“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #15:
Introduction to Equilibrium
1. Define Equilibrium,
And equilibrium
constant
Gaseous Chemical
Equilibrium
1. Can balance equations
2. Can predict direction of reaction
(H)
3. Can compute rates
d  A
m
rate  
 k A
adt
4. Can we predict what happens at end?
Our Friend for this Chapter
Dinitrogen tetroxide
A. Intermediate in:
• nitric acid & sulfuric acid production
• nitration of organic compound & explosives
• manufacture of oxidized cellulose compound (hemostatic cotton)
B. Used to bleach flour
C. Proposed as oxidizing agent in rocket propulsion nitrogen dioxide.
We studied it’s decomposition kinetics in
The preceding chapter
N 2 O4,g  2 NO2, g
Consider the reaction at 100 oC, where the initial
Pressure of dinitrogen tetroxide is 1 atm:
review
N 2 O4,g  2 NO2, g
The decomposition of dinitrogen tetroxide vs time
Is shown in the table below
1.8
N2O4, atm
1
0.6
0.35
0.22
0.22
0.22
NO2, atm
0
0.8
1.3
1.56
1.56
1.56
1.6
1.4
NO2
1.2
atm
s
0
20
40
60
80
100
1
0.8
0.6
N2O4
0.4
0.2
0
The reaction order and
Rate constant can be determined by various time/conc. plots
0
20
40
60
t, s
80
100
120
N 2 O4,g  2 NO2, g
s
0
20
40
60
80
100
N2O4, atm
1
0.6
0.35
0.22
0.22
0.22
NO2, atm
0
0.8
1.3
1.56
1.56
1.56
1.8
1/N204
1/N204
1
1.67
1.67
2.86
2.86
4.55
4.55
4.55
4.55
4.55
4.55
ln N204
0
-0.51
-1.05
-1.51
-1.51
-1.51
0
1.6
review
5
4.5
-0.2
4
1.4
-0.4
3.5
1.2
0.8
3
1/PN2O4
ln[N2O4]
atm
-0.6
1
-0.8
2.5
2
-1
0.6
1.5
0.4
-1.2
1
0.2
-1.4
0.5
0
0
-1.6
0
20
40
60
80
100
120
0
0
20
t, s
0


rate  k A
 A    A   kt
t
0
40
60
80
t, s
rate  k[ A]
1
 
 
ln At  ln Ao  kt
100
120
20
40
60
80
t, s
rate  k[ A]2
1
1

 kt
 A A0
 
100
120
1st order reaction
review
ln PN2O4 ,t  ln PN2O4 ,t  0  kt
ln PN 2 O4 ,t  ln1  kt
ln PN2O4 ,t  0  kt
1
k  0.0254
s
rate  k[ A]1
 
 
ln At  ln Ao  kt
0
-0.2
-0.4
Using the rate
Constants the
Rate at each time
Can be calculated
ln[N2O4]
-0.6
ln[N2O4] = - 0.0065 -0.0254(t)
-0.8
R2 = 0.9992
-1
-1.2
-1.4
-1.6
0
20
40
60
t, s
80
100
120
1.8
k forward
N 2 O4 , g 
 2 NO2 , g
1.6
1.4
atm
1.2
1
Conc. plot
0.8
0.03
0.03
0.6
0.4
0.025
0.025
rate  k f PN2O4
0
0
20
40
1

rate   0.0254  PN 2O4

s
0.02
0.02
rate,
rate, Atm/s
Atm/s
0.2
60
80
100
120
t, s
0.015
0.015
What do you observe?
0.01
0.01
 
0.005
0.005
rate  kb PNO2
00
00
20
20
40
40
60
60
80
80
2
rate  0.002309
100
100
 
1
PNO2
atm  s
120
120
t, s
2 NO2,g  N 2 O4, g
kbackward
t, s
1. Conc. Are “stable”
2. Rate forward = rate backward
2
“Equilibrium” = “steady state concentrations”
1.
occurs when rate forward = rate backward
k forward  0.0254
Rate forward  Ratebackward
 
k f PN2O4 , g  kb PNO2 , g
kf
kb
2.
 P 
kbackward  0.002309
NO2 , g
PN2O4 , g
 P   K
kb
PN2O4 , g
1
s  11atm
Kequilibrium 
1
0002309
.
atm  s
00254
.
NO2 , g
equilibrium
1
atms
2 NO2, g  N 2 O4,g
2
concentrations are also constant (steady state)
kf
3.
N 2 O4, g  2 NO2, g
2
2
1
s
BOTH reactions are occurring competitively
N 2 O4, g  2 NO2,g ojo
4.
An equilibrium constant describes the steady state concs
at 100 oC for any starting set of concentrations
Kequilibrium  11atm 
Experiment
time
t=0
t=equilibrium
Kequilibrium
1
N2O4
1.00
0.22
P 

 11atm 
 
2
PNO2
PN2O4
NO2
0.00
1.56
2
N2O4
0.00
0.07
NO2
NO2
1.00
1.00
0.86
0.86
3
N2O4
1.00
0.42
NO2
1.00
2.16
2
NO2 , g
PN 2O4 , g
?
Kequilibrium  11atm 
?
2
156
. atm
0.22atm
11atm  1106
. atm
?
Kequilibrium  11atm 
?
2
 086
. atm
0.07atm
11atm  1056
. atm
?
Kequilibrium  11atm 
?
11atm  1111
. atm
2
 216
. atm
0.42atm
Hamilton’s Bar: Legal occupancy: 10
The stupid
analogy
3 out
10 in
Hamilton’s appears to have an unchanged
population, but constant exchange of drinkers is
occurring. The equilibrium is 10/3.
General
FITCH Rules
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
 qq 

E  k
C1. It’s all about charge
r r 
C2. Everybody wants to “be like Mike”
 qq 
C3. Size Matters

E  k
r r 
C4. Still Waters Run Deep
Piranhas lurk
Apparent
C5. Alpha Dogs eat first
Lack of change,
1 2
Chemistry
el
1
2
1 2
el
1
2
Constant microscopic
change
The “Generic” Rule
Stoichiometry shows
up as power
aA  bB 
 cC  dD
P  P 

P  P 
c
Pressure
Of reactants
On bottom
KPr essue
C
d
D
a
A
b
B
An equilibrium constant can be written in terms of
Molarity also
Pressure
Of Products
On top
PV  nRT
aA  bB cC  dD


P
n
  molarity 
RT V
P  RTmolarity
P  P 

P  P 
c
KPr essue
C
d
a
A
K Pr essue
D
b
B
 RT C  RT D
a
b
 RT A  RT B
c
K Pr essue 
K Pr essue
d
K Pr essue
  RT  c  RT  d    C c  D d 
 
a
b 
a
b 
  RT   RT     A  B 
Kconcentration
  RT  c  RT  d 
 
a
b  K concentration
  RT   RT  
 C  c  D d

 A a  B b
( c d )  ( a b)


 RT
Kconcentration
K Pr essue
n


 RT
Kconcentration
f i
When doing problems
watch out that they
are not switching units
on you!!!!!
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #15:
Introduction to Equilibrium
Sample calculation
Of K
Calculate the Kc for the reaction at 25oC:
2 NOg  Cl2 , g 
 2 NOCl g
if the equilibrium pressures are PNOCl = 1.2 atm; PNO = 5.0x10-2 atm; and PCl2 =
3.0x10-1 atm.
DON’T
KNOW
PNOCl = 1.2 atm
PNO = 5.0x10-2 atm
PCl2 = 3.0x10-1 atm
P 

P  P
Red Herring?
Kconc
eq
1
( 2 )  ( 3)
19
. x10
  RT 
Kconcentration
atm
3
2
KP
NOCl
2
NO
Cl2
1
L  atm


19
. x103
  0.0826
 298 K

atm 
mol  K

2
12
. atm
KP 
 0.050atm 2  0.30atm
19
. x10 3
19
. x10 3 atm2
3
1
KP 

19
.
x
10
atm
atm3
K Pr essue   RT 
( c d )  ( a b)
none
Kconcentration
1
Kconcentration
1
1

K
L  atm
atm 
 concentration
 298 K 
 0.0826


mol  K
19
. x10 3
1 
L  atm

 298 K   Kconcentration
 0.0826

atm 
mol  K
Kconcentration  4.6 x10 4
1
M
Consider the Haber Reaction which “fixes” nitrogen and is immensely
N 2 , g  3H2 , g 
 2 NH3, g
important for the large scale production of ammonia used in
explosives
fertilizers
If we mix nitrogen and hydrogen
gases will we get ammonia?
N2 941 kJ/mole bond energy
3H2 432 kJ/mole
Haber found a
Catalyst to make
reaction go
N2
3H2
2NH3
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Richard AC E
Erlenmeyer
1825-1909
An alchemist
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
Amedeo Avogadro
1756-1856
Rudolph Clausius
1822-1888
Jacobus van’t Hoff
1852-1911
Niels Bohr
1885-1962
John Dalton
1766-1844
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
William Henry
1775-1836
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
Fitch Rule G3: Science is Referential
Jean Picard
1620-1682
Jacques Charles
1778-1850
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Daniel Fahrenheit
1686-1737
Max Planck
1858-1947
Rolf Sievert,
1896-1966
Blaise Pascal
1623-1662
Georg Simon Ohm
1789-1854
James Maxwell
1831-1879
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Michael Faraday
1791-1867
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
Thomas M Lowry
1874-1936
Werner Karl Linus Pauling Louis Harold Gray
1905-1965
Heisenberg 1901-1994
1901-1976
Example What is KC for the Haber Process at 127oC if
the equilibrium concentrations of gases are NH3 = 3.1x10-2 mol/L; N2 =
8.5x10-1 mol/L; and H2 = 3.1x10-3 mol/L?
KNOW
[NH3]= 3.1x10-2
[N2] = 8.5x10-1
[H2] = 3.1x10-3
DON’T KNOW
KC
RED HERRING
Temp
eq
Prison escapees
were thought to
drag red herrings
across their trail to
fool the tracking dogs.
[NH3]= 3.1x10-2
[N2] = 8.5x10-1
[H2] = 3.1x10-3
Haber Process
K=?
N 2  3H2 
 2 NH 3
NH 

K
 N  H 
2
3
2
3
2
2

 2 mol 
. x10
 31
L 
K
3

 1 mol  
 3 mol 
. x10
 8.5x10 L   31
L 
K  38
. x10
4
1
 mol 


 L 
1
4
K  38
. x10
M2
2
N 2  3H2 
 2 NH 3
NH 

K
 N  H 
Let’s reverse the reaction
2
3
2
3
2
2 NH3 
 N 2  3H 2
Kreverse
N  H 

 K' 
 NH 
2
3
2
2
3
1
K' 
K
K' 
We will compile our rules
After a few examples.
What is the first rule we have shown here?
1
1
38
. x10
M2
4
6
K '  2.6x10 M
2
An example of adding chemical reactions
SO2 ( g )  O2 ( g ) 
 SO3( g )
KSO2  2.3
NO( g )  O2 ( g )
K NO2  4.0
1
2

2( g ) 
NO
1
2
SO2 ( g )  NO2 ( g ) 
 NO( g )  SO3( g )
K SO2 








2( g )  
SO

SO3( g ) 
1
2

2( g ) 
O
K NO 2 





( g)  
NO



Krx  K SO 2 K NO 2
1
2

2( g ) 
O

NO2 ( g ) 
1

 

2 






   NO( g )   O2 ( g )  
 SO
  NO

 SO3 ( g ) 
3( g )  
( g) 







  
 Krx
 KSO2  K NO2   
1





  SO2 ( g )   NO2 ( g ) 
 NO


2 
2( g ) 



 SO
 O

  2( g )   2( g )   
 
Krx  K SO 2 K NO 2  2.3x4.0  9.2
Reversing Reactions
 C  c  D d
KC 
 A a  B b
aA  bB 
 cC  dD
cC  dD 
 aA  bB
 A  B
1
K'C 

K c  C  c  D d
a
b
Same patterns
For Kp
Summing reactions
aA  bB 
 cC  dD
K AB
cC  dD 
 eE  fF
aA  bB 
 eE  fF
KCD
 K  K 
AB
Multiplying reactions
aA  bB 
 cC  dD
K
aA  bB 
 cC  dD
K
aA  bB 
 cC  dD
K
3aA  3bB 
 3cC  3dD
KKK  K 3
CD


n aA  bB 
 cC  dD
Krx  K n
Heterogeneous Equilibria
Are equilibria that involve more than one phase
CaCO3,s 
 CaOs  CO2 , g
K P  PCO2 , g
CaOs
CaCO3,s
CaOs
CaCO3,s
The position of a heterogeneous equilibrium does not
Depend on the amounts of the pure solids or liquids present
Example Problem: Bone is called apatite and can dissolve:
2
3

Ca10  PO4  6  OH  2,s 
10
Ca

6
PO

2
OH

aq
4 ,aq
aq
What is the proper grammar for K?
Ca   PO  OH 

K
2
aq
?
10
Ca
10

K  Ca
2
aq
3
4 ,aq
6
 PO   OH 
4 6

aq
2 ,s
2

  PO  OH 
10
3
4 ,aq
6

aq
2
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #15:
Introduction to Equilibrium
Calculating Equilibrium
Concentrations; 4 examples
Kconc
1.
We can express it.
2.
We can compute it’s numerical value
3.
We can use it to:
a:
Tell if a rx will go
b. Compute [eq]
c.
Compute [eq] after some change, 
3.
Use Kconc to determine if a Rx will go:
a.
b.
Compute Q, the “reaction quotient”
if Q < Kconc, rx goes to right
if Q = Kconc, rx is at equilibrium
if Q > Kconc, rx goes to left
What is Q?
C  D 

Q
 A  B 
c
i
i
a
i
b
i
aA  bB cC  dD


C  D 


 A  B 
c
d
i=initial
KC
eq
d
eq
a
eq
b
eq
Example 1: Use old friend N2O4
What happens when 0.2mole of N2O4 and 0.2 mole of NO2 are
added together in a 4 L vol? Recall Kconc = 0.36M
Don’t KNOW
initial conc.
rx?
KNOW
mole = 0.2 mol
vol = 4 L
K = 0.36M
N 2 O4 
 2 NO2 ( g )
NO 

K  0.36 M 
N O 
2
2
2
Q<,=,>K?
Which way
will reaction go?
4
N O 
2
4 init
 NO 
2 init
0.2mol

 0.05 M
4L
0.2mol

 0.05 M
4L
 NO 
Q
N O 
2
2 init
2
4 init

 0.05 M  2
0.05 M
 0.05 M
Q  0.05 M  K  0.36 M ; rx 

right
Kconc
1.
We can express it.
2.
We can compute it’s numerical value
3.
We can use it to:
a:
Tell if a rx will go
b. Compute [eq]
c.
Compute [eq] after some change, 
Rules for Equilibrium Calculations
1.
2.
3.
4.
Write balanced reaction
Write the equilibrium expression
Calculate the initial conc., Ci
Calculate Q and determine if rx goes
to left or to right.
5. Express “equil. C” (Ceq) in terms of init C
and change, x, (Ceq =Cix)
6. Write K and put in Ceq, solve for x
7. Calculate equilibrium molarities
8****
Check your answer!!!!!
EXAMPLE 2: A BIT SIMPLE
For the system (all are gases) Kc is 0.64 at 900 K. :
CO2 , g  H2 , g 
 COg  H2 Og
Suppose we start with CO2 and H2, both at a concentration of
0.100 mol/L.
When the system reaches equilibrium, what are the concentrations
of products and reactants at 900K?
Red herrings?
1.
Balance Equation
already done
2.
CO2 , g  H2 , g 
 COg  H2 Og
Write Kc
C  D 

K 
 A  B 
CO   H O 

K 
CO   H 
c
d
eq
C
a
b
eq
eq
1
1
g
2
C
g
1
2,g
3.
Kc is 0.64 at 900 K. Suppose we
start with CO2 and H2, both at a
concentration of 0.100 mol/L.
eq
1
 0.64
2, g
Calculate Original or Initial Concentrations
[CO2]init = 0.100 mol/L
[H2]init = 0.100 mol/L
4.
Calculate Q and determine if rx goes l or r.
Q
 CO init  H2 Oinit
CO   H 
2 init
2 init

0
 0100
.  0100
. 
Q  0  Kc  0.64; rx 

right
5.
Express “equil. C” in terms of init C+ or - x
right

Q

0

K

0
.
64
;
rx



CO2 , g  H2, g  COg  H2 Og
c
CO   CO   x
H   H   x
2 eq
2 eq
2 init
2 init
Mass balance Limiting reaction
 CO eq   CO init  x
 H O   H O
2
eq
2
init
x
OR: Construct an ICE chart
Reaction:
stoichiometry
Initial conc
Change (to right)
Equil. conc
6.
K
CO2 , g  H2 , g


COg  H2 Og
1
1
1
0.100 0.100 0
-x
-x
+x
1
0
+x
0.100-x 0.100-x 0+x
0+x
Write K and put in equil Conc.
 CO eq  H2 Oeq
CO   H 
2 eq
2 eq
 x  x 


 0100


.
 x 0100
.
 x
x


 0100
.



x
2
x

K  0.64  
 0100
. 


x
2
CO2 , g  H2 , g 
 COg  H2 Og
Kc is 0.64 at 900 K. Suppose we
start with CO2 and H2, both at a
concentration of 0.100 mol/L.
x
0.8 
0100
. x
0.8 0100
.
 x  x
0.0800  08
. x x
0.0800  x  08
. x
0.0800  18
. x
0.0800
 x  0.0444
18
.
x  0.044
Reminder – x is change
In concentration
7.
x = 0.044
Calculate equilibrium molarities
stoichiometry
initial conc
change (to right)
Equil. conc
CO2
1
0.100
-x
0.100-x
H2
1
0.100
-x
0.100-x
CO
1
0
+x
0+x
H2O
1
0
+x
0+x
[CO2]eq = [H2]eq = 0.100-x = 0.05556 M
[CO]eq = [H2O]eq = 0.0444 M
8**** Check your answer!!!!!
Suggestions for ways?
K
 CO eq  H2 Oeq
CO   H 
2 eq
2 eq
?  0.0444 M  0.0444 M 
 0.64 
 0.638
 0.05556 M  0.05556 M 
OJO: When I round first and then check I get:
Kc = 0.60.
EXAMPLE PROBLEM 3: More difficult
If we add 0.1 mol of N2O4 in 1 L, what are the equilibrium
concentrations? Recall that Kc = 0.36M.
Red herrings?
1.
Balance Equation
N 2 O4 
 2 NO2 ( g )
2.
Write Kc
C  D 
NO





 0.36 M
 A  B   N O 
c
KC
eq
2
eq
a
eq
3.
d
2 , g ,eq
b
eq
2
4 , g ,eq
Calculate Original or Initial Concentrations
0.1 mol of N2O4 in 1 L  N O 
2
4 o
 NO 
2 o

01
. mol
 01
. M
1L
0.0mol

 0.00 M
1L
K  0.36 M
N 2 O4 
 2 NO2 ( g )
NO 

K  0.36 M 
N O 
2
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
2

N 2 O4
 NO 

2 o
4.
2
o

Old Friend: N2O4
4
01
. mol
 01
. M
1L
0.0mol

 0.00 M
1L
Calculate Q and determine if rx goes l or r
 
Q 
Q
 NN OO 
22
NO22
NO
init
init
22
44 init
init

 0.0 M  2
01
. M
 0.0 M  Kc ; rx 

right
If we add 0.1 mol of N2O4 in 1 L,
5.
Express “equil. C” in terms of
what are the equilibrium
orig C+ or - x
concentrations? Recall that Kc = 0.36
N 2 O4 
 2 NO2 ( g )
M.
1
stoichiometry
N2O4
NO2
0.1
-x
0.1-x
0
+ 2x
0+2x
Initial conc
Change (to right)
Equil. conc
6.
2
Write K and put in equil Conc.
K  0.36 M
N 2 O4 
 2 NO2 ( g )
NO   22xx 
NO
NO

NO 
KK00.36
.36MM

 NN OO  NN OO xx
22
22
2 2 eq
2 2 4 4 eq
and solve for x
2 2 init
init
2 2 4 4 init
init
 0  2 x 2

 01
.  x
OJO!!
Old Friend: N2O4
 NO 
K  0.36 M 
N O 
2
2 eq
2
4 eq
 NO   2 x


 N O   x
 0  2 x 2
K
 01
.  x
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
2
2 init
2
4 init
 0  2 x 2

 01
.  x
ax  bx  c  0
2
2
K 01
.  x   0  2 x
2
K 01
.  xK   2 x  4 x 2
4 x 2  xK  01
.K 0
x
b 
b 2  4ac
2a
4 x 2  Kc x  01
. Kc  0
ax 2  bx
c 0
a  4 b  Kc
b 
x
x
c  01
. Kc
b 2  4ac
2a
 0.36 
x
Kc  0.36 M
 0.36 2  4 4  0.036
2 4
OJO!!!
2


 0.36  0.36  0.5764 4 0.036
8
 0.36  0.7056  0.36  0.84
x

 0.06 and  015
.
8
8
Which is plausible?
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
x
 0.36 
0.7056
8
Initial conc
Change (to right)
Equil. conc
 0.36  0.84

 0.06 and  015
.
8
N2O4
0.1
-x
0.1-x
NO2
0
+ 2x
0+2x
No, not plausible. We would be forced to conclude:
x = -0.15
2x = [NO2] = -0.3 M
Plausible?
7.
Calculate equilibrium molarities
x = 0.060
Initial conc
Change (to right)
Equil. conc
N2O4
0.1
-x
0.1-x
01
.  0.06  0.4
Equilibrium [N2O4] = 0.04 M
Equilibrium [NO2] = 0.12 M
8**** Check your answer!!!!!
NO
0
+ 2x
0+2x
0  2 0.06  012
.
N 2 O4 
 2 NO2 ( g )
1.
.  .04 ?
Plausible: [N2O4] decreased? 01
Plausible: [NO] increased? 0  012
. ?
2.
Fits the equilibrium constant?
NO 
 012

. 
K  0.36 M 

 N O   0.04
2
2
2
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
4
2
 0.36 M
Example 4: Do Units Matter in how we approach the problem?:
For the reaction of hydrogen gas with iodine gas at room temperature the
Kp is 1x10-2. Suppose that you mix HI at 0.5, H2 at 0.01 and I2 at 0.005
atm in 5 liter volume. Calculate the equilibrium Pressures.
Beforehand: red herrings?
1.
5L volume
Balance Equation
H2 , g  I 2 , g 
 2 HI ( g )
2.
Write K
aA  bB 
 cC  dD
P  P 

P  P 
c
KPr essue
C
d
D
a
A
B
P 


P  P 
2
b
KP
HI ,eq
H2 .eq
I 2 ,eq
 1x10  2
For the reaction of hydrogen gas with iodine gas at
room temperature the Kp is 1x10-2. Suppose that
you mix HI at 0.5, H2 at 0.01 and I2 at 0.005 atm
in 5 liter volume. Calculate the equilibrium Pressures.
3.
Calculate Original or Initial Conc or Pressure
A bit of red herring = already know them.
PHI = 0.5 atminit
PH2 = 0.01 atminit
PI2 = 0.005 atminit
4.
Calculate Q and determine if rx goes l or r
Q
P 
2
HI ,init
P  P 
H2 .init
I 2 ,init
 0.5atm 2
left

 5000  K p  0.01; rx 
 01
. atm  0.005atm
OJO
5.
For the reaction of hydrogen gas with iodine gas at
room temperature the Kp is 1x10-2. Suppose that
you mix HI at 0.5, H2 at 0.01 and I2 at 0.005 atm
in 5 liter volume. Calculate the equilibrium Pressures.
Express “equil. C” in terms of orig C+ or - x
stoichiometry
Initial conc
Change (to left)
Equil. conc
6.
Kp 
H2
1
.01
+x
.01+x
Write K and put in equil Pressures
 0.5  2 x 2
(0.01  x )(0.005  x)
I2
1
.005
+x
.005+x
HI
2
0.5
-2x
0.5-2x
Kp 
 0.5  2 x 2
K p  0.01
(0.01  x )(0.005  x)
K p (0.01  x)(0.005  x)   05
.  2 x
2
K p 5x10  5  0.015x  x 2   0.25  2 x  4 x 2
5x105 K p  0.015xK p  x 2 K p  0.25  2 x  4 x 2


0.25  2 x  4 x 2  5x10 5 K p  0.015xK p  x 2 K p  0
4 x 2  2 x  0.25  5x105 K p  0.015xK p  x 2 K p 

 
 
4  K x  x2  0.015K   0.25  5x10

4 x 2  x 2 K p   2 x  0.015xK p  0.25  5x10  5 K p  0
2
p
p
5

Kp  0
 4 .01 x 2  x 2  0.00015  0.25  5x107   0
 399
.  x 2  x 2.00015   0.2499995  0
 b  b 2  4ac
x
2a
2
 399

. x  x 2.00015   0.2499995  0
2


 (  2.00015)   2.00015  4 3.99 0.2499995
x
2 3.99
x
2.00015 
4.0006  3.98992
7.98
2.00015  0102995
.
x
7.98
x  0.237739
x  0.263552
Which do we use + or -?
x  0.237739
x  0.263552
H2
I2
HI
stoichiometry
1
1
2
Initial conc
.01
.005
0.5
Change (to left)
+x
+x
-2x
Equil. conc
.01+x
.005+x
0.5-2x
If we use the second answer then we will get
PHI ,eq  0.5  2 x  05
.  2(0.263552)   0.0271
Not plausible
x  0.237739
stoichiometry
Initial conc
Change (to left)
Equil. conc
H2
1
.01
+x
.01+x
I2
1
.005
+x
.005+x
HI
2
0.5
-2x
0.5-2x
PHI ,eq  05
.  2 x  05
.  2(0.237739)  0.024523
PI2 ,eq  0.005  x  0.005  0.237739  0.242739
PH2 ,eq  0.01  x  0.01  0.237739  0.247739
P 

 0.024523
K

 0.01
 P  P   0.242739 0.247739
2
2
HI ,eq
H2 ,eq
I 2 ,eq
Checks!!!
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #15:
Introduction to Equilibrium
Making Assumptions
General
FITCH Rules
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
 qq 

E  k
C1. It’s all about charge
r r 
C2. Everybody wants to “be like Mike”
 qq 
C3. Size Matters

E  k
r r 
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
1 2
Chemistry
el
1
2
1 2
el
1
For systems with multiple reactions
Need to simply = ASSUMPTIONS
2
Example on Using Simplifications
If 1.0 mol NOCl is placed in a 2.0 L flask what are the equilibrium
concentrations of NO and Cl2 given that at 35 oC the equilibrium
constant, Kc, is 1.6x10-5 mol/L?
Red herrings:
35 oC is a red herring
Clues?
K is “small” compared to others (<<< 1) we have worked with !!!!!
Example 2
EXAMPLE 3:
Example 4:
CO2 , g  H2 , g 
 COg  H2 Og
N 2 O4 
 2 NO2 ( g )
H2 , g  I 2 , g 
 2 HI ( g )
Kc is 0.64.
Kc = 0.36M
Kp is 1x10-2.
We will define
Small in the
Next chapter!
Example on Using Simplifications
If 1.0 mol NOCl is placed in a 2.0 L flask what are the equilibrium
concentrations of NO and Cl2 given that at 35 oC the equilibrium
constant, Kc, is 1.6x10-5 mol/L?
Red herrings:
35 oC is a red herring
Clues?
K is “small” compared to others (<<< 1) we have worked with !!!!!
We are starting with reactants
1.
Balance Equation
2 NOCl 2 NO  Cl2


2. Write K
 NO 2 Cl2 
Kc 
 NOCl  2
2 NOCl 2 NO  Cl2


3.
4.
If 1.0 mol NOCl is placed in a 2.0 L flask what
are the equilibrium concentrations of NO and Cl2
given that at 35C the equilibrium constant is
1.6x10-5 mol/L?
Calculate or Initial Concentrations
[NOClinit] = 1.0 mol/2L = 0.5 M
[NOinit] = 0 M
[Cl2init ]= 0 M
Calculate Q and determine if rx goes l or
 NO  Cl   0 0
Q

 0; rx 

0.5
 NOCl 
2
g ,init
2 , g ,init
2
g .init
right
5.
Express “equil. C” in terms of orig C+ or - x
2 NOCl 
 2 NO  Cl2
stoichiometry
Initial conc
Change (to right)
Equil. conc
NOCl
2
0.5
-2x
0.5-2x
NO
2
0
+2x
0+2x
Cl2
1
0
x
0 +x
Assumptions
~0.5
2x
x
 22xx 22xx
KK
0.5
22
Assumptions are based on
NOClinit
xx
NOCl
init22
a) K is small so x is small  00.000
.000xx
sig fig
b) If x is small then sig fig 0.4999 z 
 0.5
rules eliminate x
6.
Write K and put in equil Conc


 2 x 2 x
NOClinit
 2 x 2 x 4 x 3
3
K


16
x
 0.5 2 0.25

2
K  16x 3
16
. x105  16x 3
16
. x10  5
 x3
16
10  6  x 3
3
10
6
 x
x  10 2
7.
Calculate equilibrium molarities
x  10 2
stoichiometry
Initial conc
Change (to right)
Equil. conc
NOCl
2
0.5
-2x
0.5-2x
NO
2
0
+2x
0+2x
Cl2
1
0
x
0 +x
Assumptions
~0.5
2x
x
 NOCl   0.5
g ,eq
NO   22xx  2 x10
 NO
ClCl   xx  10
2
gg,,eq
eq
22,
, gg,eq
,eq
ARE WE DONE?
2
 NOCl   0.5  2x
 NOCl   0.5  210
 NOCl   0.5  0.02
CHECK ASSUMPTIONS
g ,eq
g ,eq
2

g ,eq
0.5
 0.02
0.48
Answer
Accept error less
than 5%
Where are
The Sig Fig?
0.5
 real  estimated 
error  
100

real


 0.48 .5 
error  
100  4%

 0..48 
Are we done yet?
8**** Check your answer!!!!!
 NOCl   0.5
 NO   2 x  2 x10
Cl   x  10
g ,eq
2
g ,eq
2
2 , g ,eq
 NO 2 Cl2 
5
Kc 

16
.
x
10
M
2
 NOCl 
Kc 


2 x10
2  2 
 
2
0.5
10  2  ?
 16
. x10  5 M
Kc  16
. x10 5  16
. x10 5 M
POINT of ASSUMPTIONS
1.
Avoid using polynomial equations
2.
Can do it with small K values
3.
Assume little change (get rid of an x)
4.
Must check the assumptions
Multiple Equilibria are not solvable
With simple cubic, quadratic equations
Chemists are Lazy!
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #15:
Introduction to Equilibrium
Qualitative Predictions
For direction of Equilibrium
Chemists are ________
If we don’t have to calculate can we make
qualitative descriptions of equilibrium?
Le Châtelier’s Principle
if a change is imposed on a system at equilibrium, the position of the
equilibrium will shift in a direction that tends to reduce that change
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Richard AC E
Erlenmeyer
1825-1909
An alchemist
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
Amedeo Avogadro
1756-1856
Rudolph Clausius
1822-1888
Jacobus van’t Hoff
1852-1911
Niels Bohr
1885-1962
John Dalton
1766-1844
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
William Henry
1775-1836
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
Fitch Rule G3: Science is Referential
Jean Picard
1620-1682
Jacques Charles
1778-1850
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Daniel Fahrenheit
1686-1737
Max Planck
1858-1947
Rolf Sievert,
1896-1966
Blaise Pascal
1623-1662
Georg Simon Ohm
1789-1854
James Maxwell
1831-1879
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Michael Faraday
1791-1867
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
Thomas M Lowry
1874-1936
Werner Karl Linus Pauling Louis Harold Gray
1905-1965
Heisenberg 1901-1994
1901-1976
Effects of Changes on the System
1. Addition of inert gas does not affect the
equilibrium position.
2. Concentration: The system will shift away from
the added component.
3. Decreasing the volume shifts the equilibrium
toward the side with fewer moles.
4. Temperature: K will change depending upon the
temperature (treat the energy change as a
reactant).
LeChatlier Examples
Example: (Inert gas) Suppose we have the reaction initially at equilibrium
2SO2 , g  O2, g 
 2 SO3, g
Kc,1000 K  2.8x102
What happens when we add some N2,g, , does Q change?
SO



O SO 
2
KC
3, g ,eq
2 , g ,eq
2
2 , g ,eq
Addition of inert gas does not affect the
equilibrium position. Similarly……
Example: (Effect of Solid)
What is the effect on equilibrium in the calcination (decomposition)
CaCO3,s  CaOs  CO2, g
of limestone produced by adding a small quantity of CaCO3(s)?
CO CaO 

K 
 CO 
CaCO 
2,g
s
c
2,g
3, s
Change in quantity of solids do not affect
direction of equilibria. So addition of more
limestone is irrelevant.
Example What is the effect on equilibrium in the (decomposition)
Concentration: The system will shift

CaCO3,s  CaOs  CO2, g
toward from the removed component.
of removing some CO2,g, does Q change?

 
CO2, g ,new  CO2, g ,eq
 CO
CO 


CO 

Q
 K  CO ; rx 

x
2 , g ,eq
2 , g ,new
2 , g ,eq
right
c
x
2 , g ,eq
Example Suppose we have the reaction initially at equilibrium:
Kc,1000 K  2.8x102
2SO2 , g  O2, g 
 2 SO3, g
What happens when we add some SO3,g , does Q change?
SO   SO 
SO   xSO 
3,new
3,new
3,eq
3,eq
Q

x SO3, g ,eq

2
O SO 
2
2 , g ,eq
2 , g ,eq
 KC 


2
SO3, g ,eq
O SO 
2
2 , g ,eq
; rx 
2 , g ,eq
Concentration: The system will shift
away from the added component.
left
Example What happens if we decrease the volume of the container?
2SO2 , g  O2, g 
 2 SO3, g
2
 nSO 2
nSO3,3g,,geq,eq
2
2
SO3, g ,eq 2
 Veq 
nSO3, g ,eq
SO3, g ,eq
 Veq 
Kc 
2 
2
Kc  O
2   nO2 , g , eq   n SO2 , g , eq 2 
2 ,eq , g SO2 , g ,eq
nO2 , g ,eq nSO2 , g ,eq 
O2 ,eq , g SO2 , g ,eq
nO2 , g ,eq nSO2 , g ,eq
 Veq  Veq 
 Veq   Veq 






2



2

2
 1 
 
 Veq 
2
 1  1 
 

V
V
 eq   eq 
 1 
2
2
2

 
nSO3, g ,eq
nSO3, g ,eq
nSO3, g ,eq

 Veq 
1
Kc 

2
3 
2


nO2 , g ,eq nSO2 , g ,eq  1 
nO2 , g ,eq nSO2 , g ,eq  1   nO2 , g ,eq nSO2 , g ,eq
 
 Veq 
 Veq 
Veq
Vnew  Veq
Vnew 
x
2
2




nSO3, g ,eq
nSO3, g ,eq

 Veq


right
Q 
 Kc  

2
2  Veq ; rx 
x
 nO

 nO

 2 , g ,eq nSO2 , g ,eq 
 2 , g ,eq nSO2 , g ,eq 


























2  Veq



Q changes when we change volume because of differences in stoichiometry
Decreasing the volume (increasing P) shifts the equilibrium toward the
side with fewer moles
2SO2( g )  O

2( g ) 
2SO3( g )
Kc  2801000 K
Decrease volume, what happens?
Number of moles of reactants?
Number of moles of products?
If we form reactants we get 3 moles
If we form products we get 2 moles.
reaction moves to decrease moles
moves to the right.
13_318
LeChatlier Example: Haber Process
N 2 ( g )  3H2 ( g )
What happens when volume is decreased?


2 NH3( g )
Key:
N2
H2
To equil
NH3
7 N2
4 H2
3NH3
14
7 N2
4 H2
3NH3
14
To equil
6 N2
1H2
5NH3
12
Effect of Temperature????
Consider effect of raising temperature on two reactions:
2SO2( g )  O2( g )  2SO3( g )
 H o   180kJ
N 2( g )  O2( g )  2 NO2( g )
 H o   181kJ
What will happen?
Another way to see this is to write the rxs by considering heat
As a product or reactant. Reaction shifts away from heat:
2SO2( g )  O

2( g ) 
2SO3( g )  heat; rx 
left
N 2( g )  O2( g )  heat 2 NO2( g ) ; rx 


right
Temperature: K will change depending upon
the temperature (treat the energy change as a
reactant).
Do you see a pattern?
o


H
 K1 
1  Van’t Hoff equation (1852-1911)
forward rx  1
ln  
  
R
 K2 
 T1 T2 
 k1    E a   1 1 
ln   
  

 k 2   R   T1 T2 
Arrhenius Equation
 P1     Hvaporization   1 1 
ln   
  
R
 P2  
  T1 T2 
Clausius-Clapeyron
equation
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Richard AC E
Erlenmeyer
1825-1909
An alchemist
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
Amedeo Avogadro
1756-1856
Rudolph Clausius
1822-1888
Jacobus van’t Hoff
1852-1911
Niels Bohr
1885-1962
John Dalton
1766-1844
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
William Henry
1775-1836
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
Fitch Rule G3: Science is Referential
Jean Picard
1620-1682
Jacques Charles
1778-1850
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Daniel Fahrenheit
1686-1737
Max Planck
1858-1947
Rolf Sievert,
1896-1966
Blaise Pascal
1623-1662
Georg Simon Ohm
1789-1854
James Maxwell
1831-1879
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Michael Faraday
1791-1867
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
Thomas M Lowry
1874-1936
Werner Karl Linus Pauling Louis Harold Gray
1905-1965
Heisenberg 1901-1994
1901-1976
N 2 ( g )  3H 2 ( g ) 
 2 NH 3( g )
 H   92.2 kJ
0
K  6 x105
25C
What is K at 100 oC?
Predict first using LeChatelier’s
principle: will it get larger or
smaller? You try it!, we will compare to calc.
N 2 ( g )  3H2 ( g ) 
 2 NH3( g )  heat
 K1    H  1
1
ln

 

K
R
T
T
2 
 2
 1
o
J 

3
   92.2 x10


 6 x105 
mol 
ln 
 
J
K
2


8.31
mol  K
e
 6 x105 
ln 

 K2 
 6 x105 
7 .48628
 
 1783.405
e
 K2 
1
1



 273  25K 273  100 K 
 6 x105 
11095.07  .000675 
ln 

 7.48628



1   Kelvin 
 K2  

 Kelvin 
6 x105
 K2  336.435
1783.405
What have we learned?
1.
2.
3.
4.
5.
6.
7.
8.
Expression for K
Expression for Q
Distinguish between initial and final or
equilibrium concentrations.
How to predict direction of a reaction
How to calculate the equilibrium conc.
How to make assumptions to ease the
calculations.
How to check the calculations
How to use Le Chat.... principle to
effect of conc., pressure, volume, and
temp. on a reaction.
“A” students work
(without solutions manual)
~7 problems/night.