Ch. 15 Chemical Equilibrium

1
Ch. 15 Chemical Equilibrium

• Consider colorless 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. We now write it
  like this
• Each is constantly being formed at the same rate
  that it is being consumed. It is therefore a
  dynamic equilibrium.
• Chemical equilibrium is the point at which the
  concentrations of all species are constant.

2
The Equilibrium Constant

• Consider the following reaction
• There are two reactions going on, forward and
  reverse. The rate of each reaction can be
  expressed separately
• Ratef kfAaBb
• Rater krCcDd
• At equilibrium, Rf Rr orkfAaBb
  krCcDd
• We can rearrange this equation and combine the
  rate constants into a single constant. We end up
  with this
• where Keq is called the equilibrium constant.

3
The Equilibrium Constant

• If the substances in the reaction are gases, then
  their concentrations are proportional to their
  partial pressures according to PVnRT, so we can
  express Keq in terms of pressures as well
• (PC)c (PD)d
• (PA)a (PB)b
• (Reminder Molarity n/V P/RTand R 0.0821
  Latm/molK)
• (Changes in pressure at equilibrium will be
  explored later in the notes.)
• The actual value for Keq is determined
  experimentally at a specific temperature for a
  specific reaction.

Keq
4
Facts About Keq

• Keq does not depend on initial concentrations of
  the products or reactants since its only the
  equilibrium concentrations that we are concerned
  with!
• The larger Keq , the more products are present
  at equilibrium. If Keqgtgt 1, then products
  dominate at equilibrium and equilibrium lies to
  the right.
• Conversely, the smaller Keq , the more reactants
  are present at equilibrium. If Keq ltlt 1, then
  reactants dominate at equilibrium and the
  equilibrium lies to the left.

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The Magnitude of Equilibrium Constants
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More Facts About Keq

• The equilibrium constant of a reaction in the
  reverse direction is the inverse of the
  equilibrium constant of the reaction in the
  forward direction Keq(forward) 1/Keq(reverse)
• For example
• At 100 ºC, Keq(forward) 6.49
• At 100 ºC, Keq(reverse) 1/6.49 0.154
• Keq does not depend on the reaction mechanism.
• The value of Keq varies with temperature. (We
  will see this later.)
• The stoichiometry of a reaction that has been
  multiplied by a number changes the equilibrium
  constant. Keq gets raised to the power equal to
  that number.
• For example 2 4
  Keq (6.49)2 42.12

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More Facts About Keq

• 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.
• For example A B ? X C Keq(1) 2.0
• X B ? D Keq(2) 5.0
• A 2B ? C D Keq K1 x K2 10.0
• When this is written in terms of concentrations
  at equilibrium
• Keq10.0 CD/AB2
• Keq usually is not written with units. Why?
  The Molarity Pressures are based on a
  standard (1 M or 1 atm.)

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More Facts About Keq

• If concentrations are used in the equilibrium
  expression, Keq is sometimes written as Kc.
• When pressures are used in the expression, Keq
  it sometimes written as Kp.
• The equilibrium expression only contains the
  concentrations of gases or aqueous substances and
  NEVER solids or pure liquids. Why?
• - Consider the decomposition of calcium
  carbonate
• CaCO3(s) ?? CaO(s) CO2(g)
• - The concentrations of solids and pure liquids
  are constant. The amount of CO2 formed will
  not depend greatly on the amounts of CaO
  and CaCO3 present.
• Keq PCO2

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Note Although the concentrations of these
species are not included in the equilibrium
expression, they do participate in the reaction
and must be present for an equilibrium to be
established!
10
Changes in Equilibrium

• Le Chateliers Principle If a stress is
  applied to a system that is already at
  equilibrium, the equilibrium will shift to reduce
  the effect of the stress.
• We will now look at changing various things on a
  system at equilibrium and discuss how the
  equilibrium will shift to relieve the stress.
• (1) Changing Concentrations A B ? C
  D
• If more A is added, the rate of the forward
  reaction increases to relieve the stress. As
  this occurs C and D increase, the rate of the
  reverse reaction increases and quickly
  equilibrium is re-established. At equilibrium
  Rf Rr. The concentrations of A, B, C, D have
  changed but
• CD remains constant at a given
  temperature.
• AB

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Changes in Equilibrium

• Changing Concentrations Continued.. A B
  ? C D
• If C is removed for the system, the rate of
  the forward reaction will increase to replace the
  C that was removed. As this occurs D
  increases, and A and B decreases the rate of
  the reverse reaction increases and quickly
  equilibrium is re-established. Once again, Keq
  is the same even though the concentrations are
  different.
• (2) Pressure ONLY AFFECTS CHEMICAL REACTIONS
  WHICH INVOLVE GASES
• As the pressure on the gaseous system increases,
  the gaseous substances are compressed, their
  concentrations and of molecules/liter increase.
  The reaction, (forward or reverse), which favors
  the reduction of the number of molecules per
  liter will be favored. (Remember, if V?, then
  this would cause P?.)

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Changes in Equilibrium

• Changing pressure continued
• Example 2H2(g) O2(g) ? 2H2O(g)
• If pressure is increased by decreasing the
  volume of the gases in the container at
  equilibrium, then the forward reaction is
  favored. Why?
• - This reduces the of gas particles from 3 to
  2.
• - Note If the of gas particles on both sides
  of the equation is the same, then changing
  pressure has NO EFFECT on the equilibrium.
• If an inert gas is added, it WILL NOT change the
  equilibrium.

• (3) Temperature All chemical reactions either
  give off heat (exothermic) or take in heat
  (endothermic). An increase in temperature favors
  the endothermic reaction a decrease in
  temperature favors the exothermic reaction.
• The temperature of a reaction will change the
  value of Keq, but for now, lets just focus on
  how the equilibrium will shift.

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Changes in Equilibrium

• Changes in temperature continued
• Example H2 I2 ? 2HI ?H - 25
  kJ/mol (exothermic)
• Another way of looking at the reaction
• H2 I2 ? 2HI heat
• Lowering the temperature favors HI formation.
  (You can think of it as though we are removing
  the heat product from the equation.) Raising
  the temperature favors the reverse reaction.
• van Hoffs Law In a system at equilibrium, an
  increase in heat energy is displaced so that heat
  is absorbed.
• (4) Catalyst A catalyst increases the rate of
  both the forward and the reverse reaction by
  decreasing Ea. It has no effect on Keq but does
  cause equilibrium to be reached more quickly.

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Catalysts Changes in Equilibrium
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Predicting the Direction of a Reaction

• When given the concentrations of reactants
  products and the value of Keq, you can easily
  predict the direction of a reaction.
• Simply plug the given concentrations into the
  equilibrium equation and calculate the reaction
  quotient, Q.
• If Q gt Keq then the reverse reaction must occur
  to reach equilibrium (i.e., products are
  consumed, reactants are formed, the numerator in
  the equilibrium constant expression decreases and
  Q decreases until it equals Keq).
• If Q lt Keq then the forward reaction must occur
  to reach equilibrium. (See Practice Problems for
  an example.)

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Predicting the Direction of a Reaction
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Calculating Keq

• Practice Problem At a certain temperature, a
  1.0 L flask initially contains 0.298 moles of
  PCl3(g) and 0.0087 moles of PCl5(g). After the
  system reaches equilibrium, 0.00200 moles of
  Cl2(g) was found in the flask. Calculate the
  equilibrium concentrations of the gases in the
  flask and also the value of Keq. PCl5
  decomposes according to the following
  reactionPCl5(g)?? PCl3(g) Cl2(g)
• Step 1-- Tabulate initial and equilibrium
  concentrations (or partial pressures) given.
• This sort of table goes by the nickname
  ICEboxget it?

PCl5 PCl3 Cl2
Initial
Change
Equilibrium
0.0087 M
0.298 M
0
0.002 M
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Calculating Keq
Step 2- If an initial and equilibrium
concentration is given for a species,
calculate the change in concentration. Step 3-
Use stoichiometry on the change in concentration
line only to calculate the changes in
concentration of all species(Since our reaction
occurs in a 111 ratio, the changes for each
species is the same. Only the sign is different
on the reactant.) Step 4- Deduce the equilibrium
concentrations of all species.
PCl5 PCl3 Cl2
Initial
Change
Equilibrium
0.0087 M
0.298 M
0
0.002 M
0.002 M
- 0.002 M
0.002 M
0.0067 M
0.3 M
Step 5- Finally, plug the equilibrium
concentrations into the equilibrium equation and
solve!
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Calculating Keq
Keq PCl3Cl2/PCl5 Keq
0.30.002/0.0067 0.08955 0.09
PCl5 PCl3 Cl2
Initial
Change
Equilibrium
0.0087 M
0.298 M
0
0.002 M
0.002 M
- 0.002 M
0.002 M
0.0067 M
0.3 M

• This system of tabulating data will allow you to
  solve for equilibrium concentrations if you are
  given Keq

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Calculating Keq
Practice Problem 2-- Given this equationH2(g)
I2(g) ?? 2 HI(g) calculate all three equilibrium
concentrations when H2o I2o 0.200 M and
Kc 64.0
H2 I2 HI
Initial
Change
Equilibrium
0
0.200 M
0.200 M
- x
- x
2x
0.200 - x
0.200 - x
2x

• Heres where we use stoichiometry to calculate
  the changes in the concentrations
• Now determine the equilibrium concentrations
• Now plug them into the equilibrium expression
  and solve for x

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Calculating Keq

• Kc HI2/H2I2
• 64.0 2x2/0.200 x0.200 x
• Solving this takes some good algebra skills
  maybe even a quadratic equation will have to be
  solvedYUCK!
• Both sides are perfect squares, (done so on
  purpose), so we square root both sides to get
  8.00 (2x) / (0.200 - x)
• From there, the solution should be easier, and
  so after some cross-multiplying and dividing,
  etc x 0.160 M
• This is not the end of the solution since the
  question was asking for the equilibrium
  concentrations, so
• H2 0.200 - 0.160 0.040 MI2 0.200 -
  0.160 0.040 MHI 2 (0.160) 0.320 M
• You can check for correctness by plugging back
  into the equilibrium expression
• Kc (0.320)2 / (0.040)(0.040) 64
• Since Kc 64.0 we know that the problem was
  correctly solved.

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The End
Now we get to work on more practice problems!!