Chapter 15 Chemical Equilibrium

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Chapter 15Chemical Equilibrium
Chemistry, The Central Science, 10th
edition Theodore L. Brown H. Eugene LeMay, Jr.
and Bruce E. Bursten

• John D. Bookstaver
• St. Charles Community College
• St. Peters, MO
• ? 2006, Prentice Hall

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The Concept of Equilibriumplay the movie

• Chemical equilibrium occurs when a reaction and
  its reverse reaction proceed at the same rate.

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The Concept of Equilibrium

• As a system approaches equilibrium, both the
  forward and reverse reactions are occurring.
• At equilibrium, the forward and reverse reactions
  are proceeding at the same rate.

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A System at Equilibrium

• Once equilibrium is achieved, the amount of each
  reactant and product remains constant.

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Depicting Equilibrium

• In a system at equilibrium, both the forward and
  reverse reactions are being carried out as a
  result, we write its equation with a double arrow

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The Equilibrium Constant
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The Equilibrium Constant

• Forward reaction
• N2O4 (g) ??? 2 NO2 (g)
• Rate law
• Rate kf N2O4

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The Equilibrium Constant

• Reverse reaction
• 2 NO2 (g) ??? N2O4 (g)
• Rate law
• Rate kr NO22

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The Equilibrium Constant

• Therefore, at equilibrium
• Ratef Rater
• kf N2O4 kr NO22
• Rewriting this, it becomes

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The Equilibrium Constant

• The ratio of the rate constants is a constant at
  that temperature, and the expression becomes

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The Equilibrium Constant

• To generalize this expression, consider the
  reaction

• The equilibrium expression for this reaction
  would be

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What Are the Equilibrium Expressions for These
Equilibria?
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SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant
Expressions Write the equilibrium expression for
Kc for the following reactions
Solution Analyze We are given three equations
and are asked to write an equilibrium-constant
expression for each. Plan Using the law of mass
action, we write each expression as a quotient
having the product concentration terms in the
numerator and the reactant concentration terms in
the denominator. Each term is raised to the power
of its coefficient in the balanced chemical
equation. Solve
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The Equilibrium Constant

• Because pressure is proportional to
  concentration for gases in a closed system, the
  equilibrium expression can also be written

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Relationship between Kc and Kp

• From the ideal gas law we know that

PV nRT

• Rearranging it, we get

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Relationship between Kc and Kp

• Plugging this into the expression for Kp for
  each substance, the relationship between Kc and
  Kp becomes

Kp Kc (RT)?n
Where
?n (moles of gaseous product) - (moles of
gaseous reactant)
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Answer 0.335
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Equilibrium Can Be Reached from Either Direction

• As you can see, the ratio of NO22 to N2O4
  remains constant at this temperature no matter
  what the initial concentrations of NO2 and N2O4
  are.

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Equilibrium Can Be Reached from Either Direction

• This is the data from the last two trials from
  the table on the previous slide.

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Equilibrium Can Be Reached from Either Direction

• It does not matter whether we start with N2 and
  H2 or whether we start with NH3. We will have
  the same proportions of all three substances at
  equilibrium.

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What Does the Value of K Mean?

• If K gtgt 1, the reaction is product-favored
  product predominates at equilibrium.

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What Does the Value of K Mean?

• If K gtgt 1, the reaction is product-favored
  product predominates at equilibrium.

• If K ltlt 1, the reaction is reactant-favored
  reactant predominates at equilibrium.

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Solution Analyze We are asked to comment on the
utility of a reaction based on the magnitude of
its equilibrium constant. Plan We consider the
magnitude of the equilibrium constant to
determine whether this reaction is feasible for
the production of NO. Solve Because Kc is so
small, very little NO will form at 25C. The
equilibrium lies to the left, favoring the
reactants. Consequently, this reaction is an
extremely poor choice for nitrogen fixation, at
least at 25C.
Answer The formation of product, HI, is favored
at the lower temperature because Kp is larger at
the lower temperature.
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Manipulating Equilibrium Constants

• The equilibrium constant of a reaction in the
  reverse reaction is the reciprocal of the
  equilibrium constant of the forward reaction.

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Manipulating Equilibrium Constants

• The equilibrium constant of a reaction that has
  been multiplied by a number is the equilibrium
  constant raised to a power that is equal to that
  number.

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Manipulating Equilibrium Constants

• The equilibrium constant for a net reaction made
  up of two or more steps is the product of the
  equilibrium constants for the individual steps.

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Heterogeneous Equilibrium
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The Concentrations of Solids and Liquids Are
Essentially Constant

• Both can be obtained by dividing the density of
  the substance by its molar massand both of these
  are constants at constant temperature.

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The Concentrations of Solids and Liquids Are
Essentially Constant

• Therefore, the concentrations of solids and
  liquids do not appear in the equilibrium
  expression

Kc Pb2 Cl-2
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• As long as some CaCO3 or CaO remain in the
  system, the amount of CO2 above the solid will
  remain the same.

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for Heterogeneous Reactions
Write the equilibrium-constant expression for Kc
for each of the following reactions
Because SnO2 and Sn are both pure solids, their
concentrations do not appear in the
equilibrium-constant expression.
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Solution Analyze We are asked which of several
combinations of species can establish an
equilibrium between calcium carbonate and its
decomposition products, calcium oxide and carbon
dioxide. Plan In order for equilibrium to be
achieved, it must be possible for both the
forward process and the reverse process to occur.
In order for the forward process to occur, there
must be some calcium carbonate present. In order
for the reverse process to occur, there must be
both calcium oxide and carbon dioxide. In both
cases, either the necessary compounds may be
present initially, or they may be formed by
reaction of the other species. Solve Equilibrium
can be reached in all cases except (c) as long as
sufficient quantities of solids are present. (a)
CaCO3 simply decomposes, forming CaO(s) and
CO2(g) until the equilibrium pressure of CO2 is
attained. There must be enough CaCO3, however, to
allow the CO2 pressure to reach equilibrium. (b)
CO2 continues to combine with CaO until the
partial pressure of the CO2 decreases to the
equilibrium value. (c) There is no CaO present,
so equilibrium cant be attained because there is
no way the CO2 pressure can decrease to its
equilibrium value (which would require some of
the CO2 to react with CaO). (d) The situation is
essentially the same as in (a) CaCO3 decomposes
until equilibrium is attained. The presence of
CaO initially makes no difference.
Answers only H2(g)
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Equilibrium Calculations
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Equilibrium Calculations

• A closed system initially containing
• 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2
• At 448?C is allowed to reach equilibrium.
  Analysis of the equilibrium mixture shows that
  the concentration of HI is 1.87 x 10-3 M.
  Calculate Kc at 448?C for the reaction taking
  place, which is

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What Do We Know?
H2, M I2, M HI, M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change
At equilibrium 1.87 x 10-3
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HI Increases by 1.87 x 10-3 M
H2, M I2, M HI, M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change 1.87 x 10-3
At equilibrium 1.87 x 10-3
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Stoichiometry tells us H2 and I2decrease by
half as much
H2, M I2, M HI, M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 1.87 x 10-3
At equilibrium 1.87 x 10-3
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We can now calculate the equilibrium
concentrations of all three compounds
H2, M I2, M HI, M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 1.87 x 10-3
At equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
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and, therefore, the equilibrium constant
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Answer 1.79 ? 105
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Comment The same method can be applied to
gaseous equilibrium problems, in which case
partial pressures are used as table entries in
place of molar concentrations.
Answer 0.338
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The Reaction Quotient (Q)

• To calculate Q, one substitutes the initial
  concentrations on reactants and products into the
  equilibrium expression.
• Q gives the same ratio the equilibrium expression
  gives, but for a system that is not at
  equilibrium.

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If Q K,
the system is at equilibrium.
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If Q gt K,
there is too much product and the equilibrium
shifts to the left.
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If Q lt K,
there is too much reactant, and the equilibrium
shifts to the right.
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Answer Qp 16 Qp gt Kp, and so the reaction
will proceed from right to left, forming more SO3.
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SAMPLE EXERCISE 15.11 Calculating Equilibrium
Concentrations
For the Haber process,

at 500C. In an equilibrium mixture
of the three gases at 500C, the partial pressure
of H2 is 0.928 atm and that of N2 is 0.432 atm.
What is the partial pressure of NH3 in this
equilibrium mixture?
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Answer 1.22 atm
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from Initial Concentrations
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Le Châteliers Principle
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Le Châteliers Principle

• If a system at equilibrium is disturbed by a
  change in temperature, pressure, or the
  concentration of one of the components, the
  system will shift its equilibrium position so as
  to counteract the effect of the disturbance.

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What Happens When More of a Reactant Is Added to
a System?Play insert CD
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The Haber Process

• The transformation of nitrogen and hydrogen into
  ammonia (NH3) is of tremendous significance in
  agriculture, where ammonia-based fertilizers are
  of utmost importance.

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The Haber Process

• If H2 is added to the system, N2 will be
  consumed and the two reagents will form more NH3.

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The Haber Process

• This apparatus helps push the equilibrium to the
  right by removing the ammonia (NH3) from the
  system as a liquid.

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The Effect of Changes in Pressure
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The Effect of Changes in Temperature
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The Effect of Changes in TemperaturePlay insert
CD
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Catalysts increase the rate of both the forward
and reverse reactions.
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Equilibrium is achieved faster, but the
equilibrium composition remains unaltered.
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Solution Analyze We are given a series of
changes to be made to a system at equilibrium and
are asked to predict what effect each change will
have on the position of the equilibrium. Plan Le
Châteliers principle can be used to determine
the effects of each of these changes.
Solve (a) The system will adjust to decrease the
concentration of the added N2O4, so the
equilibrium shifts to the right, in the direction
of products.
(b) The system will adjust to the removal of NO2
by shifting to the side that produces more NO2
thus, the equilibrium shifts to the right.
(c) Adding N2 will increase the total pressure of
the system, but N2 is not involved in the
reaction. The partial pressures of NO2 and N2O4
are therefore unchanged, and there is no shift in
the position of the equilibrium.
(d) If the volume is increased, the system will
shift in the direction that occupies a larger
volume (more gas molecules) thus, the
equilibrium shifts to the right. (This is the
opposite of the effect observed in Figure 15.13,
where the volume was decreased.)
(e) The reaction is endothermic, so we can
imagine heat as a reagent on the reactant side of
the equation. Decreasing the temperature will
shift the equilibrium in the direction that
produces heat, so the equilibrium shifts to the
left, toward the formation of more N2O4. Note
that only this last change also affects the value
of the equilibrium constant, K.
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Answers (a) right, (b) left, (c) right, (d) left
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Solution Analyze We are asked to determine the
standard enthalpy change of a reaction and how
the equilibrium constant for the reaction varies
with temperature. Plan (a) We can use standard
enthalpies of formation to calculate ?H for the
reaction. (b) We can then use Le Châteliers
principle to determine what effect temperature
will have on the equilibrium constant.
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SAMPLE EXERCISE 15.14 continued
(b) Because the reaction in the forward direction
is exothermic, we can consider heat a product of
the reaction. An increase in temperature causes
the reaction to shift in the direction of less
NH3 and more N2 and H2. This effect is seen in
the values for Kp presented in Table 15.2. Notice
that Kp changes markedly with changes in
temperature and that it is larger at lower
temperatures.
Comment The fact that Kp for the formation of
NH3 from N2 and H2 decreases with increasing
temperature is a matter of great practical
importance. To form NH3 at a reasonable rate
requires higher temperatures. At higher
temperatures, however, the equilibrium constant
is smaller, and so the percentage conversion to
NH3 is smaller. To compensate for this, higher
pressures are needed because high pressure favors
NH3 formation.
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SAMPLE EXERCISE 15.14 continued
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SAMPLE INTEGRATIVE EXERCISE continued
(d) In discussing Le Châteliers principle, we
saw that endothermic reactions exhibit an
increase in Kp with increasing temperature.
Because the equilibrium constant for this
reaction increases as temperature increases, the
reaction must be endothermic. From the enthalpies
of formation given in Appendix C, we can verify
our prediction by calculating the enthalpy
change for the reaction,
The
positive sign for ?H indicates that the reaction
is endothermic.
(e) According to Le Châteliers principle, a
decrease in the pressure causes a gaseous
equilibrium to shift toward the side of the
equation with the greater number of moles of gas.
In this case there are two moles of gas on the
product side and only one on the reactant side.
Therefore, the pressure should be reduced to
maximize the yield of the CO and H2.