Download The Equilibrium Constant K

Chapter 13
Chemical Equilibrium
Section 13.1
The Equilibrium Condition
Chemical Equilibrium
 The state where the concentrations of all reactants
and products remain constant with time.
 On the molecular level, there is frantic activity.
Equilibrium is not static, but is a highly dynamic
situation.
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Section 13.2
The Equilibrium Constant
H2O+CO→H2 +CO2
Figure 13.2 shows the plot of the concentration of the reactants and products verses time.
CO and H2O are present in equal molar quantities. Since they react in 1:1 ratio, the
concentration of the two gases are always equal. Since H2 and CO2 are formed in equal
amount they are always present in the same concentration. Beyond certain time, indicated
by the dashed line in figure 13. 2. equilibrium has reaced.
Section 13.1
The Equilibrium Condition
Equilibrium Is:
 Macroscopically static
 Microscopically dynamic
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Section 13.1
The Equilibrium Condition
Changes in Concentration
N2(g) + 3H2(g)
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2NH3(g)
5
Section 13.1
The Equilibrium Condition
Chemical Equilibrium
 Concentrations reach levels where the rate of the
forward reaction equals the rate of the reverse
reaction. The system has reached equilibrium. We
will now consider the equilibrium phenomenon in
terms of the rates of opposing reactions.
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Section 13.1
The Equilibrium Condition
The Changes with Time in the Rates of Forward and Reverse
Reactions
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Section 13.1
The Equilibrium Condition
CONCEPT CHECK!
Consider an equilibrium mixture in a closed vessel
reacting according to the equation:
H2O(g) + CO(g)
H2(g) + CO2(g)
You add more H2O(g) to the flask. How does the
concentration of each chemical compare to its
original concentration after equilibrium is
reestablished? Justify your answer.
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Section 13.1
The Equilibrium Condition
CONCEPT CHECK!
Consider an equilibrium mixture in a closed vessel
reacting according to the equation:
H2O(g) + CO(g)
H2(g) + CO2(g)
You add more H2 to the flask. How does the
concentration of each chemical compare to its
original concentration after equilibrium is
reestablished? Justify your answer.
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Section 13.2
The Equilibrium Constant
Consider the following reaction at equilibrium: Guldberg and Waage proposed the law of
mass action as a general description of the equilibrium.
jA + kB
lC + mD
l
m
j
[A]
k
[B]
[C] [D]
K=




A, B, C, and D = chemical species.
Square brackets = concentrations of species at equilibrium.
j, k, l, and m = coefficients in the balanced equation.
K = equilibrium constant (given without units).
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Section 13.2
The Equilibrium Constant
Conclusions About the Equilibrium Expression
 Equilibrium expression for a reaction is the reciprocal of
that for the reaction written in reverse.
 When the balanced equation for a reaction is multiplied
by a factor of n, the equilibrium expression for the new
reaction is the original expression raised to the nth
power; thus Knew = (Koriginal)n.
 K values are usually written without units.
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Section 13.2
The Equilibrium Constant




K always has the same value at a given temperature regardless of the
amounts of reactants or products that are present initially.
For a reaction, at a given temperature, there are many equilibrium positions
but only one equilibrium constant, K.
Equilibrium concentrations will not always be the same.
The equilibrium constant which depends on the ratio of the concentrations,
remains the same.
Equilibrium position is a set of equilibrium concentrations.
There is only one equilibrium constant for a particular system at a particular
temperature, but there are an infinite number of equilibrium positions.
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Section 13.3
Equilibrium Expressions Involving Pressures





K involves concentrations.
Kp involves pressures.
PV = nRT or P=(n/V)RT = CRT,
C=n/V (the no. of moles of gas per unit volume V).
N2(g) + 3H2(g)
2NH3(g)

K = [NH3]2/[N2][H2]3

Kc = C(NH3)2/(CN2)(CH23)

Homogeneous Reactions: i) same no. of molecules on both sides of the equation
ii) Different no. of molecules on both sides of the equation
Heterogeneous Reactions: involves more than one phase.
Symbols K and Kc are used for equilibrium constant. K is used in this book.


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=
13
Section 13.3
Equilibrium Expressions Involving Pressures
Example
N2(g) + 3H2(g)
2NH3(g)
At 500 0C the value of K = 6.0 x10-2, regardless of the amounts of the gases that are
mixed to an equilibrium position.
P 

=
P P 
2
Kp
3
N
2
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H
NH3 
3
N2 H2 
2
NH
3
K =
2
14
Section 13.3
Equilibrium Expressions Involving Pressures
Example
N2(g) + 3H2(g)
2NH3(g)
Equilibrium pressures at a certain temperature:
PNH = 2.9  10 2 atm
3
PN = 8.9  10 1 atm
2
PH = 2.9  10 3 atm
2
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Section 13.3
Equilibrium Expressions Involving Pressures
Example
N2(g) + 3H2(g)
2NH3(g)
P 

=
P P 
2
Kp
NH
3
N
2
Kp =
 8.9
3
H
2
 2.9
 10
 10
1

2 2
 2.9
 10
Kp = 3.9  104
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
3 3
Section 13.3
Equilibrium Expressions Involving Pressures
The Relationship Between K and Kp
Kp = K(RT)Δn
 Δn = sum of the coefficients of the gaseous products
minus the sum of the coefficients of the gaseous
reactants.
 R = 0.08206 L·atm/mol·K
 T = temperature (in Kelvin)
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Section 13.3
Equilibrium Expressions Involving Pressures
Example
N2(g) + 3H2(g)
2NH3(g)
Using the value of Kp (3.9 × 104) from the previous
example, calculate the value of K at 35°C.
K p = K  RT 
n
3.9  10 = K  0.08206 L  atm/mol  K  308K 
4
K = 2.5  107
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 2 4 
Section 13.4
Heterogeneous Equilibria
Homogeneous Equilibria
 Homogeneous equilibria – involve the same phase:
N2(g) + 3H2(g)
2NH3(g)
HCN(aq)
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H+(aq) + CN-(aq)
19
Section 13.4
Heterogeneous Equilibria
Heterogeneous Equilibria
 Heterogeneous equilibria – involve more than one
phase:
2KClO3(s)
2KCl(s) + 3O2(g)
2H2O(l)
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2H2(g) + O2(g)
20
Section 13.4
Heterogeneous Equilibria
 The position of a heterogeneous equilibrium does not
depend on the amounts of pure solids or liquids present.
 The concentrations of pure liquids and solids are
constant.
2KClO3(s)
2KCl(s) + 3O2(g)
K =  O2 
3
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Section 13.5
Applications of the Equilibrium Constant
The Extent of a Reaction
 A value of K much larger than 1 means that at
equilibrium the reaction system will consist of mostly
products – the equilibrium lies to the right.
 Reaction goes essentially to completion.
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Section 13.5
Applications of the Equilibrium Constant
The Extent of a Reaction
 A very small value of K means that the system at
equilibrium will consist of mostly reactants – the
equilibrium position is far to the left.
 Reaction does not occur to any significant extent.
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Section 13.5
Applications of the Equilibrium Constant
CONCEPT CHECK!
If the equilibrium lies to the right, the value for K is
__________.
large (or >1)
If the equilibrium lies to the left, the value for K is
___________.
small (or <1)
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Section 13.5
Applications of the Equilibrium Constant
Reaction Quotient, Q
 Used when all of the initial concentrations are nonzero.
 Apply the law of mass action using initial concentrations
instead of equilibrium concentrations.
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Section 13.5
Applications of the Equilibrium Constant
Reaction Quotient, Q
 Q = K; The system is at equilibrium. No shift will occur.
 Q > K; The system shifts to the left.
 Consuming products and forming reactants, until
equilibrium is achieved.
 Q < K; The system shifts to the right.
 Consuming reactants and forming products, to attain
equilibrium.
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Section 13.5
Applications of the Equilibrium Constant
EXERCISE!
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)

Trial #1:
6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a certain
temperature and at equilibrium the concentration of
FeSCN2+(aq) is 4.00 M.
What is the value for the equilibrium constant for this reaction?
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Section 13.5
Applications of the Equilibrium Constant
Set up ICE Table
Fe3+(aq) + SCN–(aq)
FeSCN2+(aq)
Initial
Change
Equilibrium
6.00
10.00
0.00
– 4.00
– 4.00+4.00
2.00
6.00
4.00
FeSCN2 
4.00 M 

K =
=
3

Fe  SCN 
2.00 M 6.00 M 
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K = 0.333
28
Section 13.5
Applications of the Equilibrium Constant
EXERCISE!
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
 Trial #2:
Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq) (same temperature
as Trial #1)
Equilibrium:
? M FeSCN2+(aq)
5.00 M FeSCN2+
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Section 13.5
Applications of the Equilibrium Constant
EXERCISE!
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
 Trial #3:
Initial: 6.00 M Fe3+(aq) and 6.00 M SCN−(aq)
Equilibrium:
? M FeSCN2+(aq)
3.00 M FeSCN2+
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Section 13.6
Solving Equilibrium Problems
Solving Equilibrium Problems
1) Write the balanced equation for the reaction.
2) Write the equilibrium expression using the law of mass
action.
3) List the initial concentrations.
4) Calculate Q, and determine the direction of the shift to
equilibrium.
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Section 13.6
Solving Equilibrium Problems
Solving Equilibrium Problems
5) Define the change needed to reach equilibrium, and
define the equilibrium concentrations by applying the
change to the initial concentrations.
6) Substitute the equilibrium concentrations into the
equilibrium expression, and solve for the unknown.
7) Check your calculated equilibrium concentrations by
making sure they give the correct value of K.
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Section 13.6
Solving Equilibrium Problems
EXERCISE!
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
Fe3+
Trial #1 9.00 M
Trial #2 3.00 M
Trial #3 2.00 M
SCN5.00 M
2.00 M
9.00 M
FeSCN2+
1.00 M
5.00 M
6.00 M
Find the equilibrium concentrations for all species.
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Section 13.6
Solving Equilibrium Problems
EXERCISE!
Answer
Trial #1: [Fe3+] = 6.00 M; [SCN-] = 2.00 M; [FeSCN2+] = 4.00 M
Trial #2: [Fe3+] = 4.00 M; [SCN-] = 3.00 M; [FeSCN2+] = 4.00 M
Trial #3: [Fe3+] = 2.00 M; [SCN-] = 9.00 M; [FeSCN2+] = 6.00 M
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Section 13.6
Solving Equilibrium Problems
CONCEPT CHECK!
A 2.0 mol sample of ammonia is introduced into a
1.00 L container. At a certain temperature, the ammonia
partially dissociates according to the equation:
NH3(g)
N2(g) + H2(g)
At equilibrium 1.00 mol of ammonia remains.
Calculate the value for K.
K = 1.69
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Section 13.6
Solving Equilibrium Problems
CONCEPT CHECK!
A 1.00 mol sample of N2O4(g) is placed in a 10.0 L vessel and
allowed to reach equilibrium according to the equation:
N2O4(g)
2NO2(g)
K = 4.00 × 10-4
Calculate the equilibrium concentrations of: N2O4(g) and
NO2(g).
Concentration of N2O4 = 0.097 M
Concentration of NO2 = 6.32 × 10-3 M
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Section 13.7
Le Châtelier’s Principle
 If a change is imposed on a system at equilibrium, the
position of the equilibrium will shift in a direction that
tends to reduce that change.
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Section 13.7
Le Châtelier’s Principle
Effects of Changes on the System
1.
2.
Concentration: The system will shift away from the added component. If a component
is removed, the opposite effect occurs.
Temperature: K will change depending upon the temperature (endothermic – energy of
a reactant; exothermic – energy of a product).
N2+H2→2NH3 + 92 KJ
Exothermic reaction, and the equilibrium shift to the left. This shift decreases the
concentration of ammonia and increases the concentration of nitrogen and hydrogen.
It decreases the value of K.
556 KJ + CaCO3(s) →CaO(s) +CO2(g)
Endothermic reaction, and the equilibrium shift to the right. This shift increases the
concentration CaO and CO2 and decreases the concentration of CaCO3.
It increases the value of K.
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Section 13.7
Le Châtelier’s Principle
Effects of Changes on the System
3.
Pressure:
a) The system will shift away from the added gaseous component. If a component is
removed, the opposite effect occurs.
b) Addition of inert gas does not affect the equilibrium position. The addition of inert
gas increases the total pressure, but has no effect on the concentration or the
partial pressure of the reactant or product. Hence the added molecule do not
participate the in the reaction and cannot effect the equilibrium
c) Decreasing the volume shifts the equilibrium toward the side with fewer moles of
gas. In N2+3H2→2NH3 system will shift to the right, reducing the total number of
gaseous molecules present. The opposite is also true. If we increase the volume
system will shift to increase the volume . An increase in in volume in ammonia
synthesis system will produce a shift to the left to increase the total number of
gaseous molecules present.
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Section 13.7
Le Châtelier’s Principle
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Section 13.7
Le Châtelier’s Principle
Equilibrium Decomposition of N2O4
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