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Chemical Equilibrium

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Title: 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)
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