Chemical Equilibrium

1
Chemical Equilibrium

• The reversibility of reactions

2
Equilibrium

• Many chemical reactions do not go to
  completion. Initially, when reactants are
  present, the forward reaction predominates. As
  the concentration of products increases, the
  reverse reaction begins to become significant.

3
Equilibrium

• The forward reaction rate slows down since the
  concentration of reactants has decreased. Since
  the concentration of products is significant, the
  reverse reaction rate gets faster.
• Eventually, the
• forward reaction rate reverse reaction rate

4
Equilibrium

• forward reaction rate reverse reaction rate
• At this point, the reaction is in equilibrium.
  The equilibrium is dynamic. Both the forward and
  reverse reactions occur, but there is no net
  change in concentration.

5
Equilibrium 2NO2 ? N2O4
The concentrations of N2O4 and NO2 level off when
the system reaches equilibrium.
6
Equilibrium 2NO2 ? N2O4
The system will reach equilibrium starting with
either NO2 , N2O4, or a mixture of the two.
7
Equilibrium

• Reactions that can reach equilibrium are
  indicated by double arrows (? or ).
• The extent to which a reaction proceeds in a
  particular direction depends upon the reaction,
  temperature, initial concentrations, pressure (if
  gases), etc.

8
Equilibrium

• Scientists studying many reactions at
  equilibrium determined that for a general
  reaction with coefficients a, b, c and d
• a A b B ? c C d D
• K is the equilibrium constant for the reaction.

CcDd
K
AaBb
9
Equilibrium

• a A b B ? c C d D
• This is the equilibrium constant expression for
  the reaction. K is the equilibrium constant.
  The square brackets indicate the concentrations
  of products and reactants at equilibrium.

CcDd
K
AaBb
10
Equilibrium Constants

• The value of the equilibrium constant depends
  upon temperature, but does not depend upon the
  initial concentrations of reactants and products.
  It is determined experimentally.

11
Equilibrium Constants
12
Equilibrium Constants

• The units of K are usually omitted. The square
  brackets in the equilibrium constant expression
  indicate concentration in moles/liter.
• Some texts will use the symbol Kc or Keq for
  equilibrium constants that use concentrations or
  molarity.

13
Equilibrium Constants

• The size of the equilibrium constant gives an
  indication of whether a reaction proceeds to the
  right.

14
Equilibrium Constants

• If a reaction has a small value of K, the
  reverse reaction will have a large value of K.
• At a given temperature, K 1.3 x 10-2 for
• N2(g) 3 H2(g) ? 2 NH3(g)
• If the reaction is reversed,
• 2 NH3(g) ? N2(g) 3 H2(g)
• K 1/(1.3 x 10-2 ) 7.7 x 101

15
Equilibrium Constants

• Equilibrium constants for gas phase reactions
  are sometimes determined using pressures rather
  than concentrations. The symbol used is Kp
  rather than K (or Kc).
• For the reaction
• N2(g) 3 H2(g) ? 2 NH3(g)
• Kp (Pammonia)2
• (PN2)(PH2)3

16
Equilibrium Constants

• The numerical value of Kp is usually different
  than that for K. The values are related, since
  P nRT MRT, where M mol/liter.
• For reactions involving gases
• KC KP(RT)?n
• where n is the change in moles of gases in the
  balanced chemical reaction.

V
17
Equilibrium Constants

• The method for solving equilibrium problems
  with Kp or K is the same. With Kp, you use
  pressure in atmospheres. With K or Kc, you use
  concentration in moles/liter, or molarity.

18
Heterogeneous Equilibria

• Many equilibrium reactions involve reactants
  and products where more than one phase is
  present. These are called heterogeneous
  equilibria.
• An example is the equilibrium between solid
  calcium carbonate and calcium oxide and carbon
  dioxide.
• CaCO3(s) ? CaO(s) CO2(g)

19
Heterogeneous Equilibria

• CaCO3(s) ? CaO(s) CO2(g)
• Experiments show that the position of the
  equilibrium does not depend upon the amounts of
  calcium carbonate or calcium oxide. That is,
  adding or removing some of the CaCO3(s) or CaO(s)
  does not disrupt or alter the concentration of
  carbon dioxide.

20
Heterogeneous Equilibria

• CaCO3(s) ? CaO(s) CO2(g)

21
Heterogeneous Equilibria

• CaCO3(s) ? CaO(s) CO2(g)
• This is because the concentrations of pure
  solids (or liquids) cannot change. As a result,
  the equilibrium constant expression for the above
  reaction is
• KCO2 or Kp PCO2

22
Heterogeneous Equilibria

• The position of a heterogeneous equilibrium does
  not depend on the amounts of pure solids or
  liquids present.

23
Problem Calculation of K

• At a certain temperature, 3.00 moles of ammonia
  were placed in a 2.00 L vessel. At equilibrium,
  2.50 moles remain. Calculate K for the reaction
• N2(g) 3 H2(g) ? 2 NH3(g)

24
Problem Calculation of K

• At a certain temperature, 3.00 moles of ammonia
  were placed in a 2.00 L vessel. At equilibrium,
  2.50 moles remain. Calculate K for the reaction
• N2(g) 3 H2(g) ? 2 NH3(g)
• 1. Write the equilibrium constant expression.

25
Problem Calculation of K

• At a certain temperature, 3.00 moles of ammonia
  were placed in a 2.00 L vessel. At equilibrium,
  2.50 moles remain. Calculate K for the reaction
• N2(g) 3 H2(g) ? 2 NH3(g)
• 1. Write the equilibrium constant expression.
• K NH32/(N2H23)

26
Problem Calculation of K

• At a certain temperature, 3.00 moles of ammonia
  were placed in a 2.00 L vessel. At equilibrium,
  2.50 moles remain. Calculate K for the reaction
• N2(g) 3 H2(g) ? 2 NH3(g)
• 2. Make a table of concentrations.

27
Problem Calculation of K

• N2(g) 3 H2(g) ? 2NH3(g)

N2 H2 NH3
initial 3.00mol 2.00L
change
equili- librium 2.50mol 2.00 L
28
Problem Calculation of K

• N2(g) 3 H2(g) ? 2NH3(g)
• 3. Complete the table.

N2 H2 NH3
initial 3.00mol 2.00L
change
equili- librium 2.50mol 2.00 L
29
Problem Calculation of K

• N2(g) 3 H2(g) ? 2NH3(g)
• 3. Complete the table.

N2 H2 NH3
initial 3.00mol 2.00L
change -.50mol 2.00 L
equili- librium 2.50mol 2.00 L
30
Problem Calculation of K

• N2(g) 3 H2(g) ? 2NH3(g)
• 3. Complete the table.

N2 H2 NH3
initial 0 0 3.00mol 2.00L
change -.50mol 2.00 L
equili- librium 2.50mol 2.00 L
31
N2(g) 3H2(g) ?2NH3(g)

N2 H2 NH3
initial 0 0 3.00mol 2.00L
change .25mol 2.00 L .75mol 2.00L -.50mol 2.00 L
equili- librium 2.50mol 2.00 L
32
N2(g) 3H2(g) ?2NH3(g)

N2 H2 NH3
initial 0 0 3.00mol 2.00L
change .25mol 2.00 L .75mol 2.00L -.50mol 2.00 L
equili- librium 2.50mol 2.00 L
The equilibrium values are the sum of the initial
concentrations plus any changes that occur.
33
N2(g) 3H2(g) ?2NH3(g)

N2 H2 NH3
initial 0 0 3.00mol 2.00L
change .25mol 2.00 L .75mol 2.00L -.50mol 2.00 L
equili- librium .13M .38M 1.25M
The equilibrium values are the sum of the initial
concentrations plus any changes that occur.
34
N2(g) 3H2(g) ?2NH3(g)
N2 H2 NH3
equili- librium .13M .38M 1.25M

• 4. Substitute and solve for K.
• K NH32/(N2H23)

K(1.25)2/(.13)(.38)3 2.2 x 102
35
The Reaction Quotient

• If reactants and products of a reaction are
  mixed together, it is possible to determine
  whether the reaction will proceed to the right or
  left. This is accomplished by comparing the
  composition of the initial mixture to that at
  equilibrium.

36
The Reaction Quotient

• If the concentration of one of the reactants or
  products is zero, the reaction will proceed so as
  to make the missing component.
• If all components are present initially, the
  reaction quotient, Q, is compared to K to
  determine which way the reaction will go.

37
The Reaction Quotient

• Q has the same form as the equilibrium constant
  expression, but we use initial concentrations
  instead of equilibrium concentrations. Initial
  concentrations are usually indicated with a
  subscript zero.

38
The Reaction Quotient

• In the previous problem, we determined that K
  2.2 x 102 for the reaction
• N2(g) 3 H2(g) ? 2 NH3(g)
• Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of
  NH3 are placed in a 1.00L vessel. What will
  happen? In which direction with the reaction
  proceed?

39
The Reaction Quotient

• K 2.2 x 102 for the reaction
• N2(g) 3 H2(g) ? 2 NH3(g)
• Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of
  NH3 are placed in a 1.00L vessel. What will
  happen? In which direction with the reaction
  proceed?
• 1. Calculate Q and compare it to K.

40
The Reaction Quotient
A comparison of Q with K will indicate the
direction the reaction will go.
41
The Reaction Quotient

• K 2.2 x 102 for the reaction
• N2(g) 3 H2(g) ? 2 NH3(g)
• Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of
  NH3 are placed in a 1.00L vessel. What will
  happen? In which direction with the reaction
  proceed?
• 1. Calculate Q and compare it to K.

42
Problem

• H2(g) I2(g) ? 2HI(g) Kp 1.00 x 102
• Initially, Phydrogen Piodine 0.500 atm.
  Calculate the equilibrium partial pressures of
  all three gases.

43
Problem

• H2(g) I2(g) ? 2HI(g) Kp 1.00 x 102
• Initially, Phydrogen Piodine 0.500 atm.
  Calculate the equilibrium partial pressures of
  all three gases.
• 1. Write the Kp expression.

44
Problem

• H2(g) I2(g) ? 2HI(g) Kp 1.00 x 102
• Initially, Phydrogen Piodine 0.500 atm.
  Calculate the equilibrium partial pressures of
  all three gases.
• 1. Write the Kp expression.
• Kp (PHI)2
• (PH2)(PI2)

45
Problem

• H2(g) I2(g) ? 2HI(g) Kp 1.00 x 102
• Initially, Phydrogen Piodine 0.500 atm.
  Calculate the equilibrium partial pressures of
  all three gases.
• 2. Make a table of initial, change and
  equilibrium pressures.

46
Problem H2(g) I2(g) ? 2HI(g)
H2 I2 HI
initial .500 atm .500 atm 0
change
equili- brium
47
Problem H2(g) I2(g) ? 2HI(g)
H2 I2 HI
initial .500 atm .500 atm 0
change -x -x 2x
equili- brium
48
Problem H2(g) I2(g) ? 2HI(g)
H2 I2 HI
initial .500 atm .500 atm 0
change -x -x 2x
equili- brium .500-x .500-x 2x
49
Problem H2(g) I2(g) ? 2HI(g)
H2 I2 HI
equili- brium .500-x .500-x 2x
3. Substitute and solve.
50
Problem H2(g) I2(g) ? 2HI(g)
H2 I2 HI
equili- brium .500-x .500-x 2x

• Substitute and solve.
• Kp (PHI)2/(PH2)(PI2) 1.00 x 102
• (2x)2 1.00 x 102
• (.500-x) (.500-x)

51
Problem H2(g) I2(g) ? 2HI(g)

• 3. Substitute and solve.
• Kp (PHI)2/(PH2)(PI2) 1.00 x 102
• (2x)2 1.00 x 102
• (.500-x) (.500-x)
• Take the square root of both sides
• (2x) 10.0
• (.500-x)

52
Problem H2(g) I2(g) ? 2HI(g)

• 3. Substitute and solve.
• (2x) 10.0
• (.500-x)
• 2x 10.0 (.500-x) 5.00 -10.0x
• 12.0 x 5.00
• x .417

53
Problem H2(g) I2(g) ? 2HI(g)

• 4. Answer the question Calculate the
  equilibrium partial pressures of all three gases.
• x .417

H2 I2 HI
equili- brium .500-x .500-x 2x
54
Problem H2(g) I2(g) ? 2HI(g)

• 4. Answer the question Calculate the
  equilibrium partial pressures of all three gases.
• x .417

H2 I2 HI
equili- brium .500-x 0.083 atm .500-x 0.083 atm 2x .834 atm
55
Problem H2(g) I2(g) ? 2HI(g)

• 5. Check your answer, if possible. Does
  (PHI)2/(PH2)(PI2) 1.00 x 102 ?
• (.834)2/(.083)(.083) (.696)/(.0069) 1.0 x 102

H2 I2 HI
equili- brium .500-x 0.083 atm .500-x 0.083 atm 2x .834 atm
56
Problem N2(g) O2(g) ? 2NO(g)

• At 2200 oC, K 0.050 for the above reaction.
  Initially, 1.60 mol of nitrogen and 0.400 mol of
  oxygen are sealed in a 2.00 liter vessel.
  Calculate the concentration of all species at
  equilibrium.

57
Problem N2(g) O2(g) ? 2NO(g)

• At 2200 oC, K 0.050 for the above reaction.
  Initially, 1.60 mol of nitrogen and 0.400 mol of
  oxygen are sealed in a 2.00 liter vessel.
  Calculate the concentration of all species at
  equilibrium.
• 1. Write the equilibrium constant expression.
• K 0.050 NO2/N2O2

58
Problem N2(g) O2(g) ? 2NO(g)

• At 2200 oC, K 0.050 for the above reaction.
  Initially, 1.60 mol of nitrogen and 0.400 mol of
  oxygen are sealed in a 2.00 liter vessel.
  Calculate the concentration of all species at
  equilibrium.
• 2. Make a table of initial, change and
  equilibrium concentrations.

59
Problem N2(g) O2(g) ? 2NO(g)
N2 O2 NO
initial 1.60mol 2.00 L .400 mol 2.00 L 0
change -x -x 2x
equili- brium .800 - x .200 - x 2x
60
Problem N2(g) O2(g) ? 2NO(g)
N2 O2 NO
equili- brium .800 - x .200 - x 2x

• Substitute and solve.
• K 0.050 NO2/N2O2

61
Approaches to Solving Problems

• Some equilibrium problems require use of the
  quadratic equation to obtain an accurate
  solution. In cases where the equilibrium
  constant is quite small (roughly 10-5 or
  smaller), it is often possible to make
  assumptions that will simplify the mathematics.

62
Problem 2NOCl(g)?2NO(g) Cl2(g)

• At 35oC, K 1.6 x 10-5 for the above reaction.
  Calculate the equilibrium concentrations of all
  species present if 2.0 moles of NOCl and 1.0 mole
  of Cl2 are placed in a 1.0 liter flask.
• 1. K 1.6 x 10-5 NO2Cl2/NOCl2

63
Problem 2NOCl(g)?2NO(g) Cl2(g)

• At 35oC, K 1.6 x 10-5 for the above reaction.
  Calculate the equilibrium concentrations of all
  species present if 2.0 moles of NOCl and 1.0 mole
  of Cl2 are placed in a 1.0 liter flask.
• 2. Make a table of concentrations.

64
Problem 2NOCl(g)?2NO(g) Cl2(g)
NOCl NO Cl2
initial 2.0mol 1.00 L 0 1.0 mol 1.00 L
change -2x 2x x
equili- brium 2.0 - 2x 2x 1.0x

65
Problem 2NOCl(g)?2NO(g) Cl2(g)
NOCl NO Cl2
equili- brium 2.0 - 2x 2x 1.0x

• Substitute and solve.
• K 1.6 x 10-5 NO2Cl2/NOCl2
• 1.6 x 10-5 2x21.0 x/2.0-2x2

66
Problem 2NOCl(g)?2NO(g) Cl2(g)

• Substitute and solve.
• 1.6 x 10-5 2x2 1.0 x
• 2.0-2x2
• Since K is small, x, the amount of product
  formed, will be a small number. As a result, 1.0
  x 1.0, and 2.0-2x2.0

67
Problem 2NOCl(g)?2NO(g) Cl2(g)

• Substitute and solve.
• 1.6 x 10-5 2x2 1.0 x
• 2.0-2x2
• The expression simplifies to
• 1.6 x 10-5 2x2 1.0
• 2.02

0
0
68
Problem 2NOCl(g)?2NO(g) Cl2(g)

• Solve for x.
• 1.6 x 10-5 2x2 1.0
• 2.02
• 4x2 1.6 x 10-5 (4)/1 6.4 x 10-5
• x2 1.6 x 10-5 x 0.0040

69
Problem 2NOCl(g)?2NO(g) Cl2(g)
4. Check your assumption. x 0.0040 Does
1.0 x 1.0, and 2.0-2x2.0? 1.0 .004
1.0 2.0 2(.004) 2.0 - .0082.0

70
Validity of the Assumption
These assumptions are generally considered valid
if they cause an insignificant error. In this
case, due to significant figures, no error is
introduced. Usually, if the difference is within
5, the error is considered acceptable.

71
Problem 2NOCl(g)?2NO(g) Cl2(g)
NOCl NO Cl2
equili- brium 2.0 - 2x 2x 1.0x

• Answer the question.
• NOCl 2.0M NO 0.0080M
• Cl2 1.0M

72
Problem 2NOCl(g)?2NO(g) Cl2(g)

• Check your answer.
• NOCl 2.0M NO 0.0080M
• Cl2 1.0M
• Does NO2Cl2/NOCl2 1.6 x 10-5 ?
• (.0080)2(1.0)/(2.0)2 1.6 x 10-5

73
Le Chateliers Principle

• If a stress is applied to a system at
  equilibrium, the position of the equilibrium will
  shift so as to counteract the stress.

74
Le Chateliers Principle

• If a stress is applied to a system at
  equilibrium, the position of the equilibrium will
  shift so as to counteract the stress.
• Types of stresses
• Addition or removal of reactants or products
• Changes in pressure or volume (for gases)
• Changes in temperature

75
Fe3(aq) SCN- ? FeSCN2
76
Heat N2O4(g) ? 2 NO2(g)
77
CoCl42- 6 H2O??Co(H2O)62 4Cl1- heat
78
Le Chateliers Principle

• For reactions involving gaseous products or
  reactants, changes in pressure of volume may
  shift the equilibrium. The equilibrium will
  shift only if there is an unequal number of moles
  of gas on either side of the reaction. Reactions
  involving liquids or solids are not significantly
  affected by changes in pressure or volume.

79
Le Chateliers Principle
80
Pressure Changes
81
Le Chateliers Principle

• For the reaction
• C(s) 2 H2(g) ? CH4(g) ?Ho-75kJ
• Assuming the reaction is initially at
  equilibrium, predict the shift in equilibrium (if
  any) for each of the following changes.
• a) add carbon
• b) remove methane
• c) increase the temperature

82
Le Chateliers Principle

• The value of K will change with temperature.
  The effect of a change in temperature on K
  depends on whether the reaction is exothermic or
  endothermic.
• The easiest way to predict the result when
  temperature is changed is to write in heat as a
  product (for exothermic reactions) or a reactant
  (for endothermic reactions).

83
Le Chateliers Principle

• For the reaction
• C(s) 2 H2(g) ? CH4(g) ?Ho-75kJ
• Assuming the reaction is initially at
  equilibrium, predict the shift in equilibrium (if
  any) for each of the following changes.
• d) add a catalyst
• e) decrease the volume
• f) remove some carbon
• g) add hydrogen