CHEMICAL EQUILIBRIUM Chapter 16

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CHEMICAL EQUILIBRIUMChapter 16
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Pb2(aq) 2 Cl(aq) ? PbCl2(s)
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Properties of an Equilibrium

• Equilibrium systems are
• DYNAMIC (in constant motion)
• REVERSIBLE
• can be approached from either direction

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Pink to blue Co(H2O)6Cl2 ? Co(H2O)4Cl2 2 H2O
Blue to pink Co(H2O)4Cl2 2 H2O ? Co(H2O)6Cl2
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Chemical EquilibriumFe3 SCN- e FeSCN2
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Chemical EquilibriumFe3 SCN- e FeSCN2
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• After a period of time, the concentrations of
  reactants and products are constant.
• The forward and reverse reactions continue after
  equilibrium is attained.

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Examples of Chemical Equilibria

• Phase changes such as H2O(s) ? H2O(liq)

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Examples of Chemical Equilibria

• Formation of stalactites and stalagmites
• CaCO3(s) H2O(liq) CO2(g) ? Ca2(aq) 2
  HCO3-(aq)

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Chemical Equilibria

• CaCO3(s) H2O(liq) CO2(g) ? Ca2(aq) 2
  HCO3-(aq)
• At a given T and P of CO2, Ca2 and HCO3- can
  be found from the EQUILIBRIUM CONSTANT.

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Reaction Quotient Equilibrium Constant
See Active Figure 16.2
Product conc. increases and then becomes constant
at equilibrium
Reactant conc. declines and then becomes constant
at equilibrium
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Reaction Quotient Equilibrium Constant
At any point in the reaction H2 I2 ? 2 HI
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Reaction Quotient Equilibrium Constant
In the equilibrium region
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The Reaction Quotient, Q

• In general, ALL reacting chemical systems are
  characterized by their REACTION QUOTIENT, Q.
• a A b B ? c C d D

If Q K, then system is at equilibrium.
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THE EQUILIBRIUM CONSTANT

• For any type of chemical equilibrium of the type
• a A b B ? c C d D
• the following is a CONSTANT (at a given T)

If K is known, then we can predict concs. of
products or reactants.
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Determining K

• 2 NOCl(g) ? 2 NO(g) Cl2(g)
• Place 2.00 mol of NOCl is a 1.00 L flask. At
  equilibrium you find 0.66 mol/L of NO. Calculate
  K.
• Solution
• Set of an ICE table of concentrations
• NOCl NO Cl2
• Initial 2.00 0 0
• Change
• Equilibrium 0.66

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Determining K

• 2 NOCl(g) ? 2 NO(g) Cl2(g)
• Place 2.00 mol of NOCl is a 1.00 L flask. At
  equilibrium you find 0.66 mol/L of NO. Calculate
  K.
• Solution
• Set of an ICE table of concentrations
• NOCl NO Cl2
• Initial 2.00 0 0
• Change -0.66 0.66 0.33
• Equilibrium 1.34 0.66 0.33

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Determining K

• 2 NOCl(g) ? 2 NO(g) Cl2(g)
• NOCl NO Cl2
• Initial 2.00 0 0
• Change -0.66 0.66 0.33
• Equilibrium 1.34 0.66 0.33

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Writing and Manipulating K Expressions

• Solids and liquids NEVER appear in equilibrium
  expressions.
• S(s) O2(g) ? SO2(g)

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Writing and Manipulating K Expressions

• Solids and liquids NEVER appear in equilibrium
  expressions.
• NH3(aq) H2O(liq) ? NH4(aq) OH-(aq)

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The Meaning of K

• 1. Can tell if a reaction is product-favored or
  reactant-favored.
• For N2(g) 3 H2(g) ? 2 NH3(g)

Conc. of products is much greater than that of
reactants at equilibrium. The reaction is
strongly product-favored.
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The Meaning of K

• For AgCl(s) ? Ag(aq) Cl-(aq)
• Kc Ag Cl- 1.8 x 10-5
• Conc. of products is much less than that of
  reactants at equilibrium.
• The reaction with small K is strongly
  reactant-favored.

Ag(aq) Cl-(aq) e AgCl(s) is
product-favored.
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Product- or Reactant Favored
Product-favored K gt 1
Reactant-favored K lt 1
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The Meaning of K

• K comes from thermodynamics. (See Chapter 19)
• ?G lt 0 reaction is product favored
• ?G gt 0 reaction is reactant-favored

If K gt 1, then ?G is negative If K lt 1, then ?G
is positive
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The Meaning of K

• 2. Can tell if a reaction is at equilibrium. If
  not, which way it moves to approach equilibrium.

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The Meaning of K

• If iso 0.35 M and n 0.15 M, are you at
  equilibrium?
• If not, which way does the reaction shift to
  approach equilibrium?
• See Chemistry Now

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The Meaning of K

• All reacting chemical systems are characterized
  by their REACTION QUOTIENT, Q.

If Q K, then system is at equilibrium.
Q (2.33) lt K (2.5) Reaction is NOT at
equilibrium, so iso must become ________ and
n must ____________.
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Typical Calculations

• PROBLEM Place 1.00 mol each of H2 and I2 in a
  1.00 L flask. Calc. equilibrium concentrations.

H2(g) I2(g) ? 2 HI(g)
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H2(g) I2(g) ? 2 HI(g)Kc 55.3

• Step 1. Set up ICE table to define EQUILIBRIUM
  concentrations.
• H2 I2 HI
• Initial 1.00 1.00 0
• Change
• Equilib

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H2(g) I2(g) ? 2 HI(g)Kc 55.3

• Step 1. Set up ICE table to define EQUILIBRIUM
  concentrations.
• H2 I2 HI
• Initial 1.00 1.00 0
• Change -x -x 2x
• Equilib 1.00-x 1.00-x 2x
• where x is defined as amt of H2 and I2 consumed
  on approaching equilibrium.

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H2(g) I2(g) ? 2 HI(g)Kc 55.3

• Step 2. Put equilibrium concentrations into Kc
  expression.

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H2(g) I2(g) ? 2 HI(g)Kc 55.3

• Step 3. Solve Kc expression - take square root
  of both sides.

x 0.79 Therefore, at equilibrium
H2 I2 1.00 - x 0.21 M HI 2x
1.58 M
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Nitrogen Dioxide EquilibriumN2O4(g) e 2 NO2(g)

e
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Nitrogen Dioxide EquilibriumN2O4(g) ? 2 NO2(g)

• If initial concentration of N2O4 is 0.50 M, what
  are the equilibrium concentrations?
• Step 1. Set up an ICE table
• N2O4 NO2
• Initial 0.50 0
• Change
• Equilib

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Nitrogen Dioxide EquilibriumN2O4(g) ? 2 NO2(g)

• If initial concentration of N2O4 is 0.50 M, what
  are the equilibrium concentrations?
• Step 1. Set up an ICE table
• N2O4 NO2
• Initial 0.50 0
• Change -x 2x
• Equilib 0.50 - x 2x

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Nitrogen Dioxide EquilibriumN2O4(g) ? 2 NO2(g)

• Step 2. Substitute into Kc expression and
  solve.

Rearrange 0.0059 (0.50 - x) 4x2
0.0029 - 0.0059x 4x2 4x2 0.0059x -
0.0029 0 This is a QUADRATIC EQUATION ax2
bx c 0 a 4 b 0.0059 c
-0.0029
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Nitrogen Dioxide EquilibriumN2O4(g) ? 2 NO2(g)

• Solve the quadratic equation for x.
• ax2 bx c 0
• a 4 b 0.0059 c -0.0029

x -0.00074 1/8(0.046)1/2 -0.00074
0.027
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Nitrogen Dioxide EquilibriumN2O4(g) ? 2 NO2(g)
x -0.00074 1/8(0.046)1/2 -0.00074 0.027

• x 0.026 or -0.028
• But a negative value is not reasonable.
• Conclusion x 0.026 M
• N2O4 0.050 - x 0.47 M
• NO2 2x 0.052 M

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Solving Quadratic Equations

• Recommend you solve the equation exactly on a
  calculator or use the method of successive
  approximations
• See Appendix A.

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Writing and Manipulating K Expressions

• Adding equations for reactions
• S(s) O2(g) ? SO2(g)
• SO2(g) 1/2 O2(g) ? SO3(g)

Net equation S(s) 3/2 O2(g) ? SO3(g)
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Writing and Manipulating K Expressions

• Changing coefficients
• S(s) 3/2 O2(g) ? SO3(g)
• 2 S(s) 3 O2(g) ? 2 SO3(g)

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Writing and Manipulating K Expressions

• Changing direction
• S(s) O2(g) ? SO2(g)
• SO2(g) ? S(s) O2(g)

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Writing and Manipulating K Expressions

• Concentration Units
• We have been writing K in terms of mol/L.
• These are designated by Kc
• But with gases, P (n/V)RT conc RT
• P is proportional to concentration, so we can
  write K in terms of P. These are designated by
  Kp.
• Kc and Kp may or may not be the same.

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Writing and Manipulating K Expressions

• K using concentration and pressure units
• Kp Kc (RT)?n
• For S(s) O2(g) ? SO2(g)
• ?n 0 and Kp Kc
• For SO2(g) 1/2 O2(g) ? SO3(g)
• ?n 1/2 and Kp Kc(RT)1/2

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EQUILIBRIUM AND EXTERNAL EFFECTS

• Temperature, catalysts, and changes in
  concentration affect equilibria.
• The outcome is governed by LE CHATELIERS
  PRINCIPLE
• ...if a system at equilibrium is disturbed, the
  system tends to shift its equilibrium position to
  counter the effect of the disturbance.

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EQUILIBRIUM AND EXTERNAL EFFECTS

• Henri Le Chatelier
• 1850-1936
• Studied mining engineering.
• Interested in glass and ceramics.

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EQUILIBRIUM AND EXTERNAL EFFECTS

• Temperature change ? change in K
• Consider the fizz in a soft drink CO2(aq)
  HEAT ? CO2(g) H2O(liq)
• K P (CO2) / CO2
• Increase T. What happens to equilibrium position?
  To value of K?
• K increases as T goes up because P(CO2) increases
  and CO2 decreases.
• Decrease T. Now what?
• Equilibrium shifts left and K decreases.

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Temperature Effects on Equilibrium

• N2O4 (colorless) heat ? 2 NO2 (brown) ?Ho
  57.2 kJ

Kc (273 K) 0.00077 Kc (298 K) 0.0059
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Temperature Effects on Equilibrium
See Figure 16.8
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EQUILIBRIUM AND EXTERNAL EFFECTS
Catalytic exhaust system

• Add catalyst ? no change in K
• A catalyst only affects the RATE of approach to
  equilibrium.

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Haber-Bosch Process for NH3

• N2(g) 3 H2(g) ? 2 NH3(g) heat
• K 3.5 x 108 at 298 K

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Haber-Bosch Ammonia Synthesis
Fritz Haber 1868-1934 Nobel Prize, 1918
Carl Bosch 1874-1940 Nobel Prize, 1931
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EQUILIBRIUM AND EXTERNAL EFFECTS

• Concentration changes
• no change in K
• only the equilibrium composition changes.

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Le Chateliers Principle
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• Adding a reactant to a chemical system.

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Le Chateliers Principle
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• Removing a reactant from a chemical system.

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Le Chateliers Principle
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• Adding a product to a chemical system.

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Le Chateliers Principle
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• Removing a product from a chemical system.

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EQUILIBRIUM AND EXTERNAL EFFECTS

• Temperature, catalysts, and changes in
  concentration affect equilibria.
• The outcome is governed by LE CHATELIERS
  PRINCIPLE
• ...if a system at equilibrium is disturbed, the
  system tends to shift its equilibrium position to
  counter the effect of the disturbance.

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Butane-Isobutane Equilibrium
butane
isobutane
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Butane e Isobutane

• At equilibrium with iso 1.25 M and butane
  0.50 M. K 2.5.
• Add 1.50 M butane.
• When the system comes to equilibrium again, what
  are iso and butane?

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Butane e Isobutane

• Assume you are at equilibrium with iso 1.25 M
  and butane 0.50 M. Now add 1.50 M butane.
  When the system comes to equilibrium again, what
  are iso and butane? K 2.5
• Solution
• Calculate Q immediately after adding more butane
  and compare with K.

Q is LESS THAN K. Therefore, the reaction will
shift to the ____________.
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Butane e Isobutane

• You are at equilibrium with iso 1.25 M and
  butane 0.50 M. Now add 1.50 M butane.
• Solution
• Q is less than K, so equilibrium shifts right
  away from butane and toward isobutane.
• Set up ICE table
• butane isobutane
• Initial
• Change
• Equilibrium

0.50 1.50
1.25
- x
x
1.25 x
2.00 - x
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Butane e Isobutane

• You are at equilibrium with iso 1.25 M and
  butane 0.50 M. Now add 1.50 M butane.
• Solution

x 1.07 M At the new equilibrium position,
butane 0.93 M and isobutane 2.32 M.
Equilibrium has shifted toward isobutane.
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Le Chateliers Principle

• Change T
• change in K
• therefore change in P or concentrations at
  equilibrium
• Use a catalyst reaction comes more quickly to
  equilibrium. K not changed.
• Add or take away reactant or product
• K does not change
• Reaction adjusts to new equilibrium position

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Nitrogen Dioxide EquilibriumN2O4(g) ? 2 NO2(g)

• Increase P in the system by reducing the volume
  (at constant T).

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Nitrogen Dioxide EquilibriumN2O4(g) ? 2 NO2(g)

• Increase P in the system by reducing the volume.
• In gaseous system the equilibrium will shift to
  the side with fewer molecules (in order to reduce
  the P).
• Therefore, reaction shifts LEFT and P of NO2
  decreases and P of N2O4 increases.