Gaseous Chemical
Equilibrium
“A” students work
(without solutions manual)
~ 10 problems/night.
1. Can balance equations
2. Can predict direction of reaction
(∆H)
3. Can compute rates
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
rate = −
Office Hours Th&F 2-3:30 pm
d [ A]
m
= k [ A]
adt
4. Can we predict what happens at end?
Module #15:
Introduction to Equilibrium
1. Define Equilibrium,
And equilibrium
constant
Consider the reaction at 100 oC, where the initial
Pressure of dinitrogen tetroxide is 1 atm:
Our Friend for this Chapter
review
N 2 O4 , g → 2 NO2 , g
Dinitrogen tetroxide
The decomposition of dinitrogen tetroxide vs time
Is shown in the table below
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.
1.8
We studied it’s decomposition kinetics in
The preceding chapter
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
N 2 O4 , g → 2 NO2 , g
0
20
40
60
t, s
1
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/N204
1
1.67
2.86
4.55
4.55
4.55
1st order reaction
review
review
ln PN 2 O4 ,t = ln PN 2 O4 ,t = 0 − kt
rate = k [ A]
ln PN 2 O4 ,t = ln(1) − kt
ln PN 2 O4 ,t = 0 − kt
1
k = 0.0254
s
1
ln N204
0
-0.51
-1.05
-1.51
-1.51
-1.51
[ ]
[ ]
ln At = ln Ao − kt
0
-0.2
-0.4
1.8
0
1.6
-0.6
ln[N2O4]
4
1.4
-0.4
3.5
1.2
1/PN2O4
ln[N2O4]
3
-0.8
2.5
ln[N2O4] = - 0.0065 -0.0254(t)
-0.8
R2 = 0.9992
-1
2
-1
0.6
1.5
0.4
-1.2
1
0.2
-1.4
0.5
0
-1.6
-1.2
0
0
20
40
60
80
100
120
0
0
20
40
60
t, s
rate = k [ A]
t
80
100
40
60
80
100
120
-1.4
t, s
t, s
[ ]
20
120
rate = k [ A]2
rate = k [ A]1
0
-1.6
0
1
1
−
= kt
[ A] A0
[ ]
20
40
60
0
80
100
120
t, s
[ ]
ln At = ln Ao − kt
[ A ] = [ A ] − kt
1.8
k forward
N 2 O4, g ⎯⎯⎯⎯
⎯→ 2 NO2, g
1.6
“Equilibrium” = “steady state concentrations”
1.
occurs when rate forward = rate backward
1.4
atm
1.2
1
( )
0.6
k f PN2O4 , g = k b PNO2 , g
0.4
0.025
0.025
0.2
rate = k f PN 2O4
0
0
20
40
1⎞
⎛
rate = ⎜ 0.0254 ⎟ PN 2 O4
⎝
s⎠
0.02
0.02
60
80
100
120
t, s
kf
kb
0.015
0.015
What do you observe?
2.
0.01
0.01
(P )
=
( )
rate = k b PNO2
00
00
20
20
40
40
60
60
80
80
2
rate = 0.002309
100
100
( )
1
P
atm ⋅ s NO2
2
kb
120
120
3.
t, s
backward
2 NO2, g ⎯⎯⎯⎯
⎯→ N 2 O4, g
k
t, s
kbackward = 0.002309
PN2O4 , g
(P ) = K
=
PN2O4 , g
1
s = 11atm
Kequilibrium =
1
0.002309
atm ⋅ s
00254
.
NO2 , g
equilibrium
BOTH reactions are occurring competitively
N 2 O4, g →← 2 NO2, g ojo
1. Conc. Are “stable”
2. Rate forward = rate backward
2
1
atm⋅s
2 NO2, g ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ N 2 O4, g
2
NO2 , g
concentrations are also constant (steady state)
kf
1
N 2 O4, g ⎯⎯⎯⎯⎯⎯⎯⎯s → 2 NO2, g
2
2
0.005
0.005
k forward = 0.0254
Rate forward = Ratebackward
Conc. plot
0.8
0.03
0.03
rate,
rate, Atm/s
Atm/s
atm
-0.6
1
0.8
Using the rate
Constants the
Rate at each time
Can be calculated
5
4.5
-0.2
4.
Hamilton’s Bar: Legal occupancy: 10
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
1
N2O4
1.00
0.22
K equilibrium = 11atm =
?
Kequilibrium = 11atm =
?
(P )
(P )
The stupid
analogy
2
NO2
PN 2 O4
NO2
0.00
1.56
22
N2O4
N2O4
0.00
0.00
0.07
0.07
3
N2O4
1.00
0.42
NO2
NO2
1.00
1.00
0.86
0.86
NO2
1.00
2.16
2
3 out
NO2 , g
PN 2O4 , g
2
(156
. atm)
0.22atm
11atm = 1106
. atm
?
Kequilibrium = 11atm =
( 0.86atm) 2
?
Kequilibrium = 11atm =
0.07atm
?
Hamilton’s appears to have an unchanged
population, but constant exchange of drinkers is
occurring. The equilibrium is 10/3.
2
( 216
. atm)
0.42atm
?
11atm = 1056
. atm
11atm = 1111
. atm
The “Generic” Rule
FITCH Rules
General
10 in
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,
Stoichiometry shows
up as power
aA + bB →
← cC + dD
1 2
Chemistry
el
1
2
1 2
el
1
(P ) (P )
(P ) (P )
c
Pressure
Of reactants
On bottom
2
K Pr essue =
C
d
D
a
A
b
B
Constant microscopic
change
An equilibrium constant can be written in terms of
Molarity also
3
Pressure
Of Products
On top
PV = nRT
aA + bB →
← cC + dD
P
n
= = [molarity ]
RT V
P = RT[molarity ]
(P ) (P )
=
(P ) (P )
c
K Pr essue
⎧ ( RT ) c ( RT ) d ⎫
K Pr essue = ⎨
a
b ⎬ K concentration
⎩ ( RT ) ( RT ) ⎭
d
C
D
a
A
b
(c+ d )− (a +b)
K Pr essue = ( RT )
Kconcentration
( RT[ C]) ( RT[ D]) d
( RT[ A]) a ( RT[ B])b
Office Hours Th&F 2-3:30 pm
∆n
K Pr essue = ( RT ) f −i K concentration
⎧ ( RT ) c ( RT ) d ⎫ ⎧ [ C ] c [ D] d ⎫
K Pr essue = ⎨
a
b ⎬⎨
a
b ⎬
⎩ ( RT ) ( RT ) ⎭ ⎩ [ A] [ B] ⎭
Kconcentration
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
B
c
K Pr essue =
“A” students work
(without solutions manual)
~ 10 problems/night.
Module #15:
Introduction to Equilibrium
When doing problems
watch out that they
are not switching units
on you!!!!!
[ C ] c [ D] d
=
[ A] a [ B] b
Sample calculation
Of K
Calculate the Kc for the reaction at 25oC:
Consider the Haber Reaction which “fixes” nitrogen and is immensely
2 NOg + Cl2 , g →
← 2 NOCl g
N 2 , g + 3H 2 , g →
← 2 NH 3, g
if the equilibrium pressures are PNOCl = 1.2 atm; PNO = 5.0x10-2 atm; and PCl2 =
3.0x10-1 atm.
KNOW
PNOCl = 1.2 atm
PNO = 5.0x10-2 atm
PCl2 = 3.0x10-1 atm
(P )
=
(P ) (P
DON’T
Kconc
eq
KP
NO
Cl2
2
(12
. atm)
2
( 0.050atm) ( 0.30atm)
KP =
19
. x10 3 atm 2
= 19
. x10 3 atm −1
atm 3
gases will we get ammonia?
1
L ⋅ atm
⎛
⎞
19
. x10
= ⎜ 0.0826
⋅ 298 K ⎟
⎠
atm ⎝
mol ⋅ K
)
KP =
none
1
( 2 ) − ( 3)
19
. x10 3
= ( RT )
Kconcentration
atm
2
NOCl
2
important for the large scale production of ammonia used in
explosives
fertilizers
If we mix nitrogen and hydrogen
Red Herring?
19
. x10 3
(c+ d )− (a +b)
K Pr essue = ( RT )
Kconcentration
N2 941 kJ/mole bond energy
3H2 432 kJ/mole
−1
3
K concentration
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
Haber found a
Catalyst to make
reaction go
N2
3H2
2NH3
1
M
4
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
An alchemist
Count Alessandro G
A A Volta, 1747-1827
Amedeo Avogadro
1756-1856
John Dalton
1766-1844
Jean Picard
1620-1682
William Henry
1775-1836
Daniel Fahrenheit
1686-1737
Jacques Charles
1778-1850
Blaise Pascal
1623-1662
Georg Simon Ohm
1789-1854
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Michael Faraday
1791-1867
B. P. Emile
Clapeyron
1799-1864
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?
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
KNOW
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
Richard AC E
Erlenmeyer
1825-1909
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
Rudolph Clausius
1822-1888
Jacobus van’t Hoff
1852-1911
Niels Bohr
1885-1962
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Max Planck
1858-1947
Rolf Sievert,
1896-1966
James Maxwell
1831-1879
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
DON’T KNOW
RED HERRING
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
[NH3]= 3.1x10-2
[N2] = 8.5x10-1
[H2] = 3.1x10-3
Thomas M Lowry
1874-1936
KC
Temp
eq
Prison escapees
were thought to
drag red herrings
across their trail to
fool the tracking dogs.
Werner Karl Linus Pauling Louis Harold Gray
Heisenberg 1901-1994
1905-1965
1901-1976
Fitch Rule G3: Science is Referential
N 2 + 3H 2 →
← 2 NH 3
[NH3]= 3.1x10-2
[N2] = 8.5x10-1
[H2] = 3.1x10-3
Haber Process
K=?
N 2 + 3H 2 →
← 2 NH 3
[ NH ]
K=
[ N ][ H ]
3
2
2 NH 3 →
← N 2 + 3H 2
3
Kreverse
3
mol ⎤
⎡
. x10 −2
⎢⎣ 31
L ⎥⎦
K=
3
⎡
− 1 mol ⎤ ⎡
− 3 mol ⎤
x
x
8
.
5
10
31
.
10
⎢⎣
L ⎥⎦ ⎢⎣
L ⎥⎦
[ N ][ H ]
= K' =
[ NH ]
2
3
2
2
3
2
2
Let’s reverse the reaction
3
2
2
2
[ NH ]
[ N ][ H ]
2
K=
K = 38
. x10 4
K = 38
. x10 4
1
K' =
K
1
⎛ mol ⎞
⎜
⎟
⎝ L ⎠
2
K' =
1
38
. x10 4
1
M2
We will compile our rules
After a few examples.
What is the first rule we have shown here?
1
M2
K ' = 2.6 x10 − 6 M 2
5
Reversing Reactions
An example of adding chemical reactions
SO2 ( g ) + 2 O2 ( g ) →
← SO3( g )
K SO 2 = 2.3
1
NO2 ( g ) →
← NO( g ) + 2 O2 ( g )
cC + dD →
← aA + bB
K NO 2 = 4.0
1
SO2 ( g ) + NO2 ( g ) →
← NO( g ) + SO3( g )
Krx = K SO 2 K NO 2
K SO 2 =
⎡
⎢
⎣
SO3( g )
⎤⎡
⎥⎢
2( g ) ⎦ ⎣
SO
⎤
⎥
⎦
1
⎤2
⎥
2( g ) ⎦
O
K NO2 =
⎡
⎢
⎣
⎤⎡
1
⎤2
⎥
2( g ) ⎦
⎞
⎡
⎤⎡
⎤
⎟
⎢ SO
⎥⎢
⎥
3( g ) ⎦ ⎣ NO( g ) ⎦
⎟ = ⎡⎣
= Krx
⎤⎡
⎤
⎟ ⎢ SO2 ( g ) ⎥ ⎢ NO2 ( g ) ⎥
⎦⎣
⎦
⎠ ⎣
K CD
AB
CD
(
Multiplying reactions
Krx = K SO 2 K NO 2 = 2.3x 4.0 = 9.2
aA + bB →
← cC + dD
K
aA + bB →
← cC + dD
K
aA + bB →
← cC + dD
K
3aA + 3bB
cC + 3dD
n aA + bB →
← cC + dD
What is the proper grammar for K?
CaCO3,s →
← CaOs + CO2 , g
[Ca ] [ PO ] [OH ]
K=
2+
aq
?
K P = PCO2 , g
CaOs
Krx = K n
KKK = K 3
→
←3
−
2+
3−
Ca10 ( PO4 ) 6 ( OH ) 2 ,s →
← 10Ca aq + 6 PO4 ,aq + 2OH aq
Are equilibria that involve more than one phase
CaCO3,s
)
Example Problem: Bone is called apatite and can dissolve:
Heterogeneous Equilibria
CaOs
Same patterns
For Kp
( K )( K )
aA + bB →
← eE + fF
NO
⎛
⎞⎛ ⎡
⎤⎡
⎡
⎤
⎜
⎟ ⎜ ⎢ NO( g ) ⎥ ⎢ O
⎢ SO3 ( g ) ⎥
⎣
⎦
⎜
⎟ ⎜ ⎣ ⎡ ⎦⎣ ⎤
( KSO2 )( K NO2 ) =
1
⎜⎡
⎢ NO
⎥
⎤⎡
⎤2 ⎟⎜
2( g ) ⎦
⎣
⎝ ⎢⎣ SO2 ( g ) ⎥⎦ ⎢⎣ O2 ( g ) ⎥⎦ ⎠ ⎝
[ A] [ B]
1
=
K c [ C ] c [ D] d
b
K AB
cC + dD eE + fF
⎤
⎥
2( g ) ⎦
1
⎤2
⎥
2( g ) ⎦
K'C =
a
→
←
NO( g ) ⎥⎦ ⎢⎣ O
⎡
⎢
⎣
[ C ] c [ D] d
[ A] a [ B] b
Summing reactions
aA + bB →
← cC + dD
⎡
⎢
⎣
KC =
aA + bB →
← cC + dD
[Ca
10
10
3−
4 ,aq
6
( PO ) ( OH )
4 6
−
aq
2 ,s
2
]
CaCO3,s
[
2+
K = Ca aq
The position of a heterogeneous equilibrium does not
Depend on the amounts of the pure solids or liquids present
6
] [ PO ] [OH ]
10
3−
4 ,aq
6
−
aq
2
“A” students work
(without solutions manual)
~ 10 problems/night.
Kconc
1.
We can express it.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
2.
We can compute it’s numerical value
Office Hours Th&F 2-3:30 pm
3.
We can use it to:
a:
Tell if a rx will go
b. Compute [eq]
c.
Compute [eq] after some change, ∆
Module #15:
Introduction to Equilibrium
Calculating Equilibrium
Concentrations; 4 examples
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
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
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
What is Q?
[C ] [ D ]
[ A ] [B ]
c
Q=
i
i
b
i
i=initial
KC =
2
[C ] [ D ]
[ A ] [B ]
c
d
i
a
Don’t KNOW
initial conc.
rx?
eq
d
a
eq
2
eq
b
Q<,=,>K?
Which way
will reaction go?
eq
aA + bB →
← cC + dD
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
7
Rules for Equilibrium Calculations
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, ∆
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!!!!!
2.
d
eq
KC =
CO2 , g + H 2 , g →
← COg + H 2 Og
b
eq
1
KC
1
2
g
g
1
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
a
eq
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?
= 0.64
2,g
Calculate Original or Initial Concentrations
[CO2]init = 0.100 mol/L
[H2]init = 0.100 mol/L
Red herrings?
1.
[C ] [ D ]
[ A ] [B ]
[CO ] [ H O ]
=
[CO ] [ H ]
c
For the system (all are gases) Kc is 0.64 at 900 K. :
CO2 , g + H 2 , g →
← COg + H 2 Og
Write Kc
EXAMPLE 2: A BIT SIMPLE
Balance Equation
already done
4.
Calculate Q and determine if rx goes l or r.
Q=
[ CO] init [ H2 O]init
[
CO2
] [ ]
init
H2
init
=
0
[ 0100
. ][ 0100
. ]
Q = 0 < K c = 0.64; rx ⎯⎯⎯
⎯→
right
8
x
⎛
⎞
K = 0.64 = ⎜
⎟
⎝ 0100
.
− x⎠
5.
Express “equil. C” in terms of init C+ or - x
right
Q = 0 < Kc = 0.64; rx ⎯⎯⎯
⎯→
CO2 , g + H 2 , g →
← COg + H 2 Og
[CO ] = [CO ] − x
[H ] = [H ] − x
2 eq
2 eq
Mass balance Limiting reaction
[ CO] eq = [ CO] init + x
2 init
[ H O ] = [ H O]
2
2 init
eq
2
init
+x
0.8 =
OR: Construct an ICE chart
Reaction:
stoichiometry
Initial conc
Change (to right)
Equil. conc
6.
CO2 , g + H 2 , g
→
←
1
0
+x
0.100-x 0.100-x 0+x
0+x
K=
[CO ] [ H ]
2 eq
7.
2 eq
( x )( x )
x
⎛
=
=⎜
⎝
[ 0100
][
]
− x 0100
− x
−
.
.
0100
.
⎞
⎟
x⎠
0.0800 = x + 0.8 x
0.0800 = 18
. x
stoichiometry
initial conc
change (to right)
Equil. conc
H2
1
0.100
-x
0.100-x
CO
1
0
+x
0+x
x = 0.044
0.0800
= x = 0.0444
18
.
2
x = 0.044
Calculate equilibrium molarities
CO2
1
0.100
-x
0.100-x
x
0100
. −x
0.0800 − 0.8 x = x
Write K and put in equil Conc.
[ CO] eq [ H2 O]eq
Kc is 0.64 at 900 K. Suppose we
start with CO2 and H2, both at a
concentration of 0.100 mol/L.
0.8( 0100
.
− x) = x
COg + H 2 Og
1
1
1
0.100 0.100 0
-x
-x
+x
CO2 , g + H 2 , g →
← COg + H 2 Og
2
Reminder – x is change
In concentration
8**** Check your answer!!!!!
Suggestions for ways?
H2O
1
0
+x
0+x
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.
[CO2]eq = [H2]eq = 0.100-x = 0.05556 M
[CO]eq = [H2O]eq = 0.0444 M
9
K = 0.36 M =
Red herrings?
[N O ]
2
2 o
N 2 O4 →
← 2 NO2 ( g )
KC =
eq
2
d
2 , g ,eq
eq
a
2
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
2
2
4
=
01
. mol
= 01
. M
1L
=
0.0mol
= 0.00 M
1L
Calculate Q and determine if rx goes l or r
NO ]
[ NO
Q
Q ==
[ NN OO ]
22
b
eq
[ NO ]
[N O ]
22
22 init
init
[C ] [ D ] = [ NO ] = 0.36 M
[ A ] [B ] [ N O ]
eq
3.
4.
Write Kc
c
4 o
[ NO ]
Balance Equation
2.
Old Friend: N2O4
2
If we add 0.1 mol of N2O4 in 1 L, what are the equilibrium
concentrations? Recall that Kc = 0.36M.
1.
K = 0.36 M
N 2 O4 →
← 2 NO2 ( g )
EXAMPLE PROBLEM 3: More difficult
=
( 0.0 M ) 2
44 init
init
01
. M
= 0.0 M < K c ; rx ⎯⎯⎯
⎯→
right
4 , g ,eq
Calculate Original or Initial Concentrations
0.1 mol of N2O4 in 1 L
[N O ]
2
4 o
[ NO ]
2 o
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
01
. mol
= 01
. M
1L
=
0.0mol
= 0.00 M
1L
Old Friend: N2O4
5.
Express “equil. C” in terms of
orig C+ or - x
N 2 O4 →
← 2 NO2 ( g )
1
2 stoichiometry
N2O4
NO2
0.1
-x
0.1-x
0
+ 2x
0+2x
Initial conc
Change (to right)
Equil. conc
6.
=
[ NO ]
K = 0.36 M =
[N O ]
2
OJO!!
K=
Write K and put in equil Conc.
NO ]] ++22xx) )
[[NO
([([NO
NO ]]
KK==00.36
.36MM==
==
[ [NN OO] ] ([([NN OO ]] −−xx))
22
2 2 eq
2 2 4 4 eq
22
2 2 init
init
2 2 4 4 init
init
4 eq
([ NO ] + 2 x)
=
([ N O ] − x)
( 0 + 2 x) 2
( 01
. − x)
2
2 init
2
=
4 init
( 0 + 2 x) 2
( 01
. − x)
ax 2 + bx + c = 0
2
K ( 01
. − x) = ( 0 + 2 x)
K = 0.36 M
N 2 O4 →
← 2 NO2 ( g )
2
2 eq
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
2
K 01
. − xK = ( 2 x ) = 4 x 2
( 0 + 2 x) 2
=
( 01
. − x)
4 x 2 + xK − 01
.K=0
and solve for x
10
x=
−b ±
b 2 − 4ac
2a
Kc = 0.36 M
4 x 2 + Kc x − 01
. Kc = 0
ax 2 + bx
+c= 0
a = 4 b = Kc
x=
x=
−b ±
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
c = −01
. Kc
b 2 − 4ac
2a
− 0.36 ±
x=
( 0.36) 2 − 4( 4)( − 0.036)
2( 4)
OJO!!!
N 2 O4 →
← 2 NO2 ( g )
Calculate equilibrium molarities
x = 0.060
Initial conc
Change (to right)
Equil. conc
01
. − 0.06 = 0.4
NO2
0
+ 2x
0+2x
Plausible?
No, not plausible. We would be forced to conclude:
x = -0.15
2x = [NO2] = -0.3 M
Which is plausible?
N2O4
0.1
-x
0.1-x
N2O4
0.1
-x
0.1-x
Initial conc
Change (to right)
Equil. conc
2
− 0.36 ± ( 0.36) + 0.5764( 4)( 0.036)
x=
8
− 0.36 ± 0.7056 − 0.36 ± 0.84
x=
=
= 0.06 and − 015
.
8
8
7.
− 0.36 ± 0.7056 − 0.36 ± 0.84
=
= 0.06 and − 015
.
8
8
NO
0
+ 2x
0+2x
0 + 2( 0.06) = 012
.
1.
. → .04 ?
Plausible: [N2O4] decreased? 01
Plausible: [NO] increased? 0 → 012
. ?
2.
Fits the equilibrium constant?
[ NO ] = [ 012
. ]
[
0
N
O
[ ] .04]
2
K = 0.36 M =
2
2
Equilibrium [N2O4] = 0.04 M
Equilibrium [NO2] = 0.12 M
If we add 0.1 mol of N2O4 in 1 L,
what are the equilibrium
concentrations? Recall that Kc = 0.36
M.
8**** Check your answer!!!!!
11
4
2
= 0.36 M
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.
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.
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
5L volume
Balance Equation
H2 , g + I 2, g →
← 2 HI ( g )
2.
4.
Write K
aA + bB →
← cC + dD
K Pr essue =
( PC ) ( PD )
a
b
(P ) (P )
A
KP =
HI ,eq
H2 .eq
(P )
(P ) (P )
H2 .init
= 1x10 − 2
2
HI ,init
I 2 ,init
=
( 0.5atm) 2
left
= 5000 > K p = 0.01; rx ←⎯⎯⎯
( 01
. atm) ( 0.005atm)
I 2 ,eq
B
OJO
5.
(P )
(P ) (P )
Q=
2
d
c
Calculate Q and determine if rx goes l or r
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.
Kp =
H2
1
.01
+x
.01+x
I2
1
.005
+x
.005+x
K p = 0.01
(0.01 + x )(0.005 + x )
K p (0.01 + x )(0.005 + x ) = ( 0.5 − 2 x )
Express “equil. C” in terms of orig C+ or - x
stoichiometry
Initial conc
Change (to left)
Equil. conc
( 0.5 − 2 x ) 2
K p (5x10 − 5 + 0.015x + x 2 ) = 0.25 − 2 x + 4 x 2
HI
2
0.5
-2x
0.5-2x
5x10 − 5 K p + 0.015xK p + x 2 K p = 0.25 − 2 x + 4 x 2
(
(4 x 2 − 2 x + 0.25) − 5x10 −5 K p − 0.015xK p − x 2 K p =
(4 x − x K ) + (− 2 x − 0.015xK ) + (0.25 − 5x10 K ) = 0
(4 − K ) x − x(2 + 0.015K ) + (0.25 − 5x10 K ) = 0
2
−5
2
Write K and put in equil Pressures
( 0.5 − 2 x )
p
p
−5
2
p
Kp =
)
0.25 − 2 x + 4 x 2 − 5x10 − 5 K p + 0.015xK p + x 2 K p = 0
p
6.
2
p
p
( 4− .01) x 2 − x( 2 + 0.00015) + (0.25 − 5x10 −7 ) = 0
2
(0.01 + x )(0.005 + x )
( 3.99) x 2 − x( 2.00015) + ( 0.2499995) = 0
12
x=
− b ± b 2 − 4ac
2a
( 3.99) x 2 − x( 2.00015) + ( 0.2499995) = 0
Which do we use + or -?
x = 0.237739
2
− ( − 2.00015) ± ( − 2.00015) − 4( 3.99)( 0.2499995)
x=
2( 3.99)
x=
x=
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
2.00015 ± 4.0006 − 3.98992
7.98
2.00015 ± 0102995
.
7.98
PHI ,eq = 0.5 − 2 x = 0.5 − 2(0.263552) = − 0.0271
x = 0.237739
Not plausible
x = 0.263552
x = 0.237739
stoichiometry
Initial conc
Change (to left)
Equil. conc
H2
1
.01
+x
.01+x
I2
1
.005
+x
.005+x
“A” students work
(without solutions manual)
~ 10 problems/night.
HI
2
0.5
-2x
0.5-2x
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
PHI ,eq = 0.5 − 2 x = 0.5 − 2(0.237739) = 0.024523
PI 2 ,eq = 0.005 + x = 0.005 + 0.237739 = 0.242739
Office Hours Th&F 2-3:30 pm
PH2 ,eq = 0.01 + x = 0.01 + 0.237739 = 0.247739
( P ) = ( 0.024523)
= 0.01
( P )( P ) ( 0.242739)( 0.247739)
2
2
K=
HI ,eq
H2 ,eq
Module #15:
Introduction to Equilibrium
Checks!!!
Making Assumptions
I 2 ,eq
13
Example on Using Simplifications
General
FITCH Rules
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?
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
Clues?
K is “small” compared to others (<<< 1) we have worked with !!!!!
1 2
Chemistry
el
1
2
1 2
2
EXAMPLE 3:
Example 4:
For systems with multiple reactions
Need to simply = ASSUMPTIONS
N 2 O4 →
← 2 NO2 ( g )
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?
3.
35 oC is a red herring
Clues?
K is “small” compared to others (<<< 1) we have worked with !!!!!
4.
Kc is 0.64.
Kc = 0.36M
H2 , g + I 2, g →
← 2 HI ( g )
2 NOCl →
← 2 NO + Cl2
Example on Using Simplifications
Red herrings:
CO2 , g + H 2 , g →
← COg + H 2 Og
Example 2
el
1
35 oC is a red herring
Red herrings:
Kp is 1x10-2.
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
We are starting with reactants
2 , g ,init
2
g .init
1.
Balance Equation
2 NOCl →
← 2 NO + Cl2
2. Write K
Kc =
[ NO] 2 ⎡⎣⎢ Cl2 ⎤⎦⎥
[ NOCl ] 2
14
We will define
Small in the
Next chapter!
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
≈
K = 16 x 3
16
. x10 − 5 = 16 x 3
16
. x10 − 5
= x3
16
10 − 6 = x 3
( 2 x) 2 x
[ NOCl ]
3
2
init
x = 10 − 2
( 2 x) 2 x 4 x 3
K=
=
= 16 x 3
( 0.5) 2 0.25
7.
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
10 − 6 = x
NO ] == 22xx = 2 x10
[ NO
[[ClCl ] == xx = 10
Calculate equilibrium molarities
[ NOCl ] = 0.5 − 2 x CHECK ASSUMPTIONS
0.5
Where are
[ NOCl ] = 0.5 − 2(10 )
− 0.02
The Sig Fig?
[ NOCl ] = 0.5 − 0.02
0.48
g ,eq
−2
g ,eq
g ,eq
Answer
Accept error less
than 5%
−2
0.5
⎡ real − estimated ⎤
error = ⎢
⎥⎦ 100
real
⎣
gg,,eq
eq
⎡ 0.48− .5 ⎤
error = ⎢
100 = 4%
⎣ 0..48 ⎥⎦
−2
22,
, gg,eq
,eq
Are we done yet?
ARE WE DONE?
15
8**** Check your answer!!!!!
[ NOCl ] = 0.5
[ NO ] = 2 x = 2 x10
[Cl ] = x = 10
POINT of ASSUMPTIONS
g ,eq
−2
g ,eq
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
−2
2 , g ,eq
[ NO] 2 ⎡⎣⎢Cl2 ⎤⎦⎥
Kc =
= 16
. x10 − 5 M
[ NOCl ] 2
Kc =
⎡
⎣⎢
2 x10
−2 ⎤ 2 ⎡
⎦⎥ ⎣⎢
[0.5]
2
10 − 2 ⎤⎦⎥ ?
= 16
. x10 − 5 M
Multiple Equilibria are not solvable
With simple cubic, quadratic equations
Kc = 16
. x10 − 5 = 16
. x10 − 5 M
Chemists are Lazy!
Chemists are ________
“A” students work
(without solutions manual)
~ 10 problems/night.
If we don’t have to calculate can we make
qualitative descriptions of equilibrium?
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
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
Office Hours Th&F 2-3:30 pm
Module #15:
Introduction to Equilibrium
Qualitative Predictions
For direction of Equilibrium
16
Effects of Changes on the System
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Luigi Galvani
1737-1798
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Jabir ibn
Hawan, 721-815
Richard AC E
Erlenmeyer
1825-1909
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
An alchemist
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Rudolph Clausius
1822-1888
Niels Bohr
1885-1962
John Dalton
1766-1844
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
Jacobus van’t Hoff
1852-1911
Lawrence Henderson
1878-1942
Amedeo Avogadro
1756-1856
Jean Picard
1620-1682
William Henry
1775-1836
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
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
Michael Faraday
1791-1867
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Isaac Newton
1643-1727
Wolfgang Pauli
1900-1958
B. P. Emile
Clapeyron
1799-1864
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
Anders Celsius
1701-1744
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).
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
Heisenberg 1901-1994
1905-1965
1901-1976
Fitch Rule G3: Science is Referential
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?
LeChatlier Examples
Example: (Inert gas) Suppose we have the reaction initially at equilibrium
2 SO2 , g + O2 , g →
← 2 SO3, g
Kc ,1000 K = 2.8 x10 2
[CO ] < [CO ] [CO ] = [COx ]
2 , g ,eq
What happens when we add some N2,g, , does Q change?
[SO ]
[O ][SO ]
2 , g ,new
2
KC =
3, g ,eq
2 , g ,eq
2
2 , g ,eq
Addition of inert gas does not affect the
equilibrium position. Similarly……
s
2,g
3, s
2 , g ,eq
right
c
2 , g ,eq
What happens when we add some SO3,g , does Q change?
[SO ] > [SO ]
[SO ] = x[SO ]
3,new
of limestone produced by adding a small quantity of CaCO3(s)?
2, g
[CO ] < K = [CO ]; rx ⎯⎯⎯⎯→
x
Kc ,1000 K = 2.8 x10 2
2 SO2 , g + O2 , g →
← 2 SO3, g
CaCO3,s →← CaOs + CO2, g
[CO ][CaO ] = [CO ]
[CaCO ]
Q=
2 , g ,new
Example Suppose we have the reaction initially at equilibrium:
Example: (Effect of Solid)
What is the effect on equilibrium in the calcination (decomposition)
Kc =
2 , g ,eq
3,new
Change in quantity of solids do not affect
direction of equilibria. So addition of more
limestone is irrelevant.
3,eq
3,eq
Q=
( x[SO ])
2
[O ][SO ]
2 , g ,eq
2 , g ,eq
[SO ]
=
[O ][SO ]
2
3, g ,eq
2
> KC
3, g ,eq
2 , g ,eq
; rx ←⎯⎯⎯
left
2
2 , g ,eq
Concentration: The system will shift
away from the added component.
17
Example What happens if we decrease the volume of the container?
2 SO2 , g + O2 , g →
← 2 SO3, g
⎡ 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 =
SO
⎡ ⎢nO2 , g ,eq ⎤ ⎥⎡ ⎢nSO2 , g ,eq ⎤ ⎥
O2 ,2eq,eq, g, g SO2 ,2g, g,eq,eq
nO2 , g ,eq nSO2 , g ,eq
⎢ ⎢⎣ Veq ⎥ ⎥⎦⎢ ⎢⎣ Veq ⎥ ⎥⎦
⎢⎣ Veq ⎥⎦ ⎢⎣ Veq ⎥⎦
[[
[[
]]
][][
]]
[
2
[
][
]
[
[
[
][
][
]
]
]
]
[
[
[
][
[
]
][
]
]
Kc = 2801000 K
2
]
2
⎡ 1 ⎤
⎢ ⎥
⎢⎣ Veq ⎥⎦
2
⎡ 1 ⎤⎡ 1 ⎤
⎢
⎥⎢
⎥
⎢⎣ Veq ⎥⎦ ⎢⎣ Veq ⎥⎦
Decrease volume, what happens?
⎡ 1 ⎤
2
2
2
⎛
⎢
⎥
n SO3, 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
n
⎜
⎟ Veq
⎜
⎟
SO3 , g , eq
right
Q= ⎜
K
<
=
⎯→
⎟
⎜
2
2 ⎟ Veq ; rx ⎯⎯⎯
c
x
⎜ nO
⎟
⎜
⎟
n
n
n
⎝ 2 , g ,eq SO2 , g ,eq ⎠
⎝ O2 , g ,eq SO2 , g ,eq ⎠
[
2SO2( g ) + O2( g ) →← 2SO3( g )
2
[
[
][
]
]
Number of moles of reactants?
Number of moles of products?
⎞
⎟
2 ⎟ Veq
⎟
⎠
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.
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
13_318
LeChatlier Example: Haber Process
N 2 ( g ) + 3H 2 ( g )
What happens when volume is decreased?
→
←
Effect of Temperature????
2 NH 3( g )
Consider effect of raising temperature on two reactions:
Key:
N2
H2
NH3
2SO2( g ) + O2( g ) →← 2 SO3( g )
∆ H o = − 180kJ
N 2 ( g ) + O2 ( g ) →← 2 NO2 ( g )
∆ H o = + 181kJ
What will happen?
To equil
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 ) + O2 ( g ) →← 2SO3( g ) + heat ; rx ←⎯⎯⎯
left
7 N2
4 H2
3NH3
14
7 N2
4 H2
3NH3
14
To equil
6 N2
1H2
5NH3
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).
12
18
Do you see a pattern?
Galen, 170
⎡ K ⎤ − ∆ H forward rx ⎡ 1 1 ⎤ Van’t Hoff equation (1852-1911)
ln ⎢ 1 ⎥ =
⎢ − ⎥
R
⎣ K2 ⎦
⎣ T1 T2 ⎦
Marie the Jewess, 300
Jabir ibn
Hawan, 721-815
An alchemist
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
Jean Picard
1620-1682
Daniel Fahrenheit
1686-1737
Blaise Pascal
1623-1662
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Anders Celsius
1701-1744
o
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Luigi Galvani
1737-1798
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
⎡ k ⎤ ⎡ − Ea ⎤ ⎡ 1 1 ⎤
ln ⎢ 1 ⎥ = ⎢
⎥⎢ − ⎥
⎣ k 2 ⎦ ⎣ R ⎦ ⎣ T1 T2 ⎦
Arrhenius Equation
⎡ P ⎤ ⎡ − ∆ H vaporization ⎤ ⎡ 1 1 ⎤
ln ⎢ 1 ⎥ = ⎢
⎥⎢ − ⎥
R
⎣ P2 ⎦ ⎣
⎦ ⎣ T1 T2 ⎦
Clausius-Clapeyron
equation
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Richard AC E
Erlenmeyer
1825-1909
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
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)
Jacques Charles
1778-1850
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Max Planck
1858-1947
Rolf Sievert,
1896-1966
Georg Simon Ohm
1789-1854
James Maxwell
1831-1879
Michael Faraday
1791-1867
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
B. P. Emile
Clapeyron
1799-1864
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
∆ H 0 = − 92.2 kJ
K = 6 x105
25C
What have we learned?
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.
1.
2.
3.
N 2 ( g ) + 3H 2 ( g ) →
← 2 NH 3( g ) + heat
⎡ K1 ⎤ − ∆ H o ⎡ 1
1⎤
ln ⎢
⎥ =
⎢ −
⎥
R ⎣ T1 T2 ⎦
⎣ K2 ⎦
⎡ 6 x105 ⎤
ln ⎢
⎥ =
⎣ K2 ⎦
e
⎡ 6 x105 ⎤
ln ⎢
⎥
⎢⎣ K2 ⎥⎦
4.
5.
6.
⎡ 6 x105 ⎤
7 .48628
= 1783.405
= ⎢
⎥= e
⎣ K2 ⎦
J ⎞
⎛
− ⎜ − 92.2 x10 3
⎟
⎝
1
1
mol ⎠ ⎡
⎤
⎢⎣ 273 + 25K − 273 + 100 K ⎥⎦
J
8.31
mol − K
⎡ 6 x105 ⎤
11095.07
ln ⎢
⎥ =
1 ⎞
K
⎛
2
⎣
⎦ ⎜
⎟
⎝ Kelvin ⎠
⎡ .000675 ⎤
⎢⎣ Kelvin ⎥⎦ = 7.48628
7.
8.
6 x105
= K2 = 336.435
1783.405
19
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
Thomas M Lowry
1874-1936
Werner Karl Linus Pauling Louis Harold Gray
Heisenberg 1901-1994
1905-1965
1901-1976
Fitch Rule G3: Science is Referential
N 2 ( g ) + 3H 2 ( g ) →
← 2 NH 3( g )
Germain Henri Hess
1802-1850
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.
20