Chapter 15 Chemical Equilibrium

1
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

2
The Concept of Equilibrium

• Consider colorless frozen N2O4. At room
  temperature, it decomposes to brown NO2
• N2O4(g) ? 2NO2(g).
• At some time, the color stops changing and we
  have a mixture of N2O4 and NO2.
• Chemical equilibrium is the point at which the
  concentrations of all species are constant.

3

• The point at which the rate of decomposition
• N2O4(g) ? 2NO2(g)
• equals the rate of dimerization
• 2NO2(g) ? N2O4(g).
• is dynamic equilibrium.
• The equilibrium is dynamic because the reaction
  has not stopped the opposing rates are equal.
• Consider frozen N2O4 only white solid is
  present. On the microscopic level, only N2O4
  molecules are present.

4
The Concept of Equilibrium

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

5

• As the substance warms it begins to decompose
• N2O4(g) ? 2NO2(g)
• A mixture of N2O4 (initially present) and NO2
  (initially formed) appears light brown.
• When enough NO2 is formed, it can react to form
  N2O4
• 2NO2(g) ? N2O4(g).

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• At equilibrium, as much N2O4 reacts to form NO2
  as NO2 reacts to re-form N2O4
• The double arrow implies the process is dynamic.
• Consider
• Forward reaction A ? B Rate kfA
• Reverse reaction B ? A Rate krB
• At equilibrium kfA krB.

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• For an equilibrium we write
• As the reaction progresses
• A decreases to a constant,
• B increases from zero to a constant.
• When A and B are constant, equilibrium is
  achieved.
• Alternatively
• kfA decreases to a constant,
• krB increases from zero to a constant.
• When kfA krB equilibrium is achieved.

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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.

9
A System at Equilibrium

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

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11
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

• 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
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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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Law of mass action

• Equilibrium constant expression
• It depends on the reaction stoichiometry and NOT
  on its mechanism.
• We can not tell the reaction mechanism from the
  Equilibrium Expression
• Always products on the numerator
• Always reactants in the denominator

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What Are the Equilibrium Expressions for These
Equilibria?
20
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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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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January 11

• Manipulating Equilibrium Constants
• Heterogeneous equilibrium
• Equilibrium calculations RICE!!!
• HW 23, 27 to 41 RED

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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
35

• 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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Writing Equilibrium Constant Expressions

• The concentrations of the reacting species in the
  condensed phase are expressed in M. In the
  gaseous phase, the concentrations can be
  expressed in M or in atm.
• The concentrations of pure solids, pure liquids
  and solvents do not appear in the equilibrium
  constant expressions.
• The equilibrium constant is a dimensionless
  quantity.
• In quoting a value for the equilibrium constant,
  you must specify the balanced equation and the
  temperature.
• If a reaction can be expressed as a sum of two or
  more reactions, the equilibrium constant for the
  overall reaction is given by the product of the
  equilibrium constants of the individual reactions.

14.2
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Calculating Equilibrium Concentrations

1. Express the equilibrium concentrations of all
   species in terms of the initial concentrations
   and a single unknown x, which represents the
   change in concentration.
2. Write the equilibrium constant expression in
   terms of the equilibrium concentrations. Knowing
   the value of the equilibrium constant, solve for
   x.
3. Having solved for x, calculate the equilibrium
   concentrations of all species.

14.4
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Equilibrium Calculations
39
Equilibrium Calculations

• A closed system initially containing
• 1.000 x 10-3 M H2 and 2.000 x 10-3 M Cl2
• At 448?C is allowed to reach equilibrium.
  Analysis of the equilibrium mixture shows that
  the concentration of HCl 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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HCl Increases by 1.87 x 10-3 M
H2, M Cl2, M HCl, 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
42
Stoichiometry tells us H2 and I2decrease by
half as much
H2, M Cl2, M HCl, 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 Cl2, M HCl, 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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Example 2

• The kp for the formation of ammonia is 1.45 x 10
  -5 at 500 0C . In an equilibrium mixture of the 3
  gases at 500 0C the partial pressure of H2 is
  .928 atm and for N2 is .432 atm. Find the partial
  pressure for NH3 at equilibrium.
• Find Kc from the value of Kp

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• 1.45 x 10 -5

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January 14

• Predicting the direction of a reaction
• Le Chateliers Principle
• HW Page 660 Q 1, 4, 5, 6, 7,
• Le Chateliers Principle 51, 53, 55

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Reaction of the day

• Sodium chloride solution reacts with silver
  nitrate

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Demo

• Add NH3 to test tube and observe
• A complex ion forms

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COMPLEX IONS

• Transition metal ion with attached ligands
• Ligands are lewis bases with lone pairs of
  electrons.
• A lewis base is a substance that has an unshared
  pair of electrons in its structure
• H2O (aqua)
• NH3 (ammine)
• Cl- (chloro)
• CN- (cyano)
• OH- (hydroxo)

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Lewis acid

• Lewis acid - Electron pair acceptor
• Lewis bases electron pair donor
• Number of ligands attached to the central metal
  ion are the coordination number of a metal. Often
  is 2 times the charge of the ion.

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Transition elements

• Use d-electrons.
• When the transition element has completely filled
  d sublevel it forms white compounds and give
  colorless solutions.
• The greater the equilibrium constant of the
  formation of the complex the more stable the
  complex is.
• Bring reaction book for tomorrow!!!!

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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.
57

• A 50.0 L reaction vessel contains 1.00 mol of N2,
  3.00 mol of H2 and 0.00 mol of NH3. Will more
  ammonia be formed or it will dissociate when the
  mixture goes to equilibrium at 400 0 C?
• Kc is 0.500 at 4000C

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Strategy to solve problem

• Convert mol to molarity!!!
• Replace values in Qc
• Use the notes from last slides to decide!
• Qc 23.1

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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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Change in concentration

• Adding reactants to the system will shift the
  equilibrium to the products.
• The system will try to remove what was added by
  using it to form more products.
• As reactants are used up (and their concentration
  decrease) the, product concentrations increase.

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• Adding a reactant or product shifts the
  equilibrium away from the increase.
• Removing a reactant or product shifts the
  equilibrium towards the decrease.
• To optimize the amount of product at equilibrium,
  we need to flood the reaction vessel with
  reactant and continuously remove product (Le
  Châtelier).
• We illustrate the concept with the industrial
  preparation of ammonia.

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• Effects of Volume and Pressure Changes
• As volume is decreased pressure increases.
• Le Châteliers Principle if pressure is
  increased the system will shift to counteract the
  increase (will decrease the pressure).
• That is, the system shifts to remove gases and
  decrease pressure.
• An increase in pressure favors the direction that
  has fewer moles of gas.
• In a reaction with the same number of product and
  reactant moles of gas, pressure has no effect.

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• Effects of Volume and Pressure Changes
• An increase in pressure (by decreasing the
  volume) favors the formation of colorless N2O4.
• The instant the pressure increases, the system is
  not at equilibrium and the concentration of both
  gases has increased.
• The system moves to reduce the number moles of
  gas (i.e. the forward reaction is favored).
• A new equilibrium is established in which the
  mixture is lighter because colorless N2O4 is
  favored.

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• Increasing total pressure by adding an inert gas
  has no effect on the partial pressures of
  reactants and products, therefore it has no
  effect on the equilibrium.
• Effect of Temperature Changes
• The equilibrium constant is temperature
  dependent.
• For an endothermic reaction, ?H gt 0 and heat can
  be considered as a reactant.
• For an exothermic reaction, ?H lt 0 and heat can
  be considered as a product.

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• Adding heat (i.e. heating the vessel), favors the
  reaction that removes heat.
• An increase in temperature always favors the
  endothermic reaction! The system will react
  absorbing heat and lowering the temperature.
• Removing heat (i.e. cooling the vessel), favors
  the reaction that releases heat (and increases
  the temperature)
• A decrease in temperature will favor the
  exothermic reaction. The system will respond by
  releasing heat.

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The Effect of Changes in Temperature
68

• The forward reaction is endothermic. Heat is a
  reactant
• DH gt 0.
• Co(H2O)62 is pale pink and CoCl42- is blue.
• If a light purple room temperature equilibrium
  mixture is placed in a beaker of warm water, the
  mixture turns deep blue.
• Since ?H gt 0 (endothermic), adding heat favors
  the forward reaction, i.e. the formation of blue
  CoCl42-.

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• If the room temperature equilibrium mixture is
  placed in a beaker of ice water, the mixture
  turns bright pink.
• Removing heat favors the reaction that produces
  heat, and since ?H gt 0 the reverse reaction which
  is the formation of pink Co(H2O)62 will be
  favored.

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Le Châteliers Principle

• The Effect of Catalysis
• A catalyst lowers the activation energy barrier
  for the reaction.
• Therefore, a catalyst will decrease the time
  taken to reach equilibrium.
• A catalyst does not effect the composition of the
  equilibrium mixture.

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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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• N2 (g) 3 H2 (g) 2 NH3 (g) ?H lt 0
• How can we maximize the amount of ammonia
  produced in the Haber process, shown above?

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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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• N2 and H2 are pumped into a chamber.
• The pre-heated gases are passed through a heating
  coil to the catalyst bed.
• The catalyst bed is kept at 460 - 550 ?C under
  high pressure.
• The product gas stream (containing N2, H2 and
  NH3) is passed over a cooler to a refrigeration
  unit.
• In the refrigeration unit, ammonia liquefies not
  N2 or H2.

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• The unreacted nitrogen and hydrogen are recycled
  with the new N2 and H2 feed gas.
• The equilibrium amount of ammonia is optimized
  because the product (NH3) is continually removed
  and the reactants (N2 and H2) are continually
  being added.

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• H2O (g) CO (g) H2 (g) CO2 (g)
• Assume equilibrium conditions
• H2O 1.00M H2 0.20M
• CO 0.50 CO2 0.70M
• What will happen if
• CO is changed to 0.70M?
• H2 is changed to 0.05M

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• SO3 (g) SO2 (g) 1/2 O2 (g)
• ?H 98.9 kJ
• Determine the effect of each of the following on
  the equilibrium (direction of shift)
• What happens to the concentration of SO3 after
  each of the changes?
• A) Addition of pure oxygen gas.
• B) Compression at Constant Temperature
• C) Addition of Argon gas
• D) Decrease temperature
• Remove sulfur dioxide gas
• Addition of a catalyst

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• Calculating Equilibrium Concentrations
• The same steps used to calculate equilibrium
  constants are used.
• K is given.
• Generally, we do not have a number for the change
  in concentration line.
• Therefore, we need to assume that x mol/L of a
  species is produced (or used).
• The equilibrium concentrations are given as
  algebraic expressions.
• We solve for x, and plug it into the equilibrium
  concentration expressions.

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• Example 2 NO2 (g) N2O4 (g) Kc 0.75
• If the initial concentration of NO2 is 0.50 M and
  the initial concentration of N2O4 is 0.90 M, what
  will the equilibrium concentrations be?
• If the initial concentration of N2O4 is 1.00 M
  (no NO2 present), what will the equilibrium
  concentrations be?

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• Example
• H2 (g) FeO (s) H2O (g) Fe (s)
• Kc 5.20
• If the initial concentration of H2 is 0.50 M and
  the initial concentration of H2O is 6.50 M, what
  will the equilibrium concentrations be?
• If the initial concentration of H2 is 1.00 M (no
  H2O present), what will the equilibrium
  concentrations be?

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C(s) H2O (g) lt?CO (g) H2 (g)

• At 800 C the Keq14.1
• a) What are the eq partial pressures of each gas
  in the equilibrium mixture at this temperature if
  we start with solid carbon and 0.100 mol of H2O
  in a 1.00 L vessel

85

• B) What is the minimum amount of required to
  achieve equilibrium under these conditions?

86

• c) What is the total pressure in the vessel at
  equilibrium?

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• http//biology.clc.uc.edu/fankhauser/Labs/Microbio
  logy/Growth_Curve/Spectrophotometer.htm