16 Chemical Equilibrium
Reactions seldom go complete such that one of the reactants is
exhausted as we have discussed in stoichiometry and limiting
reagent. Instead, they are mostly reversible. For example,
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
The two-head arrow (or equal signs =) indicates that the reaction
goes both ways. Similarly,
N2O4 (colorless gas)  2 NO2 (brown gas) (show a film)
When the forward and reverse reaction rates are equal, the system is
said to be at equilibrium.
16 Chemical Equilibrium
1
The NASA Computer program CEA (Chemical Equilibrium with
Applications) calculates chemical equilibrium compositions and
properties of complex mixtures. Applications include assigned
thermodynamic states, theoretical rocket performance, ChapmanJouguet detonations, and shock-tube parameters for incident and
reflected shocks.
16 Chemical Equilibrium
2
1 mol CH4
2 mol S2
T = 1000 K
P = 1.5 atm
Countours
show G/RT
Equilibrium
occur at
minimum
G/RT
16 Chemical Equilibrium
3
A System
Scientific experiments usually
investigate a system, which is
isolated from its environment or
surrounding.
The system can be a nucleus, an
atom, a molecule, a plant, an animal,
an experimental setup, the Earth, the
solar system, the galaxy, or the
universe.
The environment or
The open
or
closed
system
The surrounding
Matter and energy can be transferred
into or out of a system.
16 Chemical Equilibrium
4
Chemical Equilibrium
– a dynamic process
Chemical equilibrium is a state of a system when reaction rates in both
directions are equal. Changes continue at the molecular level, but the
macroscopic properties stay the same. Equilibrium is a dynamic rather than
static process.
Heat is energy flowing from a high temperature object to a low temperature
object. Equilibrium is reached when the temperatures are the same.
Molecules still exchange kinetic energies during collision.
Water flows from a high potential-energy place to a low potential-energy
place. When potential energies are the same, water stops flowing. Watermolecule diffusion continues.
16 Chemical Equilibrium
5
Some Typical Equilibria
Macroscopic properties such as vapor pressure, solubility, distribution
of a substance in two immiscible liquids are typical equilibria.
Vapor pressure of a liquid is the pressure of the vapor in equilibrium
with a liquid. It is a function of temperature.
Solubility, amount in solution in equilibrium with solid, depends on the
substance and the solvent, often also on temperature.
Distribution (or equilibrium) coefficients of a substance in immiscible
liquids are useful guides for the separation of mixtures into
components.
16 Chemical Equilibrium
6
The reaction quotients
For any chemical reaction in a system,
aA+ b B  c C + d D
we can always write a quotient,
[C]c [D]d
Q = ——————
[A]a [B]b
for example,
[NO2]2
Q = ——————
[N2O4]
Know the definition of Q.
Text: 16-5 defines Q &
16-1 defines K.
(Q is reaction quotient)
for N2O4  2 NO2
16 Chemical Equilibrium
7
Expressions for reaction quotients
For the ionization of an acid,
H2SO4 = 2 H+ + SO42what is the reaction quotient expression?
Hint: The reaction quotient is
[H+]2 [SO42-]
—————— = Q
[H2SO4]
If M is used for the concentrations, what are the units for Q?
16 Chemical Equilibrium
8
More expressions for reaction quotients
2 H2 + O2  2 H2O
[H2O]2
Q = —————
[H2]2 [O2]
[H2]2 [O2]
1
Q r = ————— = —
—
[H2O]2
Q
2 H2O  2 H2 + O2
4 H2 + 2 O2  4 H2O
All relations are due to
definitions. See 16-3
[H2O]4
Q2 = ————— = Q 2
[H2]4 [O2]2
Q=Q2
Note the relationships
16 Chemical Equilibrium
9
Equilibrium constant - mass action law
When a chemical reaction in a closed system is at equilibrium, the
reaction quotient is defined as the equilibrium constant, K, which
depends on T. For a chemical reaction in a closed system,
aA+ b B  c C + d D
then
[C]c [D]d
Q = —————— = K
(at equilibrium)
[A]a [B]b
Units of K depends on units of [ ]s, M(a+b-c-d) if M is used.
If Q < K, the reaction goes forward
if Q = K, the system is at equilibrium
if Q > K, the reaction goes backward
See 16-5 & 16-3
| to increase [product]
|the law of mass action
| to decrease [product]
16 Chemical Equilibrium
10
Evaluating K
The equilibrium mixture is found to contain 0.07, 0.11, 0.03 and
0.03 moles of CO, H2, CH4 and H2O respectively for the reaction:
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
What is the equilibrium constant, K?
Solution
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
start 0.07
0.11
0.03 0.03
(0.03) (0.03)
K = —————— = 9.7
(0.07) (0.11)3
What governs the quantitative relationship
16 Chemical
Equilibrium
among equilibrium
mixtures?
11
Evaluate quantities
When 0.10 moles CO and 0.20 moles of H2 are placed in a 1-L
vessel at temperature T and allowed to come to equilibrium. The
mixture is found to contain 0.03 moles H2O for the reaction:
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
What is the equilibrium constant, K?
Solution (note the strategy please)
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
start 0.10
0.20
0
0
change - x
- 3x
+x
+x (x=0.03)
eqlbm 0.1-x 0.2-3x
0.03 0.03
[]
0.07 0.11
0.03 0.03
K = (0.03) (0.03) / (0.07) (0.11)3 = 9.7
16 Chemical Equilibrium
12
Evaluate quantities
When 0.08 moles CO, 0.14 moles of H2, and 0.08 moles of CH4
are placed in a 1-L vessel at temperature T and allowed to come
to equilibrium. The mixture is found to contain 0.01 moles H2O
for the reaction:
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
What is the equilibrium constant, K?
Solution (note the strategy please)
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
start 0.08
0.14
0.08 0
change - x
- 3x
0.08+x +x (x=0.01)
eqlbm 0.08-0.01 0.14-0.03 0.08+0.01 0.01
[]
0.07
0.11
0.09 0.01
K = (0.09) (0.01) / (0.07) (0.11)3 = 9.7
16 Chemical Equilibrium
13
Entropy
We all know that heat flow from a high-T object into a low-T object.
This type of change can be described as due to increase of entropy.
Entropy S is the amount of energy transferred, E, divided by T.
E
S = ————
or
TS = E
T
Entropy increases for spontaneous changes (reactions).
Entropy is also associated with randomness of ordering.
What causes change when no apparent energy is involved?
16 Chemical Equilibrium
14
Entropy calculation
A system consists 10.0 L of water at 300 K and 3.0 g of ice at 273 K.
Heat of fusion of ice (molar mass 18) is 6 kJ/mol. What is the
change in entropy when the system is at equilibrium.
Solution:
Since the amount of heat for melting the ice is
6 *(3/ 18) = 1.0 kJ
too little to change the temperature of the entire system appreciably
from 300 K. Thus the entropy change is
– 1/300 + 1/273 = 3.3e – 4 kJ/K = 0.33 J/K (entropy unit)
What is the molar entropy for ice melting?
+6000 J mol–1 / 273 K = 22 J mol –1 K –1 (entropy unit mol–1)
16 Chemical Equilibrium
15
Josiah W. Gibbs
(1839-1903) New Haven, Connecticut, USA
First Doctorate (1863 Yale) in Engineering of U.S.A.
Studied in Europe, influenced by Kirchhoff & Helmholtz.
1871 Professor of mathematics and physics at Yale
Unassuming in manner, genial and kindly in his intercourse with his fellowmen, never showing impatience or irritation, devoid of personal ambition of
the baser sort or of the slightest desire to exalt himself, he went far toward
realizing the ideal of the unselfish, Christian gentleman. In the minds of
those who knew him, the greatness of his intellectual achievements will
never overshadow the beauty and dignity of his life.
- Bumstead
lunar crater Gibbs
16 Chemical Equilibrium
16
Gibbs free energy, G or G
Following the definition of entropy S, Gibbs defined a function together
with enthalpy of change, H.
or
G=H–TS
now called Gibbs free energy.
G = H – T S
for differences of these quantities
Since H is negative and S is positive for spontaneous change, Gibbs
free energy is negative for all spontaneous changes, and it unified
enthalpy and entropy for changes. Gibbs free energy is the maximum
amount of available energy in any change.
Similar to Ho, Go is the Gibbs free energy at the standard T and P.
What drives physical and chemical changes?
16 Chemical Equilibrium
17
Energy and equilibrium constant K
A quantity called Gibbs free energy, G, is defined as the maximum
amount of energy from a chemical reaction for doing work.
G is negative for spontaneous reaction
G is positive for non-spontaneous reaction
The G is related to the reaction quotient Q,
G = Go + R T ln Q
When a system is at equilibrium there is no available energy,
G = Go + R T ln K = 0
Thus,
Go = – R T ln K,
ln K = – Go / R T
K = exp (– Go / R T)
16 Chemical
See 20-2 to 20-6 page 784-808
plse Equilibrium
18
More expressions for equilibrium constant
Note these relationships in light of ln K = – Go / R T
2 H2 + O2  2 H2O
[H2]2 [O2]
1
K r = ———— = ——
[H2O] 2 K
4 H2 + 2 O2  4 H2O
[H2O]2
K = —————
[H2]2 [O2]
2 H2O  2 H2 + O2
[H2O]4
K2 = ————— = K 2
[H2]4 [O2]2
K2=K
16 Chemical Equilibrium
19
An Example
Hydrogen iodide, HI, decomposes,
2 HI  H2 + I2
The amount of I2 can be determined from the intensity of the violet
color of I2. When 4.00 mol of HI was placed in a 5.00-L vessel at
458oC, the equilibrium mixture was found to contain 0.442 mol I2.
What is the value for K for the decomposition of HI at this temperature?
See 16-4 and 16-7 for this and the following examples.
16 Chemical Equilibrium
20
Analysis of Example
Hydrogen iodide, HI, decomposes,
2 HI  H2 + I2
The amount of I2 can be determined from the intensity of the violet
color of I2. When 4.00 mol of HI was placed in a 5.00-L vessel at
458oC, the equilibrium mixture was found to contain 0.442 mol I2.
What is the value for K for the decomposition of HI at this temperature?
Analysis of the problem:
[HI]0 = 4.0 mol / 5.00 L = 0.800 M
[I2]equilibrium = 0.442 mol / 5.0 L = 0.0884 M
Know your strategy 
16 Chemical Equilibrium
21
Calculate Kc
Example 14.3 The equilibrium concentration of various species are
given below their symbols,
start
change
equlbm
value
2 HI (g) = H2 (g) +
0.800
0
=?
=?
0.800-2x x
0.6232
0.0884
I2 (g)
0
= 0.0884 M from experiment
x
0.0884 M
0.0884*0.0884
Kc = ——————— = 0.0201
(0.6232)2
Skill: re-interpret question. Evaluate other quantities and K from
known start conditions and experimental results
Evaluate Go
16 Chemical Equilibrium
22
Evaluate Kc
Compare with slide 41
At 300 K and 1 atm, an equilibrium mixture contains 70% NO2 by
mass in a tube of NO2 and N2O4. Evaluate Kc for the reaction,
N2O4 = 2 NO2
The challenge (hint):
assume 100 g total mass (70 g of NO2 and 30 g N2O4)
find amount in mole for each (1.522 mol NO2; 0.326 mol N2O4)
find volume from pressure for all (45.49 L)
find concentration from volume (0.0335 M NO2; 0.00717 M)
evaluate Kc (equation dependent)
16 Chemical Equilibrium
23
Application of Kc
For
N2O4 (g) = 2 NO2 (g)
Kc = 4.20e-3 at 300 K,
What are the equilibrium concentrations if 0.100 mol of N2O4 is
placed in a 5.00-L flask?
Strategy:
write down the equation
figure out the initial condition
assume the change to be x
figure out the equilibrium concentration
use the concentration-Kc relationship
solve for x and then check the results.
N2O4 (g) = 2 NO2 (g)
0.0200-x
2x
(2x)2
———— = 4e-3 (K)
0.0200-x
4x2 + 0.0042 x – 0.0000926 = 0
-0.00463 + {(4.63e-3)2 – 4*4*(-9.265e-5)}
x = ———————————————————— = 4.27e-3
2*4
16 Chemical Equilibrium
N2O4 (g) = 2 NO2 (g)
0.0157
8.54e-3 M
24
Evaluate equilibrium concentrations of all for the
reaction condition as given in the start: (at T1)
N2 (g) + 3 H2 (g)  2 NH3 (g) K = 1e-5
start 1.0
2.0
0
change - x
- 3x
+2x (= ?)
eqlbm 1.0-x 2.0-3x
2x
(2 x)2
—————— = 1e-5
(1-x)(2.0-3x)3
Results:
[N2] = 0.995 M
[H2] = 1.99
[NH3] = 9e-3
Checking
(2*4.5e-3)2
————— = 1e-5
0.995*1.993
4x2
—————— = 1e-5
(1.0-x)(2.0-3x)3
x2 = 1e-5*1.0*2.03/4 = 2e-5
Approximations
Discuss
x = 4.5e-3 (justified)
16 Chemical Equilibrium
25
Trial and error or Newtons method
When 0.10 moles CO and 0.20 moles of H2 are placed in a 1-L
vessel at temperature T and allowed to come to equilibrium.
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g) Kc = 9.7
What are the concentrations of various species?
Solution (note the strategy please)
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
start 0.10
0.20
0
0
change - x
- 3x
+x
+x
eqlbm 0.1-x 0.2-3x
x
x (= ?)
x = Kc =
0.05* 400
0.06 15174
0.04
52
0.03 11.3
Kc = (x)2 / (0.1-x) (0.2-3x)3 = 9.7
0.029 8.21
Too difficult to solve.You need not worry.
Trial-and-error (Newton’s method) results shown>> 0.0299 9.50
*x < 0.1
16 Chemical Equilibrium
Why cannot use successive approximation?
The first x=0.44 is unreasonable.
26
Apply math skill
Exactly 0.0100, 0.0100 and 0.0200 moles of H2, I2, and HI are mixed
in a 1-L vessel, and then raised to 277 K, at which Kc = 50.3.
What concentration will increase?
Determine from Qc
What are the equilibrium concentrations of H2, I2 and HI?
Solution (summary)
H2 (g) + I2 (g) =
start 0.0100 0.0100
change
-x
-x
eqlbm [ ] 0.01-x 0.01-x
value 0.0044 0.0044
(0.02+2x)2
————— = 50.3
(0.01-x)2
2 HI
0.0200
+2x
0.02+2 x
0.0312 M
assume
x = 0.00560
(0.02+2x)
> > note > ————— = 7.09
(0.01-x)
16 Chemical Equilibrium
27
Successive approximation
For
2 H2S (g) = 2 H2 (g) + S2 (g)
Kc = 4.2e-6 at 831oC
What is [S2] when 0.0700 mol H2S is allowed to reach equilibrium in a
1.00-L vessel.
Solution:
2 H2S (g) = 2 H2 (g) + S2 (g)
0.0700-2x
2x
x
(2x)2 x
————— = 4.2e-6
x = 3(4.2e-6*0.07002/4) = 1.73e-3
(0.0700-2x)2 assume 0
(2x)2 x
————— = 4.2e-6
x = 3(4.2e-6*0.06652/4) = 1.67e-3
(0.0700-2*0.00173)2 successive inclusion
Carry out the
calculation plse
4x316 Chemical Equilibrium 3
= 4.2e-6
x = (4.2e-6*0.06672/4) = 1.67e-3
(0.07-2*0.00167)
28
Equilibria involving solvents
Since [solvent] is essentially constant, it does not appear in Kc
expression. You just have to know and apply.
CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq)
[CH3COOH] [C2H5OH]
K’c = ————————————
[CH3COOC2H5] [ H2O]
[CH3COOH] [C2H5OH]
Usually, Kc = K’c [ H2O] = ————————————
[CH3COOC2H5]
16 Chemical Equilibrium
29
Equilibria involving solvents - application
For the reaction,
CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq)
Kc = 0.25 at 25oC. Calculate [C2H5OH] at equilibrium starting with
1.00 M of ethyl acetate.
Solution:
CH3COOC2H5 (aq) + H2O (l) = CH3COOH (aq) + C2H5OH (aq)
1.00 – x
x
x= [C2H5OH ]
x2
———— = 0.25
x2 + 0.25 x – 0.25 = 0
(1.00 – x)
– 0.25 + 0.252 – 4(-0.25)
x = ———————————— = 0.39 M
2
16 Chemical Equilibrium
30
Equilibrium Calculations – a summary
Several methods have been introduced in evaluating x
Simple method
Approximation
Trial and error (Newtons method)
Recognize math relationship (taking square root on both sides)
Successive approximation
16 Chemical Equilibrium
31
Kp and Kc; Kp = Kc (R T) {c+d-(a+b)}
The equilibrium constants depends on the units used for quantities. In
general, the equilibrium constant is represented by K
When concentrations are used, Kc is the notation;
when partial pressures are used, Kp is used
Since PV = n R T,
P = (n/V) R T = [ ] R T
[ ] = (n / V) is the concentration
For the reaction a A + b B  c C + d D
PCc PDd
[C]c [D] d (R T)c + d
Kp = ———— = ———————— = Kc (R T){c+d-(a+b)}
PAa PBb
[A]a [B] b (R T)a + b
See page 635
16 Chemical Equilibrium
32
Kp more appropriate for
H2O (l) = H2O (g)
Kp = vapor pressure of water
H2O (s) = H2O (g)
Kp = vapor pressure of ice
At the triple point, it’s obvious that
vapor pressure of ice = vapor pressure of water
if these are equal, the three phases are present.
CaCO3 (s) = CaO + CO2
Kp = partial v. p. of CO2
16 Chemical Equilibrium
33
Kc and Kp conversion
For
2 H2S (g) = 2 H2 (g) + S2 (g)
What is Kp?
Kc = 4.2e-6 at 831oC
Solution:
Kp = Kc (R T){c+d – (a+b)} (recall or derive)
= Kc (R T)(2+1 – 2)
= 4.2e-6 *8.314 kJ mol – 1 K – 1 * (831+273) K
= 3.9e-2
2 H2S (g) = 2 H2 (g) + S2 (g)
0.0667
3.34e-3 1.67e-3
612 kPa 30.7
15.3
6.04 atm 0.30
0.15
Kp
confirmation
Kc = 4.2e-6 at 831oC
Kp = 3.9e-2 J mol 1 K 1 at 831oC
Kp = 3.7e-4 L atm mol 1 K 1
16 Chemical Equilibrium
34
Le Chatelier’s Principle
The system will restore its equilibrium after changing conditions.
An open bottle of water will dry eventually, because the partial pressure
never reach equilibrium (removing vapor pressure).
Increase pressure will increase the formation of NH3 in
N2 (g) + 3 H2 (g) = 2 NH3 (g)
Hof =  91.8 kJ
Increase T will decrease the formation of NH3 in this exothermic reaction.
How does [CH4] change when pressure increase for an equilibrium of
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)?
Increasing pressure tends to reduce the number of molecules.
Study sections: 16.6
16 Chemical Equilibrium
35
Le Chatelier’s Principle – cont.
Possible changes:
Temperature, T
Increase T favors forward reaction in an endothermic reaction
but favors reverse reaction in an exothermic reaction
Pressure, P and volume V
Pressure and volume are related. Decreases of volume increase the
pressure
Amounts of product and reactants
Increase amounts of reactants, products or both increases P and V
16 Chemical Equilibrium
36
Henri Louis Le Chatelier
(1850-1936)
A mining engineer
investigated structure of metals and alloys,
translated work of Gibbs into French
stated Henricus Van’t Hoff’s (Nobel Prize 1901) theory in
word and
wrote on industrial efficiency and labor management relations
Scientist would be wise to refrain from using the mathematical
equation unless he understands the theory that it represents, and can
make a statement about the theory that does not consist just in reading
the equation. L. Pauling
16 Chemical Equilibrium
37
Effect of a catalyst
There are two factors regarding the
speed of reactions: kinetics and
thermodynamics.
E
Kc and Kp are thermodynamic
parameters, related to G o. Thus,
the equilibrium constants are not
affected by a catalyst.
kinetic
thermodynamic
Catalyst only affect reaction
constant k, which is related to the
activation energy.
Reaction
See page 643-646
16 Chemical Equilibrium
38
Equilibria
- Summary
Apply stoichiometry to equilibrium mixture in a system
Write reaction quotient and equilibrium constant expression.
Rewrite Qs and Ks when the equations change.
Predict direction of reaction by comparing Q and K.
Evaluate Ks from some know conditions.
Know how energies H and G are related to equilibrium constants.
Use various methods to evaluate concentrations of an equilibrium system.
Apply Le Chatelier’s principle to predict changes in P, [ ], and T.
Separate kinetic and thermodynamic factors of reactions.
16 Chemical Equilibrium
39
Equilibrium review -1
Recall that
and
G = Go + R T ln K = 0
Go = – R T ln K, or ln K = – Go / R T
Since Go refers to the standard condition (1 atm and 273K), units for
pressures to calculate Kp should theoretically be in atm.
However, the standard condition is redefined from time to time, and
one has to be aware of the relationship here for engineering
calculations.
In Chem123, we take a simple approach to specify the units and
adhere to the unit atm for pressures.
16 Chemical Equilibrium
40
Equilibrium review – 2
Compare with slide 23
At 600K, the partial pressure of NO2 in a equilibrium mixture is 0.10 atm,
N2O4 = 2 NO2
Kp = 3.06 L atm mol 1 K 1
What is the partial pressure of N2O4?
What is the percent of dissociation of N2O4?
What percentage of molecules are NO2?
Solution:
N2O4 = 2 NO2 Kp = 3.06 L atm mol 1 K 1
x
0.10 let x be the unknown
0.102 / x = 3.06; x = 0.12/3.06 = 0.0033 atm P(N2O4)
percent dissociated = 0.05 / (0.0033+0.05) = 0.94 = 94%
% of NO2 molecule = 0.10/(0.10+0.0033) = 0.97 = 97%
16 Chemical Equilibrium
41
Equilibrium review - 3
At 1123K, Kc = 0.153 for the reaction:
C(s) + CO2 (g) = 2 CO (g)
Ho = 172.5 kJ
What is Kp?
If initially the partial pressure of CO2 was 1.50 atm, what is the total
pressure at equilibrium?
How will the equilibrium shift if temperature is decreased to 900K?
Solution:
16 Chemical Equilibrium
42
Equilibrium review - 3
At 1123K, Kc = 0.153 for the reaction:
C(s) + CO2 (g) = 2 CO (g)
Ho = 172.5 kJ
What is Kp?
If initially the partial pressure of CO2 was 1.50 atm, what is the total
pressure at equilibrium?
How will the equilibrium shift if temperature is decreased to 900K?
Solution:
R in L-atm mol-1 K-1
Kp = 0.153 (R T)1 = 0.153(0.08206*1123) = 14.1
C(s) + CO2 (g) = 2 CO (g)
1.50-x
2x
equilibrium pressure
(2x)2 / (1.50 – x) = 14.1; 4x2 + 14.1x –21.2 = 0
x = {- 14.1 + (14.12 + 4*4*21.1)} / (2*4) = 1.13 atm
total pressure = 1.50-x+2x = 1.50+x = 1.50 + 1.13 = 2.63 atm
P(CO2) = 1.50 – 1.13 = 0.37 atm; P(CO) = 2*1.13 = 2.26
P(CO) decreases at 900K (<1123K) in the endothermic reaction (+ Ho).
 shift left
43
16 Chemical Equilibrium