17.1 Liquid-vapour equilibrium
Describe the equilibrium established between a liquid and its own vapour and how it is affected by temperature changes.
Sketch graphs showing the relationship between vapour pressure and temperature and explain them in terms of kinetic theory.
State and explain the relationship between enthalpy of vaporization, boiling point and intermolecular forces.

74. Liquid- Vapor Equilibrium
If a sealed container holds a liquid some molecules at the surface of the liquid will escape and vaporize.
The gas molecules will collide with the walls of the container and exert a pressure on it.
Some molecules will collide with the surface of the liquid and condense back into the liquid phase.

75. Liquid - Vapor Equilibrium
Eventually the rate at which molecules vaporize will equal the rate at which they condense.
This dynamic equilibrium is similar to a chemical equilibrium.
The pressure of the vapor is called the VAPOR PRESSURE of the liquid.

76. Factors NOT affecting Vapor Pressure: S.A. and V
Changing the surface area of the liquid affects both rates equally so it has no overall effect on vapor pressure.
It will affect the time it takes to reach dynamic equilibrium though: The greater the surface area the faster equilibrium is reached.
The volumes of the liquid or vapor in the container will not affect vapor pressure either.

77. Factors NOT affecting Vapor Pressure: S.A. and V
In a closed system the rate of evaporation (vaporization) is constant while the rate of condensation increases with increasing concentration of vapor particles.
Equilibrium is established when the two rates are equal.

78. Factors Affecting Vapor Pressure: Temperature
Temperature changes will change the position of the equilibrium between the liquid and its vapor.
An increase in temperature will increase the vapor pressure as a higher proportion of molecules will now have enough energy to escape the liquid phase.

79. Factors Affecting Vapor Pressure: Temperature

80. Temperature
The rise of vapor pressure is not a linear relationship, it is exponential:

81. Liquid-Vapor Equilibrium
To vaporize, the surface molecules need a minimum amount of kinetic energy to overcome the intermolecular forces of the liquid molecules.
Vaporization requires energy so it is endothermic.
The amount of energy required for this phase change is called the ENTHALPY OF VAPORIZATION.

82. Enthalpy of Vaporization
Enthalpy of vaporization is defined as: the energy required to convert one mole of a substance from the liquid to the gaseous state.
∆Hvap is sometimes used to represent it.
H2O(l)  H2O(g) ∆Hvap= +40.7kJmol-1
At 373K and 101.3kPa

83. Enthalpy of Vaporization
The energy required to vaporize something is mainly needed to overcome the intermolecular forces but some work must also be done against the atmosphere.
When a substance boils, its temperature does not increase (there is no increase in K.E.) so the energy absorbed is involved in increasing potential energy by overcoming attractive forces between the particles.

84. Enthalpy of Vaporization
At a higher temperature, more molecules will have enough kinetic energy to escape the liquid phase so the rate of vaporization will be higher.
This means that the vapor pressure must also increase.

85. Boiling
A liquid will boil when its vapor pressure is equal to standard atmospheric pressure (101.3 kPa).
However a liquid will boil at a lower temperature if the pressure is also decreased.
In the same way if pressure is increased the boiling point of a liquid will increase.
This is used in pressure cookers and autoclaves to heat the substance to a higher temperature.

86. Factors Affecting Vapor Pressure: IMF’s
Substances with stronger intermolecular forces (IMF’s) require more energy to overcome those attractive forces and evaporate so they tend to have higher boiling points, higher ΔHvap and lower vapor pressures.
Substances with weak IMF’s are described as VOLATILE.

87. Factors Affecting Vapor Pressure: IMF’s
When comparing substances with similar IMF’s, the substance with the higher mass will have a higher boiling point and lower vapor pressure due to the increase in van der Waals forces.

88. Intermolecular Forces
The stronger the intermolecular forcess the higher the ∆Hvap, the lower the vapor pressure at a particular temperature and the higher the boiling point ex:

89. Raoult’s Law – NOT IB, in SAT II
If we have a mixture of liquids in a container we will also have a mixture of vapors.
We can calculate the pressure being exerted by each part of the mixture by using Raoult’s Law.
Raoult’s Law assumes that the intermolecular forces in the mixture are the same as each component on its own.

90. Raoult’s Law
Raoult’s Law states that the sum of the vapor pressures multiplied by their mole fraction equals the total vapor pressure over the mixture.
Mole fraction is just the number of moles of a component divided by the total number of moles in the mixture:mol A /mol A + mol B

91. Raoult’s Law So PA = P0A x mol A / total moles
where P0A is the pressure of A over the mixture and PA is the pressure of A over a pure sample of A.
Raoult’s Law holds true if only weak imf’s are involved. When there are dipole-dipole interactions or H bonding between the vapors there can be some deviation from this law.

92. Raoult’s Law
If the imf’s in the mixture are weaker than in the pure components then it is easier for evaporation to occur and the vapor pressure will increase.
If the imf’s in the mixture are stronger than in the pure components then it will be harder for the mixture to vaporize and the vapor pressure will be lower.