Chemical Equilibrium

1
Chemical Equilibrium
Chapter Fourteen
2
Dynamic Nature of Equilibrium

• When a system reaches equilibrium, the forward
  and reverse reactions continue to occur but at
  equal rates.

We are usually concerned with the situation after
equilibrium is reached.
After equilibrium the concentrations of reactants
and products remain constant.
3
Dynamic Equilibrium Illustrated
NaCl containing radioactive Na is added to a
saturated NaCl solution.
After a time, the solution contains radioactive
Na
NaCl dissolves and recrystallizes continuously.
and the added salt now contains some stable Na.
4
Concentration vs. Time
Beginning with 1 M H2 and 1 M I2, the HI
increases and both H2 and I2 decrease.
If we begin with only 1 M HI, the HI decreases
and both H2 and I2 increase.
Beginning with 1 M each of H2, I2, and HI, the
HI increases and both H2 and I2 decrease.
5
Regardless of the starting concentrations once
equilibrium is reached
the expression with products in numerator,
reactants in denominator, where each
concentration is raised to the power of its
coefficient, appears to give a constant.
6
The Equilibrium Constant Expression

• For the general reaction
• aA bB ? gG hH
• The equilibrium expression is

Each concentration is simply raised to the power
of its coefficient
Products in numerator.
Reactants in denominator.
7
The Equilibrium Constant

• The equilibrium constant is constant regardless
  of the initial concentrations of reactants and
  products.
• This constant is denoted by the symbol Kc and is
  called the concentration equilibrium constant.
• Concentrations of the products appear in the
  numerator and concentrations of the reactants
  appear in the denominator.
• The exponents of the concentrations are identical
  to the stoichiometric coefficients in the
  chemical equation.

8

• Example 14.1
• If the equilibrium concentrations of COCl2 and
  Cl2 are the same at 395 C, find the equilibrium
  concentration of CO in the reaction
• CO(g) Cl2(g) COCl2(g)
• Kc 1.2 x 103 at 395 C

9
The Condition of Equilibrium

• The kinetics view
• Kc (forward rate)/(reverse rate) kf/kr
• The thermodynamics view
• The equilibrium constant can be related to other
  fundamental thermodynamic properties and is
  called the thermodynamic equilibrium constant,
  Keq.
• The thermodynamic equilibrium constant expression
  uses dimensionless quantities known as activities
  in place of molar concentrations.

10
Modifying the Chemical Equation
NO22 Kc 4.67 x
1013 (at 298 K) NO2 O2
What will be the equilibrium constant K'c for the
new reaction?
NO2 O2 1 K'c
NO22
NO22

NO2 O2
1 1 2.14 x 1014
Kc 4.67 x 1013
11
Modifying the Chemical Equation (contd)
NO22 Kc 4.67 x
1013 (at 298 K) NO2 O2
What will be the equilibrium constant K"c for the
new reaction?
12
Modifying the Chemical Equation (contd)

• For the reverse reaction, K is the reciprocal of
  K for the forward reaction.
• When an equation is divided by two, K for the new
  reaction is the square root of K for the original
  reaction.
• General rule
• When the coefficients of an equation are
  multiplied by a common factor n to produce a new
  equation, we raise the original Kc value to the
  power n to obtain the new equilibrium constant.
• It should be clear that we must write a balanced
  chemical equation when citing a value for Kc.

13

• Example 14.2
• The equilibrium constant for the reaction
• ½ H2(g) ½ I2(g) HI(g)
• at 718 K is 7.07.
• (a) What is the value of Kc at 718 K for the
  reaction
• HI(g) ½ H2(g) ½ I2(g)
• (b) What is the value of Kc at 718 K for the
  reaction
• H2(g) I2(g) 2 HI(g)

14
The Equilibrium Constantfor an Overall Reaction
and were given

• Adding the given equations gives the desired
  equation.
• Multiplying the given values of K gives the
  equilibrium constant for the overall reaction.
• (To see why this is so, write the equilibrium
  constant expressions for the two given equations,
  and multiply them together. Examine the result )

15
Equilibria Involving Gases

• In reactions involving gases, it is often
  convenient to measure partial pressures rather
  than molarities.
• In these cases, a partial pressure equilibrium
  constant, Kp, is used.

Kc and Kp are related by Kp Kc
(RT)?n(gas)
where Dn(gas) is the change in number of moles of
gas as the reaction occurs in the forward
direction.
Dn(gas) mol gaseous products mol gaseous
reactants
16

• Example 14.3
• Consider the equilibrium between dinitrogen
  tetroxide and nitrogen dioxide
• N2O4(g) 2 NO2(g) Kp
  0.660 at 319 K
• (a) What is the value of Kc for this reaction?
  (b) What is the value of Kp for the reaction 2
  NO2(g) N2O4(g)? (c) If the equilibrium
  partial pressure of NO2(g) is 0.332 atm, what is
  the equilibrium partial pressure of N2O4(g)?

17
Equilibria Involving PureSolids and Liquids

• The equilibrium constant expression does not
  include terms for pure solid and liquid phases
  because their concentrations do not change in a
  reaction.
• Although the amounts of pure solid and liquid
  phases change during a reaction, these phases
  remain pure and their concentrations do not
  change.

CaO CO2 Kc
CaCO3
Kc CO2
18

• Example 14.4
• The reaction of steam and coke (a form of
  carbon) produces a mixture of carbon monoxide and
  hydrogen, called water-gas. This reaction has
  long been used to make combustible gases from
  coal
• C(s) H2O(g) CO(g) H2(g)
• Write the equilibrium constant expression for
  Kc for this reaction.

19
Equilibrium Constants When Do We Need Them and
When Do We Not?

• A very large numerical value of Kc or Kp
  signifies that a reaction goes (essentially) to
  completion.
• A very small numerical value of Kc or Kp
  signifies that the forward reaction, as written,
  occurs only to a slight extent.
• An equilibrium constant expression applies only
  to a reversible reaction at equilibrium.
• Although a reaction may be thermodynamically
  favored, it may be kinetically controlled
• Thermodynamics tells us its possible (or not)
• Kinetics tells us its practical (or not)

20

• Example 14.5
• Is the reaction CaO(s) CO2(g)
  CaCO3(s) likely to occur to any appreciable
  extent at 298 K?

21
The Reaction Quotient, Q

• For nonequilibrium conditions, the expression
  having the same form as Kc or Kp is called the
  reaction quotient, Qc or Qp.
• The reaction quotient is not constant for a
  reaction, but is useful for predicting the
  direction in which a net change must occur to
  establish equilibrium.
• To determine the direction of net change, we
  compare the magnitude of Qc to that of Kc.

22
The Reaction Quotient, Q
23

• Example 14.6
• Predict the direction of net change for
  Experiment 3 in Table 14.1.

24
Le Châteliers Principle

• When any change in concentration, temperature,
  pressure, or volume is imposed on a system at
  equilibrium, the system responds by attaining a
  new equilibrium condition that minimizes the
  impact of the imposed change.
• Analogy Begin with 100 men and 100 women at a
  dance.
• Assume that there are 70 couples dancing, though
  not always the same couples (dynamic
  equilibrium).
• If 30 more men arrive, what happens?
• The equilibrium will shift, and shortly, more
  couples will be dancing but probably not 30
  more couples.

25
Changing the Amounts ofReacting Species

• At equilibrium, Q Kc.
• If the concentration of one of the reactants is
  increased, the denominator of the reaction
  quotient increases.
• Q is now less than Kc.
• This condition is only temporary, however,
  because the concentrations of all species must
  change in such a way so as to make Q Kc again.
• In order to do this, the concentrations of the
  products increase the equilibrium is shifted to
  the right.

26
the acetic acid concentration first increases
When acetic acid (a reactant) is added to the
equilibrium mixture
27

• Example 14.7
• Water can be removed from an equilibrium
  mixture in the reaction of 1-octanol and acetic
  acid, for example, by using a solid drying agent
  that is insoluble in the reaction mixture.
  Describe how the removal of a small quantity of
  water affects the equilibrium.
• CH3(CH2)6CH2OH(soln) CH3COOH(soln)
• CH3(CH2)6CH2OCOCH3(soln)
  H2O(soln)

H
28
Heterogeneous Equilibriaand Le Chateliers
Principle

• Addition or removal of pure solids or pure
  liquids from a system at equilibrium does not
  affect the equilibrium.

29
Changing External Pressure or Volume in Gaseous
Equilibria

• When the external pressure is increased (or
  system volume is reduced), an equilibrium shifts
  in the direction producing the smaller number of
  moles of gas.
• When the external pressure is decreased (or the
  system volume is increased), an equilibrium
  shifts in the direction producing the larger
  number of moles of gas.
• If there is no change in the number of moles of
  gas in a reaction, changes in external pressure
  (or system volume) have no effect on an
  equilibrium.
• Example H2(g) I2(g) 2 HI
  equilibrium is unaffected by pressure changes.

30
Initial
When pressure is increased
to give one molecule of N2O4, reducing the
pressure increase.
two molecules of NO2 combine
31
Temperature Changes and Catalysis

• Raising the temperature of an equilibrium mixture
  shifts equilibrium in the direction of the
  endothermic reaction lowering the temperature
  shifts equilibrium in the direction of the
  exothermic reaction.
• Consider heat as though it is a product of an
  exothermic reaction or as a reactant of an
  endothermic reaction, and apply Le Châteliers
  principle.
• A catalyst lowers the activation energy of both
  the forward and the reverse reaction.
• Adding a catalyst does not affect an equilibrium
  state.
• A catalyst merely causes equilibrium to be
  achieved faster.

32

• Example 14.8
• An equilibrium mixture of O2(g), SO2(g), and
  SO3(g) is transferred from a 1.00-L flask to a
  2.00-L flask. In which direction does a net
  reaction proceed to restore equilibrium? The
  balanced equation for the reaction is
• 2 SO3(g) 2 SO2(g) O2(g)
• Example 14.9
• Is the amount of NO(g) formed from given
  amounts of N2(g) and O2(g),
• N2(g) O2(g) 2 NO(g) ?H 180.5
  kJ
• greater at high or low temperatures?

33

• Example 14.10 A Conceptual Example
• Flask A, pictured below, initially contains an
  equilibrium mixture of the reactants and products
  of the reaction
• CO(g) H2O(g) CO2(g) H2(g) ?H 41
  kJ Kc 9.03 at 698 K
• It is isolated from flask B by a closed valve.
  When the valve is opened, a new equilibrium is
  established as the contents of the two flasks
  mix. Describe, qualitatively, how the amounts of
  CO, H2, CO2, and H2O in the new equilibrium
  compare with the amounts in the initial
  equilibrium if (a) flask B initially contains
  Ar(g) at 1 atm pressure (b) flask B

initially contains 1.0 mol CO2 (c)
flask B initially contains 1.0 mol CO and the
temperature of the AB mixture is raised by 100
C. If you are uncertain of the result in any of
the three cases, explain why.
34
Determining Values of Equilibrium Constants
Experimentally

• When initial amounts of one or more species, and
  equilibrium amounts of one or more species, are
  given, the amounts of the remaining species in
  the equilibrium state and, therefore, the
  equilibrium concentrations often can be
  established.
• A useful general approach is to tabulate under
  the chemical equation
• the concentrations of substances present
  initially
• changes in these concentrations that occur in
  reaching equilibrium
• the equilibrium concentrations.
• This sort of table is sometimes called an ICE
  table Initial/Change/Equilibrium.

35

• Example 14.11
• In a 10.0-L vessel at 1000 K, 0.250 mol SO2
• and 0.200 mol O2 react to form 0.162 mol
• SO3 at equilibrium. What is Kc, at 1000 K,
• for the reaction that is shown here?
• 2 SO2(g) O2(g) 2 SO3(g)
• Example 14.12
• Consider the reaction
• H2(g) I2(g) 2 HI(g) Kc 54.3 at
  698 K
• If we start with 0.500 mol I2(g) and 0.500 mol
  H2(g) in a 5.25-L vessel at 698 K, how many moles
  of each gas will be present at equilibrium?

36
Calculating Equilibrium Quantities from Kc and Kp
Values

• When starting with initial reactants and no
  products and with the known value of the
  equilibrium constant, these data are used to
  calculate the amount of substances present at
  equilibrium.
• Typically, an ICE table is constructed, and the
  symbol x is used to identify one of the changes
  in concentration that occurs in establishing
  equilibrium.
• Then, all the other concentration changes are
  related to x, the appropriate terms are
  substituted into the equilibrium constant
  expression, and the equation solved for x.

37

• Example 14.13
• Suppose that in the reaction of Example 14.12,
  the initial amounts are 0.800 mol H2 and 0.500
  mol I2. What will be the amounts of reactants and
  products when equilibrium is attained?
• Example 14.14
• Carbon monoxide and chlorine react to form
  phosgene, COCl2, which is used in the manufacture
  of pesticides, herbicides, and plastics
• COCl2(g) CO(g) Cl2(g) Kc
  1.2 x 103 at 668 K
• How much of each substance, in moles, will there
  be at equilibrium in a reaction mixture that
  initially has 0.0100 mol CO, 0.0100 mol Cl2, and
  0.100 mol COCl2 in a 10.0-L flask?

38

• Example 14.15
• A sample of phosgene, COCl2(g), is introduced
  into a constant-volume vessel at 395 C and
  observed to exert an initial pressure of 0.351
  atm. When equilibrium is established for the
  reaction
• CO(g) Cl2(g) COCl2(g) Kp
  22.5
• what will be the partial pressure of each gas
  and the total gas pressure?

39

• Cumulative Example
• A mixture of H2S(g) and CH4(g) in the mole
  ratio 21 was brought to equilibrium at 700 C
  and a total pressure of 1.00 atm. The equilibrium
  mixture was analyzed and found to contain 9.54 x
  103 mol H2S. The CS2 present at equilibrium was
  converted, first to H2SO4 and then to BaSO4, with
  1.42 x 103 mol BaSO4 being obtained. Use these
  data to determine Kp at 700 C for the reaction
• 2 H2S(g) CH4(g) CS2(g) 4 H2(g)