Chapter 15 Principles of Chemical Equilibrium

1
Chapter 15 Principles of Chemical Equilibrium

• Dr. Peter Warburton
• peterw_at_mun.ca
• http//www.chem.mun.ca/zcourses/1051.php

2
The equilibrium state

• Chemical equilibrium is the state reached when
  the concentrations of the products and reactants
  remain constant over time. The mixture of
  reactants and products in the equilibrium state
  is the equilibrium mixture.

3
N2O4 (g) ? 2 NO2 (g)

• We have used two directional arrows (?) to show
  that this reaction does not go to completion.
• The reaction occurs both ways.

4
Pitfall

• The terms reactants and products are arbitrary.
  We must always refer to a balanced equation to be
  completely understood.
• N2O4 (g) ? 2 NO2 (g)
• reactant product
• 2 NO2 (g) ? N2O4 (g)
• reactant product

5
Each reaction occurs at its own rate as defined
by its rate law. One reaction will initially
have a faster rate than the other, and will
initially dominate the system. The other
reaction can be considered to be dominated in the
system
6

• We saw in Kinetics that the rate of a reaction
  decreases as time proceeds because the
  concentrations of the reactants decrease.
• This is what we will see in the dominant
  reaction.

7

• What will happen to the dominated reaction where
  the reactant concentrations increase?
• Its reaction rate will increase!

8

• At some point in time the rate of the forward
  reaction is THE SAME as the rate of the reverse
  reaction.
• This is a more correct means of defining
  equilibrium.

9
Initially dominant reaction slows down
Initially dominated reaction speeds up
10
Equilibrium is a dynamic process

• While no visible change is occurring,
• individual molecular events are still occurring.

At equilibrium the rates of the forward and
reverse reactions are the same, BUT are not
equal to zero.
11
The equilibrium constant expression
The final concentrations will not always be the
same because we started with different numbers of
N atoms and O atoms in the last three
experiments. But ratio of NO22 / N2O4 is
always the same!
12
Experiments 1 and 2
Start first experiment with no NO2 and 0.04 M
N2O4 Start first experiment with 0.08 M NO2 and
no N2O4
13
a A b B ? c C d D

• The concentrations of all species in an
  equilibrium mixture are related to each other
  through the equilibrium constant equation in
  terms of concentration.
• This equilibrium equation applies only to this
  specific balanced equation and the value of the
  equilibrium constant, Kc must always be stated at
  a specific temperature!

14
N2O4 (g) ? 2 NO2 (g)

• If we change the temperature then the equilibrium
  mixture will (most likely) change. This means
  the equilibrium constant will change. For this
  reaction
• Kc 1.53 at 127 ?C.
• Notice there are no units for Kc!

15
Be careful!

• Equilibrium constant equations (and therefore the
  value of K) depend on the balanced equation
  referenced.
• A B ? C D
• C D ? A B
• Kc DOES NOT EQUAL Kc, but rather
• Kc 1/Kc

16
Be careful!

• Equilibrium constant equations (and therefore the
  value of K) depend on the balanced equation
  referenced.
• A B ? C D
• 2 A 2 B ? 2 C 2 D

17
Thermodynamic equilibrium constant Keq

• The thermodynamic equilibrium constant (Keq)
  equation takes the same mathematical form as the
  equilibrium constant equation in terms of
  concentrations (Kc). However, the composition of
  the equilibrium mixture is expressed in terms of
  activities.

18
Thermodynamic equilibrium constant Keq

• Activities relate effective properties (like
  concentration or pressure) of real substances at
  given conditions in comparison to the same
  substance acting ideally at standard conditions.
• Since activities make comparisons of the same
  property type in ratio form, the property units
  cancel out and all activities are unitless.

19
Thermodynamic equilibrium constant Keq
a A b B ? c C d D
where ax X / c0 (c0 is a standard
concentration of 1 M) or ax Px / P0 (P0 is a
standard pressure of 1 bar) Note that ax 1 for
pure solids and liquids
20
Problem

• For the reaction
• CO (g) 2 H2 (g) ? CH3OH (g)
• the equilibrium concentrations of CH3OH and CO
  are found to be equal at 483 K. If Kc 14.5 at
  483 K, what is the equilibrium concentration of
  H2?

Answer H2 0.263 M
21
Problem

• For the reaction
• N2 (g) 3 H2 (g) ? 2 NH3 (g)
• Kc 1.8 x 104 at a certain temperature. What is
  the equilibrium concentration of H2 if the
  equilibrium concentrations of N2 and NH3 are
  0.015 M and 2.00 M respectively?

Answer H2 0.25 M
22
K for combined equilibria

• If we can describe an overall equilibrium
  reaction as the sum of two or more other
  equilibrium processes, then the equilibrium
  constant for the overall reaction in the
  equilibrium constants of the processes multiplied
  together.
• Krxn K1 x K2 x K3

23
Complex ions and solubility
AgCl(s) 2 NH3(aq) ? Ag(NH3)2(aq) Cl-(aq)
24
Complex ions and solubility
AgCl(s) ? Ag (aq) Cl- (aq) Ag (aq) NH3 (aq)
? Ag(NH3) (aq) Ag(NH3) (aq) NH3 (aq) ?
Ag(NH3)2 (aq)
K for the first reaction is 1.8 x 10-10 K for the
second reaction is 2.0 x 103 K for the third
reaction is 7.9 x 103
25
Complex ions and solubility
The sum of the three reactions is AgCl(s) 2
NH3(aq) ? Ag(NH3)2(aq) Cl-(aq) which will
have an equilibrium constant that is Krxn K1 x
K2 x K3 Krxn 1.8 x 10-10 x 2.0 x 103 x 7.9 x
103 Krxn 2.8 x 10-3
26
The equilibrium constant Kp

• Gas phase equilibrium constants are often
  expressed in terms of partial pressures because
  they are generally very easy to measure as a
  function of the total pressure of the system.

27
The equilibrium constant Kp

• Recall, for an ideal gas A
• PAV nART
• PA (nART) / V
• PA (nA/V) RT
• What is n / V? It is moles over volume, which is
  concentration. So nA/V is A and
• PA A RT

28
N2O4 (g) ? 2 NO2 (g)

• We can express an equilibrium constant in terms
  of partial pressures because they are related to
  concentrations!
• Again, the equilibrium constant is unitless.

29
Kc and Kp are related

• a A b B ? c C d D (all are GASES!)

30
Kc and Kp are related

• Some of the RT terms in
• will cancel each other out. In fact

31
Kc and Kp are related
Use R 0.08206 (L?atm)(K?mol)-1 because it
relates molarity to pressure at a given
temperature.
32
N2O4 (g) ? 2 NO2 (g)

• Dn (2-1) 1 so
• At 25 ?C Kc 4.64 x 10-3
• Kp Kc(RT)
• Kp (4.64 x 10-3)(0.08206)(298)
• note the lack of units and T is expressed in
  Kelvin!
• Kp 0.113 at 25 ?C

33
Problem

• In the industrial synthesis of hydrogen often the
  water-gas shift reaction is used
• CO (g) H2O (g) ? CO2 (g) H2 (g)
• What is the value of Kp at 700 K if the partial
  pressures in an equilibrium mixture at 700 K are
• 1.31 atm of CO
• 10.0 atm of water
• 6.12 atm of carbon dioxide, and
• 20.3 atm of hydrogen gas?

34
Problem answer

• Kp 9.48

35
Problem

• In the industrial synthesis of nitric acid
• 2 NO (g) O2 (g) ? 2 NO2 (g)
• If Kc 6.9 x 105 at 227 ?C,
• what is the value of Kp at this temperature?
• If Kp 1.3 x 10-2 at 1000 K,
• what is the value of Kc at this temperature?

Answers Kp at 227?C 1.7 x 104 and Kc at
1000 K 1.1
36
Heterogeneous equilibria

• Homogeneous equilibria occur in systems where all
  compounds in the equilibrium mixture are in the
  same state.
• Heterogeneous equilibria occur in systems where
  some of the chemicals of the equilibrium mixture
  are in different states.

37
CaCO3 (s) ? CaO (s) CO2 (g)

• Since one of the products is a gas, while the
  other two compounds are solids, this is a
  heterogeneous equilibrium.
• Now, if we were to express the equilibrium
  constant for this reaction, we would probably say
  
• What is the concentration of a solid, though?

38
What is the concentration of a solid or liquid?

• Concentration is moles per unit volume. Also,
  density is mass divided by volume, and molar mass
  is mass per number of moles. So, for a pure
  substance
• (mass / volume) / (mass / moles) moles / volume
  
• density / molar mass (concentration)

39
What is the concentration of a solid or liquid?

• density / molar mass (concentration)
• Since both the density and molar mass of a pure
  solid or liquid substance are constant, the
  CONCENTRATION IS CONSTANT, and does not change in
  a reaction as long as some of the solid or liquid
  exists at all times.
• This helps explain why the activities of solids
  and liquids are equal to one!

40
CaCO3 (s) ? CaO (s) CO2 (g)

• We choose not to include the concentrations of
  solids and liquids in the calculation of Kc!
• The concentrations of the solids are hidden
  inside the equilibrium constant.
• If we look at the reaction in terms of pressure,
  then
• Kp (PCO2)

41
Thermodynamic equilibrium constant Keq

• The activity of all pure solids and liquids is
  one, and so solids and liquids have no effect on
  the value of Keq

42
CaCO3 (s) ? CaO (s) CO2 (g)
43
Problem

• For each of the following reactions, write the
  equilibrium constant expression for Kc. Where
  appropriate, do the same for Kp and give the
  relationship between Kc and Kp.
• a) 2 Fe (s) 3 H2O (g) ? Fe2O3 (s) 3 H2 (g)
• b) 2 H2O (l) ? 2 H2 (g) O2 (g)
• c) SiCl4 (g) 2 H2 (g) ? Si (s) 4 HCl (g)
• d) Hg22 (aq) 2 Cl- (aq) ? Hg2Cl (s)

44
Using the equilibrium constant

• Judging the extent of a reaction The magnitude
  (size) of the constant K gives an idea of the
  extent to which reactants are converted to
  products.
• We can make general statements about the
  completeness of a given equilibrium reaction
  based on the size of the value of the equilibrium
  constant.

45

• If the equilibrium constant is very large (gt1000
  for instance), then the forward reaction is
  initially very dominant and the reaction as
  written in the balanced equation proceeds nearly
  to completion before equilibrium is reached.
• 2 H2 (g) O2 (g) ? 2 H2O (g)

46

• If the equilibrium constant is very small (lt10-3
  for instance), then the reverse reaction is
  initially very dominant and the reaction as
  written in the balanced equation barely proceeds
  at all before equilibrium is reached.
• 2 H2O (g) ? 2 H2 (g) O2 (g)

47

• If the equilibrium constant is between 10-3 and
  103, this means that the dominant reaction is not
  overpowering the other reaction and we reach
  equilibrium somewhere in between a state of no
  reaction and completeness. Appreciable
  concentrations of all species are present in the
  equilibrium mixture.
• H2 (g) I2 (g) ? 2 HI (g)

48
Predicting the direction of a reaction

• If you put known concentrations of products and
  reactants into the equilibrium constant equation
  when you know the system is NOT at equilibrium
  you would get a value that does not equal the
  equilibrium constant.
• Can we use this value to tell which reaction is
  dominant in this non-equilibrium system?

49

• We define the reaction quotient Qc (or Qp or Qeq)
  in exactly the same way we define the equilibrium
  constant Kc (or Kp or Keq).
• When the system is not at equilibrium, then
• Qc ? Kc

50

• If Qc gt Kc the reaction needs to create more
  reactants (and use up products) to get to
  equilibrium, so the reaction will be going from
  right to left.
• If Qc lt Kc the reaction needs to create more
  products (and use up reactants) to get to
  equilibrium, so the reaction will be going from
  left to right.

51
Figure
52
H2 (g) I2 (g) ? 2 HI (g)
If H2t 0.80 mol/L, I2t 0.25 mol/L, and
HIt 10.0 mol/L, then
Qc ? Kc, so the system is not at equilibrium.
Qc gt Kc, the reaction will proceed from right
to left.
53
Problem

• The equilibrium constant Kc for the reaction
• 2 NO (g) O2 (g) ? 2 NO2 (g)
• is 6.9 x 105 at 500 K. A 5.0 L reaction vessel
  at this temperature was filled with 0.060 mol of
  NO, 1.0 mol of O2, and 0.80 mol of NO2.
• a) Is the reaction mixture at equilibrium? If
  not, which direction does the reaction proceed?
• b) What is the direction of the reaction if the
  initial amounts are 5.0 x 10-3 mol of NO, 0.20
  mol of O2, and 4.0 mol of NO2?

54
Problem answer

• a) Qc 8.9 x 102. System is not at
  equilibrium, and reaction will proceed right
  since Qc lt Kc.
• b) Qc 1.6 x 107. System is not at
  equilibrium, and reaction will proceed left since
  Qc gt Kc.

55
Problem

• In an earlier problem we saw the water-gas shift
  reaction
• CO (g) H2O (g) ? CO2 (g) H2 (g)
• where we calculated the value of Kp at 700 K to
  be 9.48. If we combine equal masses of all four
  chemicals and let the system come to equilibrium,
  which chemicals will have increased in quantity
  and which will have decreased in quantity during
  the reaction?

56
Problem answer

• Qp ? 5.7 lt Kp and reaction will proceed right
  meaning we will increase CO2 (g) and H2 (g) and
  decrease CO (g) and H2O (g).

57
Altering equilibrium conditions

• We like to maximize a product yield for a
  reaction with a minimum of energy (and money)
  input.
• If a reaction doesnt go to near completion
• we must adjust experimental conditions so the
  reaction proceeds as favourably as possible!

58
Le Chataliers Principle

• Three factors can be changed to affect an
  equilibrium the concentrations of the chemicals
  involved, the pressure and/or volume of the
  system, or the temperature.

Le Chataliers Principle states that if a stress
is applied to a system at equilibrium, the system
will react in the direction that minimizes the
stress and brings the system to a NEW
equilibrium.
59
Changes in concentration

• N2 (g) 3 H2 (g) ? 2 NH3 (g)
• Kc 0.296 at 700 K.

60

• The system re-established a NEW equilibrium by
  reacting in such a way as to decrease the stress
  to the system. Since we have added a reactant
  (this is the stress), the reaction should proceed
  towards products to minimize the amount of
  extra reactant added to the system.

61
In general, if we increase the concentration of a
reactant, the reaction proceeds from reactants to
products to decrease the stress of added reactant
to our equilibrium system. If we increase the
concentration of a product, the reaction proceeds
from products to reactants to decrease the stress
of added product to our equilibrium system.
62
In the ammonia example of slide 59, before we
introduced more nitrogen (a reactant) the
reaction quotient was The system was at
equilibrium!
63
If we add 1.00 mol?L-1 nitrogen (a stress!) to
the original equilibrium system, the reaction
quotient will change and the system will no
longer be at equilibrium!
64
The reaction quotient is now less than the
equilibrium constant, meaning the reaction must
move from left to right to re-establish
equilibrium. At the new equilibrium N2 1.31
mol?L-1 H2 2.43 mol?L-1 NH3 2.36
mol?L-1
65
Note that the N2 in this new equilibrium
mixture is now lower than the 1.50 mol?L-1 we
changed the concentration to after adding N2 to
the first equilibrium mixture. The system has
reacted to minimize the stress on the system by
reducing the amount of N2 to reach a new
equilibrium!
66
Fe3 (aq) SCN- (aq) ? FeNCS2 (aq)

• a) An equilibrium mixture for this reaction is
  orange, which is the added colours of pale yellow
  Fe3 and the red FeNCS2. SCN- is colorless.

67
Fe3 (aq) SCN- (aq) ? FeNCS2 (aq)

• b) If we add FeCl3 to the solution, we see the
  mixture gets more red, meaning more FeNCS2. Why?

68
Fe3 (aq) SCN- (aq) ? FeNCS2 (aq)

• If we add KSCN to the solution, we see the
  mixture gets more red, meaning more FeNCS2.
  Why?

69
Fe3 (aq) SCN- (aq) ? FeNCS2 (aq)

• d) If we add H2C2O4 to the solution, we see the
  mixture gets more yellow, meaning less FeNCS2.
  Why?
• H2C2O4 (aq) ? 2 H (aq) C2O42- (aq)
• Fe3(aq) 3 C2O42- (aq) ? Fe(C2O4)33- (aq)

70
Fe3 (aq) SCN- (aq) ? FeNCS2 (aq)

• e) If we add HgCl2 to the solution, we see the
  mixture gets more yellow, meaning less FeNCS2.
  Why?
• HgCl2 (s) ? Hg2 (aq) 2 Cl- (aq)
• Hg2 (aq) 4 SCN- (aq) ? Hg(SCN)42- (aq)

71
Problem

• Consider the equilibrium for the water-gas shift
  reaction
• CO (g) H2O (g) ? CO2 (g) H2 (g)
• Use Le Chataliers Principle to predict how the
  concentration of H2 will change when the
  equilibrium is disturbed by
• a) Adding CO
• b) Adding CO2
• c) Removing H2O
• d) Removing CO2 also account for the change
  using the reaction quotient Qc

72
Problem

• Calcination of limestone (decomposition of
  calcium carbonate) occurs through the following
  reaction
• CaCO3 (s) ? CO2 (g) CaO (s)
• After we establish this equilibrium system in a
  constant volume container at a given temperature,
  what will be the effect on equilibrium of
• a) Doubling the amount of CO2
• b) Doubling the amount of CaO
• c) Removing half the CaCO3
• d) Removing all the CaCO3

73
Effect of changes in pressure and volume

• What happens when pressure is changed as a result
  of a change in volume?
• N2 (g) 3 H2 (g) ? 2 NH3 (g) Kc 0.296 at 700
  K
• Since PV nRT then P (nRT) / V
• an increase in the volume decreases the pressure
  of a system, or a decrease in volume increases
  the pressure of the system.

74
Pressure change due to volume change

• Say we decrease the volume (and increase the
  pressure) of our ammonia formation equilibrium
  mixture. The stress on the equilibrium is the
  increase in pressure.
• Le Chataliers Principle tells us the system will
  respond by decreasing the pressure of the system
  until a new equilibrium mixture is achieved.

75
Pressure change due to volume change

• Since the pressure is a direct result of the
  number of moles of gas (more moles in a given
  volume means more pressure), the reaction will
  proceed in the direction where the number of
  moles of gas is decreased.

76
Pressure change through a change in volume

• In general, an increase in the pressure of the
  system (caused by decreasing the system volume!)
  causes the reaction to shift to the side of the
  balanced equation with less total moles of gas.
• In general, a decrease in the pressure of the
  system (caused by increasing the system volume!)
  causes the reaction to shift to the side of the
  balanced equation with more total moles of gas.

77
Why is this the case?

• Say we reduce the volume of our ammonia
  equilibrium mixture by half.
• We have doubled the concentration of all gases!
• Reaction will shift from left to right!

78
2 SO2 (g) O2 (g) ? 2 SO3 (g)
79
Other things to note

• If the total number of moles of gas on the
  reactants side of a balanced equation
• EQUALS
• the total number of moles of gas on the products
  side of a balanced equation, then a
• pressure change due to volume change will not
  affect the equilibrium system.

80
Other things to note

• We ALWAYS talked about pressure changes in terms
  of volume change.
• If we increase the pressure by adding inert gas
  to the system, no equilibrium shift will be seen
  because the partial pressures of the gases in the
  equilibrium system have not changed!
• In other words, there is no actual stress being
  placed on the equilibrium system

81
Problem

• Does the number of moles of products increase,
  decrease, or remain the same when each of the
  following equilibria is subject to an increase in
  pressure by decreasing the volume?
• a) CO (g) H2O (g) ? CO2 (g) H2 (g)
• b) 2 CO (g) ? C (s) O2 (g)
• c) N2O4 (g) ? 2 NO2 (g)

82
Changes in temperature and equilibrium

• Our reaction for the formation of ammonia
• N2 (g) 3 H2 (g) ? 2 NH3 (g) DH? -92.2 kJ.

We see as the temperature increases, the value
of Kc decreases, so the reaction shifts towards
the reactants with increasing T
83
Is there some relationship between DH and Kc?

• Yes!
• N2 (g) 3 H2 (g) ? 2 NH3 (g) 92.2 kJ
• We can think of heat as a product in the
  reaction.
• As we increase the temperature of the system, we
  increase the concentration of this product
  and the reaction shifts from right to left
  (towards the reactants).
• The new equilibrium will have less products and
  more reactants, giving a smaller value of Kc.

84
Is there some relationship between DH and Kc?

• In an endothermic reaction, heat will be a
  reactant so increasing the temperature will
  shift the reaction from the left to the right,
  increasing the value of Kc. Overall,
• Exothermic reaction
• T ? then KC ?
• Endothermic reaction
• T ? then KC ?

85
Problem

• When air is heated at very high temperatures in
  an engine, the air pollutant nitric oxide is
  produced by the reaction
• N2 (g) O2 (g) ? 2 NO (g) DH? 180.5 kJ
• How does the equilibrium amount of NO vary with
  an increase in temperature?

86
Catalysis and equilibrium

• Since both the forward and reverse reactions pass
  through the same transition state, a catalyst
  reduces the activation energy for both the
  forward and reverse reactions, by the same
  amount.
• This increases the rates of both the forward and
  reverse reactions by the same amount!

87
(No Transcript)
88
Catalysis and equilibrium

• Another way to think of it is that a
• catalyst does not appear in the overall balanced
  equation for a reaction
• and therefore
• it wont appear in the equilibrium constant
  equation
• meaning
• no change in the equilibrium constant
• will be seen when you add a catalyst.

89
Problem

• Suppose that you have a reaction vessel
  containing an equilibrium mixture of all three
  species. Will the amount of CO increase,
  decrease, or remain the same when
• a) A platinum catalyst is added?
• b) The temperature is increased?
• c) The pressure is increased by decreasing the
  volume?
• d) The pressure is increased by adding argon gas?
  
• e) The pressure is increased by adding O2 gas?

90
Equilibrium calculations

• There are several different types of
  equilibrium-based calculations we can do
• T1) Find some equilibrium mixture data from K,
  balanced equation and other equilibrium mixture
  data (see slide 20)
• T2) Find K from equilibrium mixture data (see
  slide 33)
• T3) Find K from some initial and equilibrium
  mixture data and balanced equation
• T4) Find equilibrium mixture data from initial
  data, K, and the balanced equation

91
2 NO (g) O2 (g) ? 2 NO2 (g) Kc 6.9 x 105 at
500 K (T1)

• Say the system is at equilibrium and
• O2 1.0 mol/L and NO2 0.80 mol/L
• We can calculate NO!

92
2 NO (g) O2 (g) ? 2 NO2 (g) Kc 6.9 x 105 at
500 K (T1)

• Since concentrations are always positive, we can
  throw out the negative answer.
• The NO in the equilibrium mixture is
• 9.6 x 10-4 mol/L.
• Lets check this answer

93
2 NO (g) O2 (g) ? 2 NO2 (g) Kc 6.9 x 105 at
500 K (T1)

• You might be asked to give your final answer in
  moles (which means you need to know the volume of
  your container), or in grams (need to know the
  container volume and the molar mass)

94
Problem (T2)

• Equilibrium is established at 1405 K for the
  reaction
• 2 H2S (g) ? 2 H2 (g) S2 (g)
• in a 3.00 L reaction flask. If there are 0.11
  mol S2, 0.22 mol H2, and 2.78 mol H2S in the
  flask, what is Kc for the reaction at 1405 K?

Answer Kc 2.3 x 10-4
95
Problem (T3)

• 0.100 mol SO2 and 0.100 mol O2 are introduced
  into an evacuated 1.52 L flask at 900K. If the
  reaction is
• 2 SO3 (g) ? 2 SO2 (g) O2 (g)
• and 0.0916 mol of SO3 are found at equilibrium
  then what is Kp for the reaction at 900 K?

Answer Kp 2.2 x 10-2
96
Finding K
97
Problem (T4) 1

• The H2/CO ratio in mixtures of carbon monoxide
  and hydrogen (called synthesis gas) is increased
  by the water-gas shift reaction
• CO (g) H2O (g) ? CO2 (g) H2 (g)
• which has an equilibrium constant Kc 4.24 at
  800 K. Calculate the equilibrium concentrations
  of all species at 800 K if only CO and H2O are
  present initially at concentrations of 0.150
  mol/L.

98
Problem (T4) 1 (ICE tables)

• Using a balanced equation, we create a table of
• Initial concentrations,
• the Change in concentrations (based on unknown
  quantities related by the stoichiometry of the
  balanced equation),
• and Equilibrium concentrations (sum of initial
  concentration and change in concentration).
• We can substitute our Equilibrium concentrations
  into our equilibrium constant expression.
• NOTE If our system data are given as pressures,
  we do exactly the same thing, but with pressures.

99
Problem (T4) 1

• Taking the square root of both sides

100
Problem (T4) 1
If we put both values of x back into all our
Equilibrium concentration expressions, well see
one value of x will give at least one negative
equilibrium concentration. This isnt
physically possible! Throw that value of x out
and use the other.
101
Problem (T4) 1

• We can check our results by inserting these
  equilibrium concentrations into the equilibrium
  equation.
• Our result is (within rounding error) the
  equilibrium constant we were given, so our
  answers for the equilibrium concentrations are
  correct.

102
Problem (T4) 2

• The equilibrium constant Kp is 2.44 at 1000 K for
  the reaction
• C(s) H2O (g) ? CO (g) H2 (g)
• What are the equilibrium partial pressures of
  H2O, CO, and H2 if the initial partial pressures
  are PH2O 1.20 atm, PCO 1.00 atm, and PH2
  1.40 atm?

103
Problem (T4) 2

• Since this question starts with both reactants
  and products in the initial mixture, it makes
  sense to first check the reaction quotient to see
  in which direction the reaction is going to occur
  to reach equilibrium
• Since Q lt K we expect to lose reactants and gain
  products to get to equilibrium. This tells us
  the signs of the changes that are occurring.

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Problem (T4) 2

• While this reaction quotient calculation step
  isnt absolutely necessary,
• if we perform this and assign the correct signs
  to our pressure changes, then we will find that
• any negative value of x we calculate
• will not be physically possible.

105
Problem (T4) 2

• Rearranging, we get

106
Problem (T4) 2
107
Problem (T4) 2

• Our equilibrium partial pressures must all be
  positive.
• This only occurs for x 0.30 atm
• (the negative x value wont work because we have
  used the reaction quotient to help set up our ICE
  table)
• At equilibrium
• PH2O (1.20 atm 0.30 atm) 0.90 atm,
• PCO (1.00 atm 0.30 atm) 1.30 atm, and
• PH2 (1.40 atm 0.30 atm) 1.70 atm.

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Problem (T4) 2

• We should check our answer
• which is the equilibrium constant we were given,
  within rounding errors.

109
Problem (T4) 3

• In a basic aqueous solution, chloromethane
  undergoes a substitution reaction in which Cl- is
  replaced by an OH-
• CH3Cl (aq) OH- (aq) ? CH3OH (aq) Cl- (aq)
• The equilibrium constant Kc is 1 x 1016.
  Calculate the equilibrium concentrations of all
  species in a solution prepared by mixing equal
  volumes of 0.1 mol/L CH3Cl and 0.2 mol/L NaOH.

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Problem (T4) 3 answers

• CH3OH 0.05 mol/L,
• Cl- 0.05 mol/L
• OH- 0.05 mol/L,
• CH3Cl 5 x 10-18 mol/L.

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