Chemical%20Equilibrium

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
When first added, the air has no liquid
vapor. The vapor pressure of the liquid starts to
populate the vapor With molecules
The rate of evaporation does not change, but the
rate of condensation continuously increases Until
the rate of evaporation the rate of
condensation
5
Reactions Go Both Ways
2 NO2 (g) ? N2O4 (g)
? Brown Colorless
N2O4 (g) ? 2 NO2 (g)
? Colorless Brown
6
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) 2HI (g)
• All species are in solution.
• HC2H2O2 (aq) H (aq) C2H3O2- (aq)

9
Chemical equilibrium

• 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

• 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

• 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

• 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
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 krNO22
• Isolating the concentrations and rate constants
  together
• NO22/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, thats moles per liter of solution

19
Law of Mass Action

• pA qB ? rC tD
• K Cr Dt Ap Bq
• molarity
• K equilibrium constant

Products over
reactants
20
Law of Mass Action

• 4 NH3(g) 7O2 (g) ? 4NO2 (g) 6H2O (g)
• K Cr Dt NO24 H2O6
  Ap Bq NH34 O27

21
Equilibrium Constant
Y Reactant ? X Product (g)

• Keq productsx reactantsy
• 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 NH32 N2 H23
K 6.02 x 10-2
23
N2(g) 3H2(g) ? 2NH3(g)
K NH32 N2 H23
K 6.02 x 10-2
K 6.02 x 10-2
24
N2(g) 3H2(g) ? 2NH3(g)
K NH32 N2 H23
K 6.02 x 10-2
K 6.02 x 10-2
K 6.02 x 10-2
25
Equilibria

• It doesnt 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 gt 1 indicate that forming products is
  favored. Keq lt 1 indicate that reactants dont
  react easily.

27

• Equilibrium Constant, Keq

• 3H2(g)N2(g) 2NH3(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) 3H2(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 2 or 9.52

29

• Equilibrium Constant, Keq

• NH3(g) 1.5H2(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.

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) 2H2S (g)

• Kp Kc (RT)Dng
• T 700 273 973 K
• R 0.08206
• Dng ( 2 ) - ( 2 1) -1

34
Partial pressureequilibrium constants

• Kp Kc (RT)Dng
• 1.10 x 107 (0.08206 ) (973 K)
• 1.378 x105

-1


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 dont 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 (lt0.1M).
• 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) 2 NH3(g) CO2(g) H2O(g)

• Kc NH32 CO2 H2O
• (NH4)2CO3 is a pure solid so is not included Kc
• We use NH32 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 soup

• Ka

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) 2H2O (g)
• Kc 2.9 x 1031 H2O2
• H22 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) 2HI (g)

44
Determining equilibrium constants

• H2 (g) I2 (g) 2HI (g)
• Assume that we started with the following initial
  concentrations at 425.4oC.
• H2 (g) 0.00500 M
• I2 (g) 0.01250 M
• HI (g) 0.00000 M
• At equilibrium, we determine that the
  concentration of iodine is 0.00772 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
• 0.00772 M
• That means that 0.00478 mol I2 reacts (0.01250M -
  0.00772M) to produce HI in 1.00 L of solution.

46
Determining equilibrium constants

• I2 (g) 0.00772 M
• HI (g) 0.00478 M
• 0.00956 M
• H2 (g) 0.00500 M - 0.00478 M
• 0.00022 M
• Kc
  54

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.

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 lt Kc Net forward reaction will occur.
• Q Kc No change, at equilibrium.
• Q gt Kc Net reverse reaction will occur.

49
Reaction quotient example

• For an earlier example
• H2 (g) I2 (g) 2HI (g)
• we determined the Kc to be 54 at 425.4 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

• Q
• 5.73
• Since Q is lt Kc, the system is not in
  equilibrium and will proceed in the forward
  direction.

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 10-10 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 0.095 0.000 0.000
• Change in conc. - X X X
• due to reaction
• Equilibrium
• Concentration, M (0.095 -X) X X
• Kc

54
Equilibrium calculation example

• Rearrangement gives

X2 2.2 x 10-10 X - 2.09 x 10-11 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.

56
Equilibrium calculation example
X2 2.2 x 10-10 X - 2.09 x 10-11 0 a
b c
X

• X 9.1 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 0.095 - X 0.095 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)62 4Cl1- CoCl42- 6H2O

• 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 Chateliers principle

• Co(H2O)62 4Cl1- CoCl42- 6H2O
• Note ?H value

• Add HCl (adds Cl1-)

• Add heat

• Add acetone

• (removes H2O)

• (removes Cl1- )

• Add Ag1

• Add ice

• What other stresses could be placed on this
  system to change the color back and forth between
  pink and blue?

64
Le Chateliers 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) 2NH3 (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 Châteliers 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 Châteliers Principle

• Change in Reactant or Product Concentrations

68

• Le Châteliers 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 Châteliers Principle

• Change in Reactant or Product Concentrations

• The catalyst bed is kept at 460 - 550 ?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 Châteliers 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 Châteliers 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) 2HI (g)
• N2O2 (g) 2NO2 (g)

• Unaffected by pressure

• Increased pressure, shift to left

73

• Le Châteliers 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 Châteliers 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 Châteliers Principle

• Effect of Temperature Changes
• The equilibrium constant is temperature dependent.

• For an endothermic reaction, ?H gt 0 and heat can
  be considered as a reactant.
• For an exothermic reaction, ?H lt 0 and heat can
  be considered as a product.
• Adding heat (i.e. heating the vessel) favors away
  from the increase
• if ?H gt 0, adding heat favors the forward
  reaction,
• if ?H lt 0, adding heat favors the reverse
  reaction.

76

• Le Châteliers Principle

• Effect of Temperature Changes
• Removing heat (i.e. cooling the vessel), favors
  towards the decrease
• if ?H gt 0, cooling favors the reverse reaction,
• if ?H lt 0, cooling favors the forward reaction.
• for which DH gt 0.
• Co(H2O)62 is pale pink and CoCl42- is blue.

77

• Le Châteliers Principle

• Effect of Temperature Changes

78

• Le Châteliers 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 gt 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 gt 0, removing heat favors the reverse
  reaction which is the formation of pink
  Co(H2O)62.

• Co

79

• Le Châteliers 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.