Chapter 6: Chemical Equilibrium
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
The Equilibrium Condition
The Equilibrium Constant
Equilibrium Expressions Involving Pressures
The Concept of Activity
Heterogeneous Equilibria
Applications of the Equilibrium Constant
Solving Equilibrium Problems
LeChatelier’s Principle
Equilibria Involving Real Gases
1
Chemical Equilibrium
So far, we have assumed that a chemical reaction goes to
completion as written.
CO(g) + H2O (g) →
CO2 (g)
+ H2 (g)
In general, this is not correct. Instead, a stable state of the
system in reached, which includes both reactants and
products at constant concentrations. It is called the
equilibrium state, or simply “equilibrium”.
CO(g) + H2O (g)
⇌
CO2 (g)
+
H2 (g)
2
Molecular Picture of
Establishment of Equilibrium
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
AFTER IT STOPS CHANGING
INITIAL
2 CO(g) + 2 H2O(g) + 5 CO2(g) + 5 H2(g)
7 CO(g) + 7 H2O(g) + 0 CO2(g) + 0 H2(g)
3
Concentration vs. Time
CO(g) + H2O(g) → CO2(g) + H2(g)
Ratio depends
on temperature
4
Characteristics of Chemical
Equilibrium States
•
Reaching equilibrium requires that reactions
occur.
Chemical reactions occur via collisions of reacting molecules.
The energy released during collisions breaks bonds in
reactant molecules, allowing the atoms to rearrange to form
a product.
•
Once equilibrium is reached, there is no
macroscopic evidence of further change.
•
Reached through dynamic balance of
forward and reverse reaction rates.
5
CHEMICAL EQUILIBRIUM
2 NO2 (g)
N2O4 (g)
Dimerization of nitrogen dioxide
Decomposition of dinitrogen tetroxide
“Chemical equilibrium is the state at which the
concentrations of all reactants and products remain
the same”
“ At chemical equilibrium the rate of formation and
decomposition remains the same”
“ When equilibrium favors product, equilibrium is
shifted to the right”
6
Reaching Equilibrium on the
Macroscopic and Molecular Levels
DECOMPOSITION
N2O4 (g)
2 NO2 (g)
Colorless
Brown
7
Reaction of 2 NO2(g) ↔ N2O4(g)
DIMERIZATION
2 NO2 (g)
BROWN
N2O4 (g)
COLORLESS
8
Consider the following equation:
5
CO(g) + H2O(g)
CO2(g) + H2(g)
Which of the following must be true at equilibrium?
1.
2.
3.
4.
5.
[CO2] = [H2] because they
are in a 1:1 mole ratio in
the balanced equation.
The total concentration of
reactants is equal to the
total concentration of the
products.
The total concentration of
the reactants is greater than
the total concentration of
the products.
The total concentration of
the products is greater than
the total concentration of
the reactants.
None of these is true.
9
The Equilibrium Constant, K
K - equilibrium constant is based on
LAW of MASS ACTION
jA + kB
lC+mD
where A, B, C, D are chemical species and j, k, l, m are coefficients
K=
[C]l [D]m
[A]j [B]k
The “apparent units” for K are
concentration units raised to
some power = l+m–(j+k)
NOTE: conc. of chemical species are at EQUILIBRIUM!!
K depends only on T , not on initial concentrations.
10
KINETICS of CHEMICAL REACTIONS
EQUILIBRIUM
Changes of
concentrations
of reactant
and product
during chemical
reaction.
Equilibrium
concentrations will
yield K value.
11
TIME
EQUILIBRIUM CONSTANT
ratio of products to reactants
depends only on TEMPERATURE
versus…
Chemical equilibrium (the equilibrium position)
a dynamic situation
can be affected by temperature, pressure, and/or
initial concentration of reactants and products.
12
Initial and Equilibrium Concentrations for the
N2O4-NO2 System at 100°C
Initial
[N2O4]
Equilibrium
Ratio
[NO2]
[N2O4]
[NO2]
[NO2]2 [N2O4]
0.1000
0.0000
0.0491
0.1018
0.211
0.0000
0.1000
0.0185
0.0627
0.212
0.0500
0.0500
0.0332
0.0837
0.211
0.0750
0.0250
0.0411
0.0930
0.210
=K
Experiments carried out at a certain temperature, but with different
initial concentrations, yield the same value for the EQUILIBRIUM
13
CONSTANT, K.
Reaching Equilibrium from Different Starting Points
CO(g) + 2 H2(g) ⇌ CH3OH(g)
K=
[CH 3OH ]
2
[CO][H 2 ]
In Exp 3, all three
species start at the
same
concentration.
14
Types of Equilibrium Constants
Dissociation constant (acids, bases, salts; Chs 7/8)
Solubility product (precipitation reactions; Ch 8)
Stability constant (complexation reactions; Ch 8)
Standard potentials (redox reactions; Ch 11)
15
Equilibrium Constants Big and Small
Equilibrium values can be wide ranging, both big and small!!!
Small K
N2 (g) + O2 (g) ⇌ 2 NO(g)
K = 1 x 10 –30
Essentially only reactants at equilibrium.
Large K
2 CO(g) + O2 (g) ⇌ 2 CO2 (g)
K = 2.2 x 1022
Essentially only products at equilibrium.
Intermediate K
2 BrCl(g) ⇌ Br2 (g) + Cl2 (g)
K=5
Comparable amounts of products and reactants at
equilibrium.
16
Equilibrium arises through dynamic balance
between forward and reverse reactions
k1
→ NO (g) + NO(g)
Forward:
2NO2(g)
Reverse:
NO3(g) + NO(g)
3
k -1
→ 2NO (g)
2
Forward rate = k1[NO2][NO2] = k1[NO2]2
Reverse rate = k-1[NO3][NO]
k1 and k-1 reflect probabilities that one collision leads to reaction
17
Kinetic Approach to Equilibrium
k1 [NO2 ] = k!1 [NO][NO3 ]
2
k1 [NO][NO3 ]
=
=K
2
k!1
NO
[ 2]
The RATE of forward reaction decreases as the reaction
progresses since the initial concentration of reactants
18
decreases over time.
Fig 6.5
Fe
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
K=
[NH3]2
[N2][H2]3
Due to reaction stoichiometry, hydrogen concentration
decreases three times as fast as nitrogen concentration.
19
Table 6.1 K has the same value regardless of the
initial concentrations of reactants.
Experiment
Initial
Concentrations
Equilibrium
Concentrations
K=
[NH3]2
[N2] [H2]3
I
[N2]o= 1.00 M
[H2]o = 1.00 M
[NH3]o = 0
[N2] = 0.921 M
[H2] = 0.763 M
[NH3] = 0.157 M
6.02 x 10-2
L2/mol2
II
[N2]o = 0
[H2]o = 0
[NH3]o = 1.00 M
[N2] = 0.399 M
[H2] = 1.197 M
[NH3] = 0.203 M
6.02 x 10-2
L2/mol2
III
[N2]o = 2.00 M
[H2]o = 1.00 M
[NH3]o = 3.00 M
[N2] = 2.59 M
[H2] = 2.77 M
[NH3] = 1.82 M
6.02 x 10-2
L2/mol2
20
Calculation of the Equilibrium Constant from
Equilibrium Concentrations
At 454 K, the following reaction takes place:
3 Al2Cl6(g) ⇌ 2 Al3Cl9(g)
At equilibrium, the concentration of Al2Cl6(g) is 1.00 M and of
Al3Cl9(g) is 1.02 x 10-2 M.
Calculate the equilibrium constant at 454 K.
Strategy: Substitute values into K
K=
[Al3Cl9]2 (1.02 x 10-2 M)2
=
= 1.04 x 10-4 M-1
3
3
[Al2Cl6]
(1.00 M)
21
Determining Equilibrium Concentrations from K
Methane can be made by reacting carbon disulfide with
hydrogen gas:
CS2 (g) + 4 H2 (g) ⇌ CH4 (g) + 2 H2S (g)
K for this reaction is 27.8 (L/mol)2 at 900oC. At equilibrium
the reaction mixture in a 4.70 L flask is: 0.250 mol CS2, 1.10
mol of H2, and 0.45 mol of H2S. How many moles of
methane are formed?
Plan:
Calculate the
equilibrium
concentrations
Use moles and V
Concentration of
methane
Use K and equilibrium
expression
Convert
molarity to
moles
22
What is the equilibrium constant
expression for this reaction?
CS2 (g) + 4 H2 (g) ⇌ CH4 (g) + 2 H2S (g)
[CH ][ H S ]
[CS ][ H ]
[CH ][ H S ]
2. K=
[CS ][ H ]
[CS ][ H ]
3. K= CH H S
[ ][ ]
1. K=
4
2
2
2
2
4
2
2
2
4
2
2
4
2
[CS ][ H ]
4. K=
[CH ][ H S ]
4
2
2
4
2
2
23
Determining Equilibrium Concentrations from K
0.250 mol
[CS2] =
= 0.0532 M
4.70 L
[H2] =
1.10 mol
= 0.234 M
4.70 L
0.450 mol
[H2S] =
= 0.0957 M
4.70 L
K=
[CH4] = K
[CH4][H2S]2
= 27.8 L2/mol2
[CS2][H2]4
(0.0532 M)(0.234 M)4 = 0.484 M
[CS2][H2]4
2
=
27.8
(L/mol)
(0.0957 M)2
[H2S]2
moles CH4 = 0.484 M x 4.70 L = 2.27 moles
24
Using and Rearranging K
CS2(g) + 3 O2(g) ⇌ CO2(g) + 2 SO2(g)
K=
[CO2][SO2]2
[CS2][O2]3
The equilibrium expression for a reaction written in reverse is
the reciprocal of that for the original expression.
CO2(g) + 2 SO2(g) ⇌ CS2(g) + 3 O2(g)
[CS2][O2]3
1
K’ =
=
2
[CO2][SO2] K
If the original reaction is multiplied by a factor n, the new
equilibrium constant is the original raised to the power n.
2 CS2(g) + 6 O2(g) ⇌ 2 CO2(g) + 4 SO2(g)
K’’ =
[CO2] 2[SO2]4
= K2
[CS2] 2[O2]6
25
Like Example 6.1 - I
The following equilibrium concentrations were observed for the
reaction between CO and H2 to form CH4 and H2O at 927oC.
CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g)
[CO] = 0.613 mol/L
[H2] = 1.839 mol/L
[CH4] = 0.387 mol/L
[H2O] = 0.387 mol/L
a) Calculate the value of K at 927oC for this reaction.
b) Calculate the value of the equilibrium constant at 927oC for:
H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g)
c) Calculate the value of the equilibrium constant at 927oC for:
1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g)
Solution:
a) Based on the balanced reaction above:
K=
[CH4] [H2O]
(0.387 mol/L) (0.387 mol/L)
= 0.0393 L2/mol2
=
26
[CO] [H2]3
(0.613 mol/L) (1.839 mol/L)3
K=0.0393 L2/mol2 for the reaction:
CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g)
What is the value of the equilibrium constant for each of
these reactions under the same conditions?
K’: H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g)
K’’: 1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g)
1. K’=-0.0393 ; K’’= 0.3400
2. K’=-0.0393 ; K’’= 6.07x10-5
3. K’=-0.0393 ; K’’= 0.0131
4. K’=25.44 ; K’’= 0.3400
5. K’=25.44 ; K’’= 6.07x10-5
6. K’=25.44 ; K’’=0.0131
27
Like Example 6.1 - II
b) Calculate the value of the equilibrium constant at 927oC for:
H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g)
K’ =
[CO] [H2]3
[H2O] [CH4]
=
(0.613 mol/L) (1.839 mol/L)3
(0.387 mol/L) (0.387 mol/L)
= 25.45 mol2/L2
Shortcut if K is known: K’ is just the reciprocal of K from part a:
1
1
K’ =
=
= 25.45 mol2/L2
K
0.0393 L2/mol2
c) Calculate the value of the equilibrium constant at 927oC for:
1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g)
K” =
[H2O]1/3 [CH4]1/3
[CO]1/3 [H2]
=
(0.387mol/L)1/3 (0.387 mol/L)1/3
(0.613 mol/L)1/3 (1.839 mol/L)
Shortcut if K is known: K” = (K)1/3 = (0.0393L2/mol2)1/3
28
Summary: Some Characteristics of the
Equilibrium Expression
The equilibrium expression for a reaction written in reverse is the
reciprocal of that for the original reaction. Thus Knew = 1/Koriginal
When the balanced equation for a reaction is multiplied by a
factor n, the equilibrium expression for the new reaction is the
original expression raised to the nth power. Thus Knew = (Koriginal)n
The apparent units for K are determined by the powers of the
various concentration terms. The (apparent) units for K therefore
depend on the reaction being considered.
29
Expressing K with Pressure Units
For gases, PV=nRT can be rearranged to give: P = n RT
V
n
P
=
V
RT
or:
n
= Molar concentration; R is a
V
constant, so if we keep the T
constant, the molar concentration
is directly proportional to the
pressure.
Example: 2 NO (g) + O2 (g) ⇌ 2 NO2 (g)
K=
[NO2]2
[NO]2 [O2]
Using concentrations
and
Kp =
P 2NO2
P 2NO PO2
Using partial pressures
30
Relationship between Kc and Kp
jA + kB
Kc =
[C]l [D]m
[A]j [B]k
lC + mD
Kp =
PlC PmD
PjA PkB
For an equilibrium between gaseous compounds, there is a
relationship between the equilibrium constants Kc and KP.
Kp = Kc (RT)Δn
Example:
Δn = (l+m) – (j+k)
for gaseous components only
2 NO (g) + O2 (g) ⇌ 2 NO2 (g)
Δn = (l) – (j+k) = 2 - (2+1) = 2 – 3 = -1
So, Kp = Kc
RT
31
For which of the following reactions
does Kc = Kp at the same temperature?
Kp = Kc (RT) Δn
Δn = (l+m) – (j+k)
for gaseous components only
1.
2.
3.
4.
N2(g) + 3H2(g)
2NH3(g)
CaCO3(s)
CaO(s) + CO2(g)
CO(g) + H2O(g)
CO2 (g) + H2(g)
3Fe(s) + H2O(g)
Fe3O4(s) + 4H2(g)
32
Use of Activities in Equilibrium Constant Expressions
For Pressures:
Activity (ith component) = ai =
Pi
Pref
where Pi = partial pressure of the ith gaseous component
and Pref = 1 atm (exactly)
For Concentrations:
Activity (ith component) = ai =
Mi
Mref
This is why K has
“apparent units” rather
than actual units – they
cancel out when you
include the reference
states.
where Mi = molarity of the ith component
and Mref = 1 M (exactly)
33
For pure solids and liquids, activity = 1.0
CaCO3 (s) ⇌ CaO (s) + CO2 (g)
KP = PCO2 (the 2 solids do NOT appear in KP!)
The position of an equilibrium involving a pure solid or liquid
does not depend upon the amounts of pure solid or liquids present.
The activity of a pure solid or liquid is always equal to 1, so the
34
“concentration” does not appear in the K expression.
Equilibrium involving pure solids or liquids
Always set their activities to 1
Example: NH4NO2(s) ⇌ N2(g) + 2 H2O(g)
The equilibrium constant for this reaction would normally be
expressed as:
[N2][H2O]2
K=
[NH4NO2]
However, a pure solid or liquid retains the same activity during
the reaction. Thus, we set the activity of NH4NO2(s) to 1.
K = [N2][H2O]2
or
Kp = (PN2)(PH2O)2
35
EXTENT OF REACTION
• value of K >> 1 means that at equilibrium the system will
consist mainly of products
• value of K << 1 means that the system consists mainly of
reactants –the chemical reaction does not occur to much
extent
• K and the time required to reach equilibrium are NOT
directly related
• Rates of chemical reactions are temperature dependent and
so are the equilibrium positions and constants
36
The Reaction Quotient, Q
Consider the reaction:
aA(g) + bB(g) ⇌ cC(g) +dD(g)
The reaction quotient, Q is defined as:
[C]tc [D]td
Q=
[A]ta [B]tb
Q has same form as K, but the concentrations are the actual
concentrations at any time (t) rather than the concentrations
after equilibrium is reached.
37
If Q<K, which direction will
the reaction shift?
aA(g) + bB(g) ⇌ cC(g) +dD(g)
[C]tc [D]td
Q=
[A]ta [B]tb
1. Products (right)
2. Reactants (left)
3. No shift (equilibrium)
38
Equilibrium Constant, K
vs.
Reaction Quotient, Q
While the equilibrium constant describes equilibrium
concentrations, the reaction quotient uses initial concentrations
to determine in which direction a reaction will proceed.
• if Q<K, the system will produce products
(forward reaction; shifts to the right)
• if Q=K, the system is in equilibrium
(no change in concentrations of products or reactants)
• if Q>K , the system will consume products
(reverse reaction; shifts to the left)
39
Reaction Direction and the
Relative Sizes of Q and K
“Excess reactants
initially”
“Excess products
initially”
40
N2O4(g) ⇌ 2NO2(g)
[NO2 ]
Q=
[N 2O4 ]
2
TIME
41
Ways of Expressing Q and K values
Form of Q
Form of Chemical Equation
Forward reaction: A
B
Reversed reaction: B
A
[B]
[A]
Q = 1 = [A]
Qfwd [B]
Qfwd =
Reaction as sum of two steps:
(1) A
C
(2) C
B
Q1 =
Q2 =
Value of K
Kfwd =
[B]eq
[A]eq
1
K=
Kfwd
[C]
[A]
[B]
Qoverall = Q1 x Q2
Koverall = K1 x K2
[C]
Reaction with pure solid or liquid component:
A(s )
B
Q = [B]
K = [B]
42
Writing the Reaction Quotient from the Balanced Equation
Problem: Write the reaction quotient for each of the following
reactions:
(a) The thermal decomposition of potassium chlorate:
KClO3 (s) ⇌ KCl(s) + O2 (g)
(b) The combustion of butane in oxygen:
C4H10 (g) + O2 (g) ⇌ CO2 (g) + H2O(g)
Plan: We first balance the equations, then construct the
reaction quotient.
Solution:
(a)2 KClO3 (s)
2 KCl(s) + 3 O2 (g)
(b) 2 C4H10 (g) + 13 O2 (g)
Qc =
Qc =
[O2]3
8 CO2 (g) + 10 H2O(g)
[CO2]8 [H2O]10
Remember to use initial
43
[C4H10]2 [O2]13 concentrations!
Predict Reaction Direction using Q
Consider the following reaction:
CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g)
1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S and 2.00 mol H2
are mixed in a 250 mL vessel at 960 °C. If K = 0.036 at
this temperature, calculate Q and figure out in which
direction will the reaction proceed in order to reach
equilibrium?
Plan:
Calculate actual
concentrations
Calculate Q
Compare to K
Use moles and
vessel volume
44
Predict Reaction Direction using Q
[CH4]0= 1.00 mol/0.250 L = 4.00 M
[H2S]0 = 2.00 mol/0.250 L = 8.00 M
[CS2]0 = 1.00 mol/0.250 L = 4.00 M
[H2]0 = 2.00 mol/0.250 L = 8.00 M
K = 0.036
Calculate the value of Q:
Q=
[CS2]0[H2]04
(4.00 M)(8.00 M)4
=
[CH4]0[H2S]02 (4.00 M)(8.00 M)2
= 64
Comparing Q and K: Q > K, so the reaction goes to the left.
[reactants] increase and [products] decrease
45
Example 6.2 - I
For the synthesis of ammonia at 500oC, the equilibrium constant is
6.0 x 10-2 L2/mol2. Predict the direction in which the system will
shift to reach equilibrium in each of the following cases.
a) [NH3]0 = 1.0 x 10-3 M
[N2]0= 1.0 x 10-5 M
[H2]0=2.0 x 10-3 M
b) [NH3]0 = 2.00 x 10-4 M
[N2]0= 1.50 x 10-5 M [H2]0= 3.54 x 10-1 M
c) [NH3]0 = 1.0 x 10-4 M
[N2]0= 5.0 M
[H2]0= 1.0 x 10-2 M
Solution:
a) First we calculate Q:
[NH3]02
(1.0 x 10-3 mol/L)2
Q=
=
[N2]0[H2]03
(1.0 x 10-5 mol/L)(2.0 x 10-3 mol/L)3
= 1.3 x 107 L2/mol2
Since K = 6.0 x 10-2 L2/mol2, Q is much greater than K. For the system
to attain equilibrium, [products] must decrease and [reactants] must
increase.
N2 (g) + 3 H2 (g)
2 NH3 (g)
46
Predict the direction of the reaction for the
following conditions:
[NH3]0 = 2.00 x 10-4 M [N2]0= 1.50 x 10-5 M [H2]0= 3.54 x 10-1 M
Q=
[NH3]02
[N2]0[H2]03
1. Right (products)
2. Left (reactants)
3. Equilibrium (no
shift)
47
Example 6.2 - II
K = 6.0 x 10-2 L2/mol2
b) We calculate the value of Q:
Q=
[NH3]02
=
(2.00 x 10-4 mol/L)2
[N2]0[H2]03
(1.50 x 10-5 mol/L) (3.54 x 10-1 mol/L)3)
= 6.01 x 10-2 L2/mol2
In this case Q = K, so the system is at equilibrium..
N2 (g) + 3 H2 (g) =
2 NH3 (g)
c) The value of Q is:
Q=
[NH3]02
[N2]0[H2]03
=
= 2.0 x 10-3 L2/mol2
(1.0 x 10-4 mol/L)2
(5.0 mol/L) (1.0 x 10-2 mol/L)3
N2 (g) + 3 H2 (g)
2 NH3 (g)
Here Q is less than K, so the system will shift to the right, attaining
equilibrium by increasing [product] and decreasing [reactants]:
48
More ammonia!
Calculating Equilibrium Pressures and Concentrations
from K and Initial Conditions
Consider the equilibrium:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
0.250 mol CO and 0.250 mol H2O are placed in a 125 mL
flask at 900 K. What is the composition of the equilibrium
mixture if K = 1.56?
The original reactant concentrations are:
[CO]0 = [H2O]0 = 0.250 mol/ 0.125 L = 2.00 M
[CO2]0 = [H2]0 = 0 mol/ 0.125 L = 0 M
Q = 0. Therefore, Q < K so reactants are consumed and
products made.
49
Calculating Equilibrium Pressures and
Concentrations from K and Initial Conditions
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Construct the reaction table:
often called an “I.C.E.” table.
Conc. (M)
CO(g)
H2O(g)
CO2(g)
H2(g)
Init. [i]
2.00
2.00
0
0
Change
-x
-x
+x
+x
2.00 - x
2.00 - x
x
x
= x times coefficient
= neg. for reactants
Equil. [i]
= Init.+Change
50
Calculating Equilibrium Pressures and
Concentrations from K and Initial Conditions
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Construct the reaction table:
Conc. (M)
CO(g)
often called an “I.C.E.” table.
H2O(g)
CO2(g)
H2(g)
Init. [i]
2.00
2.00
0
0
Change
-x
-x
+x
+x
2.00 - x
2.00 - x
x
x
= x times coefficient
= neg. for reactants
Equil. [i]
= Init.+Change
51
Calculating Equilibrium Pressures and Concentrations
from K and Initial Conditions
Substitute [i]eq into the equilibrium expression:
K=
[CO2][H2]
=
[CO][H2O]
(x)(x)
x2
=
(2.00 - x)(2.00 – x)
(2.00 – x) 2
= 1.56
Taking the square root of both sides:
x
= 1.56 = ± 1.25
2.00 - x
Since only the positive root is meaningful, ignore the negative root:
1.25 =
x
2.00 – x
x = 1.11 M
52
Calculating Equilibrium Pressures and
Concentrations from K and Initial Conditions
Calculating equilibrium concentrations:
[CO] = [H2O] = 2.00 – x = 2.00 M – 1.11 M = 0.89 M
[CO2] = [H2] = x = 1.11 M
Check the results:
K=
[CO2][H2]
(1.11 M)(1.11 M)
=
= 1.56
[CO][H2O]
(0.89 M)(0.89 M)
53
Solving Equilibrium Problems
1) Write the balanced equation for the reaction.
2) Write the equilibrium expression K.
3) List the initial concentrations, [i]initial.
4) Calculate Q and determine the direction of shift to equilibrium.
5) Define change in [i] to reach equilibrium:
Change = x times the stoich. coefficient
6) Express equilibrium concentrations in terms of x.
7) Substitute the equilibrium concentrations into the equilibrium
expression for K.
8) Solve for x and calculate [i]eq for all i.
9) Check the solution by calculating K and making sure it is
identical to the original K.
54
Calculate Equilibrium Partial Pressures
Using Quadratic Equation
The reaction between nitrogen and oxygen to form nitric
acid proceeds according to the following reaction:
N2 (g) + O2 (g) ⇌ 2 NO(g)
K = 4.1x10-4 at 2000K
Initially 0.500 moles of N2 and 0.860 moles of O2 are put into a
2.00 L vessel. Calculate the concentrations of all the species at
equilibrium.
Initial concentration of N2 = 0.500 mol/2.00 L = 0.250 M
Initial concentration of O2 = 0.860 mol/2.00 L = 0.430 M
Initial concentration of NO = 0
Q = 0, therefore the reaction proceeds to the right
Construct the reaction table.
55
Calculate Equilibrium Partial Pressures
Using Quadratic Equation Solver
N2 (g) + O2 (g) ⇌ 2NO(g)
Molarity
N2(g)
O2(g)
NO(g)
Initial
0.250
0.430
0
Change
-x
-x
+2x
Equil.
0.250-x
0.430-x
2x
56
Calculate Equilibrium Partial Pressures
Using Quadratic Equation Solver
N2 (g) + O2 (g) ⇌ 2NO(g)
Molarity
N2(g)
O2(g)
NO(g)
Initial
0.250
0.430
0
Change
-x
-x
+2x
Equil.
0.250-x
0.430-x
2x
57
Calculate Equilibrium Partial Pressures
Using Quadratic Equation Solver
Substituting the equilibrium concentrations from the I.C.E. table into
the equilibrium expression:
K=
[NO]2
[N2][O2]
=
(2x)2
(0.250-x)(0.430-x)
= 4.10 x 10-4
This expression simplifies to:
4.00 x2 + 2.78 x10-4 x - 4.41x10-5 = 0
This is a quadratic equation of the general form:
ax2 + bx + c = 0
where the (two) roots can be obtained from the quadratic formula.
! b ± b 2 ! 4ac
x=
2a
APPENDIX A1.4
58
Calculate Equilibrium Partial Pressures
Using Quadratic Equation Solver
We obtain two possible solutions:
x = -3.35x10-3 and x = 3.28x10-3
Since only the positive root leads to all positive concentrations, we
ignore the negative root.
Calculating equilibrium concentrations:
[NO] = 2x = 6.56x10-3 M
[N2] = 0.250 M – x = 0.247 M
[O2] = 0.430 M – x = 0.427 M
Check:
[NO]2
K=
[N2][O2]
=
(6.56x10-3)2
(0.247)(0.427)
= 4.10x10-4
59
Using the Quadratic Formula to Solve for the Unknown
Given the Reaction between CO and H2O:
Concentration (M)
CO(g)
Initial
Change
Equilibrium
Kc =
2.00
-x
2.00-x
[CO2][H2]
[CO][H2O]
=
quadratic equation:
4.68 +
H2O(g)
CO2(g)
1.00
-x
1.00-x
(x) (x)
(2.00-x)(1.00-x)
We rearrange the equation:
x=
+
=
+
0
+x
x
x2
x2 - 3.00x + 2.00
H2(g)
0
+x
x
= 1.56
0.56x2 - 4.68x + 3.12 = 0
ax2 + bx + c = 0
! b ± b 2 ! 4ac
x=
2a
(-4.68)2 - 4(0.56)(3.12)
2(0.56)
= 7.6 M
and 0.73 M
[CO] = 1.27 M
[H2O] = 0.27 M
[CO2] = 0.73 M
[H2] = 0.73 M
60
Solving equilibrium problems with
simplifying assumptions
Phosgene decomposes into CO and Cl2 when heated according
to the equation.
COCl2 (g)⇌ CO (g) + Cl2 (g)
K = 8.3x10-4 M at 360oC
Calculate the concentration of all species at equilibrium if
5.00 moles of phosgene are placed into a 10.0 L flask.
[COCl2] =
5.00 mol
10.0 L
= 0.500 M
61
Solving equilibrium problems with
simplifying assumptions
COCl2 (g)⇌ CO (g) + Cl2 (g)
K = 8.3x10-4 M at 360oC
Molarity
CO(g)
COCl2 (g)
Cl2 (g)
Init.
0.500
0
0
Change
-x
+x
+x
Equil.
0.500 - x
x
x
62
Solving equilibrium problems with
simplifying assumptions
COCl2 (g)⇌ CO (g) + Cl2 (g)
K = 8.3x10-4 M at 360oC
Molarity
CO(g)
COCl2 (g)
Cl2 (g)
Init.
0.500
0
0
Change
-x
+x
+x
Equil.
0.500 - x
x
x
63
Solving equilibrium problems with
simplifying assumptions
K=
[CO][Cl2]
[COCl2]
=
x2
0.500 - x
= 8.3x10-4
For K to be so small, (0.500 – x) >>x2, so 0.500-x ≈ 0.500
x2
0.500 - x
≈
x2
0.500
= 8.3x10-4
x = 2.037x10-2 → 2.04x10-2
Exact solution is x = 1.996x10-2 → 2.00x10-2
Approximation gives error of 2%.
This ASSUMPTION can ONLY be made for SMALL VALUES of K!!
Use the 5% rule – if the error is less than 5%, the assumption is valid.
(if x < 5% of [ ]0, then Δ[ ] < 5%...assumption is valid) 64
Problem: Hydrogen iodide decomposes at moderate temperatures by
the reaction:
2 HI(g)
H2 (g) + I2 (g)
When 4.00 mol HI were placed in a 5.00 L vessel at 458oC, the
equilibrium mixture was found to contain 0.442 mol I2. What is the
value of Kc ?
Plan: First we calculate the molar concentrations of HI and I2, and then
substitute them into the equilibrium expression to find the value of Kc.
1.
2.
3.
4.
5.
6.
0.00977
0.0122
0.0154
0.0201
0.0549
0.0627
65
Calculating K from Concentration Data–I
Problem: Hydrogen iodide decomposes at moderate temperatures by the
reaction:
2 HI
H
+I
(g)
2 (g)
2 (g)
When 4.00 mol HI were placed in a 5.00 L vessel at 458 °C, the equilibrium
mixture was found to contain 0.442 mol I2. What is the value of Kc ?
Plan: First we calculate the molar concentrations of HI and I2, and then
substitute them into the equilibrium expression to find the value of Kc.
Solution:
Starting conc. of HI =
4.00 mol = 0.800 M
5.00 L
Equilibrium conc. of I2 =
Conc. (M)
Initial
Change
Equilibrium
2HI(g)
0.800
- 2x
0.800 - 2x
0.442 mol = 0.0884 M
5.00 L
H2 (g)
I2 (g)
0
+x
x
0
+x
x = 0.0884
66
Calculating K from Concentration Data–II
[I2] = x = [H2] = 0.0884 M
[HI] = (0.800 – (2 * 0.0884)) M = 0.623 M
Kc =
[H2] [I2]
[HI]2
=
( 0.0884)(0.0884)
(0.623)2
= 0.0201
The equilibrium constant for the decomposition of hydrogen iodide at
458oC is only 0.0201, meaning that the decomposition does not
proceed very far under these temperature conditions.
67
Le Châtelier’s Principle
“If a change in conditions (a “stress”) is imposed on a system
at equilibrium, the equilibrium position will shift in a direction
that tends to reduce that change in conditions.”
A + B
C + D + Energy
For example:
In the reaction above, if more A or B is added you will force the reaction
to produce more product; if either is removed, the reaction will shift to
form more reactants.
If C or D is added you will force the reaction to form more reactants, if
either is removed, the reaction will shift to form more products.
If it is heated you will get more reactants, and if cooled, more products.
68
Equilibrium Position
Factors that control the position of
chemical equilibria are:
• Concentration
• Pressure
• Temperature
NOTE: catalysts influence rate, not position of equilibrium
69
The Effect of a Change in Concentration–I
Given an equilibrium equation such as :
CH4 (g) + NH3 (g)
CH4 (g) + NH3 (g)
HCN(g) + 3 H2 (g)
Forces equilibrium to
produce more product.
Add NH3
CH4 (g) + NH3 (g)
Remove NH3
HCN(g) + 3 H2 (g)
HCN(g) + 3 H2 (g)
Forces the reaction equilibrium to go back
to the left and produce more of the reactants.
70
The Effect of a Change in Concentration–II
CH4 (g) + NH3 (g)
HCN(g) + 3 H2 (g)
Forces equilibrium to go
toward the reactants.
CH4 (g) + NH3 (g)
Add H2
HCN(g) + 3 H2 (g)
Forces equilibrium to make more
produce and replace the lost HCN.
Remove HCN
71
The Effect of a Change in Concentration
• If a gaseous reactant or
product is added to a
system at equilibrium, the
system will shift in a
direction to reduce the
concentration of the
added component.
• If a gaseous reactant or
product is removed from a
system at equilibrium, the
system will shift in a
direction to increase the
concentration of the
removed component.
NH4NO2(s) ⇌ N2(g) + 2 H2O(g)
[ N ][ H O ]
K=
[ NH NO ]
2
2
2
4
2
K = [ N 2 ][ H 2 O ]
2
72
Consider the following reaction:
2 H2S(g) + O2(g) ⇌ 2 S(s) + 2 H2O(g)
Which of the following will not result in an
equilibrium shift to the right?
1.
2.
3.
4.
5.
Adding H2S
Removing H2O
Removing S
Adding O2
All will shift the
equilibrium to the
right
73
The Effect of a Change in Concentration
Consider the following reaction:
2 H2S(g) + O2(g) ⇌ 2 S(s) + 2 H2O(g)
What happens to:
(a) [H2O] if O2 is added?
The reaction proceeds to the right so H2O increases.
(b) [H2S] if O2 is added?
Some H2S reacts with the added O2 to move the reaction to
the right, so [H2S] decreases.
74
The Effect of a Change in Concentration
2 H2S(g) + O2(g) ⇌ 2 S(s) + 2 H2O(g)
(c) [O2] if H2S is removed?
The reaction proceeds to the left to re-form H2S, more O2
is formed as well, O2 increases.
(d) [H2S] if S(s) is added?
S is a solid, so its activity does not change. Thus, [H2S] is
unchanged.
75
The Effect of a Change in Pressure (Volume)
Pressure changes mainly involve gases since liquids and solids
are nearly incompressible. For gases, pressure changes can occur in
three ways:
 Changing the concentration of a gaseous component
 Adding an inert gas (one that does not take part in the reaction)
 Changing the volume of the reaction vessel
When a system at equilibrium that contains a gas undergoes a change in
pressure as a result of a change in volume, the equilibrium position shifts
to reduce the effect of the change.
 If the volume decreases (increased partial pressure), the total number of
gas molecules decreases (reaction shifts to side with fewer moles of gas).
 If the volume increases (decreased partial pressure), the total number of
gas molecules increases (reaction shifts to side with more moles of gas).
76
Figure 6.9: Brown NO2(g) and colorless
N2O4(g) at equilibrium in a syringe
2 NO2 (g) ⇌ N2 O4 (g)
Brown
Colorless
Source: Ken O’Donoghue
77
The Effect of a Change in Pressure (Volume)
How would you change the total pressure or volume in the
following reactions to increase the yield of the products:
(a) CaCO3(s) ⇌ CaO(s) + CO2 (g)
The only gas is the product CO2. To move the reaction to the right,
increase the volume.
(b) S(s) + 3 F2 (g) ⇌ SF6 (g)
With 3 moles of gas on the left and only one on the right, we
increase the pressure (decrease volume) to form more SF6.
(c) Cl2(g) + I2(g) ⇌ 2 ICl (g)
The number of moles of gas is the same on both sides of the
equation, so a change in pressure or volume will have no effect
on the equilibrium.
78
The Effect of a Change in Temperature
Only temperature changes will alter the equilibrium constant, and that
is why we always specify the temperature when giving the value of K.
The best way to look at temperature effects is to realize that temperature
is a component of the equation, the same as a reactant or product.
O2 (g) + 2 H2 (g)
Energy + 2 H2O(g)
2 H2O(g) + Energy
= Exothermic
2 H2 (g) + O2 (g) = Endothermic
A temperature increase favors the endothermic direction and a
temperature decrease favors the exothermic direction.
79
Exothermic vs Endothermic Reactions
Exothermic
Endothermic
Role of energy in the
reaction:
Released as a
product
Absorbed as a
reactant
Effect of increasing
temperature:
Equilibrium shifts
toward the
reactants
Decreases
Equilibrium shifts
toward the
products
Increases
Change in K because of
increased T:
Change in concentration at constant T changes the
equilibrium position, but not the value of K (ratio of
products to reactants).
80
How does an increase in temperature affect the
equilibrium concentration of the substance in bold
and K for the following reactions:
a) CaO(s) + H2O (l) ⇌ Ca(OH)2 (aq) + energy
b) CaCO3 (s) + energy ⇌ + CaO(s) + CO2 (g)
c) SO2 (g) + energy ⇌ S(s) + O2(g)
1.
2.
3.
4.
a) ↓↓, b) ↑↑, c) ↓↑
a) ↓↑, b) ↑↓, c) ↓↓
a) ↑↓, b) ↓↑, c) ↑↑
a) ↑↑, b) ↓↓, c) ↑↓
81
Temperature and K
How does an increase in temperature affect the equilibrium
concentration of the substance in bold and K for the following
reactions:
(a) CaO(s) + H2O (l) ⇌ Ca(OH)2 (aq) + energy
Increasing T shifts the system to the left, where it absorbs energy.
[Ca(OH)2] and K decrease.
(b) CaCO3 (s) + energy ⇌ + CaO(s) + CO2 (g)
Increasing T shifts the system to the right, where it absorbs energy.
[CO2] and K increase.
(c) SO2 (g) + energy ⇌ S(s) + O2(g)
Increasing T shifts the system to the right, where it absorbs energy.
[SO2] will decrease and K increases.
82
Effect of Various Stresses on an Equilibrium System
Stress
Net Direction of Reaction
Concentration:
Increase [reactant]
Decrease [reactant]
Effect on Value of K
Toward formation of products
Toward formation of reactants
None
None
Pressure (1/Volume) of closed container, holding reaction at constant T:
Decrease P
Toward formation of larger
amount (mol) of gas unless Δn = 0
None
Increase P
Toward formation of smaller
amount (mol) of gas unless Δn = 0
None
Temperature - adding energy in the form of heat (opposite if T decreases):
Increase T
Toward formation of products if energy
K increases
is a “reactant” (endothermic reaction)
with inc T
Toward formation of reactants if energy
is a “product” (exothermic reaction)
Catalyst added:
None
K decreases
with inc T
rates of fwd & rev reactions increase equally None
83