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Chemical Equilibrium

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Title: Chemical Equilibrium


1
Section 14 Chemical Equilibrium
2
Chemical Equilibrium
  • Previously we have assumed that chemical
    reactions results in complete conversion of
    reactants to products
  • A B C D
  • No A or B remaining or possibly an excess of A
    or B but not both and eventually reaction stops

Many chemical reactions do not completely convert
reactants to products. Stop somewhere between no
rxn and complete rxn. A B C D some
left some formed reversible (both
directions)
3
The concentration of all species level off to
some individual constant value that persists
indefinitely as long as no stress is added to
system. The state that is reached when the
concentrations of R and P remain a constant in
time is called state of chemical equilibrium and
the mixture is considered an equilibrium
mixture. Individual molecules are continually
reacting even though the overall composition of
the reaction does not change. There is a
exchange going on A B forms C D but then C
D breaks down into A B at the same rate
therefore, concentration of each component
remains a constant and the rate of exchange is
constant at equilibrium. Reaction has not
stopped just the rates of the forward and
reverse reactions are equal.
4
Chemical Equilibrium
  • Therefore, many reactions do not go to completion
    but rather form a mixture of products and
    unreacted reactants, in a dynamic equilibrium.
  • A dynamic equilibrium consists of a forward
    reaction, in which substances react to give
    products, and a reverse reaction, in which
    products react to give the original reactants.
  • Chemical equilibrium is the state reached by a
    reaction mixture when the rates of the forward
    and reverse reactions have become equal.

5
Graphically we can represent this A B C
D The concentrations and reaction rate (less
collisions, less component) of A and B decreases
over time as the concentrations and reaction rate
of C and D increases (more collisions, more
component) over time until the rates are equal
and the concentrations of each components reaches
a constant. This occurs at what we call
equilibrium -- Rf Rr. If the rates are equal,
then there must be a relationship to show
this.
6
Figure Catalytic methanation reaction
approaches equilibrium. CO 3 H2 CH4
H2O
7
If we assume these reactions are elementary rxns
(based on collisions), we can write the rate laws
directly from the reaction A B
C D Rf kf AB
For the reverse reaction we have, C D
A B Rr kr CD We know at
equil that Rf Rr therefore, we can set these
two expressions as equal kf AB kr
CD Rearrange to put constants on one side we
get
8
Constant divided by constant just call a new
constant K. This ratio is given a special name
and symbol called equilibrium constant K relating
to the equilibrium condition at a certain
temperature (temp dependent) for a particular
reaction relating conc of each component. This
is basically a comparison between forward and
reverse reaction rates. At equilibrium, the
ratio of conc of species must satisfy K.
9
The Equilibrium Constant
  • Every reversible system has its own position of
    equilibrium- K- under any given set of
    conditions.
  • The ratio of products produced to unreacted
    reactants for any given reversible reaction
    remains constant under constant conditions of
    pressure and temperature. If the system is
    disturbed, the system will shift and all the
    concentrations of the components will change
    until equilibrium is re-established which occurs
    when the ratio of the new concentrations equals
    what -- "K". Different constant conc but ratio
    same as before.
  • The numerical value of this ratio is called the
    equilibrium constant for the given reaction, K.

10
The Equilibrium Constant
  • The equilibrium-constant expression for a
    reaction is obtained by multiplying the equil
    concentrations ( or partial pressures) of
    products, dividing by the equil concentrations
    (or partial pressures) of reactants, and raising
    each concentration to a power equal to its
    coefficient in the balanced chemical equation.
  • The molar concentration of a substance is denoted
    by writing its formula in square brackets for aq
    solutions. For gases can put Pa - atm. As long
    as use M or atm, K is unitless.
  • Temp dependent any changes, ratio must equal K
    when equil established

11
The Equilibrium Constant
  • The equilibrium constant, K, is the value
    obtained for the equilibrium-constant expression
    when equilibrium concentrations (not just any
    conc but equil conc) are substituted.
  • A large K indicates large concentrations of
    products at equilibrium.
  • A small K indicates large concentrations of
    unreacted reactants at equilibrium.

12
The Equilibrium Constant
  • Do same set up for all equil equations (future
    chapters), just different subscript describing
    the reaction in question i.e. acid hydrolysis -
    acid water - Ka, Kb, Kc, Kp, Ksp
  • Kc is based on conc (M) and Kp is based on
    pressures (atm). There is a difference between Kc
    and Kp and to jump between these two, there is an
    equation to use. Let's show how to jump
    between.

HW 14
13
The Equilibrium Constant, Kp
  • In discussing gas-phase equilibria, it is often
    more convenient to express concentrations in
    terms of partial pressures rather than molarities.

14
The Equilibrium Constant, Kp
  • The equilibrium-constant expression in terms of
    partial pressures becomes (same way but partial
    pressures instead of M)

15
The Equilibrium Constant, Kp
  • In general if you need to jump between K's, the
    numerical value of Kp differs from that of Kc.

16
A Problem to Consider
  • Consider the reaction
  • Kc for the reaction is 280 at 1000. K . Calculate
    Kp for the reaction at this temperature.

17
A Problem to Consider
  • Consider the reaction at 1000. K and Kc 280
  • We know that

18
A Problem to Consider
  • Consider the reaction
  • Since

19
The Equilibrium Constant
  • The law of mass action states that the value of
    the equilibrium constant expression K is constant
    for a particular reaction at a given temperature,
    whenever equilibrium concentrations are
    substituted.
  • When at equil, conc ratio equals K and concs are
    constant but if disturbed, values change until
    equil reached, different constant conc but same
    ratio of K).

20
Obtaining Equilibrium Constants for Reactions
  • Equilibrium concentrations for a reaction must be
    obtained experimentally and then substituted into
    the equilibrium-constant expression in order to
    calculate Kc.

21
Obtaining Equilibrium Constants for Reactions
  • Consider the reaction below
  • Suppose we started with initial concentrations of
    CO and H2 of 0.100 M and 0.300 M, respectively.
  • Obviously shift to right, decrease CO and H2 and
    increase CH4 and water until equil established.

22
Obtaining Equilibrium Constants for Reactions
  • When the system finally settled into equilibrium
    we determined the equilibrium concentrations to
    be as follows.

23
Obtaining Equilibrium Constants for Reactions
  • If we substitute the equilibrium concentrations,
    we obtain

24
Obtaining Equilibrium Constants for Reactions
  • Regardless of the initial concentrations (whether
    they are reactants or products or mixture), the
    law of mass action dictates that the reaction
    will always settle into an equilibrium where the
    equilibrium-constant expression equals Kc.

25
Obtaining Equilibrium Constants for Reactions
  • As an example, lets repeat the previous
    experiment, only this time starting with initial
    concentrations of products (note if only
    products to start shift left until equil
    established)
  • CH4initial 0.2000 M and H2Oinitial
    0.2000 M
  • Obviously shift to left decrease CH4 and H2O and
    increase CO and H2 until equil established.

26
Obtaining Equilibrium Constants for Reactions
  • We find that these initial concentrations result
    in the following equilibrium concentrations.

27
Obtaining Equilibrium Constants for Reactions
  • Substituting these values into the
    equilibrium-constant expression, we obtain the
    same result.
  • Whether we start with reactants or products at
    any initial conc, the system establishes the same
    ratio.

28
A Problem to Consider
  • Applying Stoichiometry to an Equilibrium Mixture
    (basic setup for future problems).
  • What is the composition of the equilibrium
    mixture if it contains 0.080 mol NH3 at
    equilibrium?

29
  • Using the information given, set up the following
    table. (ratio works for atm, M, mols, etc.)

Initial, no
Change, Dn
Equil, neq
  • This procedure for many types of problems,
    however in this problem given equil quantity of
    NH3 therefore, figure out rest.
  • The equilibrium amount of NH3 was given as 0.080
    mol. Therefore, 2x 0.080 mol NH3 (x 0.040
    mol).

30
A Problem to Consider
  • Using the information given, set up the following
    table.

Starting 1.000 3.000 0
Change
Equilibrium
Equilibrium amount of N2 Equilibrium amount of
H2 Equilibrium amount of NH3
HW 15
31
Equilibrium Constant for the Sum of Reactions
  • Similar to the method of combining reactions that
    we saw using Hesss law in Chapter 6, we can
    combine equilibrium reactions whose K values are
    known to obtain K for the overall reaction.
  • With Hesss law, when we reversed reactions
    (change sign) or multiplied them prior to adding
    them together (mult by factor). We had to
    manipulate the DHs values to reflect what we had
    done.
  • The rules are a bit different for manipulating K.

32
Equilibrium Constant for the Sum of Reactions
  • If you reverse a reaction, invert the value of K.
    (reciprocal rule, 1/K)
  • If you multiply/divide each of the coefficients
    in an equation by the same factor (2, 1/2, ),
    raise Kc to the same power (2, 1/2, ). (known
    as coefficient rule, Kn)
  • When you finally combine (that is, add) the
    individual equations together, take the product
    of the equilibrium constants to obtain the
    overall K.(rule of multiple equilibria, K1 x K2 x
    K3 KT)

33
Equilibrium Constant for the Sum of Reactions
  • For example, nitrogen and oxygen can combine to
    form either NO(g) or N2O (g) according to the
    following equilibria.

Kc 6.4 x 10-16
(1)
Kc 2.4 x 10-18
(2)
overall
Kc ?
34
(1)
(2)
overall
HW 16
35
Heterogeneous Equilibria
  • A heterogeneous equilibrium is an equilibrium
    that involves reactants and products in more than
    one phase. Up to now all our reactions have been
    homogeneous - all gases or aqueous solutions.
  • The equilibrium of a heterogeneous system is
    unaffected by the amounts of pure solids or
    liquids present, as long as some of each is
    present.
  • The concentrations of pure solids and liquids are
    always considered to be 1 activity and
    therefore, do not appear in the equilibrium
    expression. Solids and pure liquids have no
    effect on conc or pressure.

36
Heterogeneous Equilibria
  • Consider the reaction below.

HW 17
37
Predicting the Direction of Reaction
  • How could we predict the direction in which a
    reaction at non-equilibrium conditions will shift
    to reestablish equilibrium? Remember did example
    with only reactants, obviously had to go right
    and if only products obviously must go left but
    what if have some of R and P?
  • To answer this question, substitute the current
    concentrations into the reaction quotient
    expression and compare it to Kc.
  • The reaction quotient, Qc, is an expression that
    has the same form as the equilibrium-constant
    expression but whose concentrations are not
    necessarily at equilibrium.

38
Predicting the Direction of Reaction
  • For the general reaction

39
Predicting the Direction of Reaction
  • For the general reaction
  • If Qc Kc, then the reaction is at equilibrium.
  • If Qc gt Kc, the reaction will shift left toward
    reactants until equil reached.
  • If Qc lt Kc, the reaction will shift right toward
    products until equil reached.

40
A Problem to Consider
  • Consider the following equilibrium.
  • A 50.0 L vessel contains 1.00 mol N2, 3.00 mol
    H2, and 0.500 mol NH3. Is the sytem at equil? If
    not, in which direction (toward reactants or
    toward products) will the system shift to
    reestablish equilibrium at 400 oC?
  • Kc for the reaction at 400 oC is 0.500.

41
A Problem to Consider
  • First, calculate concentrations from moles of
    substances.

42
A Problem to Consider
  • First, calculate concentrations from moles of
    substances.

43
A Problem to Consider
  • First, calculate concentrations from moles of
    substances.

HW 18
44
Calculating Equilibrium Concentrations
  • Once you have determined the equilibrium
    constant, K, for a reaction, you can use it to
    calculate the concentrations of substances in the
    equilibrium mixture.

45
Calculating Equilibrium Concentrations
  • For example, consider the following equilibrium
    mixture.
  • Suppose a gaseous mixture contained 0.30 mol CO,
    0.10 mol H2, 0.020 mol H2O, and an unknown amount
    of CH4 per liter at equilibrium.
  • What is the concentration of CH4 at equilibrium
    in this mixture? The equilibrium constant Kc
    equals 3.92.
  • Note the amounts given are equil amounts
    therefore can plug into K eq.

46
Calculating Equilibrium Concentrations
  • First, calculate concentrations from moles of
    substances.

47
Calculating Equilibrium Concentrations
  • First, calculate concentrations from moles of
    substances.

48
Calculating Equilibrium Concentrations
  • First, calculate concentrations from moles of
    substances.

49
Calculating Equilibrium Concentrations
  • Suppose we begin a reaction with known amounts of
    starting materials and want to calculate the
    quantities at equilibrium.

50
Calculating Equilibrium Concentrations
  • Consider the following equilibrium.
  • Suppose you start with 1.000 mol each of carbon
    monoxide and water in a 50.0 L container.
    Calculate the molarity of each substance in the
    equilibrium mixture at 1000 oC.
  • Kc for the reaction is 0.58 at 1000 oC.
  • Which way will reaction shift?

51
Calculating Equilibrium Concentrations
  • First, calculate the initial molarities of CO and
    H2O.

52
Calculating Equilibrium Concentrations
  • First, calculate the initial molarities of CO and
    H2O.
  • The starting concentrations of the products are
    0.
  • We must now set up a table of concentrations
    (starting, change, and equilibrium expressions in
    x).

53
Calculating Equilibrium Concentrations
  • The equilibrium-constant expression is
  • Let x be the moles per liter of product formed.

o 0.0200 0.0200 0 0
D
eq
54
  • Solving for x.

Starting 0.0200 0.0200 0 0
Change
Equilibrium
55
Calculating Equilibrium Concentrations
  • Solving for x.

Starting 0.0200 0.0200 0 0
Change -x -x x x
Equilibrium 0.0200-x 0.0200-x x x
56
Calculating Equilibrium Concentrations
  • Solving for x.

Starting 0.0200 0.0200 0 0
Change -x -x x x
Equilibrium 0.0200-x 0.0200-x x x
57
Calculating Equilibrium Concentrations
  • Solving for equilibrium concentrations.

Starting 0.0200 0.0200 0 0
Change -x -x x x
Equilibrium 0.0200-x 0.0200-x x x
  • If you substitute for x in the last line of the
    table you obtain the following equilibrium
    concentrations. If plug into Keq, should equal K
    or close to it because of sign fig for a check

HW 19
58
Calculating Equilibrium Concentrations
  • The preceding example illustrates the three steps
    in solving for equilibrium concentrations.
  1. Set up a table of concentrations (starting,
    change, and equilibrium expressions in x).
  2. Substitute the expressions in x for the
    equilibrium concentrations into the
    equilibrium-constant equation.
  3. Solve the equilibrium-constant equation for the
    values of the equilibrium concentrations.

59
Another example If the initial pressure of C is
1.0 atm, what would be the partial pressures of
each species at equil.
A B
2C
Kp 9.0
Po 0 0 1.0
DP
Peq
HW 20
60
Another example If the initial pressure of C is
0.10 atm and A and B are 1.00 atm, what would be
the partial pressures of each species at equil.
Kp 0.016
2C A
B
Po 0.10 1.00 1.00
DP
Peq
HW 21
61
Calculating Equilibrium Concentrations
  • In some cases it is necessary to solve a
    quadratic equation to obtain equilibrium
    concentrations - not a perfect square.
  • The next example illustrates how to solve such an
    equation.

62
Calculating Equilibrium Concentrations
  • Consider the following equilibrium.
  • Suppose 1.00 atm H2 and 2.00 atm I2 are placed in
    a 1.00-L vessel. What are the partial pressures
    of all species when it comes to equilibrium at
    458 oC?
  • Kp at this temperature is 49.7.

63
Calculating Equilibrium Concentrations
  • The concentrations of substances are as follows.

Po 1.00 2.00 0
DP
Peq
64
Calculating Equilibrium Concentrations
  • The concentrations of substances are as follows.

Po 1.00 2.00 0
DP -x -x 2x
Peq 1.00-x 2.00-x 2x
  • Substituting our equilibrium concentration
    expressions gives

65
Calculating Equilibrium Concentrations
  • Solving for x.

Starting 1.00 2.00 0
Change -x -x 2x
Equilibrium 1.00-x 2.00-x 2x
  • Because the right side of this equation is not a
    perfect square (sq over sq), you must solve the
    quadratic equation.

66
(No Transcript)
67
Calculating Equilibrium Concentrations
  • However, x 2.33 gives a negative value to 1.00
    - x (the equilibrium concentration of H2), which
    is not possible.

Obviously if you get a negative and positive
number, take the positive number and if two
positive numbers, take the smaller number.
However, if you neglect the shift if all
components are present (you don't do Q - not only
initial reactants or products) and select it
incorrectly, you will end up with either a and
- or two -'s. In this case the negative
number will be correct or the smaller of the two
negative numbers will be correct. Bottom line
examine numbers carefully when selecting answer.
68
Calculating Equilibrium Concentrations
  • Solving for equilibrium concentrations.

Starting 1.00 2.00 0
Change -x -x 2x
Equilibrium 1.00-x 2.00-x 2x
  • If you substitute 0.934 for x in the last line of
    the table you obtain the following equilibrium
    concentrations.

HW 22
69
Le Chateliers Principle
  • Obtaining the maximum amount of product from a
    reaction depends on the proper set of reaction
    conditions which gets us to
  • Le Chateliers principle states that when a
    system in a chemical equilibrium is disturbed by
    a change of temperature, pressure, or
    concentration, the equilibrium will shift in a
    way that tends to counteract this change.

70
Removing Products or Adding Reactants
  • If a chemical system at equilibrium is disturbed
    by adding a gaseous or aqueous species (not solid
    or liquid R or P), the reaction will proceed in
    such a direction as to consume part of the added
    species. Conversely, if a gaseous or aqueous
    species is removed (complex or escape gas), the
    system shifts to restore part of that species.
    This shift will occur until equilibrium is
    re-established.
  • A B C D add - shift to opposite
    side, remove - shift towards that
    side.

71
Effects of Pressure Change
  • A pressure change caused by changing the volume
    of the reaction vessel can affect the yield of
    products in a gaseous reaction only if the
    reaction involves a change in the total moles of
    gas present
  • Ex. N2O4 (g) 2NO2 (g)
  • Suppose system is compressed by pushing down a
    piston (decrease volume of space), which way
    would the shift be that would benefit and use the
    the available space wisely?
  • Shift to smaller number of gas molecules to pack
    more efficiently and relieve the increase in
    pressure due to piston coming down.

72
Effects of Pressure Change
  • Basically the reactants require less volume (that
    is, fewer moles of gaseous reactant) and by
    decreasing the volume of the reaction vessel by
    increasing the pressure, the rxn would shift the
    equilibrium to the left (toward reactants) until
    equil is established.

73
Effects of Pressure Change
  • Literally squeezing the reaction (increase P)
    will cause a shift in the equilibrium toward the
    fewer moles of gas.
  • Reducing the pressure in the reaction vessel by
    increasing its volume would have the opposite
    effect.
  • Decrease P, increase V, shift larger mols of gas
    (L S not compressible)
  • Increase P, decrease V, shift to smaller mols of
    gas
  • In the event that the number of moles of gaseous
    product equals the number of moles of gaseous
    reactant, vessel volume/pressure will have no
    effect on the position of the equilibrium no
    advantage to shift one way over the other.

74
SO2 (g) 1/2 O2 (g) lt--gt SO3 (g) N2 (g)
3 H2 (g) lt--gt 2NH3 (g) N2 (g) O2 (g) lt--gt
2NO (g) C (s) H2O(g) lt--gt CO (g) H2
(g)
75
Effect of Temperature Change
  • Temperature has a significant effect on most
    reactions.
  • Reaction rates generally increase with an
    increase in temperature. Consequently,
    equilibrium is established sooner.
  • However, when you add or remove
    reactants/products or change pressure, the result
    is that the system establishes new conc of
    species but ratio still equal to same K at that
    temperature. For temp changes, the numerical
    value of the equilibrium constant Kc varies with
    temperature. K temp dependent.

76
Effect of Temperature Change
  • Lets look at heat as if it were a product in
    exothermic reactions and a reactant in
    endothermic reactions.
  • We see that increasing the temperature is related
    to adding more product (in the case of exothermic
    reactions) or adding more reactant (in the case
    of endothermic reactions).
  • This ultimately has the same effect as if heat
    were a physical entity.

77
Effect of Temperature Change
  • For example, consider the following generic
    exothermic reaction.
  • Increasing temperature would be like adding more
    product, causing the equilibrium to shift left.
  • Since heat does not appear in the
    equilibrium-constant expression, this change
    would result in a smaller numerical value for Kc
    (numerator smaller and den larger)

78
Effect of Temperature Change
  • For an endothermic reaction, the opposite is true.
  • Increasing temperature would be analogous to
    adding more reactant, causing the equilibrium to
    shift right.
  • This change results in more product at
    equilibrium, and a larger numerical value for Kc
    (larger numerator, smaller den)

79
Effect of Temperature Change
  • In summary
  • For an endothermic reaction (DH positive) the
    amounts of products are increased at equilibrium
    by an increase in temperature (Kc is larger at
    higher temperatures).
  • For an exothermic reaction (DH is negative) the
    amounts of reactants are increased at equilibrium
    by an increase in temperature (Kc is smaller at
    higher temperatures).

80
Effect of a Catalyst
  • A catalyst is a substance that increases the rate
    of a reaction but is not consumed by it.
  • It is important to understand that a catalyst has
    no effect on the equilibrium composition of a
    reaction mixture.
  • A catalyst merely speeds up the attainment of
    equilibrium but does not cause it shift one way
    or the other just get to direction is was going
    faster.

81
Consider the system I2 (g) 2I (g)
DH 151 kJ Suppose the system is at
equilibrium at 1000oC. In which direction will
rxn occur if a.) I atoms are added? b.) the
system is compressed? c.)the temp is increased?
d.)effect increase temp has on K? e.) add
catalyst?
HW 23
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