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```					                   Chemical Equilibrium

Concept of
Dynamic Equilibrium

Heterogeneous                                Homogeneous
Equilibrium Constant
Equilibrium                                  Equilibrium

Applications

Le Chatelier’s
Principle

Temperature        Reactants/Products     Pressure/Volume
11/25/2011
The Concept of Equilibrium

• Consider colorless frozen N2O4. At room temperature, it
decomposes to brown NO2:
N2O4(g)  2NO2(g).
• At some time, the color stops changing and we have a
mixture of N2O4 and NO2.
• Chemical equilibrium is the point at which the
concentrations of all species are constant.
The Concept of Equilibrium
The Concept of Equilibrium

• At equilibrium, as much N2O4 reacts to form NO2 as NO2
reacts to re-form N2O4:
N2O4(g)     2NO2(g)
• The double arrow implies the process is dynamic.
• Consider
Forward reaction: A  B Rate = kf[A]
Reverse reaction: B  A Rate = kr[B]
• At equilibrium kf[A] = kr[B].
NO2 – equilibrium
The Concept of Equilibrium
The Equilibrium Constant

• For a general reaction in the gas phase
aA + bB           cC + dD
the equilibrium constant expression is
c d
PC PD
Keq 
a b
PA PB

where Keq is the equilibrium constant.
The Equilibrium Constant

• For a general reaction
aA + bB       cC + dD
the equilibrium constant expression for everything in
solution is

K eq 
Cc Dd
A a Bb
where Keq is the equilibrium constant.
The Equilibrium Constant
N2O4(g)        2NO2(g)

2
P
K eq 
NO2

PN 2O 4
The Equilibrium Constant
The Magnitude of Equilibrium Constants
The Equilibrium Constant
The Direction of the Chemical Equation and Keq

• In the reverse direction:
2NO2(g)       N2O4(g)

P                 1
N 2O 4
K eq          0.155 
P2               6.46
NO 2
The Equilibrium Constant
Other Ways to Manipulate Chemical
Equations and Keq Values
• The reaction
2N2O4(g)          4NO2(g)
has
P4
NO 2
Keq 
P2
N 2O 4

which is the square of the equilibrium constant for
N2O4(g)         2NO2(g)
The Equilibrium Constant
Other Ways to Manipulate Chemical
Equations and Keq Values
• Equilibrium constant for the reverse direction is the
inverse of that for the forward direction.
• When a reaction is multiplied by a number, the
equilibrium constant is raised to that power.
• The equilibrium constant for a reaction which is the sum
of other reactions is the product of the equilibrium
constants for the individual reactions.
The Equilibrium Constant
Given at 700K:
(i) H2(g) + I2(g)  2 HI(g)                    Keq = 54.0
(ii) N2(g) + 3 H2(g)  2 NH3(g)                Keq’ = 1.04x10-4

Determine Keq’’ at 700K for:
2 NH3(g) + 3 I2(g)  6 HI(g) + N2(g)          Answer: 1.51x109

Unit(s) of the Equilibrium Constant
Dimensionless because:
 Partial Pressures divided by Reference Pressure of 1 atm.
 Molarities divided by Reference Concentration of 1 Molar.
Given at 700K:
(i) H2(g) + I2(g)  2 HI(g)            Keq = 54.0
(ii) N2(g) + 3 H2(g)  2 NH3(g)        Keq’ = 1.04x10-4
Determine Keq’’ at 700K for:
2 NH3(g) + 3 I2(g)  6 HI(g) + N2(g)

Heterogeneous Equilibria

• When all reactants and products are in one phase, the
equilibrium is homogeneous.
• If one or more reactants or products are in a different
phase, the equilibrium is heterogeneous.
• Consider:
CaCO3(s)        CaO(s) + CO2(g)
– experimentally, the amount of CO2 does not seem to depend on
the amounts of CaO and CaCO3. Why?
Heterogeneous Equilibria

• The concentration of a solid or pure liquid is its density
divided by molar mass.
• Neither density nor molar mass is a variable, the
concentrations of solids and pure liquids are constant.
• For the decomposition of CaCO3:
[CaO]
K eq            [CO 2 ]  constant  [CO 2 ]
[CaCO3 ]
Heterogeneous Equilibria

• We ignore the concentrations of pure liquids and pure
solids in equilibrium constant expressions.
• The amount of CO2 formed will not depend greatly on
the amounts of CaO and CaCO3 present.

• Examples:
CO 2 ( g )  H 2 ( g )  CO(g)  H 2 O(l)

Sn(s)  2 H  (aq)  Sn 2 (aq)  H 2 (g)
Calculating Equilibrium Constants

• Proceed as follows:
– Tabulate initial and equilibrium concentrations (or partial
pressures) given.
– If an initial and equilibrium concentration is given for a species,
calculate the change in concentration.
– Use stoichiometry on the change in concentration line only to
calculate the changes in concentration of all species.
– Deduce the equilibrium concentrations of all species.
• Usually, the initial concentration of products is zero.
(This is not always the case.)
Calculating Equilibrium Constants

1. A mixture of H2(g) and N2(g) is allowed to attain equilibrium at 472oC. At
equilibrium, the partial pressures of the gases were determined to be 7.38
atm H2, 2.46 atm N2 and 0.166 atm NH3. Calculate Keq .

2. Enough ammonia is dissolved in 5.00 L of water at 25oC to produce a
solution that is 0.0124 M in ammonia. The solution is then allowed to come
to equilibrium. Analysis of the equilibrium mixture shows that the
concentration of OH- is 4.64x10-4 M. Calculate Keq at 25oC for the reaction.

3. Gaseous sulfur trioxide decomposes to gases of sulfur dioxide and oxygen
at high temperature in a sealed container. Initially the vessel is charged at
1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium, the
SO3 partial pressure is 0.200 atm. Calculate the value of Keq at 1000 K.
A mixture of H2(g) and N2(g) is allowed to attain equilibrium at 472oC. At equilibrium,
the partial pressures of the gases were determined to be 7.38 atm H2, 2.46 atm N2
and 0.166 atm NH3. Calculate Keq .

Enough ammonia is dissolved in 5.00 L of water at 25oC to produce a solution that is 0.0124 M in
ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture
shows that the concentration of OH- is 4.64x10-4 M. Calculate Keq at 25oC for the reaction.

Gaseous sulfur trioxide decomposes to gases of sulfur dioxide and oxygen at high temperature in a
sealed container. Initially the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500
atm. At equilibrium, the SO3 partial pressure is 0.200 atm. Calculate the value of Keq at 1000 K.

Applications of Equilibrium Constants

Predicting the Direction of Reaction
• We define Q, the reaction quotient, for a general reaction
aA + bB          cC + dD
as
c d
PC PD
Q
a b
PA PB

• Q = K only at equilibrium.
Applications of Equilibrium Constants

Predicting the Direction of Reaction
• If Q > K then the reverse reaction must occur to reach
equilibrium (i.e., products are consumed, reactants are
formed, the numerator in the equilibrium constant
expression decreases and Q decreases until it equals K).
• If Q < K then the forward reaction must occur to reach
equilibrium.
Applications of Equilibrium Constants

Calculating Equilibrium Constants
•   The same steps used to calculate equilibrium constants
are used.
•   Generally, we do not have a number for the change in
concentration line.
•   Therefore, we need to assume that x mol/L of a species is
produced (or used).
•   The equilibrium concentrations are given as algebraic
expressions.
Applications of Equilibrium Constants

Predicting the Direction of Reaction

1. At 1000 K the value of Keq for the reaction 2SO3(g)  2SO2(g) + O2(g) is 0.338 .
Calculate the value for Q, and predict the direction in which the reaction will
proceed toward equilibrium if the initial partial pressures are 0.16 atm for SO3,
0.41 atm for SO2 , and 2.5 atm for O2 .

2. At 448oC, Keq is 51 for: H2(g) + I2(g)  2HI(g) . Predict how the reaction will
proceed to reach equilibrium at 448oC if we start with 2.0x10-2 mol HI , 1.0x10-2
mol H2 , and 3.0x10-2 mol I2 in a 2.00-L container.
[Hint: Calculate partial pressures using Ideal Gas Law. Calculate Q.]
[Answer: Q=1.33 Reaction proceeds from left to right (i.e. more products)]
Applications of Equilibrium Constants

Calculating Equilibrium Concentrations

1. At 500 K the reaction: PCl5(g)  PCl3(g) + Cl2(g) , has Keq = 0.497 . In an
equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that
of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium
mixture?

2. A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448oC. The
value of the Keq for the reaction: H2(g) + I2(g)  2HI(g) , at 448oC is 50.5.
What are the partial pressures of H2 , I2 , and HI in the flask at equilibrium?
[Hint: Solve for initial partial pressures using Ideal Gas law. Set up
initial/change/equil’m table and define x as change in partial pressures of
reactants. Solve for x in Keq expression using quadratic formulation.]
[Answer: x = 55.3 atm; H2 = 3.90 atm, I2 = 63.1 atm, HI = 111 atm.]
2.   A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448oC. The value of the Keq for the reaction:
H2(g) + I2(g)  2HI(g) , at 448oC is 50.5. What are the partial pressures of H2 , I2 , and HI in the flask at equilibrium?
[Hint: Solve for initial partial pressures using Ideal Gas law. Set up initial/change/equil’m table and define x as change in
partial pressures of reactants. Solve for x in Keq expression using quadratic formulation.]
[Answer: x = 55.3 atm; H2 = 3.90 atm, I2 = 63.1 atm, HI = 111 atm.]
Le Châtelier’s Principle
Consider the production of ammonia
N2(g) + 3H2(g)         2NH3(g)
• As the pressure increases, the amount of ammonia
present at equilibrium increases.
• As the temperature decreases, the amount of ammonia at
equilibrium increases.
• Can this be predicted?
Le Châtelier’s Principle: if a system at equilibrium is
disturbed, the system will move in such a way as to
counteract the disturbance.
Le Châtelier’s Principle
Change in Reactant or Product
Concentrations
• Consider the Haber process
N2(g) + 3H2(g)          2NH3(g)
• If H2 is added while the system is at equilibrium, the
system must respond to counteract the added H2 (by Le
Châtelier).
• The system must consume the H2 and produce products
until a new equilibrium is established.
• So, [H2] and [N2] will decrease and [NH3] increases.
Le Châtelier’s Principle
Change in Reactant or Product
Concentrations
•   Adding a reactant or product shifts the equilibrium away
from the increase.
•   Removing a reactant or product shifts the equilibrium
towards the decrease.
•   To optimize the amount of product at equilibrium, we
need to flood the reaction vessel with reactant and
continuously remove product (Le Châtelier).
•   We illustrate the concept with the industrial preparation
of ammonia.
Le Châtelier’s Principle
Effects of Volume and Pressure Changes
•   As volume is decreased pressure increases.
•   Le Châtelier’s Principle: if pressure is increased the
system will shift to counteract the increase.
•   That is, the system shifts to remove gases and decrease
pressure.
•   An increase in pressure favors the direction that has
fewer moles of gas.
•   In a reaction with the same number of product and
reactant moles of gas, pressure has no effect.
Le Châtelier’s Principle
Effects of Volume and Pressure Changes
N2O4(g)        2NO2(g)
• An increase in pressure (by decreasing the volume)
favors the formation of colorless N2O4.
• The instant the pressure increases, the system is not at
equilibrium and the concentration of both gases has
increased.
• The system moves to reduce the number moles of gas
(i.e. backward reaction is favored).
Le Châtelier’s Principle
Effect of Temperature Changes

• The equilibrium constant is temperature dependent.
• For an endothermic reaction, H > 0 and heat can be
considered as a reactant.
• For an exothermic reaction, H < 0 and heat can be
considered as a product.
Le Châtelier’s Principle
Effect of Temperature Changes

• Adding heat (i.e. heating the vessel) favors away from
the increase:
– if H > 0, adding heat favors the forward reaction,
– if H < 0, adding heat favors the reverse reaction.

• Removing heat (i.e. cooling the vessel), favors towards
the decrease:
– if H > 0, cooling favors the reverse reaction,
– if H < 0, cooling favors the forward reaction.
Le Châtelier’s Principle
Effect of Temperature Changes
• Consider
Cr(H2O)62+(aq) + 4Cl-(aq)                    CoCl42-(aq) + 6H2O(l)
for which H > 0.
– Co(H2O)62+ is pale pink and CoCl42- is blue.
– If a light purple room temperature equilibrium mixture is placed in a beaker
of warm water, the mixture turns deep blue.
– Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the
formation of blue CoCl42-.
– If the room temperature equilibrium mixture is placed in a beaker of ice
water, the mixture turns bright pink.
– Since H > 0, removing heat favors the reverse reaction which is the
formation of pink Co(H2O)62+.
Le Châtelier’s Principle
The Effect of Catalysis

• A catalyst lowers the activation energy barrier for the
reaction.
• Therefore, a catalyst will decrease the time taken to reach
equilibrium.
• A catalyst does not affect the composition of the
equilibrium mixture.
Chemical Equilibrium

Concept of
Dynamic Equilibrium

Heterogeneous                                  Homogeneous
Equilibrium Constant
Equilibrium                                    Equilibrium

c d
Keq 
PC PD           Applications
K eq 
Cc Dd
a b
PA PB                                     A a Bb
Le Chatelier’s
Principle

Temperature        Reactants/Products        Pressure/Volume

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