CHEMICAL EQUILIBRIUM Chapter 16

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Chemistry and Chemical Reactivity 6th
Edition John C. Kotz Paul M. Treichel
Gabriela C. Weaver
CHAPTER 16 Principles of Reactivity Chemical
Equilibria
Lecture written by John Kotz as modified by
George Rhodes
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Chapter Goals p-788

• Understand the nature and characteristics of
  chemical equilibria
• Understand the significance of the equilibrium
  constant K, and the reaction quotient Q
• Understand how to use K in quantitative studies
  of chemical equilibria

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CHEMICAL EQUILIBRIUM
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

Pink to blue Co(H2O)6Cl2 ---gt Co(H2O)4Cl2 2
H2O
Blue to pink Co(H2O)4Cl2 2 H2O ---gt Co(H2O)6Cl2
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Chemical EquilibriumFe3 SCN- ? FeSCN2
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Chemical EquilibriumFe3 SCN- ? FeSCN2

• 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
Screen 16.4 Active Figure 16.3
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 a 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) ? 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, page 929
• ?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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Chemistry In Action
Life at High Altitudes and Hemoglobin Production
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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?

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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) ? 2 NO2(g)

?
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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 0.50 - 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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Method of successive approximations

• For example, if
• We could use the quadradic equation to give
  x3.4x10-4 or the method of successive
  approximations
• Assume that x is so small that (0.0010-x) is
  still 0.0010
• Then substitute and calculate x again. If the
  value doesnt change much, it is valid.

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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 ManipulatingK 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 ---gt 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
Figure 16.8
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EQUILIBRIUM AND EXTERNAL EFFECTS
Catalytic exhaust system

• Add catalyst ---gt 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

• Adding a reactant to a chemical system.

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Le Chateliers Principle

• Removing a reactant from a chemical system.

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Le Chateliers Principle

• Adding a product to a chemical system.

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Le Chateliers Principle

• Removing a product from a chemical system.

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Butane-Isobutane Equilibrium
butane
isobutane
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Butane ? 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 ? 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 ? 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 ? 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.

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Le Châteliers Principle
Change
Shift Equilibrium
Concentration
yes
no
Pressure
yes
no
Volume
yes
no
Temperature
yes
yes
Catalyst
no
no
14.5
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Heisenberg

• Q How many Heisenbergs does it take to change a
  light bulb? A If you know the number, you don't
  know where the light bulb is.
• Q How many physical chemists does it take to
  change a light bulb? A If the light bulb is a
  perfect sphere, one. The solution for a light
  bulb of arbitrary shape is left as an exercise to
  the reader.
• Q How many experimental physicists does it take
  to change a light bulb? A They don't replace the
  bulbs, they repair them.

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Exam 1 18 MC 4 problems

• Rate expressions write
• Effects of catalysts, T and concentration
  changes
• Given rate data write rate expressions
• Write rate constants for different Rx orders
• Determine exponents from rate laws
• Label axes for rate data (x,y)
• Collision theory what does it mean?
• Activation energy what does it mean?
• Gas phase equilibrium what does it mean?
• Equilibrium expressions be able to write
• When does KpKc?

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More

• Difference between Q and K?
• Le Chatelier what happens to reactants and
  products if?
• Write expression for Kp or Kc given the reaction
  expression
• Do half life calculations, especially those
  involving carbon dating
• Calculate the equilibrium constant for a specific
  reaction involving the dissolution of a solid
• Calculate the equilibrium constant for a specific
  reaction in the gas phase
• Be able to employ the ICE format

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Problems

• K for the conversion of butane and isobutane
  2.5 _at_25C. Put 0.017 mol butane in a flask. What
  will the equiliberium concentrations be?
• butane Û isobutane
• Initial (M) 0.034 0Change (M) x xEquilibriu
  m (M) 0.034 x xisobutane x 0.024
  Mbutane 0.034 M x 0.010 M

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Problems 2

• Ammonium Iodide dissociates into HI and NH3
  according to the equation
• NH4I(s) Û NH3 (g) HI (g), what is Kp if the
  pressure in the flask is 705 mm when the partial
  pressures are in atm?
• Ptotal PNH3 PHIPNH3 PHI 1/2(0.928
  atm) 0.464 atmKP (0.464)2 0.215