Chapter Fourteen

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Chapter Fourteen

• Chemical Equilibrium

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Contents

• Dynamic Nature of Equilibrium
• Equilibrium Constant Expression
• Modifying Equilibrium Constant Expressions
• Le Châteliers Principle
• 5. Some Illustrative Equilibrium Calculations

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1. Dynamic Nature Of Equilibrium

• Definitions
• Chemical equilibrium At the condition that the
  concentrations of reactants and products no
  longer changes with time, it is chemical
  equilibrium.
• Physical equilibrium The equilibrium between two
  phases, that is according to physical processes.
• Dynamic equilibrium The rate of forward and
  rate of reverse reactions are equal, called
  dynamic equilibrium.

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2) Dynamic Equilibrium Illustrated
Radioactive NaCl
Saturated NaCl(aq)
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3) Concentration vs. Time Graph for reversible
Reaction
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2. Equilibrium Constant Expression
1) Heuristic approach (trial and error)
They are constant
K is independent of initial concentrations!!
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2) Kinetic view of equilibrium

• In chemical kinetics view, the equilibrium
  constant of a reaction can be expressed as a
  ratio of the rate constants of forward and
  reverse reactions
• Kc (forward rate)/(reverse rate) kf/kr
• The expression of chemical equilibrium according
  to the coefficients of the balance equation,
  independent to the kinetic mechanism.

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Solution
(A) ratef kfAB2 rater krAB2 at
equilibrium, ratef rater
(B)
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Pressure equilibrium constant (KP, using partial
pressure, atm for p)
Conversion between KC and KP KP KC(RT)?n
KC(0.0821 T)?n ?n moles of gaseous Products -
moles of gaseous Reactants In this case ?n
(cd) (ab)
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• 4) Equilibria involving solids and liquids
• The equilibrium constant expression does not
  include terms of pure solid and liquid phases,
  i.e., pure solid and liquid their concentrations
  do not change.
• Examples
• C(s) H2O(g) CO(g) H2(g)

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3. Modifying Equilibrium Constant Expressions

• 1) General description
• When a chemical reaction with the equilibrium
  constant, Kc, is reversed, the reverse reaction
  has the equilibrium constant 1/Kc.
• When the coefficients of an equation are
  multiplied by a common factor n to produce a new
  equation, the original Kc value is raised to the
  power n to obtain the new equilibrium constant.
• When the equations for individual reactions are
  added to obtain an overall equation, their
  equilibrium constants are multiplied to obtain
  the equilibrium constant for the overall
  reaction.
• The balanced chemical equation must be written
  when citing a value for Kc.

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2) Necessity Of Equilibrium Constants

• A very large numerical value of K signifies that
  a reaction goes to completion.
• A very small numerical value of K signifies that
  the forward reaction occurs only to a very slight
  extent.
• An equilibrium constant expression applies only
  to a reversible reaction ( ) at equilibrium.

• Although a reaction may be thermodynamically
  favored, it may kinetically controlled.
• Thermodynamic equilibrium constant expression
  uses dimensionless unit known as activities
  (detailed in Chapter 17).

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3) The Reaction Quotient, Q

• For non-equilibrium (instantaneous) conditions,
  the expression having the same form as K is
  called the reaction quotient, Q.
• The Q value is not constant for a reaction, but
  is useful for predicting the direction in which a
  net change must occur to establish equilibrium.
• If Q lt K, a net change occurs in the forward
  direction, that is, from left to right.
• If Q gt K, a net change proceeds in the reverse
  direction, that is, from right to left.
• If Q K, the equilibrium is reached.

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Example 14.1 If the equilibrium concentrations of
Cl2 and COCl2 are the same at 395 C, find the
equilibrium concentration of CO in the reaction
Solution
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Solution (a)
(b)
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Solution (a)
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(b)
(c)
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Example
Given 2 NO(g) O2(g) 2 NO2(g) KC
4.67x1013 (at 298 K) 1. What is the equilibrium
constant at 298 K for following reaction? 2
NO2(g) 2 NO(g) O2(g) Ans
2. What is the equilibrium constant at 298 K for
following reaction? NO2(g) NO(g) ½
O2(g) Ans
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Example
Given
2 N2O(g) 1/2 O2(g) 2 NO(g) KC
1.7x1013 2 NO(g) O2(g) ? 2 NO2(g) KC
4.67x1013 What is the equilibrium constant (KC)
for the following reaction 2 N2O(g) 3/2 O2(g)
2 NO2(g) Solution 2 N2O(g) 1/2 O2(g)
2 NO(g) KC 1.7x1013 2 NO(g) O2(g)
2 NO2(g) KC 4.67x1013 2 N2O(g) 3/2 O2(g)
2 NO2(g) KC 1.7x1013x 4.67x1013 7.9
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Example 14.6
2HI(g) H2(g) I2(g) KC 1.84x102 In
which direction would a net change occur if the
initial conditions in the reaction were HI
H2 I2 1.00 M? Solution
Q gt KC, Shift from right to left
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4. Le Châteliers Principle

• 1) Definition
• When a change (concentration, temperature,
  pressure, or volume) is imposed on a system at
  equilibrium, the system responds by attaining a
  new equilibrium condition that minimizes the
  impact of the imposed change.

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• Changing The Amounts Of Reacting Species
• At equilibrium, Q K.
• If the concentration of one of the reactants is
  increased or one of the products is removed,
  resulted in Q lt K, the concentrations of all
  species must change to make Q once again Kc.
• If Q lt K, the reaction is pushed (shift) to the
  right.

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Example 14.7 Describe how the removal of a small
quantity of water affects the equilibrium CH3(CH2
)6CH2OH(soln) CH3COOH(soln)
CH3(CH2)6CH2OCOCH3(soln) H2O(soln)
H
Answer equilibrium shifts to the right.
Example Describe how the adding a small quantity
of CaCO3(s) affects the equilibrium
CaCO3(s) CaO(s) CO2(g)
KpP CO2
Answer No effect
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• Changing external pressure or volume in gaseous
  equilibria
• When the external pressure is increased (or
  system volume is reduced, i.e., concentration
  increased), an equilibrium shifts in the
  direction producing the smaller number of moles
  of gas.
• If there is no change in the number of moles of
  gas in a reaction, changes in external pressure
  (or system volume) have no effect on an
  equilibrium.
• Changes in gas pressure by adding inert gas in a
  constant volume, the partial pressure of the
  reactants and products do not change, no effect
  on the equilibrium.

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Example 14.8 An equilibrium mixture of SO2(g),
O2(g), and SO3(g) is transferred from a 1.00-L
flask to a 2.00-L flask. In which direction does
a net reaction proceed to restore equilibrium?
Solution equilibrium shifts to the right.
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4) Changing The Equilibrium Temperature

• The equilibrium constant (K) depend on
  temperature

• For endothermic reaction (?Ho gt 0), increase T,
  increase K, shift right for restoring the
  equilibrium.
• For exothermic reaction (?Ho lt 0), increase T,
  decrease K, shift left for restoring the
  equilibrium.

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Example 14.9 Is the amount of NO(g) formed from
given amounts of N2(g) and O2(g), greater at high
or low temperatures?
Solution Endothermic reaction, raising the
temperature, shift right, increasing
concentration of NO(g)
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5) Adding A Catalyst

• The role of a catalyst is to change the mechanism
  of a reaction to lower activation energy.
• Equilibrium is a function only of the states of
  the reactants and products, not the reaction
  path.
• Catalyst dose not effect the equilibrium.

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• Some Illustrative Equilibrium Calculations
• Determining the equilibrium constants
• Calculating equilibrium quantities from
  equilibrium constant

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Example 14.11 In a 10.0-L vessel at 1000 K, 0.250
mol SO2 and 0.200 mol O2 react to form 0.162 mol
SO3 at equilibrium. What is Kc, at 1000 K, for
the reaction that is shown here?
Solution
Initial
Equilibrium
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Example 14.12
Consider the reaction H2(g) I2(g)
2HI(g) KC 54.3 at 698 K If we start with
0.500 mol H2(g) and 0.500 mol I2(g) in a 5.25-L
vessel at 698 K, how many moles of each gas will
be present at equilibrium? Solution
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Example 14.13
Consider the reaction H2(g) I2(g)
2HI(g) KC 54.3 at 698 K If the initial
amounts are 0.800 mol H2 and 0.500 mol I2 in a
5.25-L vessel at 698 K. What will be the amounts
of reactants and products when equilibrium is
attained? Solution
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x 0.0874, not 0.179, because the I2 cannot be
greater than 0.0952 M, the initial concentration.
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End of Chapter 14