CHE 115
Chemical Equilibrium
IV. Chemical Equilibrium
A) Law of Chemical Equilibrium
Concentration expressed as molarity
Molarity of A
' [A] '
moles of A
volume in liters
Consider the hypothetical reaction
A (g)
!
% B (g)
C (g)
units:
mole
L
' M
% D (g)
Compound A is allowed to react with B and the concentrations of the reactants and products are
plotted as a function of time.
Note: After eleven seconds the concentrations do not change with time.
Chemical equilibrium is the condition that exists when the concentrations of the reactants and
products no longer change with time. Chemical equilibrium is a dynamic state. It occurs when
the rate of the forward reaction (A + B 6 C + D) equals the rate of the reverse reacion (A + B
7 C + D).
A (g) + B (g)
C (g) + D (g)
indicates equilibrium
1
CHE 115
Chemical Equilibrium
At equilibrium there is a specific relationship between the concentrations of the reactants and
products.
Consider three experiments for the reaction
A (g) + B (g)
Ÿ
C (g) + D (g)
in which the initial concentrations of A and B vary from experiment to experiment.
Table of Equilibrium Concentrations
[A]eq
[B]eq
[C]eq
3.00
2.00
1.00
9.60
10.0
4.00
0.500
3.00
0.500
Exp. #
1
2
3
[D]eq
1.00
4.00
0.500
The following relationship fits the data in each experiment.
[C]eq [D]eq
[A]eq [B]eq
' 0.167 ' constant ' Kc
For the general reaction
m A (g) + n B (g) +
㠟
p C (g) + q D (g) +
ã
Kc is defined
[C]peq [D]qeq
[A]meq [B]neq
ã
' Kc
ã
The magnitude of the constant Kc depends on the nature of the reaction and the temperature.
The units of Kc depend on the specific reaction.
Example: A + 3 B
Kc '
Example: A + B
Ÿ2C
[C]2eq
[A]eq [B]3eq
Ÿ
units:
ï
mole 2
L „
ï
mole 4
L „
C + D
2
'
ð
L
mole
2
ë
CHE 115
Chemical Equilibrium
Kc '
[C]eq [D]eq
[A]eq [B]eq
units:
ï
mole 2
L „
mole 2
L „
ï
' unitless
Terms:
When K < 1, then the equilibrium is said to lie to the left.
When K > 1, then the equilibrium is said to lie to the right.
B) Equilibrium Calculations
Example #1: Initially, 0.2660 mole of PCl3 and 0.3340 mole of Cl2 are placed in a 2.00 L
container at 300EC. If the moles of PCl5 are found experimentally to be 0.0660 at equilibrium,
calculate Kc for the reaction
PCl3 + Cl2 Ÿ PCl5
at 300EC.
where [A]in is the initial (at time = 0) molarity of A and [A]eq is the equilibrium molarity of A
Important observations from the above plot
3
CHE 115
Chemical Equilibrium
• Initial moles of a substance + change in moles of substance = equilibrium moles of substance
Example:
initial moles of PCl5 + change in moles of PCl5
0.0 moles of PCl5 +
=
equilibrium moles of PCl5
0.0660 moles of PCl5 = 0.0660 moles of PCl5
• The changes in number of moles of reactants and products are related by mole ratios.
These observations can be utilized in an Equilibrium Table to solve equilibrium problems.
Equilibrium Table
Change in moles
of PCl5 = (equilibrim
PCl3
initial:
+
0.2660
change: -0.0660
equil:
Cl2
PCl5
0.3340
0.0
mole
0.0660
mole
0.0660
mole
-0.0660
0.2000
moles PCl5) - (initial
moles of PCl5)
0.2680
Negative sign
indicates moles
of Cl2 are
Change in moles of Cl2 =
consumed
(change in moles PCl5)(1 mole Cl2 /1 mole PCl5 )
Equilibrium moles of Cl2 = (initial moles Cl2) + (change in moles Cl2)
Substituting the equilibrium moles into the equilibrium expression
Kc
'
0.0660 mole
[PCl5]eq
[PCl3]eq [Cl2]eq
'
ï
2.00 L
0.200 mole
ï
2.00 L
' 2.46
„
0.268 mole
„
ï
2.00 L
L
mole
„
Example #2: Initially, 3.00 moles of PCl3 and 3.00 moles of Cl2 are placed in a 5.00 L container
at 300EC. Calculate the equilibrium concentration of PCl5, [PCl5]eq. Kc is 2.46 L/mole.
Unknown: [PCl5]eq , Let x = moles of PCl5 at equilibrium.
Knowns: initial moles of PCl3 = 3.00 mole, initial moles of Cl2 = 3.00 mole, volume of container =
4
CHE 115
Chemical Equilibrium
5.00L, Kc = 2.46 L/mole
Concepts: chemical equilibrium, molarity
Substitute known values into equilibrium table.
initial
change
equil
PCl3
3.00
+
Ÿ
Cl2
3.00
PCl5
0 mole
mole
x mole
Use the concepts presented above to fill in the table.
PCl3
initial
3.00
change
-x
equil
3.00 - x
+
Ÿ
Cl2
3.00
-x
3.00 - x
PCl5
0 mole
x mole
x mole
Substituting equilibrium moles into the equilibrium expression
Kc '
x
[PCl5]eq
[PCl3]eq [Cl2]eq
'
ï
3.00 & x
ï
' 2.46
5.00 L „
5.00 L
3.00 & x
„
ï
5.00 L
L
mole
„
2.46 x 2 & 19.76x % 22.14 ' 0
For the quadratic equation
ax2 + bx + c = 0
the solutions are
x '
& b ± b 2 & 4 ac
2a
Substituting
x '
& b ± b 2 & 4 ac ' & (& 19.76 ) ± (& 19.76)2 & 4(2.46) (22.14)
2a
2 (2.46)
x ' 1.35, 6.67
Although both roots (1.35 and 6.67) are mathematically correct, only one root satisfies the physical
constraints of the problem and that root is 1.35. If root 6.67 is used, then the moles of PCl3 and Cl2 at
equilibrium would be - 3.67 (3.00 - 6.67) and the equilibrium concentrations of PCl3 and Cl2 would be 0.734 M. A negative or zero equilibrium concentration for reactants or products is physically impossible.
5
CHE 115
Chemical Equilibrium
x ' 1.35 mole ' moles of PCl5 at equilibrium
[PCl5]eq '
[PCl3]eq '
x
'
5.00 L
(3.00 & x)
'
5.00L
1.35mole
5.00 L
(3.00 & 1.35)
'
5.00 L
[Cl2]eq '
1.65mole
5.00 L
1.65mole
5.00 L
The degree of dissociation, , of B
B
Ÿ
2A
is defined as
' change in moles of B
initial moles of B
and the percent dissociation is ( )(100).
Example #3: If 3.00 moles of B are placed in a 1.00 L flask at 25EC, calculate the degree of
dissociation, , of B. Kc for B at 25EC is 0.100 mole/L.
Unknown: change in moles of B. Let x = change in moles of B
Knowns: initial moles of B = 3.00 mole, volume of flask = 1.00 L, Kc = 0.100 mole/L
Concepts: chemical equilibrium, degree of dissociation
Substitute known values into the equilibrium table.
initial
change
equil
B
3.00
-x
Ÿ
2A
0
mole
mole
mole
The reaction must go to the right in order to produce A. Moles of B will be consumed and thus,
the change in moles of B will be -x.
Use the concepts presented above to fill in the table.
B
initial
3.00
change
-x
equil
3.00 - x
Ÿ
2A
0
2x
2x
mole
mole
mole
Substitute equilibrium moles into the equilibrium expression.
6
CHE 115
Chemical Equilibrium
Kc '
2x
[A]2eq
'
[B]eq
ï
2
' 0.100
1.00 L „
3.00 & x
ï
1.00 L
mole
L
„
4.00x 2 % 0.100 x & 0.300 ' 0
x ' 0.261, & 0.287
x = 0.261 mole = change in moles of B
' change in moles of B '
initial moles of B
x
'
3.00 mole
0.261mole
' 8.70@10&2
3.00 mole
C) Factors Affecting Equilibrium Concentrations at Constant Temperature
Le Châtelier’s principle - when a system is in equilibrium , a change in any one of the factors upon
which the equilibrium depends will cause the equilibrium to shift in such a way as to diminish the
effect of the change.
To determine if the reaction
a A (g) + b B (g)
Ÿ
c C (g) + d D (g)
will go to the left or to the right to reestablish the equilibrium, Q is defined as
Q
'
[C]cin [D]din
[A]ain [B]bin
If Q < Kc , then the reaction will go to the right to reestablish the equilibrium. Moles of reactants
A and B will be consumed and moles of products C and D will be produced.
If Q > Kc , then the reaction will go to the left to reestablish the equilibrium. Moles of products C
and D will be consumed and moles of reactants A and B will be produced.
1) Addition or Removal of Moles of Reactants or Products
Example #4: If 1.00 mole of PCl5 is added to the equilibrium in Example #2, what will be
the NEW equilibrium concentration of PCl5?
Unknown: new equilibrium molarity of PCl5 after the addition of 1.00 mole of PCl5
Knowns: equilibrium moles of PCl3, Cl2, and PCl5 (1.65, 1.65, and 1.35 mole) before the
addition of 1.00 mole of PCl5 (see Example #2); Kc = 2.46 L/mole; volume of the
container = 5.00 L
7
CHE 115
Chemical Equilibrium
Concepts: chemical equilibrium, Q, molarity
Step #1: Evaluate Q and determine whether the reaction will go to the left or
right to reestablish the equilibrium.
initial
PCl3
1.65
+
Ÿ
Cl2
1.65
PCl5
1.35 + 1.00 mole
The moles in the table are the equilibrium values from Example #2 and the 1.00 mole of
PCl5 added.
Q '
(2.35 mole)
[PCl5]in
'
[PCl3]in [Cl2]in
(1.65 mole)
5.00 L
ï
Q ' 4.31
5.00 L
ï
' 4.31
„
(1.65 mole)
„
5.00 L
ï
L
mole
„
L
L
> 2.46
' Kc
mole
mole
Since Q > Kc, the reaction will go to the left to reestablish the equilibrium. Moles of
PCl5 will be consumed and moles of PCl3 and Cl2 will be produced.
Step #2: Solve the equilibrium problem.
Let x = change in moles of PCl5. Since the reaction goes to the left to reestablish the
equilibrium, moles of PCl5 will be consumed and the entry for PCl5 in the row labeled
“change” of the equilibrium table will be -x.
PCl3
initial
1.65
change
x
equil
1.65 + x
+
[PCl5]eq
'
Kc '
[PCl3]eq [Cl2]eq
Ÿ
Cl2
1.65
x
1.65 + x
PCl5
2.35
mole
-x
mole
2.35 - x mole
2.35 & x
ï
5.00 L
1.65 % x
ï
5.00 L
' 2.46
„
1.65 % x
„
ï
5.00 L
„
0.492x 2 & 2.624 x % 1.010 ' 0
x = 0.361 mole = change in moles of PCl5
8
L
mole
CHE 115
Chemical Equilibrium
[PCl5]eq '
(2.35 mole & x)
'
5.00 L
(2.35mole & 0.36 mole)
'
5.00 L
1.99mole
5.00 L
2) Change in Volume of the Container
Example #5: If the volume of the container in Example #2 is decreased from 5.00 L to
2. 00 L, will the moles of PCl5 at the NEW equilibrium be greater or less than the moles
of PCl5 in Example #2?
Unknown: new equilibrium moles of PCl5 after the volume of the container in Example #2 is
reduced to 2.00 L
Knowns: equilibrium moles of PCl3, Cl2, and PCl5 in Example #2, new volume = 2.00 L, Kc =
2.46 L/mole
Concept: Q, molarity
Step #1
PCl3
1.65
initial
Q '
+
Ÿ
Cl2
1.65
PCl5
1.35 mole
(1.35 mole)
[PCl5]in
'
[PCl3]in [Cl2]in
ï
ï
Q ' 0.992
2.00 L
(1.65 mole)
2.00 L
' 0.992
„
(1.65 mole)
„
ï
2.00 L
L
mole
„
L
L
< 2.46
' Kc
mole
mole
Since Q < Kc, the reaction will go to the right to reestablish the equilibrium. Moles of PCl3 and
Cl2 will be consumed and moles of PCl5 will be produced. The moles of PCl5 at the NEW
equilibrium will be greater than the moles of PCl5 in Example #2.
D) Dependency of K on Temperature
Plot of ln(Kp) versus 1/T for the reaction
Ÿ
N2O4 (g)
9
2 NO2 (g)
CHE 115
Chemical Equilibrium
Equation for a straight line is
y
' (slope) x % (intercept)
ln(Kp)
'
& H 1
ð
R
ëð
T
ë
% C
where C is a constant.
Consider K2 at T2 and K1 at T1 for a specific reaction, then
10
CHE 115
Chemical Equilibrium
& H 1
'
ln(Kp)2
ëð
T2
% C
ë
& H 1
& ln(Kp)1 '
ð
ln(Kp)2
R
ð
R
ð
& ln(Kp)1 '
ëð
ð
T1
% C
ë
ë
& H 1 & 1
R
ëð
T2
T1
ë
Rearranging
ln(Kp)
'
2
H T2 & T1
R ë ð T2 T1 ë
ð
% ln(Kp)1
When H > 0, heat is absorbed and the reaction is said to be endothermic.
When H < 0, heat is evolved and the reaction is said to be exothermic.
Example: If Kp = 0.09823 at 20EC and 0.4554 at 40EC, calculate H for the reaction.
Ÿ
N2O4 (g)
ln(Kp)
2
'
2 NO2 (g)
H T2 & T1
% ln(Kp)1
R
T2 T1
ð
ëð
ë
Rearranging
H ' R ln
(Kp)
ð
(Kp)
T2 T1
2
ëð
1
T2 & T1
ë
' ( 8.314 KJ ) ln (0.4554) (313 K)(293 K ' 5.85 @ 104 J
ð
E) Entropy and Free Energy
11
(0.09823)
ëð
20K
ë
CHE 115
Chemical Equilibrium
The thermodynamic function entropy, S, is a measure of the disorder or randomness of a system.
Entropy is a state function.
The second law of thermodynamics states that an increase in entropy of the universe is associated
with every spontaneous process. The change in entropy of the universe is the sum of the changes
in the entropy of the system and surroundings.
'
Suniv
Ssys
% Ssurr
(1)
For a process to proceed spontaneously Suniv must be a positive number. If Suniv is a negative
number, then the process will proceed spontaneously in the reverse direction and when Suniv = 0,
then the system is at equilibrium.
The change in the entropy of the surroundings, Ssurr, is dependent on the heat that is transferred
between the system and the surroundings. For a constant pressure and constant temperature
process
qsurr
'
Ssurr
T
'
& (qp )sys
T
' & H
T
(2)
As an example consider the reaction
N2 (g) + 3 H2 (g)
6
HE = - 92.4kJ,
2 NH3 (g)
SE = - 198.5 J/K
The change in the entropy of the surroundings with the formation of two moles of NH3 is
Ssurr
& (& 92.4kJ)
' & H '
T
1000 J
ï
1 kJ
„
' 310 J
K
298 K
and the change in the entropy of the universe is
Suniv
'
Ssys
% Ssurr ' &198.5 J % 310 J ' 111 J
K
K
K
The entropy of the system, SE = - 198.5 J/K, decreases primarily because there are fewer
product molecules than reactant molecules. The reduction in number of molecules results in a
lowering of the positional disorder. Although SE for the reaction is a negative number, Suniv >
0 due to the large increase in the entropy of the surroundings and the reaction proceeds
spontaneously to the right.
For a constant pressure and constant temperature process, eq 2 can be substituted into eq 1.
Suniv
'
Ssys
% & H '
T
Multiply eq 3 by - T and rearrange.
12
S
% & H
T
(3)
CHE 115
Chemical Equilibrium
& T Suniv '
H
&T S
(4)
Gibbs free energy, G, is defined as
G
/ H & TS
For a constant pressure and constant temperature process
G
'
G
' & T Suniv
H
&T S
(5)
Comparing eqs 4 and 5
A constant pressure and constant temperature process will be spontaneous when Suniv > 0 or
< 0.
Example: Calculate
G E at 25EC for the following reaction?
H2 (g)
GE '
6
2 H (g)
HE = 436 kJ,
HE & T SE ' 436kJ & (298K) (98.5 J )
K
SE = 98.5 J/K
1 kJ
ð
1000 J ë
' 407 kJ
F) Reaction Free Energy
For the generic reaction
a A (g) + b B (g)
6
c C (g) + d D (g)
the reaction free energy, G, is defined as
Partial Pressure of Gas C
Standard Reaction
Free Energy
)G
=
)Go
Reaction Free
Energy
Partial Pressure of Gas D
Absolute
Temp
+
RT ln
P cP d
C D
P aP b
A
B
Ideal Gas
Constant
Partial Pressure of Gas B
Partial Pressure of Gas A
A plot of G for the above reaction versus the extent of reaction yields
13
G
CHE 115
Chemical Equilibrium
.
The degree of reaction is a measure of the extent to which the reaction has advanced from the left
to right.
The slope of the above curve at any point is equal to the reaction free energy, G. When the
degree of reaction is 0, only the reactants A and B are present, G < 0, and the reaction will
proceed spontaneously to the right. When the degree of reaction is 1.0, only the products are
present, G > 0, and the reaction will proceed spontaneously to the left. When the degree of
reaction is 0.16, G = 0 and the reaction is at equilibrium.
At equilibrium, G = 0 and
G
' 0 '
GE
% R T ln
Kp is defined as
14
(PC ) c
eq
ð
(PA )
a
(PD ) d
eq
(PB ) b ë
eq
eq
(6)
CHE 115
Chemical Equilibrium
'
Kp
(PC ) c
eq
ð
(PA )
a
(PD ) d
eq
(7)
(PB ) b ë
eq
eq
Substituting eq 7 into eq 6 and rearranging
' & RTln(Kp)
GE
or
' exp &RTGE
Kp
ð
ë
Relationship between Kc and Kp
For the generic reaction
6
a A (g) + b B (g)
'
Kp
(PC ) c
eq
ð
(PA )
a
(PD ) d
c C (g) + d D (g)
eq
(PB ) b ë
eq
eq
Assuming ideal gas behavior
'
PA
nA
ï
V
' [A] R T ; PB ' [B] R T ; PC ' [C] R T ; PD ' [D] R T
RT
„
Substituting
Kp
'
(PC ) c
(PD ) d
(PA ) a
(PB ) b
eq
ð
eq
eq
eq
ë
'
( [C]R T) ceq ( [D]R T) deq
ð
([A] RT ) aeq ([B] RT ) beq ë
'
[C] ceq [D] deq
ð
[A] aeq [B] beq ë
and
Kp
' Kc (R T )
n
where n = c + d - a - b.
G) Standard Reaction Free Energy and Standard Absolute Entropy
1) GE
15
(R T )c % d & a& b
CHE 115
Chemical Equilibrium
The free energy change for the formation of a compound in its standard state from its
elements (most stable form) in their standard states is call the standard free energy of
formation, GEf .
Example:
N2 (g) + 3 H2 (g)
6
GEf = - 33.2 kJ
2 NH3 (g)
By definition GEf for elements in their most stable form is zero.
A table of GEf is found in the text. See example below.
GE, from the standard free energies of
To calculate the standard reaction free energy,
formation
GEreaction
'
j
product
GEf (product)
&
j
reactant
GEf (reactant)
where , called the mole number, is the stoichiometric coefficient in the thermochemical
equation.
2) SE
The third law of thermodynamics states that the entropy for all pure, perfectly ordered,
crystalline substances is zero at absolute zero(0 K). Consequently the absolute entropies,
SE, of all compounds and elements in their most stable form can be determined. Some of
these entropies are list in a table of thermodynamic data found in the text.
Example: Table of Selected Thermodynamic Data
GEf (kJ/mole)
- 394.4
- 237.2
0.0
203.3
Substance
CO2 (g)
H2O (l)
H2 (g)
H (g)
SE (J/mole-K)
213.6
70.0
130.5
115
To calculate the change in the standard entropy for a reaction
SEreaction
'
j
product
SE (product)
16
&
j
reactant
SE (reactant)