CHEMICAL
EQUILIBRIUM
Chapter 16
ALL BOLD NUMBERED PROBLEMS
Properties of an
Equilibrium
Equilibrium systems are
• DYNAMIC (in constant motion) & REVERSIBLE
2
3
Cool
Heat
Blue
Pink
Blue to pink
Cool
CoCl42- (aq) + 6 H2O (l)  Co(H2O)62+ (aq)
Pink to blue Heat
Co(H2O)62+ (aq)  CoCl42- (aq) + 6 H2O (l)
4
Chemical Equilibrium
Fe3+ + SCN-
FeSCN2+
+
Fe(H2O)63+ + SCN-
colorless
Fe(SCN)(H2O)52+ + H2O
red-orange
Chemical Equilibrium
Fe3+ + SCN-
FeSCN2+
• After a period of time, the concentrations of
reactants and products are constant.
• The forward and reverse reactions continue
after equilibrium is attained.
5
6
Examples of Chemical
Equilibria
Phase changes such as
H2O(s)
H2O(liq)
7
Examples
of
Chemical
Acid Base Equilibria
Equilibria
+Na2CO3
H3O+(aq) + 2CO32-(aq)
+CO2
OH-(aq) + 2HCO3-(aq)
Color
Red
Purple
Violet
Blue
pH
2
4
6
8
Blue-Green
10
Green
12
8
Examples
of
Chemical
Equilibria
Formation of stalactites (ceiling) and
stalagmites (floor)
CaCO3(s) + H2O(liq) + CO2(g)
Ca2+(aq) + 2 HCO3-(aq)
9
Chemical
Equilibria
CaCO3(s) + H2O(liq) + CO2(g)
Ca2+(aq) + 2 HCO3-(aq)
At a given T and P of CO2, [Ca2+] and
[HCO3-] can be found from the
EQUILIBRIUM CONSTANT.
THE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the
type
aA + bB
c C + d D the following
is a CONSTANT (at a given T).
conc. of products
K =
[C]c [D]d
[A]a [B]b
equilibrium constant
conc. of reactants
If K is known, then we can predict
concentrations of products or reactants.
10
Determining K
2 NOCl(g)
2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl in a 1.00 L flask.
At equilibrium you find 0.66 mol/L of NO.
Calculate K.
Solution
Set up a table of concentrations:
[NOCl]
[NO]
[Cl2]
Initial
2.00
0
0
Change
Equilibrium
0.66
11
Determining K
2 NOCl(g)
2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask.
At equilibrium you find 0.66 mol/L of NO.
Calculate K.
Solution
Set of a table of concentrations:
[NOCl]
[NO]
[Cl2]
Initial
2.00
0
0
Change
+0.33
+0.66
-0.66
0.33
Equilibrium
0.66
1.34
12
13
Determining K
2 NOCl(g)
Initial
Change
Equilibrium
2 NO(g) + Cl2(g)
[NOCl]
[NO]
[Cl2]
2.00
0
0
-0.66
+0.66
+0.33
1.34
0.66
0.33
K=
K=
[NO]2 [Cl2 ]
[NOCl]2
2
[NO] [Cl2 ]
[NOCl]2
2
=
(0.66) (0.33)
(1.34)2
= 0.080
14
Writing and Manipulating K
Expressions
Solids and liquids
NEVER appear in
equilibrium expressions.
S(s) + O2(g)
[SO 2 ]
K=
[O 2 ]
SO2(g)
15
Writing and Manipulating K
Expressions
Solids and liquids NEVER
appear in equilibrium
expressions.
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
[NH 4+ ][OH - ]
K=
[NH3 ]
16
Writing and Manipulating K
Expressions
Changing coefficients
K=
S(s) + 3/2 O2(g)
SO3(g)
2 S(s) + 3 O2(g)
2 SO3(g)
Knew =
Knew =
2
[SO3 ]
[O2 ]3
= (Kold )2
[SO 3 ]
[O 2 ]3/2
2
[SO3 ]
[O2 ]3
Writing and Manipulating K
Expressions
Changing direction
S(s) + O2(g)
SO2(g)
SO2(g)
S(s) + O2(g)
K=
[SO 2 ]
[O 2 ]
[O 2 ]
K new =
[SO 2 ]
[O 2 ]
1
K new =
=
[SO 2 ]
K old
17
Writing and Manipulating K
Expressions
Adding equations for reactions
S(s) + O2(g)
SO2(g)
SO2(g) + 1/2 O2(g)
K1 = [SO2] / [O2]
SO3(g)
K2 = [SO3] / [SO2][O2]1/2
NET EQUATION
S(s) + 3/2 O2(g)
K net =
[SO 3 ]
SO3(g)
= K1 • K 2
[O 2 ] 3/2
18
Manipulating K Expressions
19
• The equilibrium expression is tied to the
equation from which it is written.
• Changing the equation changes the
expression and thus the numerical value of K.
• Three general cases need to be considered.
1. If the equation is multiplied by a number, “a”,
then the K is raised to the “a” power.
2. If the equation is reversed, then the new K is
the reciprocal of the old K.
3. If two equations are added, the new K is the
product of the two old K's.
• Using these rules, new K's can be derived for
the modified equations.
SAMPLE QUESTIONS
Writing and Manipulating K
Expressions
Concentration Units
We have been writing K in terms of mol/L.
These are designated by
Kc .
But with gases, P = (n/V)•RT = [ ] • RT
P is proportional to concentration, so we can
write K in terms of P.
These are designated by Kp.
Kc and Kp may or may not be the same.
Kp = Kc(RT)Dn
20
Writing and Manipulating K
Expressions
Practice Problems
• Write the equation and the Kc expression
for the formation of two moles of gaseous
ammonia from the elements in the
standard state.
• Kc for this reaction at 25oC is 3.5x108.
• Calculate Kp for this reaction.
• Calculate Kc for the reaction forming the
elements from one mole of ammonia gas.
21
22
The Meaning of K
We can tell if a reaction is product-favored
or reactant-favored.
For
N2(g) + 3 H2(g)
2 NH3(g)
Kc =
[NH3 ]2
[N2 ][H2 ]3
8
= 3.5 x 10
Concentration of products is much greater
than that of reactants at equilibrium.
The reaction is strongly product-favored.
23
The Meaning of K
For AgCl(s)
Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 1.8 x 10-5
Concentration of products is
much less than that of
reactants at equilibrium.
The reaction is strongly
reactant-favored.
Ag+(aq) + Cl-(aq)
AgCl(s)
is product-favored.
24
Comparing Q and K
• The relative magnitudes of Q and
K tell us which direction the
reaction will proceed to reach
equilibrium.
• If Q<K, not at equilibrium and
Reactants ------> Products.
• If Q=K, the system is at
equilibrium.
• If Q>K, not at equilibrium and
Products ------> Reactants.
Using Q
Can tell if a reaction is at equilibrium.
If not, which way it moves to approach
equilibrium.
n-butane
H H H H
H—C—C—C—C—H
H H H H
K =
[iso]
[n]
iso-butane
H H H
H—C—C—C—H
H
H
H C H
H
= 2.5
25
26
Using Q
n-butane
iso-butane
H H H H
H—C—C—C—C—H
H H H H
H H H
H—C—C—C—H
H
H
H C H
K =
[iso]
[n]
H
= 2.5
If [iso] = 0.35 M and [n] = 0.15 M, are you
at equilibrium?
Which way does the reaction “shift” to
approach equilibrium?
Using Q
27
In general, all reacting chemical systems are
characterized by their REACTION QUOTIENT, Q.
product concentrations
Q =
reactant concentrations
If Q = K, then system is at equilibrium.
[iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?
conc. of iso
0.35
Q =
=
= 2.3
conc. of n
0.15
Q (2.3) < K (2.5)
Reaction is NOT at equilibrium, so [Iso] must
larger and [n] must ____________.
decrease
become ________
SAMPLE QUESTIONS
28
Using K
PROBLEM: Place 1.00 mol each of H2 and I2
in a 1.00 L flask. Calculate the equilibrium
concentrations.
H2(g) + I2(g)
[HI] 2
Kc =
= 55.3
[H2 ][I2 ]
2 HI(g)
29
H2(g) + I2(g)
2 HI(g), Kc = 55.3
Step 1. Set up table to define
EQUILIBRIUM concentrations.
[H2]
[I2]
[HI]
Initial
1.00
1.00
0
Change
-x
-x
+2x
Equilib
1.00-x
1.00-x
2x
where x is defined as amount of H2 and I2
consumed on approaching equilibrium.
30
H2(g) + I2(g)
2 HI(g), Kc = 55.3
Step 2. Put equilibrium concentrations
into Kc expression.
2
[2x]
Kc =
= 55.3
[1.00 - x][1.00 - x]
Step 3. Solve Kc expression - take
square root of both sides.
2x
7.44 =
1.00 - x
31
H2(g) + I2(g)
2 HI(g), Kc = 55.3
Step 3. Solve Kc expression - take
square root of both sides.
2x
7.44 =
1.00 - x
x = 0.788
Therefore, at equilibrium
[H2] = [I2] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M
32
Sample Problem
At a certain temperature 8.00 atm of
H2S(g) comes to equilibrium with H2(g)
and S2(g). The equilibrium pressure of
the sulfur gas is 0.60 atm.
Write the equation for the
decomposition of the hydrogen sulfide,
calculate the equilibrium pressure for
each gas, and calculate KP.
Solution
33
Nitrogen Dioxide
Equilibrium
N2O4(g)
2 NO2(g)
Nitrogen Dioxide Equilibrium
N2O4(g)
2 NO2(g)
2
[NO2 ]
Kc =
= 0.0059 at 298 K
[N2O4 ]
If initial concentration of N2O4 is 0.50 M, what
are the equilibrium concentrations?
Step 1. Set up an equilibrium table
[N2O4]
[NO2]
Initial
0.50
0
Change
+2x
-x
Equilib
0.50 - x
2x
34
35
Nitrogen Dioxide Equilibrium
N2O4(g)
2 NO2(g)
Step 2. Substitute into Kc expression and solve.
[NO2 ]2
(2x)2
Kc = 0.0059 =
=
[N2O 4 ]
(0.50 - x)
Rearrange:
0.0059 (0.50 - x) = 4x2
0.0029 - 0.0059x = 4x2
4x2 + 0.0059x - 0.0029 = 0
This is a QUADRATIC
ax2 + bx + c = 0
a = 4
b = 0.0059
EQUATION
c = - 0.0029
Nitrogen Dioxide Equilibrium
N2O4(g)
2 NO2(g)
36
Solve the quadratic equation for x.
ax2 + bx + c = 0
a = 4
b = 0.0059
c = - 0.0029
x =
x =
-b 
-0.0059 
b
2
- 4ac
2a
2
(0.0059)
2(4)
- 4(4)(-0.0029)
x = - 0.00074 ± 1/8(0.046)1/2 = - 0.00074 ± 0.027
37
Nitrogen Dioxide Equilibrium
N2O4(g)
2 NO2(g)
x =
-0.0059 
2
(0.0059)
2(4)
- 4(4)(-0.0029)
x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027
x = 0.026 or -0.028
But a negative value is not reasonable.
Conclusion
[N2O4] = 0.50 - x = 0.47 M
[NO2] = 2x = 0.052 M
More Sample Problems
EQUILIBRIUM AND
EXTERNAL EFFECTS
• Temperature, catalysts, and changes in
concentration affect equilibria.
• The outcome is governed by
LE CHATELIER’S
PRINCIPLE
• “...if a system at equilibrium is
disturbed, the system tends to shift
its equilibrium position to counter
the effect of the disturbance.”
38
39
EQUILIBRIUM AND
EXTERNAL EFFECTS
Henri Le Chatelier
1850-1936
Studied mining
engineering.
Interested in glass
and ceramics.
40
Figure 16.6
50o C
0o C
NO2 / N2O4 is temperature dependent.
41
EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature change --->
change in K
• Consider the fizz in a soft drink
CO2(g) + H2O(liq)
CO2(aq) + heat
• Decrease T. What happens to equilibrium
position? To value of K?
• K = [CO2] / P (CO2)
K increases as T goes down because
[CO2] increases and P(CO2) decreases.
• Increase T. Now what?
• Equilibrium shifts left and K decreases.
42
Temperature Effects
on Equilibrium
N2O4 (colorless) + heat
2 NO2 (brown)
DHo = + 57.2 kJ
Kc =
2
[NO2 ]
[N2O 4 ]
Kc = 0.00077 at 273 K
Kc = 0.0059 at 298 K
43
EQUILIBRIUM AND EXTERNAL EFFECTS
• Add catalyst --->
no change in K
• A catalyst only affects the RATE of
approach to equilibrium.
Catalytic exhaust system
44
NH3
Production
Fritz Haber, 1909
• N2(g) + 3 H2(g)
2 NH3(g)
• K = 3.5 x 108 at 298 K
• K = 0.16 at 723 K
45
EQUILIBRIUM AND EXTERNAL EFFECTS
Concentration changes
no change in K
only the position of
equilibrium changes
46
Le Chatelier’s Principle
Adding a “reactant” to a chemical system.
47
Le Chatelier’s Principle
Removing a “reactant” from a chemical system.
48
Le Chatelier’s Principle
Adding a “product” to a chemical system.
49
Le Chatelier’s Principle
Removing a “product” from a chemical system.
50
Figure 16.7
(a) 7 n-butane molecules
(b) 7 iso-butane are added
(c) What are the new equilibrium concentration?
51
ButaneIsobutane
Equilibrium
butane
[isobutane]
K =
= 2.5
[butane]
isobutane
52
Butane
Isobutane
Assume you are at equilibrium with [iso] = 1.25 M
and [butane] = 0.50 M. Now add 1.50 M butane.
When the system comes to equilibrium again,
what are [iso] and [butane]? K = 2.5
butane
isobutane
Butane
Isobutane
53
Assume you are at equilibrium with [iso] = 1.25 M
and [butane] = 0.50 M. Now add 1.50 M butane.
When the system comes to equilibrium again,
what are [iso] and [butane]? K = 2.5
Solution
Calculate Q immediately after adding more
butane and compare with K.
[isobutane]
1.25
Q =
= 0.63
=
[butane]
0.50 + 1.50
Q is LESS THAN K. Therefore, the reaction will
Right
(products)
shift to the ____________.
Butane
Isobutane
You are at equilibrium with [iso] = 1.25 M and
[butane] = 0.50 M. Now add 1.50 M butane.
Solution
Q is less than K, so equilibrium shifts right —
away from butane and toward isobutane.
Set up concentration table
Initial
Change
Equilibrium
[butane]
[isobutane]
0.50 + 1.50
1.25
-x
+x
2.00 - x
1.25 + x
54
55
Butane
Isobutane
You are at equilibrium with [iso] = 1.25 M and
[butane] = 0.50 M. Now add 1.50 M butane.
Solution
[isobutane]
1.25 + x
K = 2.50 =
=
[butane]
2.00 - x
x = 1.07 M
At the new equilibrium position,
[butane] = 0.93 M and [isobutane] = 2.32 M.
Equilibrium has shifted toward
isobutane.
56
Nitrogen Dioxide Equilibrium
N2O4(g)
2 NO2(g)
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
Increase P in the system
by reducing the volume.
57
Nitrogen Dioxide Equilibrium
N2O4(g)
2 NO2(g)
2
[NO2 ]
Kc =
= 0.0059 at 298 K
[N2O4 ]
Increase P in the system by reducing the
volume.
In gaseous system the equilibrium will shift to
the side with fewer molecules (in order to
reduce the P).
Therefore, reaction shifts LEFT and P of NO2
decreases and P of N2O4 increases.
58
Sample Problem
Write the equation for the formation of two mole of
ammonia gas from the elements in the standard
state. DH for this reaction is
-92kJ. Predict the
change in K and the direction of shift of an
equilibrium mixture when:
3 H2 (g) + N2 (g) <--> 2 NH3 (g) + 92kJ
1. The temperature is increased.smaller, left
2. The PT is increased by adding He.none, right
3. The total volume is increased.none, left
4.Hydrogen gas is removed. none, left
5.Nitrogen gas is added.none, right
6.The temperature is decreased.larger, right
7.The total volume is decreased. none, right
Practice Problems
59
1.
2 NO (g) + O2 (g) <--> 2 NO2 (g)
a) Write the Kp expression.
b) Write the Kc expression.
c) If Kc = 0.50 at 25oC, calculate Kp.
d) Determine Kc for the following reaction:
4 NO (g) + 2 O2 (g) <--> 4 NO2 (g)
e) If 2.5 moles of NO, O2, and NO2 are placed in
a 5.0 L container, is the system at equilibrium?
If not, will the reaction proceed to the left or to
the right?
f) If 2.5 moles of NO and O2 are placed in a 5.0 L
container at 350oC, 1.1 moles of NO2 are
present at equilibrium. Calculate Kc.
Practice Problems
60
2. 2.5 moles of carbon dioxide is placed in a 0.50 L
flask.
CO2 (g) <--> C (s) + O2 (g) Kc = 1.82 at 25oC
Calculate all equilibrium concentrations.
3. Calculate all equilibrium concentrations if the
initial
concentration of Cl2 is .100.
Cl2 (g) <--> 2 Cl (g)
Kc = 0.036
4. Calculate the equilibrium pressure of NO2 if the
initial
pressure of N2O4 is 1.0 atm at 25o C.
N2O4 (g) <--> 2 NO2 (g)
Kc = 0.11
61
Practice Problems
5. The initial concentrations of CO2 and H2 are
0.100 M.
Calculate all equilibrium concentrations.
CO2 (g) + H2 (g) <--> CO (g) + H2O (g) Kc = 0.64
6. BaCO3(s) + 2H+(aq) <--> H2O(l) + CO2(g) +Ba2+(aq)
Explain the effects of the following on the given
equation:
a) adding HCl
b) adding BaCl2
c) adding H2O
d) adding BaCO3
e) adding NaOH
62
Practice Problems Answers
1. a)
Kp =
P2 NO2
P2 NO P O2
c) 0.02
e) left
2. 1.8, 3.2
4. 1.1 atm
6. a) right
e) right
b) left
b)
Kc =
[NO2]2
[NO]2[O2]
d) 0.25
f) 1.6
3. 0.074, 0.052
5. 0.044, 0.044,
0.056, 0.056
c) none
d) none
The End!!!
63
Sample Questions
Write the expression.
1.
2 H2 (g) + O2 (g)
Kc =
[H2O]2
[H2]2[O2]
<-->
2 H2O (g)
64
Sample Questions
Write the expression.
2.
4 H2 (g) + 2 O2 (g) <-->
Kc =
[H2O]4
[H2]4[O2]2
4 H2O (g)
65
Sample Questions
Write the expression.
3.
2 H2O (g) <--->
Kc =
2 H2 (g) + O2 (g)
[H2] 2 [O2]
[H2O]2
66
Sample Questions
1.
2 H2 (g) + O2 (g) <-->
If
Kc =
[H2O]2
[H2]2 [O2]
2 H2O (g)
= 5.0
2.
4 H2 (g) + 2 O2 (g) <--> 4 H2O (g)
[H2O]4
Kc =
= 25
[H2]4 [O2]2
3.
2 H2O (g)
<--->
2 H2 (g) + O2 (g)
[H2]2 [O2]
Kc =
= 0.20
[H2O]2
67
Sample Questions
Write the equation and the Kc expression for the
formation of two moles of gaseous ammonia from
the elements in the standard state.
3 H2 (g) + N2 (g) <--> 2 NH3 (g)
Kc =
[NH3] 2
[H2] 3 [N2]
68
Sample Questions
Kc for this reaction at 25oC is 3.5x108.
Calculate Kp for this reaction.
3 H2 (g) + N2 (g) <--> 2 NH3 (g)
Kp = Kc (RT)Dn
Kp = 3.5 x 108 (0.0821x 298)-2
Kp = 5.8 x 105
69
Sample Questions
Calculate Kc for the reaction forming the
elements from one mole of ammonia gas.
NH3 (g) <--> 3/2 H2 (g) + 1/2 N2 (g)
Kc =
[H2] 3/2 [N2]1/2
[NH3]
1
=
(3.5 x 108)1/2
= 5.3 x 10-5
70
The Meaning of Q
• Q, Reaction Quotient
• Same expression as the
equilibrium expression.
• Non-equilibrium concentration
numbers may be used.
• Used to determine if a system is at
equilibrium and if not, which
direction it will move to reach
equilibrium.
71
Sample Questions
1. If [NO2] = 0.050 and [N2O4] = 0.050, is the
system at equilibrium? If not, in which
direction does the reaction move to
come to equilibrium?
N2O4 (g)
Q=
<-->
[NO2] 2
[N2O4]
2 NO2 (g)
Kc = 0.36
(0.050) 2
=
(0.050)
= 0.050
Q < Kc forward direction
(to right, products)
72
Sample Questions
2. If [NO2] = 0.50 and [N2O4] = 0.40, is the
system at equilibrium? If not, in which
direction does the reaction move to
come to equilibrium?
N2O4 (g) <--> 2 NO2 (g)
Kc = 0.36
Q=
[NO2] 2
[N2O4]
(0.50) 2
=
(0.40)
= 0.62
Q > Kc reverse direction
(to left, reactants)
73
Sample Questions
3. If [NO2] = 0.50 and [N2O4] = 0.69, is the
system at equilibrium? If not, in which
direction does the reaction move to
come to equilibrium?
N2O4 (g) <--> 2 NO2 (g)
Kc = 0.36
Q=
[NO2] 2
[N2O4]
(0.50) 2
=
(0.69)
= 0 .36
Q = Kc equilibrium
Determining K
2 NOCl(g)
2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At
equilibrium you find 0.66 mol/L of NO.
Calculate K. (Review of slides 9-11.)
Solution
Set up a table of concentrations:
[NOCl]
[NO]
[Cl2]
Initial
2.00
0
0
+0.66
+ 0.33
Change
- 0.66
Equilibrium
0.66
1.34
0.33
74
75
Determining K
2 NOCl(g)
Before
Change
Equilibrium
2 NO(g) + Cl2(g)
[NOCl]
[NO]
[Cl2]
2.00
0
0
-0.66
+0.66
+0.33
1.34
0.66
0.33
K=
K=
[NO]2 [Cl2 ]
[NOCl]2
2
[NO] [Cl2 ]
[NOCl]2
2
=
(0.66) (0.33)
(1.34)2
= 0.080
76
Sample Questions
1. PCl5 dissociates to produce PCl3 and
Cl2. If 2.00 moles of PCl5 is placed in a
1.00 L flask it is found that 10.0% of the
original PCl5 had dissociated, at
equilibrium. Calculate Kc.
PCl5
<-->
PCl3
+
Cl2
2.00
0
0
- 0.200
+ 0.200
+0 .200
1.80
Kc =
0.200
[PCl3][Cl2]
[PCl5]
=
0.200
(.200)(.200)
(1.80)
= 0.0222
77
Sample Questions
2. If 1.50 moles of H2O are placed in a
3.00 L flask 1.20 moles of H2O remain
at equilibrium. Calculate Kc.
2 H 2O
<-->
2 H2
+
O2
Can we use
only Moles?...
0.500
0
0
- 0.100
+ 0.100
+ 0.050
0.400
0.100
0.050
No….reason is all K’s are relative, that is
2(.050)
since all reactions
are
in different
[H2]2 [O
2] done(.100)
Kc =moles and size
= of the container
= 0.0031
containers,
2
2
[H2O]
(0.400)
determine the K.
The End!!!
78
Sample Questions
3. When 0.020 mole N2O and 0.0560 mole
O2 are placed in a 1.00 L flask it is
found that there is 0.020 mole NO2 at
equilibrium. Calculate Kc.
2 N 2 O + 3 O2
<-->
4 NO2
0.020
0.0560
0
- 0.010
- 0.015
+0 .020
0.010
Kc =
0.041
[NO2]4
[N2O]2 [O2]3
0.020
=
(0.020)4
(.010)2(.041)3
= 23
79
Sample Questions
4. Some solid ammonium hydrogen sulfide
is placed in a flask at 298 K. At
equilibrium, the pressure was 0.660 atm.
Calculate Kp and Kc.
NH4HS
<-->
NH3 + H2S
some
0
0
- some
+ 0.330
+ 0.330
some - some
0.330
0.330
Kp = PNH3PH2S = (0.330)2 = 0.109
Kp = Kc(RT)2 0.109 = Kc (0.0821x298)2
Kc = 0.000182
80
Sample Problem
2 H2S (g)
8.00
- 1.20
<-->
6.80
Kp =
2 H2 (g)
0
+ 1.20
+
1.20
PH22 PS2
PH2S2
=
S2 (g)
0
+ 0.60
0.60
(1.20)2 (0.60)
(6.80)2
= 0.019
81
Sample Problems
1. Calculate the equilibrium pressure of
CO if 25 g COBr2 is placed in a 10.0 L
flask at 298 K.
CO (g) + Br2 (g) <--> COBr2 (g) Kp = 0.19
25 g
mole
= 0.13 mole
187.8 g
P = nRT/V
P = (0.13 mol)(0.0821 L atm/molK)(298K)/10.0 L)
P = 0.32 atm
82
Sample Problems
1. Calculate the equilibrium pressure of
CO if 25 g COBr2 is placed in a 10.0 L
flask at 298 K.
CO (g) + Br2 (g) <--> COBr2 (g) Kp = 0.19
0
0
+x
+x
x
x
Kp =
PCOBr2
PCOPBr2
0.32
-x
0.32 - x
=
(0.32 - x)
x2
= 0.19
x = 0.30atm = PCO
83
Sample Problem
2. Calculate all equilibrium concentrations
if 0.500 moles of HI is placed in a 2.00 L
flask.
2 HI
<-->
H2 +
I2
Kc = 0.018
0.250
0
0
- 2x
+x
+x
0.250 - 2x
[H2][I2]
Kc =
[HI]2
x
x
=
x2
(0.250 -
2x)2
= 0.018
x = 0.026 M = [H2] = [I2]
[HI] = 0.20 M
84
Sample Problem
3. Calculate all equilibrium concentrations
when the initial concentrations of SO2 and
NO2 are 0.0500.
SO2 + NO2 <--> NO + SO3
Kc = 85.0
0.0500
- x
0.0500 - x
Kc =
0.0500
- x
0.0500 - x
[NO][SO3]
[SO2][NO2]
=
0
+x
0
+x
x
x
x2
(0.0500 -
x)2
= 85.0
x = 0.0451 M = [NO] = [SO3]
[SO2] = [NO2] = 0.0049 M