CHAPTER 13:
CHEMICAL
EQUILIBRIUM
A
Reversible
Reaction
Can Occur In Either Direction
CO(
g
) + H
2O(
g
)
CO
2(
g
) + H
2(
g
)
The forward reaction
proceeds to the right.
The reverse reaction
proceeds to the left.
A System is At
Equilibrium
When
the Rate of the Forward Reaction
Equals the Rate of the Reverse Reaction
H
2+ I
22HI
H
2+ I
22HI
H
2+ I
22HI
The Net Concentrations of Reactants and Products
Do NOT Change At Equilibrium
Equilibrium
Is Established When
the
Rate of the Forward Reaction
Equals
the Rate of the Reverse Reaction
H
2+ I
22HI
What can be said about concentrations at
equilibrium?
When you open the valve on the flasks above, the hydrogen and iodine gases will react according to the equilibrium equation below. By visually observing the flasks, how do
What can be said about concentrations at
equilibrium?
Forward reaction:
Reverse reaction:
What can be said about concentrations at
equilibrium?
Forward & reverse
reactions at the same time:
A system is at
equilibrium
when the rates of the
forward and reverse reactions are the same.
Both the forward and reverse reactions are
elementary
reactions, so we can write their rate
laws from the balanced equation.
rate forward =
k
f[N
2O
4] and rate reverse =
k
r[NO
2]
2N
2O
4(
g
)
⇌
2NO
2(
g
)
N
2
O
4
(
g
)
⇌
2NO
2
(
g
)
Starting with N2O4
N
2
O
4
(
g
)
⇌
2NO
2
(
g
)
Starting with NO2
Important Things to Remember
About Equilibrium
1.
Equilibrium is a dynamic state – both the
forward and reverse reactions continue to
occur, although there is no net change in
reactant and product concentrations over time.
2.
At equilibrium, the rates of the forward and
reverse reactions are equal.
The Equilibrium Constant,
K
rate forward = rate reverse kf[N2O4] eq = kr[NO2]2eq
The subscript “eq” denotes a concentration at equilibrium.
Rearranging
The ratio of two constants (kf/kr) is also a constant:
equilibrium expression
N2O4(
g
)
⇌
2NO2(
g
)
2
k k
2 eq f r 2 4 eq
NO N O
2
Kc 2 eq 2 4 eq NO N O
equilibrium constant
The
Equilibrium Constant
,
K
,
Relates the Equilibrium Concentrations
of Reactants and Products.
The value of Keq varies with temperature and units are usually omitted.
a
A +
b
B
c
C +
d
D
equilibrium
constant
=
K
=
[C]
c[D]
d[A]
a[B]
b=
[products]
[reactants]
Brackets, [ ], are used to symbolize concentration in
moles
per
liter
(mol/L).
When Writing an Equilibrium Constant,
K
, the
Coefficient
Becomes the
Exponent
N
2(
g
) + O
2(
g
)
2 NO(
g
)
equilibrium
constant
=
K
=
[N
2] [O
2]
[NO]
24NH
3+ 3O
22N
2+ 6H
2O
When Writing an Equilibrium Constant,
K
, the
Coefficient
Becomes the
Exponent
K
eq=
N
2[ ]
2H
2O
[
]
6NH
3[
]
4O
2[ ]
3The Size of
K
Indicates Whether Reactants or
Products are Favored at Equilibrium
•
When
K
is around 1
(0.01 <
K
< 100),
[products]
[reactants]
Both are similar
in magnitude.
both reactants
and
products
are present in
similar amounts
.
The Size of
K
Indicates Whether Reactants or
Products are Favored at Equilibrium
•
When
K
is much greater than 1,
•
When
K
is much
less than 1
,
[products]
[reactants]
The numerator is larger.
[products]
[reactants]
The denominator is larger.
equilibrium lies to the
right
and favors the
products
.
The Size of
K
Indicates Whether Reactants or
Products are Favored at Equilibrium
Very small
Very large
Reaction proceeds hardly at all
Reaction proceeds nearly to completion Appreciable concentrations
of both reactants and products are present at equilibrium
K
eq10–3 1 103
2 H
2(
g
) + O
2(
g
)
2 H
2O(
g
)
K
= 2.9 x 10
82The product is favored because
K
> 1
.
The equilibrium lies to the
right
.
How To Calculate the Equilibrium Constant for
a Reaction
Given the following information, calculate K for the
reaction between the reactants A
2and B
2.
[A
2] = 0.25 M
[B
2] = 0.25 M
[AB] = 0.50 M
A
2+ B
22 AB
How To Calculate the Equilibrium Constant for
a Reaction
STEP 1:
Write the expression for the equilibrium constant from
the balanced equation.
A
2+ B
22 AB
[AB]
2[A
2][B
2]
K
=
How To Calculate the Equilibrium Constant for
a Reaction
STEP 2:
Substitute the given concentrations in the equilibrium
expression and calculate K.
[AB]
2[A
2][B
2]
K
=
=
[0.50]
2
[0.25][0.25]
=
0.25
0.0625
=
4.0
The units of
K
are usually omitted.
Use Equilibrium Concentrations and the Equilibrium Constant
Expression to Calculate the Value of the Equilibrium Constant
Calculate the value of the equilibrium constant for
the reaction:
If the concentrations at equilibrium are
[A] = 2.0 M
[B] = 1.5 M
[C] = 0.010 M
A + 2B
C
K
eq=
[ ]
C
A
[ ]
[ ]
B
2=
0.010
(
)
2.0
( )
( )
1.5
2= 0.0022
Use Equilibrium Concentrations and the Equilibrium Constant
Expression to Calculate the Value of the Equilibrium Constant
Carbonyl chloride (COCl
2), also called phosgene, is a highly
poisonous gas that was used on the battlefield in World War I. It is
produced by the reaction of carbon monoxide with chlorine gas:
CO(
g
) + Cl
2(
g
)
COCl
2(
g
)
The equilibrium concentrations are:
[CO] = 0.012
M
[Cl
2] = 0.054
M
[COCl
2] = 0.14
M
When the species in a reversible chemical reaction are not all
in the same phase, the equilibrium is
heterogeneous
.
CaCO
3(s)CaO
(s)+ CO
2(g)¢
K
=
[
CO
2]
[
CaO
]
CaCO
3[
]
K
eq=
[
CO
2]
Pure Solids and Pure Liquids do not appear in
Equilibrium Expressions
(
s
)
not included
in the equilibrium expression
(
l
)
not included
in the equilibrium expression
(
g
) included in the equilibrium expression
(
aq
) included in the equilibrium expression
CO
2(g)+ C
(s)2CO
(g)K
eq=
[ ]
CO
2
CO
2[
]
K
eq=
1
Hg
22+[
]
[ ]
Cl
- 2Hg
22+(aq)+ 2Cl
(aq)Hg
2Cl
2(s)Worked Example 15.3
Strategy Use the law of mass action to write the equilibrium expressions for each reaction. Only gases and aqueous species appear in the expression. Write equilibrium expressions for each of the following reactions:
(a) CaCO3(s) ⇌ CaO(s) + CO2(g)
(b) Hg(l) + Hg2+(aq) ⇌ Hg 22+(aq)
(c) 2Fe(s) + 3H2O(l) ⇌ Fe2O3(s) + 2H2(g)
(d) O2(g) + 2H2(g) ⇌ 2H2O(l)
Solution (a) Kc = [CO2] (b) Kc =
[Hg22+]
[Hg2+]
(d) Kc =
1 [O2][H2]2
(c) Kc = [H2]2
Think About It Like writing equilibrium expressions for homogeneous equilibria, writing equilibrium expressions for heterogeneous equilibria becomes second nature if you practice. The importance of developing this skill now cannot be overstated. Your ability to understand the principles and to solve many of the problems in and Chapters 16 to 19 depends on your ability to write equilibrium expressions correctly and easily.
When a chemical equation is reversed,
the new equilibrium expression is
the reciprocal of the original equilibrium expression.
C + D A + B
K
eq=
C
[ ]
[ ]
D
A
[ ]
[ ]
B
A + B C + D
¢
K
eq=
A
[ ]
[ ]
B
C
[ ]
[ ]
D
=
1
K
eqWhen a chemical equation is
multiplied
by a constant,
the concentrations of the new equilibrium expression
are raised to a power equal to the constant.
2A + 2B 2C + 2D
A + B C + D
¢
K
eq=
C
[ ]
2D
[ ]
2A
[ ]
2B
[ ]
2=
C
[ ]
[ ]
D
A
[ ]
[ ]
B
æ
è
çç
ö
ø
÷÷
2=
K
2eqK
eq=
C
[ ]
[ ]
D
A
[ ]
[ ]
B
Worked Example 15.4
Strategy Begin by writing the equilibrium expressions for the reactions that are given. Then, determine the relationship of each equation’s equilibrium expression to the equilibrium expression of the original equations, and make the corresponding change to the equilibrium constant for each.
Kc = and Kc =
The following reactions have the indicated equilibrium constants at 100°C:
(1) 2NOBr(g) ⇌ 2NO(g) + Br2(g) Kc = 0.014
(2) Br2(g) + Cl2(g) ⇌ 2BrCl(g) Kc = 7.2
Determine the value of Kc for the following reactions at 100°C:
(a) 2NO(g) + Br2(g) ⇌ 2NOBr(g) (d) 2NOBr(g) + Cl2(g) ⇌ 2NO(g) + 2BrCl(g)
(b) 4NOBr(g) ⇌ 4NO(g) + 2Br2(g) (e) NO(g) + BrCl(g) ⇌ NOBr(g) + Cl2(g)
(c) NOBr(g) ⇌ NO(g) + Br2(g)
1 2 1
2
[NO]2[Br 2]
[NOBr]2
[BrCl]2
[Br2][Cl2]
Worked Example 15.4 (cont.)
Solution (a) This equation is the reverse of original equation 1. Its equilibrium expression is the reciprocal of that for the original equation:
Kc = = 1/0.014 = 71
(b) This is original equation 1 multiplied by a factor of 2. Its equilibrium expression is the original expression squared:
Kc = = (0.014)2 = 2.0×10-4
(c) This is the original equation 1 multiplied by . Its equilibrium expression is the square root of the original
Kc = or Kc = = (0.014)1/2 = 0.12
[NOBr]2
[NO]2[Br 2]
[NO]2[Br 2]
[NOBr]2
2
1 2
[NO]2[Br 2]
[NOBr]2
1/2
2 2 2
] [
] [ ] [
NOBr Br NO
Worked Example 15.4 (cont.)
Solution (d) This is the sum of the original equations 1 and 2. Its equilibrium expression is the product of the two individual expressions:
Kc = = (0.014)(7.2) = 0.10
(e) Probably the simplest way to analyze this reaction is to recognize that it is the reverse of the reaction in part (d), multiplied by . Its equilibrium expression is the square root of the reciprocal of the expression in part (d):
Kc = = (1/0.10)1/2 = 3.2
[NO]2[BrCl]2
[NOBr]2[Cl 2]
[NO]2[BrCl]2
[NOBr]2[Cl 2]
1/2
1 2
Worked Example 15.4 (cont.)
Think About It The magnitude of an equilibrium constant reveals whether products or reactants are favored, so the reciprocal relationship between Kc values
of forward and reverse reactions should make sense. A very large Kc value means
that products are favored. In the reaction of hydrogen ion and hydroxide ion to form water, the value of Kc is very large, indicating that the product, water, is
favored.
H+(aq) + OH-(aq) ⇌ H
2O(l) Kc = 1.0×1014 (at 25°C)
Simply writing the equation backward doesn’t change the fact that water is the predominate species. In the reverse reaction, therefore, the favored species is on the reactant side:
H2O(l) ⇌ H+(aq) + OH-(aq) Kc = 1.0×10-14 (at 25°C)
As a result, the magnitude of Kc should correspond to reactants being favored;
that is, it should be very small.
When an equilibrium expression contains only gases,
we can write an alternative form of the expression in which
the concentrations of gases are expressed
in partial pressures (atm).
we can write as
K
c=
or
K
P=
The relationship between
K
c and
K
Pcan be expressed as
where Δ
n
= moles of gaseous products – moles of gaseous reactants.
N
2O
4(
g
)
⇌
2NO
2(
g
)
[NO
2]
2[N
2O
4]
(
P
NO2)
2P
N2O4K
P=
K
c[(0.08206 L∙atm/K∙mol)×
T
]
ΔnEquilibrium Expressions Involving Pressures
PV
=
nRT
P
=
n
V
æ
è
ç
ö
ø
÷
RT
=
CRT
C = moles of gas per volume
Equilibrium Expressions Involving Pressures
K
eq=
Z
[ ]
4A
[ ]
[ ]
B
=
C
Z4C
AC
B=
P
ZRT
æ
è
ç
ö
ø
÷
4
P
ART
æ
è
ç
ö
ø
÷
P
BRT
æ
è
ç
ö
ø
÷
A + B
4Z
C
=
P
RT
Equilibrium Expressions Involving Pressures
=
P
ZRT
æ
è
ç
ö
ø
÷
4
P
ART
æ
è
ç
ö
ø
÷
P
BRT
æ
è
ç
ö
ø
÷
=
P
Z4P
A( )
( )
P
B´
1
RT
æ
è
ç
ö
ø
÷
4
1
RT
æ
è
ç
ö
ø
÷
1
RT
æ
è
ç
ö
ø
÷
Equilibrium Expressions Involving Pressures
K
eq=
P
Z4P
A( )
( )
P
B´
1
RT
æ
è
ç
ö
ø
÷
2
K
eq=
K
p´
( )
RT
-2
K
pEquilibrium Expressions Involving Pressures
K
p=
K
eq´
( )
RT
2
K
eq=
K
p´
( )
RT
-2
Where Dn = sum of product coefficients – sum of reactant coefficients
Worked Example 15.5
Strategy Write equilibrium expressions for each equation, expressing the concentrations of the gases in partial pressures.
Write KP expressions for (a) PCl3(g) + Cl2(g) ⇌ PCl5(g),
(b) O2(g) + 2H2(g) ⇌ 2H2O(l), and (c) F2(g) + H2(g) ⇌ 2HF(g).
Solution(a) All the species in this equation are gases, so they will all appear in the KP expression.
KP =
(b) Only the reactants are gases. KP =
(c) All species are gases. KP =
(PPCl5)
(PPCl3)(PCl2)
1 (PO2)(PH2)2
(PHF)2
(PF2)(PH2)
Think About It It isn’t necessary for every species in the reaction to be a gas– only those species that appear in the equilibrium expression.
Worked Example 15.6
Strategy Use KP = Kc[(0.08206 L∙atm/K∙mol)×T]Δn. Be sure to convert
temperature in degrees Celsius to kelvins. Using Δn = moles of gaseous products – moles of gaseous reactants, Δn = 2(NO2) – 1(N2O4) = 1. T = 298K.
The equilibrium constant, Kc, for the reaction
N2O4(g) ⇌ 2NO2(g)
is 4.63×10-3 at 25°C. What is the value of K
P at this temperature.
Solution
KP = Kc ×T
= (4.63×10-3)(0.08206 × 298)
= 0.113 0.08206 L∙atm
K∙mol
Think About It Note that we have essentially disregarded the units of R and T so that the resulting equilibrium constant, KP, is unitless.
The
Reaction Quotient
,
Q
, has the same form as the
Equilibrium Constant, K, except that the concentrations are
not the equilibrium concentrations
(usually
initial concentrations
are used to calculate Q).
c d
a b
Q
cC D
A B
a
A +
b
B
⇌
c
C +
d
D
The Equilibrium Constant
The value of the reaction quotient,
Q,
changes as the reaction
progresses
N2O4(
g
)
⇌
2NO2(
g
)
To predict the direction in which a reaction will proceed
when we start with a mixture of reactants and products,
calculate the value of the reaction quotient Q
and compare it to the value of the equilibrium constant K
There are three possibilities:
1.
Q
<
K
The ratio of initial concentrations of products to
reactants is too small. To reach equilibrium, reactants must be
converted to products. The
system proceeds in the forward
direction
.
2.
Q
=
K
the initial concentrations are equilibrium concentrations.
The
system is at equilibrium
.
3.
Q > K
the ratio of initial concentrations of products to reactants
is too large. To reach equilibrium products must be converted
to reactants. The
system proceeds in the reverse direction
.
Worked Example 15.7
Strategy Use the initial concentrations to calculate Qc, and then compare Qc
with Kc.
Qc = = = 0.61
At 375°C, the equilibrium constant for the reaction
N2(g) + 3H2(g) ⇌ 2NH3(g)
is 1.2. At the state of a reaction, the concentrations of N2, H2, and NH3 are
0.071 M, 9.2×10-3M, and 1.83×10-4M, respectively. Determine whether the
system is at equilibrium, and if not, determine in which direction it must proceed to establish equilibrium.
[NH3]i2
[N2]i[H2]i3
(1.83×10-4)2
(0.071)(9.2×10-3)3
Strategy The calculated value of Qc is less than Kc. Therefore, the reaction is
not at equilibrium and must proceed to the right to establish equilibrium. Think About It In proceeding to the right, a reaction consumes reactants and produces more products. This increases the numerator in the reaction quotient and decreases the denominator. The result is an increase in Qc until it is equal to Kc, at which point equilibrium
will be established.
Equilibrium concentrations can be calculated from initial
concentrations if the equilibrium constant is known.
Kc = 24.0 (200°C)
Initial concentration (M) 0.850 0 Change in concentration (M)
Equilibrium concentration (M)
–x
0.850 – x +x
x
cis-Stilbene trans-Stilbene
cis-Stilbene ⇌ trans-Stilbene
Equilibrium concentrations can be calculated from initial
concentrations if the equilibrium constant is known.
Initial concentration (M) 0.850 0 Change in concentration (M)
Equilibrium concentration (M)
–x
0.850 – x
+x
x
cis-Stilbene ⇌ trans-Stilbene
Use the equilibrium concentrations, defined in terms of
x
, in the
equilibrium expression:
trans
K cis c
-stilbene -stilbene
24.0 0.850
x x
Equilibrium concentrations can be calculated from initial
concentrations if the equilibrium constant is known.
Initial concentration (M) 0.850 0 Change in concentration (M)
Equilibrium concentration (M)
–0.816
0.850 – 0.816
+0.816
0.816 cis-Stilbene ⇌ trans-Stilbene
Calculate the equilibrium concentrations of
cis-
and
trans-
stilbene:
[
cis
-stilbene] = (0.850 –
x
)
M
= (0.850 – 0.816)
M
= 0.034
M
[
trans
-stilbene] =
x
M
= 0.816
M
Calculate the value of the equilibrium constant for the reaction
if 1.20 mol of A and 1.70 mol of B are dissolved in 1.00 L of solution, whereupon 0.100 mol of C is produced at equilibrium.
A + 2B
C + 2D
Example Problem 1
A + 2B
C
+ 2D
Initial concentration (M)
Change due to reaction (M)
Equilibrium concentration (M)
1.20 1.70 0.000 0.000
0.100
Calculate the value of the equilibrium constant for the reaction
if 1.20 mol of A and 1.70 mol of B are dissolved in 1.00 L of solution, whereupon 0.100 mol of C is produced at equilibrium.
A + 2B
C + 2D
Example Problem 1
A + 2B
C
+ 2D
Initial concentration (M)
Change due to reaction (M)
Equilibrium concentration (M)
1.20 1.70 0.000 0.000
0.100
0.100 0.200 –0.200
–0.100
Calculate the value of the equilibrium constant for the reaction
if 1.20 mol of A and 1.70 mol of B are dissolved in 1.00 L of solution, whereupon 0.100 mol of C is produced at equilibrium.
A + 2B
C + 2D
Example Problem 1
A + 2B
C
+ 2D
Initial concentration (M)
Change due to reaction (M)
Equilibrium concentration (M)
1.20 1.70 0.000 0.000
0.100
0.100 0.200 –0.200
–0.100
1.10 1.50 0.200
Example Problem 1
A + 2B
C
+ 2D
Initial concentration (M)
Change due to reaction (M)
Equilibrium concentration (M)
1.20 1.70 0.000 0.000
0.100
0.100 0.200 –0.200
–0.100
1.10 1.50 0.200
K
eq=
[ ]
C
[ ]
D
2
A
[ ]
[ ]
B
2=
0.100
(
)
(
0.200
)
21.10
(
)
(
1.50
)
2= 0.00162
A + B
2Z
For the reaction shown below, Keq = 4.0 x 10–4.Calculate the equilibrium concentration of Z if 0.500 mol of A and 0.500 mol of B are dissolved in 1.00 L of solution and allowed to come to equilibrium.
A + B
2Z
Example Problem 2
Initial concentration (M)
Change due to reaction (M)
Equilibrium concentration (M)
0.500 0.500 0.000
A + B
2Z
For the reaction shown below, Keq = 4.0 x 10–4.Calculate the equilibrium concentration of Z if 0.500 mol of A and 0.500 mol of B are dissolved in 1.00 L of solution and allowed to come to equilibrium.
A + B
2Z
Example Problem 2
Initial concentration (M)
Change due to reaction (M)
Equilibrium concentration (M)
0.500 0.500 0.000
–x –x 2x
0.500–x 0.500–x 2x
A + B
2Z
Example Problem 2
Initial concentration (M)
Change due to reaction (M)
Equilibrium concentration (M)
0.500 0.500 0.000
–x –x 2x
0.500–x 0.500–x 2x
K
eq=
Z
[ ]
2A
[ ]
[ ]
B
=
2
x
( )
20.500
-
x
(
)
(
0.500
-
x
)
= 4.0
´
10
-4
Example Problem 2
K
eq=
[ ]
Z
2
A
[ ]
[ ]
B
=
2
x
( )
20.500
-
x
(
)
(
0.500
-
x
)
= 4.0
´
10
-4
2
x
( )
20.500
-
x
(
)
2= 4.0
´
10
-42
x
0.500
-
x
æ
è
ç
ö
ø
÷
2
= 4.0
´
10
-4Example Problem 2
2
x
0.500
-
x
æ
è
ç
ö
ø
÷
2
= 4.0
´
10
-42
x
0.500
-
x
= 4.0
´
10
-4
= 0.020
2
x
= 0.500
(
-
x
)
(
0.020
)
= 0.0100
-
0.020
x
2
x
= Z
[ ]
= 0.0100 M
Worked Example 15.8
Strategy Insert the starting concentrations that we know into the equilibrium table:
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) + I2(g) ⇌ 2HI(g)
is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and I2?
Initial concentration (M) 0.240 0.240 0
Change in concentration (M)
Equilibrium concentration (M)
H2 + I2 ⇌ 2HI
Worked Example 15.8 (cont.)
Solution We define the change in concentration of one of the reactants as x. Because there is no product at the start of the reaction, the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium. According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by twice that amount (2x). Combining the initial concentration and the change in concentration for each species, we get expressions (in terms of x) for the equilibrium concentrations.
Initial concentration (M) 0.240 0.240 0
Change in concentration (M) –x –x +2x
Equilibrium concentration (M) 0.240 – x 0.240 – x 2x
Worked Example 15.8 (cont.)
Solution Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x.
Kc =
54.3 = =
=
x = 0.189
Use the calculated value of x, we can determine the equilibrium concentration of each species as follows:
[H2] = (0.240 – x) M = 0.051 M
[I2] = (0.240 – x) M = 0.051 M
[HI] = 2x = 0.378 M [HI]2
[H2][I2]
(2x)2
(0.240 – x)(0.240 – x) (2x)2
(0.240 – x)2
3 .
54 2x
0.240 – x
Think About It Always check your answer by inserting the calculated concentrations into the equilibrium expression:
The small difference between the calculated Kc and the one given in
the problem statement is due to rounding. = 54.9 ≈ Kc
[HI]2
[H2][I2]
(0.378)2
(0.051)2
=
Worked Example 15.9
Strategy Using the initial concentrations, calculate the reaction quotient, Qc,
and compare it to the value of Kc (given in the problem statement of Worked
Example 15.8) to determine which direction the reaction will proceed to establish equilibrium. Then, construct an equilibrium table to determine the equilibrium concentrations.
For the same reaction and temperature as in Worked Example 15.8, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: [H2] = 0.00623 M, [I2] = 0.00414 M, [HI] = 0.0424 M.
[HI]2
[H2][I2]
(0.0424)2
(0.00623)(0.00414)
= = 69.7
Worked Example 15.9 (cont.)
Strategy Therefore, Qc > Kc, so the system will proceed to the left (reverse) to
reach equilibrium. The equilibrium table is
Initial concentration (M) 0.00623 0.00414 0.0424
Change in concentration (M)
Equilibrium concentration (M)
H2 + I2 ⇌ 2HI
Worked Example 15.9 (cont.)
Solution Because we know the reaction must proceed from right to left, we know that the concentration of HI will decrease and the concentrations of H2 and
I2 will increase. Therefore, the table should be filled in as follows:
Next, we insert these expressions for the equilibrium concentration into the equilibrium expression and solve for x.
Initial concentration (M) 0.00623 0.00414 0.0424
Change in concentration (M) +x +x –2x
Equilibrium concentration (M) 0.00623 + x 0.00414 + x 0.0424 – 2x
H2 + I2 ⇌ 2HI
[HI]2
[H2][I2]
Kc =
(0.0424 – 2x)2
(0.00623 + x)(0.00414 + x) 54.3 =
Worked Example 15.9 (cont.)
Solution It isn’t possible to solve this equation the way we did in Worked Example 15.8 (by taking the square root of both sides) because the concentrations of H2 and I2 are unequal. Instead, we have to carry out the multiplications.
54.3(2.58×10-5 + 1.04×10-2x + x2) = 1.80×10-3 – 1.70×10-1x + 4x2
Collecting terms we get
50.3x2 + 0.735x – 4.00×10-4 = 0
This is a quadratic of the form ax2 + bx + c = 0. The solution for the quadratic
equation [Appendix 1] is
x =
Here we have a = 50.3, b = 0.735, and c = -4.00×10-4, so
x =
a ac b b 2 4 2 ) 3 . 50 ( 2 ) 10 00 . 4 )( 3 . 50 ( 4 ) 735 . 0 ( 735 .
0 2 4
Worked Example 15.9 (cont.)
Solution x = 5.25×10-4 or x = –0.0151
Only the first of these values, 5.25×10-4, makes sense because concentration
cannot be a negative number. Using the calculated value of x, we can determine the equilibrium concentration of each species as follows:
[H2] = (0.00623 + x) M = 0.00676 M
[I2] = (0.00414 + x) M = 0.00467 M
[HI] = (0.0424 – 2x) M = 0.0414 M
Think About It Checking this result gives
[HI]2
[H2][I2]
(0.0414)2
(0.00676)(0.00467)
= = 54.3
Worked Example 15.10
Strategy Construct an equilibrium table to determine the equilibrium partial pressures.
A mixture of 5.75 atm of H2 and 5.75 atm of I2 is contained in a 1.0-L vessel at 430°C. The equilibrium constant (KP) for the reaction
H2(g) + I2(g) ⇌ 2HI(g)
at this temperature is 54.3. Determine the equilibrium partial pressures of H2, I2, and HI.
Initial partial pressure (atm) 5.75 5.75 0
Change in partial pressure (atm) –x –x +2x
Equilibrium partial pressure (atm) 5.75 – x 5.75 – x 2x
H2 + I2 ⇌ 2HI
Worked Example 15.10 (cont.)
Solution Setting the equilibrium expression equal to KP,
54.3 =
Taking the square root of both sides of the equation gives
=
The equilibrium partial pressures are PH2 = PI2 = 5.75 – 4.52 = 1.23 atm, and
PHI = 9.04 atm.
(2x)2
(5.75 – x)2
3 .
54 2x
5.75 – x 2x 5.75 – x 7.369(5.75 – x) = 2x
42.37 – 7.369x = 2x 42.37 = 9.369x
x = 4.52 7.369 =
Think About It Plugging the calculated partial pressures into the equilibrium expression gives
The small difference between this result and the equilibrium constant given in the problem is due to rounding.
= 54.0 (PHI)2
(PH2)(PI2)
(9.04)2
(1.23)2
=
Le Châtelier
’
s Principle
is a General Rule
Used to Explain the Effect of a
Change in Reaction Conditions on Equilibrium
If a chemical system at equilibrium is disturbed or
stressed, the system will react in a direction that
counteracts the disturbance or relieves the stress.
The direction of shift is usually stated as
“
to the left
”
or
“
to the right.
”
Le Chatelier’s Principle
states that when a stress is
applied to a system at equilibrium, the system will
respond by shifting in the direction that minimizes the
effect of the stress.
Stress refers to any of the following:
The addition of a reactant or product.
The removal of a reactant or product.
A change in volume of the system, resulting in a change
in concentration or partial pressure of the reactants and
products.
A change in temperature.
What Happens When the
Concentration
of a
Reactant or Product is
Changed
?
2 CO(
g
) + O
2(
g
)
2 CO
2(
g
)
What happens if [CO(
g
)] is
increased
?
•
The concentration of O
2(
g
) will
decrease
.
•
The concentration of CO
2(
g
) will
increase
.
What Happens When the
Concentration
of a
Reactant or Product is
Changed
?
2 CO(
g
) + O
2(
g
)
2 CO
2(
g
)
What happens if [CO
2(
g
)] is
increased
?
•
The concentration of CO(
g
) will
increase
.
What Happens When the
Concentration
of a
Reactant or Product is
Changed
?
•
What happens if a product is
removed
?
•
The concentration of ethanol will
decrease
.
•The concentration of the other product (C
2H
4)
will
increase
.
Application of Le Châtelier
’
s Principle
Hydrogen sulfide (H
2S) is a contaminant commonly found in
natural gas. It is removed by reaction with oxygen to produce
elemental sulfur.
2 H
2S(
g
) + O
2(
g
)
2 S(
s
) + 2 H
2O(
g
)
For each of the following, determine whether the equilibrium
will shift to the right, shift to the left, or neither.
a)
Addition of O
2(
g
)
b)
Removal of H
2S(
g
)
c)Removal of H
2O(
g
)
d)
Addition of S(
s
)
What Happens When the
Temperature
of a
Reaction is
Changed
?
•
When the temperature is
increased
, the reaction
that
absorbs heat is favored
.
•
An
endothermic
reaction absorbs heat, so increasing
the temperature favors the forward reaction.
What Happens When the
Temperature
of a
Reaction is
Changed
?
•
Conversely, when the
temperature is decreased
,
the
reaction that adds heat is favored
.
•
An
exothermic
reaction releases heat, so
increasing
the temperature
favors the
reverse
reaction.
Factors That Affect Chemical Equilibrium
Changes in volume and concentration do not change the value of the
equilibrium constant.
A change in temperature can alter the value of the equilibrium
constant.
Heat + N2O4(
g
)
⇌
2NO2(
g
)
Δ
H
°
= 58.0
kJ/mol
Because the processes is endothermic, adding heat shifts the equilibrium
toward products
Factors That Affect Chemical Equilibrium
For any endothermic reaction, heat is a reactant:
Adding heat shifts the reaction towards products,
K
cincreases
Removing heat shifts the reaction towards reactants,
K
cdecreases.
For any exothermic reaction, heat is a product:
Adding heat shifts the reaction towards reactants,
K
cdecreases
Removing heat shifts the reaction towards products,
K
cincreases
heat + reactants
⇌
products
Δ
H
°
> 0
kJ/mol
What Happens When the
Pressure
of a
Reaction is
Changed
?
•
When
pressure increases
, equilibrium shifts in
the
direction that decreases the number of moles
in order to decrease pressure.
What Happens When the
Pressure
of a
Reaction is
Changed
?
•
When
pressure decreases
, equilibrium shifts in the
direction that increases the number of moles
in
order to increase pressure.
Worked Example 15.11
Strategy Use Le Châtelier’s principle to predict the direction of shift in each case. Remember that the position of the equilibrium is only changed by the addition or removal of a species that appears in the reaction quotient expression.
Because sulfur is a solid, it does not appear in the expression.
Hydrogen sulfide (H2S) is a contaminant commonly found in natural gas. It is removed by reaction with oxygen to produce elemental sulfur.
2H2S(g) + O2(g) ⇌ 2S(s) + 2H2O(g)
For each of the following scenarios, determine whether the equilibrium will shift to the right, shift to the left, or neither: (a) addition of O2(g), (b) removal of H2S(g), (c) removal of H2O(g), and (d) addition of S(s).
[H2O]2
[H2S][O2]
Qc =
Worked Example 15.11 (cont.)
Solution
Changes in concentration of any of the other species will cause a change in the equilibrium position. Addition of a reactant or removal of a product that appears in the expression Qc will shift the equilibrium to the right:
2H2S(g) + O2(g) 2S(s) + 2H2O(g)
Removal of a reactant or addition of a product that appears in the expression Qc will shift the equilibrium to the left:
2H2S(g) + O2(g) 2S(s) + 2H2O(g)
(a) Shift to the right (d) No change
[H2O]2
[H2S][O2]
Qc =
addition addition
removal
removal removal
addition
(b) Shift to the left (c) Shift to the right Think About It In each case, analyze the effect the change will have on the value of Qc. In part (a), for example, O2 is added, so its
concentration increases. Looking at the reaction quotient expression, we can see that a larger concentration of oxygen corresponds to a larger overall denominator – giving the overall fraction a smaller value. Thus, Q will temporarily be smaller that K and the reaction will have to shift to the right, consuming some of the added O2
(along with some of the H2S in the mixture) to reestablish
equilibrium.
When
volume decreases
, equilibrium shifts in the
direction that decreases the number of moles of gas
in order to decrease pressure.
N
2O
4(
g
)
⇌
2NO
2(
g
)
Equilibrium mixture:[N2O4] = 0.643 M [NO2] = 0.0547 M
2 c
2 4
NO
N O
. .
. K
2 2
3
0 0547 4 65 10 0 643
2
c c
2 4
NO
N O
. .
.
Q K
2 2
3
0 1094 9 31 10 1 286
Volume decreases by half, concentrations are initially
doubled: [N2O4] = 1.286 M [NO2] = 0.1094 M
The reaction shifts to the left. N2O4(g) ⇌ 2NO2(g)
Worked Example 15.12
Strategy Determine which direction minimized the number of moles of gas in the reaction. Count only moles of gas.
For each reaction, predict in what direction the equilibrium will shift when the volume of the reaction vessel is decreased.
(a) PCl5(g) ⇌ PCl3(g) + Cl2(g)
(b) 2PbS(s) + 3O2(g) ⇌ 2PbO(s) + 2SO2(g)
(c) H2(g) + I2(g) ⇌ 2HI(g)
Solution (a) We have 1 mole of gas on the reactant side and 2 moles of gas on the product side, so it will shift to the left.
(b) 3 moles of gas on the reactant side and 2 moles of gas on the product side, so it will shift to the right.
(c) 2 moles of gas on each side, so no shift.
Summary of Le Chatelier’s Principle
Summary of Le Chatelier’s Principle
Component added
Shift to
opposite
side
Component removed
Shift to
same
side
A + B C
+ D
Application of Le Châtelier
’
s Principle
For an exothermic reaction, indicate which direction the
equilibrium will shift for each of the indicated changes:
N
2(g)+ O
2(g)2NO
(g)First where do we write
heat
?
Application of Le Châtelier
’
s Principle
i.
Some N
2is removed.
ii. The temperature is decreased.
iii. Some NO is added.
iv. Some O
2is removed.
v. A catalyst is added.
vi. The temperature is increased and some O
2is removed.
N
2(g)+ O
2(g)2NO
(g)+
heat
Application of Le Châtelier
’
s Principle
For an exothermic reaction, indicate which direction the
equilibrium will shift for each of the indicated changes:
i.
Decreasing the concentration of H
2.
ii. Increasing the concentration of C
6H
6.
iii. Decreasing the temperature.
iv. Increasing the pressure by decreasing the volume of the
container.
C
6H
6(g)+ 3H
2(g)C
6H
12(g)+
heat
Le Châtelier
’
s Principle and Respiration
Hb + O
2HbO
2Lung [O
2] high
Le Châtelier
’
s Principle and Respiration
Hb + O
2HbO
2Muscle [O
2] low
Reaction shifts left
More Applications of Le Châtelier
’
s Principle
Co(H
2O)
62+(aq)+ 4Cl
–(aq)+ heat 4CoCl
42–(aq)+ 6H
2O
(l)Room temperature
More Applications of Le Châtelier
’
s Principle
Co(H
2O)
62+(aq)+ 4Cl
–(aq)+ heat 4CoCl
42–(aq)+ 6H
2O
(l)Heating the solution
Reaction shifts rightMore Applications of Le Châtelier
’
s Principle
Co(H
2O)
62+(aq)+ 4Cl
–(aq)+ heat 4CoCl
42–(aq)+ 6H
2O
(l)Cooling the solution
Reaction shifts leftCoCl
42-+ 6H
2O
⇌
Co(H
2O)
62++ 4Cl
-+ heat
blue
pink
How does a catalyst effect equilibrium?
•
Catalysts act by lowering the activation energy of a reaction,
which occurs to the same extent for both the forward and
reverse reactions.