Chapter 16
Principles of Chemical
Reactivity: Equilibria
Jeffrey Mack
California State University,
Sacramento
Chemical Equilibrium: A Review
• All chemical reactions are reversible, at least in
principle.
• The concept of equilibrium is fundamental to
chemistry.
• The general concept of equilibrium was introduced
in Chapter 3 to explain the limited dissociation of
weak acids.
• The goals of this and the following chapter will be to
consider chemical equilibria in quantitative terms.
• The extent to which that equilibrium lies (product
favored verses reactant favored) will be discussed.
Equilibrium
• At some point in time during the
progress of a reaction, if the
concentration of the reactants and
products remains constant,
equilibrium is said to be achieved.
• The concentrations are NOT equal.
Equilibrium
Moving
towards
equilibrium
Equilibrium
established
Properties of Chemical Equilibria
Equilibrium systems are said to
be:
• Dynamic (in constant motion)
• Reversible
• Equilibrium can be approached
from either direction.
Pink to blue
Co(H2O)6Cl2  Co(H2O)4Cl2 + 2 H2O

Blue to pink
Co(H2O)4Cl2 + 2 H2O  Co(H2O)6Cl2

Chemical Equilibrium
Fe3+(aq) + SCN (aq)  Fe(SCN)2+ (aq)
• After a period of time, the concentrations of reactants
and products are constant.
• The forward and reverse reactions continue after
equilibrium is attained.
• They are equal and opposite in rate.
Examples of Chemical Equilibria
Phase changes such as
H2O(s) vs. H2O(liq)
The Equilibrium Constant
When a general chemical reaction
aA  bB
cC  dD
is at equilibrium, the equilibrium constant is given by:
c
d
[C] [D]
K=
[A]a [B]b
If K > 0 then the reaction is said to be product
favored.
If K < 0 then the reaction is said to be reactant
favored.
The Equilibrium Constant
Product-favored
K>1
Reactant-favored
K<1
The Equilibrium Constant
For the formation of HI(g) the
equilibrium constant is given
by:
2
[HI]
K=
[H2 ][I2 ]
N2O4 (g)
2NO2 (g)
equilibrium
equilibrium
Start with N2O4
Start with NO2 & N2O4
equilibrium
Start with NO2
11
Disturbing a Chemical Equilibrium
The equilibrium between reactants and products may
be disturbed in three ways:
(1) by changing the temperature
(2) by changing the concentration of a reactant
(3) by changing the volume (for systems involving
gases)
A change in any of these factors will cause a system
at equilibrium to shift back towards a state of
equilibrium.
This statement is often referred to as Le Chatelier’s
principle.
Disturbing a Chemical Equilibrium
Effect of the Addition or Removal of a
Reactant or Product
• If the concentration of a reactant or product is
changed from its equilibrium value at a given
temperature, equilibrium will be reestablished
eventually.
• The new equilibrium concentrations of
reactants and products will be different, but
the value of the equilibrium constant
expression will still equal K
Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the
system adjusts in such a way that the stress is partially offset
as the system reaches a new equilibrium position.
• Changes in Concentration
N2 (g) + 3H2 (g)
2NH3 (g)
Equilibrium
shifts left to
offset stress
Add
NH3
14
Le Châtelier’s Principle
• Changes in Concentration continued
Remove
Add
Remove
Add
aA + bB
cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
Decrease concentration of product(s)
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
left
right
right
left
15
Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase
Equilibria
• For a reaction that involves gases, what happens to
equilibrium concentrations or pressures if the size of
the container is changed? (Such a change occurs,
for example, when fuel and air are compressed in
an automobile engine.)
• To answer this question, recall that concentrations
are in moles per liter. If the volume of a gas
changes, its concentration therefore must also
change, and the equilibrium composition can
change.
Le Châtelier’s Principle
• Changes in Volume and Pressure
A (g) + B (g)
C (g)
Change
Shifts the Equilibrium
Increase pressure
Decrease pressure
Increase volume
Decrease volume
Side with fewest moles of gas
Side with most moles of gas
Side with most moles of gas
Side with fewest moles of gas
• When the number of gas moles on either side is
the same, there is no effect.
17
Disturbing a Chemical Equilibrium
Effect of Temperatue Changes on Gas-Phase
Equilibria
Consider the reaction of nitrogen and oxygen to form
nitric oxide:
N2 (g)  O2 (g)
2NO(g)
rHo  180.6 kJ
As the temperature of the reaction is increased, the
equilibrium constant increases.
K
4.5  1031
6.7  1010
1.7  103
Temperature (K)
298
900
2300
Why?
Disturbing a Chemical Equilibrium
Effect of Temperature Changes on Gas-Phase
Equilibria
Lets write the reaction in this manner:
N2 (g)  O2 (g)  180.6 kJ
2NO(g)
Notice that energy is included as a reactant!
Disturbing a Chemical Equilibrium
Effect of Temperature Changes on Gas-Phase
Equilibria
Lets write the reaction in this manner:
N2 (g)  O2 (g)  180.6 kJ
2NO(g)
As temperature (Energy) is increased, equilibrium
shifts to the right, favoring products.
Disturbing a Chemical Equilibrium
Effect of Temperatue Changes on Gas-Phase
Equilibria
Lets write the reaction in this manner:
N2 (g)  O2 (g)  180.6 kJ
2NO(g)
As the concentration of products increases so does
the value of the equilibrium constant.
Disturbing a Chemical Equilibrium
Effect of Temperatue Changes on Gas-Phase
Equilibria
This explains the increase in the equilibrium constant
with increasing temperature.
NO]
[
K=
[N2 ][O2 ]
2
K
4.5  1031
6.7  1010
1.7  103
Temperature (K)
298
900
2300
Disturbing a Chemical Equilibrium
Effect of Temperature Changes on Gas-Phase
Equilibria
Conclusion:
• Increasing the temperature of an endothermic
reaction favors the products, equilibrium shifts to
the right.
• Increasing the temperature of an exothermic
reaction favors the reactants, equilibrium shifts to
the left.
• Lowering temperature results in the reverse effects.
Temperature Effects on Equilibrium
N2O4 (colorless) + heat
rHo = + 57.2 kJ
Kc =
[NO2 ]2
[N2O4 ]
Kc (273 K) = 0.00077
Kc (298 K) = 0.0059
2 NO2 (brown)
Le Châtelier’s Principle
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift
25
equilibrium.
Le Chatelier’s Principle Practice
N2 (g) + 3H2 (g)
K = 3.5  108
2NH3 (g)
rxnHo  91.8 kJ
Change
Adding more N2(g)
Removing H2(g)
Decreasing the container volume
Increasing the container temperature
Increasing the container volume
Decreasing the container temperature
adding a catalyst
adding argon to the container
Reaction Shift
Right
Left
Right
Left
Left
Right
no effect
no effect
Le Châtelier’s Principle - Summary
Change Equilibrium
Constant
no
Change
Shift Equilibrium
Concentration
yes
Pressure
yes*
no
Volume
yes*
no
Temperature
yes
yes
Catalyst
no
no
*Dependent on relative moles of gaseous reactants and products
27
Writing Equilibrium Constant
Expressions
• In an equilibrium constant expression, all
concentrations are reported as equilibrium
values.
• Product concentrations appear in the numerator,
and reactant concentrations appear in the
denominator.
• Each concentration is raised to the power of its
stoichiometric balancing coefficient.
• Values of K are dimensionless.
• The value of the constant K is particular to the
given reaction at a specific temperature.
Writing Equilibrium Constant
Expressions
Reactions Involving Solids
• So long as a solid is present in the course of
a reaction, its concentration is not included in
the equilibrium constant expression.
S(s)  O2 (g)
• Equilibrium constant:
[SO2 ]
K=
[O2 ]
SO2 (g)
Writing Equilibrium Constant
Expressions
Reactions in Solution
• If water is a participant in the chemical
reaction, its concentration based on
magnitude is considered to remain constant
throughout.
NH3 (aq)  H2O(l)

4

NH (aq)  OH (aq)
• Equilibrium constant:
+
[NH4 ][OH ]
K=
[NH3 ]
Writing Equilibrium Constant
Expressions
Reactions Involving Gases: Kc and Kp
• Concentration data can be used to calculate
equilibrium constants for both aqueous and
gaseous systems.
• In these cases, the symbol K is sometimes given
the subscript “c” for “concentration,” as in Kc.
• For gases, however, equilibrium constant
expressions can be written in another way: in
terms of partial pressures of reactants and
products.
Writing Equilibrium Constant
Expressions
Reactions Involving Gases: Kp
H2 (g)  I2 (g)
2HI(g)
2
HI
P
Kp =
PH2 ´ PI2
• Notice that the basic form of the equilibrium
constant expression is the same as for Kc. In some
cases, the numerical values of Kc and Kp are the
same. They are different when the numbers of
moles of gaseous reactants and products are
different.
Writing Equilibrium Constant
Expressions
Reactions Involving Gases: Kp & Kc
K p = K c (RT )
Dn
R = the gas constant
T = the absolute temperature
n = (mols gas product)  mols gas reactant)
• The general relationship between Kp and Kc is
derived in chapter 26, pa726.
• When the number of gas mole is equivalent on
either side of the chemical equation, the two
equilibrium constants are the same value.
The Reaction Quotient, Q
In general, ALL reacting chemical systems are
characterized by their REACTION QUOTIENT,
Q.
aA  bB
cC  dD
If Q = K, then system is at equilibrium.
The Reaction Quotient, Q
• If Q < K then the system is heading towards
equilibrium: There are more reactants than
products as expected at equilibrium. The
reaction is said to be headed to the “right”.
• If Q > K the system has gone past
equilibrium. There are more products than
reactants as expected at equilibrium. The
reaction is said to be headed to the left.
• If Q = K then the system is at equilibrium.
K=
[C]c[D]d
aA + bB
cC + dD
[A]a[B]b
Equilibrium Will
K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
36
The equilibrium concentrations for the reaction between carbon
monoxide and molecular chlorine to form COCl2 (g) at 740C are
[CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate
the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g)
COCl2 (g)
[COCl2]
0.14
=
= 220
Kc =
[CO][Cl2]
0.012 x 0.054
Kp = Kc(RT)n
n = 1 – 2 = -1
R = 0.0821
T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
37
Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s)
[CaO][CO2]
Kc′ =
[CaCO3]
Kc = [CO2] = Kc′ x
[CaCO3]
[CaO]
CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kp = PCO2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
38
Determining the Equilibrium
Constant
• The value of a reaction’s equilibrium
constant is determined by measuring the
concentrations of the reactants and products
when a system is at equilibrium.
• The equilibrium constant can also determined
by looking at the changes in concentrations
as a system achieves equilibrium.
• This is know as an “ICE” table.
Determining the Equilibrium
Constant
•
•
•
•
ICE tables:
I = Initial concentration
C = change in concentration
E = concentrations at equilibrium
Reactants
I
C
E
Products
Determining K
2NOCl(g)
2NO(g) + Cl2 (g)
2.00 mol of NOCl is added to 1.00 L flask. At
equilibrium the concentration of NO(g) is found to be
0.66 mol/L. What is the value of K?
Determining K
2NOCl(g)
2NO(g) + Cl2 (g)
2.00 mol of NOCl is added to 1.00 L flask. At
equilibrium the concentration of NO(g) is found to be
0.66 mol/L. What is the value of K?
[NOCl]
Initial
Change
Equilibrium
[NO]
[Cl2]
Determining K
2NOCl(g)
2NO(g) + Cl2 (g)
2.00 mol of NOCl is added to 1.00 L flask. At
equilibrium the concentration of NO(g) is found to be
0.66 mol/L. What is the value of K?
Initial
Change
Equilibrium
[NOCl]
2.00
[NO]
0
[Cl2]
0
Determining K
2NOCl(g)
2NO(g) + Cl2 (g)
2.00 mol of NOCl is added to 1.00 L flask. At
equilibrium the concentration of NO(g) is found to be
0.66 mol/L. What is the value of K?
Initial
Change
[NOCl]
2.00
[NO]
0
[Cl2]
0
-0.66
+0.66
0.33
Equilibrium
note the reaction
stoichiometry
Determining K
2NOCl(g)
2NO(g) + Cl2 (g)
2.00 mol of NOCl is added to 1.00 L flask. At
equilibrium the concentration of NO(g) is found to be
0.66 mol/L. What is the value of K?
[NOCl]
2.00
[NO]
0
[Cl2]
0
Change
-0.66
+0.66
+0.33
Equilibrium
1.34
0.66
0.33
Initial
Determining K
2NOCl(g)
2NO(g) + Cl2 (g)
[NOCl]
2.00
[NO]
0
[Cl2]
0
Change
-0.66
+0.66
+0.33
Equilibrium
1.34
0.66
0.33
Initial
[NO]2 [Cl2 ]
K=
[NOCl]2
(0.66)2 (0.33)
=
= 0.080
2
(1.34)
Equilibrium Concentrations from K
H2 (g)  I2 (g)
2HI(g)
1.00 mol each of H2 and I2 in a 1.00 L flask. What are
the equilibrium concentrations of all species. Kc =
55.3?
Equilibrium Concentrations from K
H2 (g)  I2 (g)
2HI(g)
1.00 mol each of H2 and I2 in a 1.00 L flask. What are
the equilibrium concentrations of all species. Kc =
55.3?
Initial
Change
Equilibrium
[H2]
1.00
[I2]
1.00
[HI]
0
Equilibrium Concentrations from K
H2 (g)  I2 (g)
2HI(g)
1.00 mol each of H2 and I2 in a 1.00 L flask. What are
the equilibrium concentrations of all species. Kc =
55.3?
Initial
Change
Equilibrium
[H2]
1.00
[I2]
1.00
[HI]
0
-x
-x
+ 2x
Equilibrium Concentrations from K
H2 (g)  I2 (g)
2HI(g)
1.00 mol each of H2 and I2 in a 1.00 L flask. What are
the equilibrium concentrations of all species. Kc =
55.3?
Initial
Change
Equilibrium
[H2]
1.00
[I2]
1.00
[HI]
0
-x
-x
+ 2x
1.00 - x
1.00 - x
+2x
Where x is defined as amount of H2 and I2
consumed on approaching equilibrium in moles.
Equilibrium Concentrations from K
H2 (g)  I2 (g)
2HI(g)
1.00 mol each of H2 and I2 in a 1.00 L flask. What are
the equilibrium concentrations of all species. Kc =
55.3?
Initial
Change
Equilibrium
2
[HI]
Kc =
[H2 ] ´ [I2 ]
[H2]
1.00
[I2]
1.00
[HI]
0
-x
-x
+ 2x
1.00 - x
1.00 - x
+2x
Equilibrium Concentrations from K
H2 (g)  I2 (g)
2HI(g)
1.00 mol each of H2 and I2 in a 1.00 L flask. What are
the equilibrium concentrations of all species. Kc =
55.3?
Initial
Change
Equilibrium
2
[H2]
1.00
[I2]
1.00
[HI]
0
-x
-x
+ 2x
1.00 - x
1.00 - x
+2x
2
[HI]
(2x)
Kc =
=
= 55.3
[H2 ] ´ [I2 ] (1.00 - x) ´ (1.00 - x)
Equilibrium Concentrations from K
1.00 mol each of H2 and I2 in a 1.00 L flask. What are
the equilibrium concentrations of all species. Kc =
55.3?
[HI]2
(2x)2
Kc =
=
= 55.3
[H2 ] ´ [I2 ] (1.00 - x) ´ (1.00 - x)
2x
= 55.3
1.00 - x
x = 0.788
[H2 ] = [I2 ] = 0.21 M
[HI] = 2x = 1.58 M
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
Initial
Change
Equilibrium
[N2O4]
0.50
[NO2]
0
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
Initial
Change
Equilibrium
[N2O4]
0.50
[NO2]
0
-x
+2x
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
Initial
Change
Equilibrium
[N2O4]
0.50
[NO2]
0
-x
+2x
0.50 - x
2x
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
Initial
Change
Equilibrium
[N2O4]
0.50
[NO2]
0
-x
+2x
0.50 - x
2x
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
4x + 0.0059x - 0.0030 = 0
2
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
4x + 0.0059x - 0.0030 = 0
2
Solving this requires the Quadratic Equation:
ax 2 + bx + c = 0
x =
-b ±
b 2 - 4ac
2a
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
x 
0.0059 
(0.0059)2  4  (4)  (-0.0030)
2  (4)
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
x =
x =
-0.0059 ±
(0.0059)2 - 4 ´ (4) ´ (-0.0030)
2 ´ (4)
0.046
- 0.00074 ±
= -0.00074 ± 0.027
8
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
x = 0.027 or 0.028
The negative value is not reasonable which gives the
equilibrium values of concentration to be:
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
x = 0.027 or 0.028
The negative value is not reasonable which gives the
equilibrium values of concentration to be:
x = 0.027 M
[N2O4] = 0.50 M  x = 0.47 M
[NO2] = 2x = 0.054 M
Equilibrium Concentrations from K
Consider the following reaction: N2O4 (g) 2NO2 (g)
If initial concentration of N2O4 is 0.50 M, what are the
equilibrium concentrations?
x = 0.027 or 0.028
The negative value is not reasonable which gives the
equilibrium values of concentration to be:
x = 0.027 M
[N2O4] = 0.50 M  x = 0.47 M
[NO2] = 2x = 0.054 M
The results are in agreement with the magnitude of K: The
reactant concentration is favored due to the small value of K
Approximations in Equilibrium
Concentrations
• What happens when the change in
concentration for a reaction is much less than
the initial concentration of the reactants?
• It turns out that when K  100 < [A]0, then the
quadratic equation is not required.
A
C][B] (x )(x )
[
K=
=
[A ] [A ]0 - x
C+B
reduces to
since [A]0  x  [A]0
(x )(x )
[A ]0
Approximations in Equilibrium
Concentrations
• Let’s look at what happens to the previous
problem if the initial concentration of N2O4(g)
is doubled to 1.00 M.
Approximations in Equilibrium
Concentrations
Consider the following reaction: N2O4 (g)
2NO2 (g)
If initial concentration of N2O4 is 1.00 M, what are the
equilibrium concentrations?
K c = 0.0059
K c ´ 100 = 0.59 << 1.00
Approximations in Equilibrium
Concentrations
Consider the following reaction:
If initial concentration of N2O4 is 1.00 M, what are the
equilibrium concentrations?
Solving for the equilibrium concentration using the
quadratic equation yields:
Approximations in Equilibrium
Concentrations
Consider the following reaction:
If initial concentration of N2O4 is 1.00 M, what are the
equilibrium concentrations?
Solving for the equilibrium concentration using the
quadratic equation yields:
x = 0.038 M, [N2O4] = 0.096 & [N2O] = 0.076
Approximations in Equilibrium
Concentrations
Consider the following reaction:
If initial concentration of N2O4 is 1.00 M, what are the
equilibrium concentrations?
Solving for the equilibrium concentration using the
approximation, ([N2O4]0  x) = [N2O4]0 also yields:
x = 0.038 M, [N2O4] = 0.096 & [N2O] = 0.076
The two results are in agreement.
More About Balanced Equations &
Equilibrium Constants
Multiplying the coefficients of a reaction:
3
O2 (g)
2
vs.
2S(s) + 3O2 (g)
S(s) +
SO3 (g)
2SO3 (g)
The second reaction is 2 the first.
[SO3 ]
K1 =
[O2 ]3 / 2
[SO3 ]2
2
K2 =
= (K1 )
3
[O2 ]
The general relationship is:
K new = (K old )
n
When “n” is the multiplication factor.
More About Balanced Equations &
Equilibrium Constants
Reversing a reaction
2S(s) + 3O2 (g)
2SO3 (g)
vs.
2SO3 (g)
2S(s) + 3O2 (g)
The second reaction is the reverse the first.
[SO3 ]2
K1 =
[O2 ]3
[O2 ]3
-1
K2 =
= (K1 )
2
[SO3 ]
The general relationship is:
K new = (K old )
-1
The equilibrium constant is the inverse of the original.
More About Balanced Equations &
Equilibrium Constants
Adding reactions at equilibrium:
S(s) + O2 (g)
SO2 (g)
1
SO2 (g) + O2 (g)
2
Net.
[SO2 ]
K1 =
[O2 ]
3
S(s) + O2 (g)
2
K2 =
[SO3 ]
[SO2 ][O2 ]
1
2
SO3 (g)
SO3 (g)
K net =
[SO3 ]
3
[O2 ]
2
= K1 ´ K 2
K new = K1 ´ K 2 ´ K 3 ...
The new equilibrium constant is the product of the individual.
Butane-Isobutane
Equilibrium
butane
[isobutane]
K=
= 2.5
[butane]
isobutane
Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] =
0.50 M. K = 2.50
If 1.50 M butane is added to the system, what is the
mew concentration of each when the system returns
to equilibrium?
Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
Butane
Isobutane K = 2.50
If 1.50 M butane is added to the system, what is the new
concentration of each when the system returns to equilibrium?
Let’s first calculate the value of Q to see which direction the
reaction will take to reestablish equilibrium.
Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
Butane
Isobutane K = 2.50
If 1.50 M butane is added to the system, what is the new
concentration of each when the system returns to equilibrium?
Let’s first calculate the value of Q to see which direction the
reaction will take to reestablish equilibrium.
isobutane]
1.25
[
Q=
=
= 0.625
[butane] 0.50 + 1.50
Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
Butane
Isobutane K = 2.50
If 1.50 M butane is added to the system, what is the new
concentration of each when the system returns to equilibrium?
Let’s first calculate the value of Q to see which direction the
reaction will take to reestablish equilibrium.
isobutane]
1.25
[
Q=
=
= 0.625
[butane] 0.50 + 1.50
Q < K: There are more reactants than products as expected at
equilibrium. The reaction shift to the “right”. Reactants must
react to form products.
Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
Butane
Isobutane K = 2.50
If 1.50 M butane is added to the system, what is the new
concentration of each when the system returns to equilibrium?
Let’s now calculate the changes in concentration by setting up
an “ICE” table.
[butane]
Initial
Change
Equilibrium
[isobutane]
Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
Butane
Isobutane K = 2.50
If 1.50 M butane is added to the system, what is the new
concentration of each when the system returns to equilibrium?
Let’s now calculate the changes in concentration by setting up
an “ICE” table.
Initial
Change
Equilibrium
[butane]
0.5 +1.50
[isobutane]
1.25
Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
Butane
Isobutane K = 2.50
If 1.50 M butane is added to the system, what is the new
concentration of each when the system returns to equilibrium?
Let’s now calculate the changes in concentration by setting up
an “ICE” table.
Initial
Change
Equilibrium
[butane]
0.5 +1.50
[isobutane]
1.25
x
+x
Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
Butane
Isobutane K = 2.50
If 1.50 M butane is added to the system, what is the new
concentration of each when the system returns to equilibrium?
Let’s now calculate the changes in concentration by setting up
an “ICE” table.
Initial
Change
Equilibrium
[butane]
0.5 +1.50
[isobutane]
1.25
x
+x
2.00  x
1.25 + x
Butane-Isobutane Equilibrium
Butane
Initial
Change
Equilibrium
Isobutane K = 2.50
[butane]
0.5 +1.50
[isobutane]
1.25
x
+x
2.00  x
1.25 + x
Using the value of K, determine the change in concentration,
“x”.
Butane-Isobutane Equilibrium
Butane
Initial
Change
Equilibrium
Isobutane K = 2.50
[butane]
0.5 +1.50
[isobutane]
1.25
x
+x
2.00  x
1.25 + x
Using the value of K, determine the change in concentration,
“x”.
isobutane] 1.25 - x
[
K=
=
= 2.5
[butane] 2.00 - x
Butane-Isobutane Equilibrium
Butane
Initial
Change
Equilibrium
Isobutane K = 2.50
[butane]
0.5 +1.50
[isobutane]
1.25
x
+x
2.00  x
1.25 + x
Using the value of K, determine the change in concentration,
“x”.
isobutane] 1.25 - x
[
K=
=
= 2.5
[butane] 2.00 - x
X = 1.07
Butane-Isobutane Equilibrium
Butane
Initial
Change
Equilibrium
Isobutane K = 2.50
[butane]
0.5 +1.50
[isobutane]
1.25
x
+x
2.00  x
1.25 + x
The new concentrations for butane and isobutane are:
[butane] = 0.93 M
[isobutane] = 2.32 M
Equilibrium is shifted to the right, forming more products.
Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase
Equilibria
Consider the following reaction:
2NO2 (g)
N2O4 (g)
K = 1.70  10
2
@298K
At Equilibrium, the concentrations are:
[N2O4] = 0.0280 M & [NO2] = 0.0128 M
What would happen if the total volume of the system
was suddenly doubled?
Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase
Equilibria
What would happen if the total volume of the system was
suddenly doubled?
2NO2 (g)
N2O4 (g)
K = 1.70  102 @298K
½ [N2O4] = ½  0.0280M = 0.0140M
½ [NO2] = ½  0.0128M = 0.00640M
Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase
Equilibria
What would happen if the total volume of the system was
suddenly doubled?
2NO2 (g)
N2O4 (g)
K = 1.70  102 @298K
½ [N2O4] = ½  0.0280M = 0.0140M
½ [NO2] = ½  0.0128M = 0.00640M
Calculate Q:
N2O4 ] (0.0140)
[
Q=
=
= 342
2
2
[NO2 ] (0.00640)
Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase
Equilibria
What would happen if the total volume of the system was
suddenly doubled?
2NO2 (g)
N2O4 (g)
K = 1.70  102 @298K
Q = 342 > K: Therefore some products must shift to reactants
(left) to reestablish equilibrium.
2NO2 (g)
N2O4 (g)
¬¾¾¾¾¾¾¾¾¾¾¾¾¾¾
Increase the volume of the container
N2O4 must be converted to NO2 to reach equilibrium
Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase
Equilibria
Conclusion:
• Increasing the volume of a container favors the side
of equilibrium with the greatest number of gas
moles.
• Decreasing the volume favors the side with the
least number of moles.
• When the number of gas moles on either side is the
same, there is no effect.
Chemical Kinetics and Chemical Equilibrium
A + 2B
kf
kr
ratef = kf [A][B]2
AB2
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2]
kf
[AB2]
= Kc =
kr
[A][B]2
94
Modifying the Chemical Equation (cont’d)
Consider the reaction:
2 NO(g) + O2(g)
2 NO2(g)
[NO2]2
Kc = –––––––––
=
4.67
x 1013 (at 298 K)
[NO]2 [O2]
Now consider the reaction:
NO2(g)
NO(g) + ½ O2(g)
What will be the equilibrium constant K"c for the new reaction?
[NO] [O2]1/2
1 1/2
K"c = –––––––––
= –––
[NO2]2
Kc
=
2.14 x 10–14 = 1.46 x 10–7
95
•
Calculate K of reversed reaction, ½ of a reaction or doubled
Example:
2SO2(g) + O2(g)  2SO3(g)
4SO2(g) + 2O2(g)  4SO3(g)
SO2(g) + ½ O2(g)  SO3(g)
2SO3(g)  O2(g) + 2SO2(g)
Example: 2CO(g) + O2(g)  2CO2(g) K = 2.75 x 1020 @1000K
CO2(g)  CO(g) + ½ O2(g) K = 6.03 x 10-11 @1000K
Summarize:
If the coef in the reaction is:
Doubled
Halved
Reversed in sign
Multiplied by a constant n
Then K is:
Squared
Square root
Inverted
Raised to the nth power
96