Chapter 31
Chemical
Equilibrium
1
Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved.
Chemical Equilibrium
Previously we have assumed that chemical
reactions results in complete conversion of
reactants to products:
A + B
C + D
No A or B remaining or possibly an excess of A or B but
not both and eventually reaction stops.
Many chemical reactions do not completely convert
reactants to products. Stop somewhere between no rxn and
complete rxn. A + B
C + D
A + B C + D
some left
some formed
A + B C + D
reversible (both directions) exchange, constant conc.,
Ratef = Rater “equilibrium”
2
Chemical Equilibrium
• Therefore, many reactions do not go to
completion but rather form a mixture of
products and unreacted reactants, in a
dynamic equilibrium.
– A dynamic equilibrium consists of a forward
reaction, in which substances react to give products,
and a reverse reaction, in which products react to
give the original reactants.
– Chemical equilibrium is the state reached by a
reaction mixture when the rates of the forward
and reverse reactions have become equal.
4
Graphically we can represent this
A + B
C +D
The concentrations and reaction rate (less collisions, less component)
of A and B decreases over time as the concentrations and reaction rate
of C and D increases (more collisions, more component) over time
until the rates are equal and the concentrations of each components
reaches a constant. This occurs at what we call equilibrium -- Rf = Rr.
If the rates are equal, then there must be a relationship to show this.
5
3H2 + CO
CH4 + H2O
Rf=Rr
3H2 + CO

CH4 + H2O rate decreases over time
CH4 + H2O  3H2 + CO rate increases over time
Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New
York, NY, 2005.
If we assume these reactions are elementary rxns (based on
collisions), we can write the rate laws directly from the
reaction:
A + B
C+ D
Rf = kf [A][B]
For the reverse reaction we have,
C +D
A+ B
Rr = kr [C][D]
We know at equil that Rf = Rr ;therefore, we can set these two
expressions as equal
kf [A][B] = kr [C][D]
Rearrange to put constants on one side we get
kf
[C ][ D]
K

k r [ A][ B ]
7
Constant divided by constant just call a new constant K. This ratio is
given a special name and symbol called equilibrium constant K
relating to the equilibrium condition at a certain temperature (temp
dependent) for a particular reaction relating conc of each component.
This is basically a comparison between forward and reverse reaction
rates. At equilibrium, the ratio of conc of species must satisfy K.
kf
[C ][ D]
K

k r [ A][ B ]
8
The Equilibrium Constant
• Every reversible system has its own “position of
equilibrium”- K- under any given set of conditions.
– The ratio of products produced to unreacted reactants
for any given reversible reaction remains constant under
constant conditions of pressure and temperature. If the
system is disturbed, the system will shift and all the
concentrations of the components will change until
equilibrium is re-established which occurs when the ratio of
the new concentrations equals "K". Different constant conc
but ratio same as before.
– The numerical value of this ratio is called the equilibrium
constant for the given reaction, K.
9
The Equilibrium Constant
• The equilibrium-constant expression for a reaction is
obtained by multiplying the equil concentrations ( or partial
pressures) of products, dividing by the equil concentrations (or
partial pressures) of reactants, and raising each concentration to
a power equal to its coefficient in the balanced chemical
equation.
aA  bB
c
cC  dD
d
[C ] [ D]
Kc 
a
b
[ A] [ B ]
c
d
( PC ) ( PD )
Kp 
( PA ) a ( PB ) b
The molar concentration of a substance is denoted by writing its formula in
square brackets for aq solutions. For gases can put Pa - atm. As long as use M
or atm, K is unitless; liquids and solids = 1; setup same for all K’s (Ka, Kb, etc.)
Temp dependent; any changes, ratio will still equal K when equil established
10
31.1 The Equilibrium Constant, K
• The law of mass action states that the value of the
equilibrium constant expression K is constant for a
particular reaction at a given temperature, whenever
equilibrium concentrations are substituted.
• When at equil, conc ratio equals K and concs are
constant but if disturbed, values change until equil
reached, different constant conc but same ratio of K).
11
Obtaining Equilibrium Constants
for Reactions
• Equilibrium concentrations for a reaction
must be obtained experimentally and
then substituted into the equilibriumconstant expression in order to
calculate Kc.
12
Obtaining Equilibrium Constants
for Reactions
• Consider the reaction below
CO(g )  3 H 2 (g)
0.100 M
0.300 M
CH 4 (g)  H 2O(g)
0
0
Suppose we started with initial concentrations of
CO and H2 of 0.100 M and 0.300 M, respectively.
Obviously shift to right, decrease CO and H2 and
increase CH4 and water until equil established.
13
Obtaining Equilibrium Constants
for Reactions
CO(g )  3 H 2 (g)
CH 4 (g)  H 2O(g)
– When the system finally settled into equilibrium we
determined the equilibrium concentrations to be as
follows.
Reactants
[CO]eq
= 0.0613 M
[H2] eq
= 0.1839 M
Products
[CH4] eq = 0.0387 M
[H2O] eq
= 0.0387 M
14
Obtaining Equilibrium Constants
for Reactions
CO(g )  3 H 2 (g)
Kc 
CH 4 (g)  H 2O(g)
[CH 4 ]eq [ H 2O]eq
[CO ]eq [ H 2 ]3eq
– If we substitute the equilibrium concentrations,
we obtain:
(0.0387)(0.0387)
Kc 
 3.93
3
(0.0613)(0.1839)
15
Obtaining Equilibrium Constants
for Reactions
CO(g )  3 H 2 (g)
CH 4 (g)  H 2O(g)
– Regardless of the initial concentrations (whether
they are reactants or products or mixture), the law
of mass action dictates that the reaction will
always settle into an equilibrium where the
equilibrium-constant expression equals Kc.
16
Obtaining Equilibrium Constants
for Reactions
CO(g )  3 H 2 (g)
0
0
CH 4 (g)  H 2O(g)
0.2000 M
0.2000 M
As an example, let’s repeat the previous
experiment, only this time starting with initial
concentrations of products (note: if only products
to start shift left until equil established):
[CH4]initial = 0.2000 M and [H2O]initial = 0.2000 M
Obviously shift to left decrease CH4 and H2O and
increase CO and H2 until equil established.
17
Obtaining Equilibrium Constants
for Reactions
CO(g )  3 H 2 (g)
CH 4 (g)  H 2O(g)
– We find that these initial concentrations result in
the following equilibrium concentrations.
Reactants
Products
[CO] eq = 0.0990 M
[CH4] eq = 0.1010 M
[H2] eq = 0.2970 M
[H2O] eq = 0.1010 M
18
Obtaining Equilibrium Constants
for Reactions
CO(g )  3 H 2 (g)
CH 4 (g)  H 2O(g)
[CH 4 ]eq [ H 2O]eq
Kc 
[CO ]eq [ H 2 ]3eq
– Substituting these values into the equilibriumconstant expression, we obtain the same result.
(0.1010)(0.1010)
Kc 
 3.93
3
(0.0990)(0.2970)
– Whether we start with reactants or products at any
initial conc, the system establishes the same ratio
19
at same temperature.
The Equilibrium Constant
• The equilibrium constant, K, is the value obtained
for the equilibrium-constant expression when
equilibrium concentrations (not just any conc but
equil conc) are substituted.
– A large K, K>1, indicates large concentrations of
products at equilibrium.
– A small K, K<1, indicates large concentrations of
unreacted reactants at equilibrium.
HW 14
code: horse
20
The Equilibrium Constant
– Do same set up for all equil equations (future chapters), just
different subscript describing the reaction in question: i.e.
acid hydrolysis - acid + water - Ka, Kb, Kc, Kp, Ksp
– Kc is based on conc (M) and Kp is based on pressures (atm).
There is a difference between Kc and Kp and to jump
between these two, there is an equation to use. Let's show
how to jump between.
21
31.1.1 The Equilibrium Constant in
Terms of Pressure, Kp
• In discussing gas-phase equilibria, it is often
more convenient to express concentrations in
terms of partial pressures rather than
molarities.
– It can be seen from the ideal gas equation (PV =nRT) that
the partial pressure of a gas is proportional to its molarity
– P proportional to M - n/V ; therefore handled the same way.
n
P  ( )RT  MRT
V
22
The Equilibrium Constant, Kp
– Consider the reaction below.
CO(g )  3 H 2 (g)
CH 4 (g)  H 2O(g)
– The equilibrium-constant expression in terms of partial
pressures becomes (same way but partial pressures instead
of M):
Kp 
PCH 4 PH 2O
PCO ( PH 2 )
3
Kc 
[CH 4 ] [ H 2 O]
3
[CO ] [H 2 ]
23
The Equilibrium Constant, Kp
• In general if you need to jump between
K's, the numerical value of Kp differs
from that of Kc.
K p  K c ( RT )
Dng
where Dng is the sum of the moles of gaseous
products in a reaction minus the sum of the moles
of gaseous reactants.
24
A Problem to Consider
• Consider the reaction
2SO 2 (g )  O 2 (g )
2 SO 3 (g)
– Kc for the reaction is 280 at 1000. K . Calculate Kp
for the reaction at this temperature.
25
A Problem to Consider
• Consider the reaction at 1000. K and Kc
= 280
2SO 2 (g )  O 2 (g )
2 SO 3 (g)
– We know that
K p  K c ( RT )
Dng
From the equation we see that Dng = 2 – 3 = -1. We
can simply substitute the given reaction temperature
and the value of R (0.0821 L.atm/mol.K) to obtain Kp.
26
A Problem to Consider
• Consider the reaction
2SO 2 (g )  O 2 (g )
– Since
2 SO 3 (g)
K p  K c ( RT )
K p  280 (0.0821
Latm
mol K
Dng
 1000 K)  3.4
-1
27
A Problem to Consider
• Applying Stoichiometry to an Equilibrium Mixture
(basic setup for future problems).
– Suppose we place 1.000 mol N2 and 3.000 mol H2
in a reaction vessel at 450 oC and 10.0
atmospheres of pressure. The reaction is
N 2 ( g )  3H 2 ( g )
2NH 3 (g)
2
[NH 3 ]
Kc 
3
[N 2 ][H 2 ]
– What is the composition of the equilibrium mixture
if it contains 0.080 mol NH3 at equilibrium?
28
• Using the information given, set up the
following table. (ratio works for atm, M, mols,
etc.)
N 2 ( g )  3H 2 ( g )
Initial, no 1.000
Change,
-x
Dn
Equil, neq 1.000 - x
3.000
- 3x
3.000 - 3x
2NH 3 (g)
0
+ 2x
2x = 0.080 mol
– This procedure for many types of problems, however in
this problem given equil quantity of NH3 ;therefore, figure
out rest.
– The equilibrium amount of NH3 was given as 0.080 mol.
Therefore, 2x = 0.080 mol NH3 (x = 0.040 mol).
29
A Problem to Consider
• Using the information given, set up the
following table.
N 2 ( g )  3H 2 ( g )
Initial
1.000
3.000
Change
-x
-3x
Equilibrium 1.000 - x 3.000 - 3x
2NH 3 (g)
0
+2x
2x = 0.080 mol
x = 0.040 mol
Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2
Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2
Equilibrium amount of NH3 = 2x = 0.080 mol NH3
HW 15
30
code: rhino
31.2 Changing the Chemical
Equation
• Similar to the method of combining reactions
that we saw using Hess’s law in
thermochemistry, we can combine equilibrium
reactions whose K values are known to
obtain K for the overall reaction.
– With Hess’s law, when we reversed reactions
(change sign) or multiplied them prior to adding them
together (mult by factor). We had to manipulate the
DH’s values to reflect what we had done.
– The rules are a bit different for manipulating K.
31
Equilibrium Constant for the Sum
of Reactions
1.
2.
3.
If you reverse a reaction, invert the value of K.
(reciprocal rule, 1/K)
If you multiply/divide each of the coefficients in an
equation by the same factor (2, 1/2, …), raise Kc to the
same power (2, 1/2, …). (known as coefficient rule, Kn)
When you finally combine (that is, add) the individual
equations together, take the product of the equilibrium
constants to obtain the overall K.(rule of multiple
equilibria, K1 x K2 x K3… = KT)
32
Equilibrium Constant for the Sum
of Reactions
• For example, nitrogen and oxygen can
combine to form either NO(g) or N2O (g)
according to the following equilibria.
same and multiply by factor of 2
(1)
1
1
N 2 ( g )  O 2 (g )
2
2
(2)
N 2 (g )  12 O 2 (g )
NO(g) Kc = 6.4 x 10-16
N 2O(g)
Kc = 2.4 x 10-18
reverse and multiply by factor of 1
– Using these two equations, we can obtain K for
the formation of NO(g) from N2O(g):
overall
N 2O(g )  12 O 2 (g )
2 NO(g)
Kc = ? 33
1
1
N 2 ( g )  O 2 (g )
2
2
NO(g) Kc = 6.4 x 10-16
N 2 (g )  12 O 2 (g )
N 2O(g)
(1)
N 2 (g )  O 2 (g )
(2)
N 2O(g)
overall
Kc = 2.4 x 10-18
2 NO(g)
Kc = (6.4 x 10-16)2 =
4.1 x 10-31
1
Kc =

N 2 (g)  12 O 2 (g )
N 2O(g )  12 O 2 (g )
31
2.4  10-18
2 NO(g)
= 4.2 x 1017
K c (overall)  (4.110 )  (4.2 10 )  1.7 10
17
HW 16
code: ally
34
13
31.1.2 Heterogeneous Equilibria
• A heterogeneous equilibrium is an equilibrium that
involves reactants and products in more than one
phase. Up to now all our reactions have been
homogeneous - all gases or aqueous solutions.
– The equilibrium of a heterogeneous system is
unaffected by the amounts of pure solids or liquids
present, as long as some of each is present.
– The concentrations of pure solids and liquids are
always considered to be “1 activity” and therefore, do
not appear in the equilibrium expression. Solids and
pure liquids have no effect on conc or pressure.
35
Heterogeneous Equilibria
• Consider the reaction below.
C ( s)  H 2O(l )
CO(g)  H 2 (g)
K c  [CO ][ H 2 ]
K p  PCO PH 2
code: mike
HW 1736
31.3
Direction of Reaction
• How could we predict the direction in which a
reaction at non-equilibrium conditions will shift to
reestablish equilibrium? Remember did example
with only reactants, obviously had to go right and if
only products obviously must go left but what if have
some of R and P?
– To answer this question, substitute the current
concentrations into the reaction quotient
expression and compare it to Kc.
– The reaction quotient, Qc, is an expression that
has the same form as the equilibrium-constant
expression but whose concentrations are not
necessarily at equilibrium.
37
Predicting the Direction of Reaction
• For the general reaction
aA  bB
cC  dD
the Qc expresssion would be (i=initial):
c
d
[C]i [D]i
Qc 
a
b
[ A ]i [B]i
Kc 
d
[C ]ceq [ D]eq
a
eq
b
eq
[ A] [ B]
38
Predicting the Direction of Reaction
• For the general reaction
aA  bB
cC  dD
– If Qc = Kc, then the reaction is at equilibrium.
– If Qc > Kc, the reaction will shift left toward reactants until equil
reached.
c
d
[C]i [D]i
Qc 
[ A ]ai [B]bi
– If Qc < Kc, the reaction will shift right toward products until equil
reached.
[C]c [D]d
Qc 
i
i
[ A ]ai [B]bi
39
A Problem to Consider
– Consider the following equilibrium.
N 2 ( g )  3H 2 ( g )
2NH 3 (g)
– A 50.0 L vessel contains 1.00 mol N2, 3.00 mol
H2, and 0.500 mol NH3. Is the sytem at equil? If
not, in which direction (toward reactants or
toward products) will the system shift to
reestablish equilibrium at 400 oC?
– Kc for the reaction at 400 oC is 0.500.
40
A Problem to Consider
First, calculate concentrations from moles
of substances.
N 2 ( g )  3H 2 ( g )
2NH 3 (g)
Next, plug into Q expression
2
[ NH 3 ]i
Qc 
3
[ N 2 ]i [ H 2 ]i
41
A Problem to Consider
N 2 ( g )  3H 2 ( g )
2NH 3 (g)
Note: you cannot just look at number of mols or molarity to decide direction of reaction
because must account for the size of K.
– Substituting these concentrations into the reaction
quotient gives:
2
(0.0100)
Qc 
 23.1
3
(0.0200)(0.0600)
42
A Problem to Consider
N 2 ( g )  3H 2 ( g )
0.0200 M
0.0600 M
2NH 3 (g)
0.0100 M
– Because Qc = 23.1 is greater than Kc = 0.500,
the reaction will go to the left (toward
reactants - consume products, produce
reactants) as it approaches equilibrium.
HW 18
code: josh
43
31.4
Calculating K and Equilibrium
Quantities
• Once you have determined the
equilibrium constant, K, for a reaction,
you can use it to calculate the
concentrations of substances in the
equilibrium mixture.
44
Calculating Equilibrium
Concentrations
– For example, consider the following equilibrium
mixture.
CO( g )  3 H 2 (g)
CH 4 (g)  H 2O(g)
– Suppose a gaseous mixture contained 0.30 mol CO,
0.10 mol H2, 0.020 mol H2O, and an unknown
amount of CH4 per liter at equilibrium.
– What is the concentration of CH4 at equilibrium in this
mixture? The equilibrium constant Kc equals 3.92.
– Note: the amounts given are equil amounts; therefore
can plug into K eq.
45
Calculating Equilibrium
Concentrations
– First, calculate concentrations from moles of substances
then plug into K expression.
CO ( g )
 3 H 2 (g)
Kc 
CH 4 (g)  H 2 O(g)
[CH 4 ]eq [ H 2O]eq
3
[CO ]eq [ H 2 ]
eq
46
Calculating Equilibrium
Concentrations
CO(g )  3 H 2 (g)
0.30 M
0.10 M
CH 4 (g)  H 2O(g)
??
0.020 M
– Substituting the known concentrations and the value
of Kc gives:
[CH 4 ](0.020)
3.92 
(0.30)(0.10) 3
47
Calculating Equilibrium
Concentrations
CO(g )  3 H 2 (g)
0.30 M
CH 4 (g)  H 2O(g)
0.10 M
??
0.020 M
– You can now solve for [CH4].
3
(3.92)(0.30)(0.10)
[CH 4 ] 
 0.059 M
(0.020)
– The concentration of CH4 in the equil mixture is
0.059 mol/L.
48
31.4.1 Calculating Equilibrium Quantities
from Initial Values (Perfect Square)
• Suppose we begin a reaction with
known amounts of starting materials
and want to calculate the quantities at
equilibrium.
49
Calculating Equilibrium
Concentrations
– Consider the following equilibrium.
-
CO(g )  H 2O(g )
+
CO 2 (g)  H 2 (g)
0
0
• Suppose you start with 1.000 mol each of carbon
monoxide and water in a 50.0 L container.
Calculate the molarity of each substance in the
equilibrium mixture at 1000 oC.
• Kc for the reaction is 0.58 at 1000 oC.
Which way will reaction shift?
50
Calculating Equilibrium
Concentrations
– First, calculate the initial molarities of CO and H2O.
CO(g )  H 2O(g )
CO 2 (g)  H 2 (g)
0
0
51
Calculating Equilibrium
Concentrations
CO(g )  H 2O(g )
0.0200 M
0.0200 M
CO 2 (g)  H 2 (g)
0M
0M
• The starting concentrations of the products are 0.
• We must now set up a table of concentrations (initial, change,
and equilibrium expressions in x, ICE table).
52
Calculating Equilibrium
Concentrations
– The equilibrium-constant expression is:
[CO 2 ][H 2 ]
Kc 
[CO][H 2O]
– Let x be the moles per liter of product formed.
CO(g )  H 2O(g )
CO 2 (g)  H 2 (g)
Initial
0.0200
0.0200
0
0
Change
Equil
-x
0.0200-x
- x
0.0200-x
+ x
x
+ x
x
53
– Solving for x.
CO(g )  H 2O(g )
CO 2 (g)  H 2 (g)
Initial
0.0200 0.0200
Change
-x
-x
Equilibrium 0.0200-x 0.0200-x
0
+x
x
0
+x
x
[CO 2 ][H 2 ]
Kc 
[CO][H 2O]
– Substituting the values for equilibrium concentrations, we
get:
( x )( x )
0.58 
(0.0200  x )(0.0200  x )
54
Calculating Equilibrium
Concentrations
– Solving for x.
CO(g )  H 2O(g )
CO 2 (g)  H 2 (g)
Initial
0.0200 0.0200
Change
-x
-x
Equilibrium 0.0200-x 0.0200-x
0
+x
x
0
+x
x
2
x
0.58 
2
(0.0200  x )
55
–
Calculating Equilibrium
Concentrations
Solving for x.
CO(g )  H 2O(g )
Initial
0.0200 0.0200
Change
-x
-x
Equilibrium 0.0200-x 0.0200-x
– Taking the square root of both sides
because perfect square, we get:
2
x
0.58 
(0.0200  x) 2
CO 2 (g)  H 2 (g)
0
+x
x
0
+x
x
x
0.76 
(0.0200  x)
0.76(0.0200  x)  x
0.0152  0.76 x  x
0.0152  1.76 x
0.0152
x
 0.0086M 56
1.76
Calculating Equilibrium
Concentrations
– Solving for equilibrium concentrations.
CO(g )  H 2O(g )
Initial
0.0200 0.0200
Change
-x
-x
Equilibrium 0.0200-x 0.0200-x
CO 2 (g)  H 2 (g)
0
+x
x
0
+x
x
– If you substitute for x in the last line of the table you obtain the
following equilibrium concentrations. If plug into Keq, should
equal K or close to it because of sign fig for a check
[CO]eq = 0.0200 – x = 0.0200 – 0.0086 = 0.0114 M
[H2O]eq = 0.0200 – x = 0.0200 – 0.0086 = 0.0114 M
[CO2]eq = x = 0.0086 M
code:
[H2]eq = x = 0.0086 M
katie
HW 19 57
Calculating Equilibrium
Concentrations
• The preceding example illustrates the
three steps in solving for equilibrium
concentrations.
1. Set up a table of concentrations (initial, change, and
equilibrium expressions in x – ICE table).
2. Substitute the expressions in x for the equilibrium
concentrations into the equilibrium-constant equation.
3. Solve the equilibrium-constant equation for the values
of the equilibrium concentrations.
58
Another example: If the initial pressure of C is 1.0 atm, what
would be the partial pressures of each species at equil.
Kp = 9.0
Initial, Po
Change, DP
Equil, Peq
HW 20
code: chris
A
0
+x
x
+
B
2C
0
1.0
- 2x
+x
x
1.0-2x
59
Another example: If the initial pressure of C is 0.10 atm and A
and B are 1.00 atm, what would be the partial pressures of each
species at equil.
2C
A
+ B
Kp = 0.016
Initial, Po
0.10
1.00
1.00
Change, DP
+ 2x
- x
- x
1.00 – x
1.00 - x
Equil, Peq 0.10 + 2x
Q > K therefore, shift left
HW 21
code: three
60
31.4.2 Calculating Equilibrium Quantities
from Initial Values (Quadratic Formula)
• In some cases it is necessary to solve a
quadratic equation to obtain equilibrium
concentrations - not a perfect square.
• The next example illustrates how to
solve such an equation.
61
Calculating Equilibrium
Concentrations
– Consider the following equilibrium.
H 2 (g )  I 2 (g )
2HI(g)
• Suppose 1.00 atm H2 and 2.00 atm I2 are placed in
a 1.00-L vessel. What are the partial pressures of
all species when it comes to equilibrium at 458 oC?
• Kp at this temperature is 49.7.
62
Calculating Equilibrium
Concentrations
– The concentrations of substances are as follows.
H 2 (g )  I 2 (g )
Initial, Po
1.00
Change, DP - x
Equil, Peq 1.00 - x
2.00
-x
2.00 – x
2HI(g)
0
+ 2x
2x
63
Calculating Equilibrium
Concentrations
– The concentrations of substances are as follows.
H 2 (g )  I 2 (g )
Po
DP
Peq
1.00
-x
1.00-x
2.00
-x
2.00-x
2HI(g)
0
+2x
2x
– Substituting our equilibrium concentration expressions
2
gives:
(2x )
Kp 
 49.7
(1.00  x )(2.00  x )
64
Calculating Equilibrium
Concentrations
– Solving for x.
H 2 (g )  I 2 (g )
Initial
Change
Equilibrium
1.00
-x
1.00-x
2HI(g)
2.00
-x
2.00-x
0
+2x
2x
– Because the right side of this equation is not a perfect square (sq over
sq), you must solve the quadratic equation.
ax 2  bx  c  0
 b  b 2  4ac
x
2a
65
(2x ) 2
Kp 
 49.7
(1.00  x )(2.00  x )
49.7(1.00-x)(2.00-x) = (2x)2
49.7(2.00 - 3.00x +
x2)
=
4x2
99.4 - 149.1x + 49.7x2 = 4x2
45.7x2 - 149.1x + 99.4 = 0
a
b
c
ax 2  bx  c  0
 b  b 2  4ac
x
2a
 (149.1)  (149.1) 2  4(45.7)(99.4)
x
2(45.7)
 (149.1)  (22230.81)  18170.32
x
91.4
 (149.1)  4060.49
x
91.4
149.1  63.72
x
91.4
85.38
x 
 0.934
91.4
212.82
x 
 2.33
91.4
66
Calculating Equilibrium
Concentrations
– However, x = 2.33 gives a negative value to 1.00 - x (the
equilibrium concentration of H2), which is not possible.
Only x  0.934 possible
If you do the shift correctly, you may get a negative and
positive number, take the positive number and if two positive
numbers, take the smaller number. However, if you neglect the
shift and select it incorrectly, you will end up with either a +
and - # or two -#'s. In this case the negative number will be
correct or the smaller of the two negative numbers will be
correct. Bottom line examine numbers carefully when
selecting answer and selecting proper shift makes life easier
(use Q when necessary).
67
Calculating Equilibrium
Concentrations
• Solving for equilibrium pressures.
H 2 (g )  I 2 (g )
Initial
Change
Equilibrium
1.00
-x
1.00-x
2.00
-x
2.00-x
2HI(g)
0
+2x
2x
– If you substitute 0.934 for x in the last line of the table
you obtain the following equilibrium concentrations.
HW 22
code: two
68
31.5 Effect of Changing the Reaction
Conditions Upon Equilibrium
• Obtaining the maximum amount of
product from a reaction depends on the
proper set of reaction conditions which
gets us to:
– Le Chatelier’s principle states that when a
system in a chemical equilibrium is disturbed by a
change of temperature, pressure, or
concentration, the equilibrium will shift in a way
that tends to counteract this change until
equilibrium is established again.
69
31.5.1 Adding or Removing a
Species
• If a chemical system at equilibrium is disturbed by
adding a gaseous or aqueous species (not solid
or liquid R or P), the reaction will proceed in such
a direction as to consume part of the added
species. Conversely, if a gaseous or aqueous
species is removed (complex or escape gas), the
system shifts to restore part of that species. This
shift will occur until equilibrium is re-established.
• A + B
C + D add - shift to opposite side,
remove - shift towards that side.
70
31.5.2 Compression or Expansion
• A pressure change caused by changing the volume of the
reaction vessel can affect the yield of products in a gaseous
reaction; only if the reaction involves a change in the total moles
of gas present
• Ex. N2O4 (g)
2NO2 (g)
• Suppose system is compressed by pushing down a piston
(decrease volume of space), which way would the shift be that
would benefit and use the available space wisely?
• Shift to smaller number of gas molecules to pack more
efficiently and relieve the increase in pressure due to piston
coming down.
71
Effects of Pressure Change
• Basically the reactants require less volume
(that is, fewer moles of gaseous reactant) and
by decreasing the volume of the reaction vessel
by increasing the pressure, the rxn would shift
the equilibrium to the left (toward reactants)
until equil is established.
72
Effects of Pressure Change
• Literally “squeezing” the reaction (increase P) will cause a shift
in the equilibrium toward the fewer moles of gas.
• Reducing the pressure in the reaction vessel by increasing its
volume would have the opposite effect.
• Decrease P, increase V, shift larger mols of gas (L &S
not compressible)
• Increase P, decrease V, shift to smaller mols of gas
• In the event that the number of moles of gaseous
product equals the number of moles of gaseous
reactant, vessel volume/pressure will have no effect
on the position of the equilibrium; no advantage to
shift one way over the other.
73
SO2 (g) + 1/2 O2 (g)
SO3 (g)
Increase P?
Shift to right toward smaller mols gas
N2 (g) + 3 H2 (g)
2NH3 (g)
Decrease P?
Shift to left toward larger mols gas
N2 (g) + O2 (g)
2NO (g)
Decrease P?
No shift
C (s) + H2O(g)
CO (g) + H2 (g)
Increase P?
Shift to left toward smaller mols gas; note
that C is a solid.
74
31.5.3
Temperature
• Temperature has a significant effect on
most reactions.
– Reaction rates generally increase with an increase
in temperature. Consequently, equilibrium is
established sooner.
– However, when you add or remove
reactants/products or change pressure, the result
is that the system establishes new conc of species
but ratio still equal to same K at that temperature.
For temp changes, the numerical value of the
equilibrium constant Kc varies with
75
temperature. K temp dependent.
Effect of Temperature Change
• Let’s look at “heat” as if it were a
product in exothermic reactions and
a reactant in endothermic reactions.
A + B  C + D + heat
A + B + heat  C + D
exo
endo
• We see that increasing the temperature is
related to adding more product (in the case of
exothermic reactions) or adding more
reactant (in the case of endothermic
reactions).
• This ultimately has the same effect as if heat
76
were a physical entity.
Effect of Temperature Change
• For example, consider the following generic
exothermic reaction.
reactants
products " heat" (DH is negative)
• Increasing temperature would be like adding
more product, causing the equilibrium to
shift left.
• Since “heat” does not appear in the equilibriumconstant expression, this change would result
in a smaller numerical value for Kc
(numerator smaller and den larger)
77
Effect of Temperature Change
• For an endothermic reaction, the opposite is
true.
" heat"reactants
products ( DH is positive)
• Increasing temperature would be analogous to
adding more reactant, causing the
equilibrium to shift right.
• This change results in more product at
equilibrium, and a larger numerical value for
Kc (larger numerator, smaller den)
78
Effect of Temperature Change
• In summary:
– For an endothermic reaction (DH positive) the
amounts of products are increased (shift right) at
equilibrium by an increase in temperature (Kc is
larger at higher temperatures).
– For an exothermic reaction (DH is negative) the
amounts of reactants are increased (shift left) at
equilibrium by an increase in temperature (Kc is
smaller at higher temperatures).
Note: opposite occurs when lower temperature.
79
Effect of a Catalyst
• A catalyst is a substance that increases
the rate of a reaction but is not
consumed by it.
– It is important to understand that a catalyst has
no effect on the equilibrium composition of a
reaction mixture.
– A catalyst merely speeds up the attainment of
equilibrium but does not cause it shift one way or
the other, just get to direction is was going faster.
80
Consider the system
I2 (g)
2I (g)
DH = 151 kJ
Suppose the system is at equilibrium at 1000oC. In which direction will rxn
occur if
a.) I atoms are added?
Add gas products, shift left until equil established.
b.) the system is compressed?
Increase P, decrease V, shift towards smaller mols of gas; therefore to the left until equil.
c.)the temp is increased?
Endo reaction, therefore similar to adding reactants and rxn shifts to the right
d.)effect increase temp has on K?
Producing more products (shift right) and less reactants, therefore larger K
e.) add catalyst?
No effect, if reaction wasn’t at equil just getting there faster.
code: reaction
HW 23
81