Chapter 13: Chemical Equilibrium

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Chapter 13 Chemical Equilibrium
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• In previous chapters we have generally assumed
  that chemical reactions result in complete
  conversion of reactants to products
• Many reactions do not go to completion!!
• Example

N2O4
NO2
double arrow
Reversible reaction the product molecules
recombine to form reactant molecules Reactants
and products are at equilibrium
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• How far does a reaction proceed toward completion
  before it reaches a state of chemical
  equilibrium?
• 2. Chemical equilibrium
• a) The state reached when the concentrations of
  reactants and products remain constant over time
• b) A state in which the concentration of
  reactants and products no longer change (net)
• The reaction does NOT stop!!!

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• 3. Equilibrium mixture
• A mixture of reactants and products in the
  equilibrium state
• What are we interested in?
• a) What is the relationship between the
  concentration of reactants and products in an
  equilibrium mixture?
• b) How can we determine equilibrium
  concentrations from initial concentrations?
• c) What factors can be exploited to alter the
  composition of an equilibrium mixture?

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N2O4 2NO2
No matter with which component we start, either
N2O4 or NO2, their concentrations at equilibrium
are the same
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The Equilibrium State

• The two experiments demonstrate that the
  interconversion of N2O4 and NO2 is reversible and
  that the same equilibrium state is reached
  starting from either substance.
• 1. This is why we use a ? instead of ?
• 2. Since both NO2 and N2O4 are products and
  reactants we will call the chemical on the left
  reactants and on the right products.
• 3. All chemical reactions are reversible     

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The Equilibrium State

• We call a reaction irreversible when it proceed
  nearly to completion
• a. Equilibrium mixture contains almost all
  products and almost no reactants
• b. Reverse reaction is too slow to be detected
• 5. In an equilibrium state the reaction does not
  stop at particular concentrations of reactants
  and products, the rates of the forward and
  reverse reactions become equal.
• Important reaction does not stop!!!

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The Equilibrium State

• 6. Chemical equilibrium is a dynamic state in
  which forward and reverse reactions continue at
  equal rates so that there is no net conversion of
  reactants to products

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The Equilibrium Constant KC At equilibrium
rate for forward reaction rate for reverse
reaction aA bB cC
dD rate forward kf Aa Bb rate reverse
kr Cc Dd At equilibrium kr Cc
Dd kf Aa Bb kf Cc Dd --
------------ kc kr Aa
Bb
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The Equilibrium Constant, Kc

• General equation aA bB ? cC dD
•  
• The substances in the equilibrium equation must
  be gases or molecules and ions in solution,
• NO SOLIDS! NO PURE LIQUIDS!
•  
• Kc units are omitted but you must say at what
  temperature!

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NO22 (0.0125)2 Kc ---------
---------- 4.64 x 10-3 N2O4
0.0337
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The Equilibrium Constant, Kc

• a A bB ? cC dD
•  
• Kc Cc Dd
• Aa Bb
•  
• If we write the equation in the reverse
  direction
• cC dD ? aA bB
•  
• Kc Aa Bb 1
• Cc Dd Kc

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Example 2

•  Write the equilibrium equation for each of the
  following reaction 
• a)      N2(g) 3 H2(g) ? 2 NH3(g)
• b) 2 NH3(g) ? N2(g) 3 H2(g)

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• The oxidation of sulfur dioxide to give sulfur
  trioxide is an important step in the industrial
  process for synthesis of sulfuric acid. Write
  the equilibrium equation for each of the
  following reactions
• a)      2 SO2(g) O2(g) ? 2 SO3(g)
• b) 2 SO3(g) ? 2 SO2(g) O2(g)
• The following equilibrium concentrations were
  measured at 800 K
• SO2 3.0 x 10-3 M O2 3.5 x 10-3 M
• SO3 5.0 x 10-2 M
•  
• Calculate the equilibrium constant at 800 K a
  and b

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The Equilibrium Constant Kp

• For gas-phase reactions use partial pressure
  rather than
• concentrations.
• Kp equilibrium constant with respect to partial
  pressures of reactants and products
• a A bB ? cC dD
•  
• Kp (PC)c (PD)d
• (PA)a (PB)b 
• Relationship between Kc and Kp
• Kp Kc(RT)?n ?n(cd) (ab)

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Example 4

• In the industrial synthesis of hydrogen, mixtures
  of CO and H2O are enriched in H2 by allowing the
  CO to react with steam. The chemical equation
  for this so-called water-gas shift reaction is
•  
• CO(g) H2O(g) ? CO2(g) H2(g)
•  
• What is the value of Kp at 700 K if the partial
  pressures in an equilibrium mixture at 700 K are
  1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of
  CO2, and 20.3 atm H2?

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Example 5

• Nitric oxide reacts with oxygen to give nitrogen
  dioxide, an important reaction in the Ostwald
  process for the industrial synthesis of nitric
  acid
•  
• 2 NO(g) O2(g) ? 2 NO2(g)
•  
• If Kc 6.9 x 105 _at_ 227?C, what is the value of
  Kp _at_ 227?C?
• b) If Kp 1.3 x 10-2 _at_ 1000 K, what is the
  value of Kc _at_ 1000 K?

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Heterogeneous Equilibria

• So far we have been talking about homogeneous
  equilibria,
• in which all reactants and products are in single
  phase
• (gas or solution).
• Heterogeneous equilibria are those in which
  reactants and
• products are present in more than one phase
• Solids and pure liquids
• molar concentrations are constant
• Concentration of solids and pure liquids are not
  included
• when writing the equilibrium equation for any
  heterogeneous
• reaction.
• The concentrations are constant and included in
  value of
• equilibrium constant

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CaCO3(s) CaO(s)
CO2(g) Limestone Lime CaO
CO2 Kc ---------------- CaCO3 CaO
and CaCO3 are solids They have no concentration
in moles/Liter These values dont change, they
are constant Kc CO2 General rule
concentrations of pure solids and liquids are
not included in equilibrium equations
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Example 6

• For each of the following reactions, write the
  equilibrium constant expression for Kc
•  
• a)   2 Fe(s) 3 H2O(g) ? Fe2O3(s) 3
  H2(g)
• b) 2 H2O(l) ? 2 H2(g) O2(g)
• c) SiCl4(g) 2 H2(g) ? Si(s) 4 HCl(g)
  
• d) Hg22(aq) 2 Cl-(aq) ? Hg2Cl2(s)

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Using the Equilibrium Constant

• Knowing the value of the equilibrium constant for
  a chemical reaction lets us
• 1. Judge the extent of the reaction
• 2. Predict the direction of a reaction mixture
• 3. Calculate the equilibrium concentrations from
  any initial concentrations

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Using the Equilibrium Constant

• The numerical value of the equilibrium constant
  for a reaction indicates the extent to which
  reactants are converted to products
• Large value for Kc gt 103 reaction proceeds
  essentially to 100 (mostly products)
• Small value for Kc lt 10-3 reaction proceeds
  hardly at all before equilibrium is reached
  (mostly reactants)
• If a reaction has an intermediate value of 10-3
  ltKclt 103
• a.  Appreciable concentrations of both reactants
  and products are present in the equilibrium
  mixture

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Judge the extent of the reaction

• 1. Kc gt 103 100 (mostly products)
• 2. Kc lt 10-3 (mostly reactants)
• 3. 10-3 ltKclt 103 both reactants and products

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Predicting the direction of Reaction

• Reaction Quotient Qc
•       1. Not necessarily equilibrium
  concentrations, at some time, t, snapshot of
  reaction
•       2. As time passes, Qc changes toward the
  value of Kc
•       3. When the equilibrium state is reached
  Qc Kc
•       4. Qc allows us to predict the direction
  of reaction by comparing the values of Kc and Qc
•        a) If Qclt Kc, net reaction goes
  from left to right, (reactant to products)
•       b) If Qc gt Kc, net reaction goes
  from right to left, (products to reactants)
•       c) If Qc Kc, no net reaction
  occurs

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Predicting the direction of Reaction

• H2(g) I2 (g) 2HI(g)
  Kc57.0 at 700K
• Suppose that we have a mixture of H2(g), I2 (g)
  HI(g) at 700K with
• H2t 0.1M
• I2t 0.20M
• HI 0.40M.
• Predict the direction of the reaction.

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Example 7

• The equilibrium constant for the reaction
•  
• 2 NO(g) O2(g) ? 2 NO2(g)
•  
• is 6.9 x 105 _at_ 500 K. A 5.0 L reaction vessel
  at this temperature was filled with 0.060 mol of
  NO, 1.0 mol O2, and 0.80 mol NO2.
• a) Is the reaction mixture at equilibrium? If
  not, in which direction does the net reaction
  proceed?
• b) What is the direction of the net reaction if
  the initial amounts are 5.0 x 10-3 mol of NO,
  0.20 mol of O2 and 4.0 mol of NO2?

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Factors that Alter the Composition of an
Equilibrium Mixture

• One of the principal goals of chemical synthesis
  is to maximize the conversion of reactants to
  products while minimizing the expenditure of
  energy.
• 1. Can be achieved if the reaction goes nearly
  to completion at mild temperatures and
  pressures.
• 2. If the equilibrium mixture is high in
  reactants and poor in products, the experimental
  conditions must be changed.
• 3. Several factors can be exploited to alter
  the composition of an equilibrium mixture.
• A.       The concentration of reactants or
  products
•      B. The pressure and volume
• C. The temperature

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Le Chateliers Principle

• If a stress is applied to a reaction mixture at
  equilibrium, net reaction occurs in the direction
  that relieves the stress
•       1. Stress means a change in the
  concentration, pressure, volume, or temperature
  that disturbs the original equilibrium
•       2. Reaction then occurs to change the
  composition of the mixture until a new state of
  equilibrium is reached
• 3. The direction that the reaction takes
  (reactants to products or products to reactants)
  is the one that reduces the stress

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Altering an Equilibrium Mixture Changes in
Concentration  

• In general, when an equilibrium is disturbed by
  the addition or removal of any reactant or
  product, Le Chateliers principle predicts that
• 1.  The concentration stress of an added
  reactant or product is relieved by net reaction
  in the direction that consumes the added
  substance
• 2. The concentration stress of a removed
  reactant or product is relieved by net reaction
  in the direction that replenishes the removed
  substance

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Example 8

• Consider the equilibrium for the water-gas shift
  reaction
•  
• CO(g) H2O(g) ? CO2(g) H2(g)
•  
• Use Le Chateliers principle to predict how the
  concentration of H2 will change and what
  direction the reaction will flow when the
  equilibrium is disturbed by
•  
• 1.      Adding CO
• 2.      Adding CO2
• 3.      Removing H2O
• 4.      Removing CO2
•  

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Altering an Equilibrium Mixture Changes in
Pressure and Volume

• In general Le Chateliers Principle predicts
  that
• 1. An increase in pressure by reducing the
  volume will bring about net reaction in the
  direction that decreases the number of moles of
  gas
• 2. A decrease in pressure by enlarging the
  volume will bring about net reaction in the
  direction that increases the number of moles of
  gas.

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Example 9

• Which direction will the reaction flow if the
  following equilibria is subjected to an increase
  in pressure by decreasing the volume?
• 1.      CO(g) H2O(g) ? CO2(g) H2(g)
•  
• 2.      2 CO(g) ? C(s) CO2(g)
•  
• 3.      N2O4(g) ? 2 NO2(g)

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Altering the Equilibrium Mixture Changes in
Temperature

• In general, the temperature dependence of the
  equilibrium constant depends on the sign of ?H?
  for the reaction
• 1.The equilibrium constant for an exothermic
  reaction (negative ?H?) decreases as the
  temperature increases
• 2. The equilibrium constant for an endothermic
  reaction (positive ?H?) increases as the
  temperature increases.
• 3. ?H? standard enthalpy of reaction, enthalpy
  change measured under standard conditions
• 4. Standard conditions most stable form of a
  substance at 1 atm pressure and at a specified
  temperature, usually 25?C 1 M concentration for
  all substances

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Altering the Equilibrium Mixture Changes in
Temperature

• Le Chateliers Principle says that if heat is
  added to an equilibrium mixture (increasing the
  temperature) net reaction occurs in the direction
  that relieves the stress of the added heat.
• 1.      For an endothermic reaction heat is
  absorbed by reaction in the forward direction.
  The equilibrium shifts to the right at the
  higher temperatures, Kc increases with increasing
  temperature
• 2.      For an exothermic heat is absorbed by
  net reaction in the reverse direction, so Kc
  decreases with temperature, and the reaction
  would flow to the left (reactants)

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Example 10

• When air is heated at very high temperatures in
  an automobile engine, the air pollutant nitric
  oxide is produced by the reaction
•  
• N2(g) O2(g) ? 2 NO(g) ?H? 180.5 kJ
•  
• 1. How does the equilibrium amount of NO vary
  with an increase in temperature?
• 2. What direction is the net reaction flowing?

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The Effect of a Catalyst on Equilibrium

• A catalyst increases the rate of a chemical
  reaction by making available a new, lower-energy
  pathway for conversion of reactants to products.
• 1.  Since the forward and reverse reaction pass
  through the same transition state, a catalyst
  lowers the activation energy for both
• 2.   The rates of the forward and reverse
  reactions increase by the same factor
• 3. Catalyst accelerates the rate at which
  equilibrium is reached
• 4. Catalyst does not affect the composition of
  the equilibrium mixture

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The Effect of a Catalyst on Equilibrium
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Example 11

• A platinum catalyst is used in automobile
  catalytic converters to hasten the oxidation of
  carbon monoxide
• 2 CO(g) O2(g) ? 2 CO2(g) ?H? -566
  kJ
•  
• Suppose that you have a reaction vessel
  containing an equilibrium mixture. Will the
  amount of CO increase, decrease, or remain the
  same when
• A platinum catalyst is added
• The temperature is increased
• The pressure is increased by decreasing the
  volume
• The pressure is increased by adding argon gas
• The pressure is increased by adding O2 gas

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The Link Between Chemical Equilibrium and
Chemical Kinetics

• A B ? C D
•  
• Assuming that the forward and reverse reactions
  occur in a single bimolecular step, elementary
  reactions, we can write the following rate laws
•  
• Rate of forward reaction kf A B
• Rate of reverse reaction kr C D
• When t0 C D 0
• As A and B are converted to C and D the rate of
  the forward reaction decreases and the rate of
  the reverse reaction is increasing, until they
  are equal, chemical equilibrium
•  
• kf A B kr C D

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The Link Between Chemical Equilibrium and
Chemical Kinetics

• kf C D
• kr A B
• The right side of this equation is the
  equilibrium constant expression for the forward
  reaction, which equals the equilibrium constant
  Kc
•  
• Kc C D
• A B
•  
• Therefore the equilibrium constant is simply the
  ratio of the rate constants for the forward and
  reverse reactions
•  
• Kc kf
• kr

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Example 12

• Nitric oxide emitted from the engines of
  supersonic transport planes can contribute to the
  destruction of stratospheric ozone
•  NO(g) O3(g) ? NO2(g) O2(g)
•  
• This reaction is highly exothermic (?E -200
  kJ), and its equilibrium constant Kc is 3.4 x
  1034 at 300 K
• Which rate constant is larger, kf or kr?
• The value of kf at 300 K is 8.5 x 106 M-1 s-1.
  What is the value of kr at the same temperature?
• A typical temperature in the stratosphere is 230
  K. Do the values of kf, kr, and Kc increase or
  decrease when the temperature is lowered from 300
  K to 230 K?

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• Calculating Equilibrium Concentrations
• The equilibrium constant Kc for the reaction
• H2(g) I2(g) 2HI
  (g)
• 57.0 at 700K. If 1.00mol of H2 is allowed to
  react with 1.00mol of I2 in a 10.0L reaction
  vessel at 700K, what are the concentrations of
  H2, I2 and HI at the equilibrium? What is the
  composition of the equilibrium mixture in moles.

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Calculating Equilibrium Concentrations (P.542)
What is the concentration of NO in an
equilibrium mixture of gaseous NO, O2 and NO2 at
500 K that contains 0.01M O2 and 0.05M NO2 ? At
500k, the Kc for the reaction 2NO(g) O2 (g)
2NO2 (g) is 690000
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• Example 3 The Equilibrium constant Kp for the
  following reaction at 1000k is 0.259
• FeO(s) CO(g) Fe(s) CO2
  (g)
• What are the equilibrium partial pressures of CO
  and CO2 at 1000K if the initial partial pressure
  are Pco 1.000atm and Pco2 0.500atm?

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