Chemical Equilibrium

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Chapter 15

• Chemical Equilibrium

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15.1
3
15.1
4
Cold Temp
Hot Temp
15.1
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15.1
6
15.1
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15.2 Law of Mass Action

• Derived from rate laws by Guldberg andWaage
  (1864)
• For a balanced chemical reactionin equilibrium
• a A b B ? c C d D
• Equilibrium constant expression (Keq)

Cato Guldberg Peter Waage (1836-1902)
(1833-1900)
or

• Keq is strictly based on stoichiometry of the
  reaction (is independent of the mechanism).
• Units Keq is considered dimensionless (no units)

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Relates Kc to Kp
15.2
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For Example

• Write the equilibrium expression Kc for the
  following reactions

15.2
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Example 2

• In the synthesis of ammonia from nitrogen and
  hydrogen,
• Kc 9.60 at 300C
• N2(g) 3H2(g) 2NH3(g)
• Calculate Kp for this reaction at this
  temperature

Hint..you will need to use..
15.2
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What does the Equilibrium constant tell us??
15.2
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Looking at reversible reactions

• So far we have only written the expression
  forwards, but it can also be written equally
  backwards!
• The constants are the recipricals of each other

Kc 0.212
Kc 4.72
15.2
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15.3 Types of Equilibria

• Homogeneous all components in same phase
  (usually g or aq)
• N2 (g) H2 (g) ? NH3 (g)

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2
1
Fritz Haber(1868 1934)
15.3
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• Heterogeneous different phases
• CaCO3 (s) ? CaO (s) CO2 (g)
• Definition What we use

• Concentrations of pure solids and pure liquids
  are not included in Keq expression because their
  concentrations do not vary, and are already
  included in Keq

• Even though the concentrations of the solids or
  liquids do not appear in the equilibrium
  expression, the substances must be present to
  achieve equilibrium.

15.3
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For Example

• Write equilibrium-constant expressions for Kc and
  Kp for each of the following reactions

Click for answers
15.3
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15.4 Calculating Equilibrium Constants

• Steps to use ICE table
• I Tabulate known initial and equilibrium
  concentrations of all species in equilibrium
  expression
• C Determine the concentration change for the
  species where initial and equilibrium are known
• Use stoichiometry to calculate concentration
  changes for all other species involved in
  equilibrium
• E Calculate the equilibrium concentrations

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• Ex Enough ammonia is dissolved in 5.00 L of
  water at 25ºC to produce a solution that is
  0.0124 M ammonia. The solution is then allowed
  to come to equilibrium. Analysis of the
  equilibrium mixture shows that OH1- is 4.64 x
  10-4 M. Calculate Keq at 25ºC for the reaction
• NH3 (aq) H2O (l) ? NH41 (aq) OH1- (aq)

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NH3 (aq) H2O (l) ? NH41 (aq) OH1- (aq)
X
0.0124 M
0 M
0 M
X
- x
x
x
X
0.0119 M
4.64 x 10-4 M
4.64 x 10-4 M
x 4.64 x 10-4 M
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• Ex A 5.000-L flask is filled with 5.000 x 10-3
  mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC.
  The value of Keq is 1.33. What are the
  concentrations of each substance at equilibrium?
• H2 (g) I2 (g) ? 2 HI (g)

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H2 (g) I2 (g) ? 2 HI (g)
1.000x10-3 M
2.000x10-3 M
0 M
- x M
- x M
2x M
(1.000x10-3 x) M
(2.000x10-3 x) M
2x M
4x2 1.33x2 (-3.000x10-3)x 2.000x10-6 0
-2.67x2 3.99x10-3x 2.66x10-6 Using quadratic
eqn x 5.00x10-4 or 1.99x10-3 x
5.00x10-4 Then H25.00x10-4 M I21.50x10-3
M HI1.00x10-3 M
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15.6 Le Chateliers Principle
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Changes that do not affect Keq

• Concentration
• Upon addition of a reactant or product,
  equilibrium shifts to re-establish equilibrium by
  consuming part of the added substance.
• Addition of solids/liquids do not appreciably
  shift the system and can be ignored. But they are
  still made/consumed!
• Upon removal of reactant or product, equilibrium
  shifts to re-establish equilibrium by producing
  more of the removed substance.

• Ex Co(H2O)62 (aq) 4 Cl1- ? CoCl42- (aq) 6
  H2O (l)
• Add HCl, temporarily inc forward rate

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Volume, with a gas present (T is constant)

• Upon a decrease in V (thereby increasing
  P),equilibrium shifts to reduce the number of
  moles of gas.
• Upon an increase in V (thereby decreasing
  P),equilibrium shifts to produce more moles of
  gas.
• Ex N2 (g) 3 H2 (g) ? 2 NH3 (g)
• If V of container is decreased, equilibrium
  shifts right.
• XN2 and XH2 dec
• XNH3 inc

Since PT also inc, KP remains constant.
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3. Pressure, but not Volume

• Usually addition of a noble gas, p. 560
• Avogadros law adding more non-reacting
  particles fills in the empty space between
  particles.
• In the mixture of red and blue gas particles,
  below, adding green particles does not stress the
  system, so there is no Le Châtelier shift.

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Catalysts

• Lower the activation energy of both forward and
  reverse rxns, therefore increases both forward
  and reverse rxn rates.
• Increase the rate at which equilibrium is
  achieved, but does not change the ratio of
  components of the equilibrium mixture (does not
  change the Keq)

Ea, uncatalyzed
Ea, catalyzed
Energy
Rxn coordinate
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Change that does affect Keq

• Temperature consider heat as a part of the
  reaction
• Upon an increase in T, endothermic reaction is
  favored (equilibrium shifts to consume the extra
  heat)
• Upon a decrease in T, equilibrium shifts to
  produce more heat.

• Effect on Keq
• Exothermic equilibria Reactants ? Products
  heat
• Inc T increases reverse reaction rate which
  decreases Keq
• Endothermic equilibria Reactants heat ?
  Products
• Inc T increases forward reaction rate increases
  Keq

• Ex Co(H2O)62 (aq) 4 Cl1- ? CoCl42- (aq) 6
  H2O (l) DH?
• Inc T temporarily inc forward rate
• Dec T temporarily inc reverse rate

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Effect of Various Changes on Equilibrium
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Effect of Pressure on Equilb.
Disturbance Direction of Reaction Effect of K
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Effect of Temperature on Equilb
Disturbance Direction of Reaction Effect of K
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For Example

• For the following reaction
• 5 CO(g) I2O5(s) I2(g) 5 CO2(g)
  1175 kJ
• for each change listed, predict the equilibrium
  shift and the effect on the indicated quantity.