1
2
Chemical Equilibrium
• The state where the concentrations of all
reactants and products remain constant
with time.
CHEMICAL
EQUILIBRIUM
Chapter 13
Properties of an Equilibrium
• On the molecular level, there is frantic
activity. Equilibrium is not static, but is a
highly dynamic situation.
3
4
Chemical Equilibrium
Equilibrium systems are
• DYNAMIC (in constant
motion)
• REVERSIBLE
• can be approached from
either direction
Fe3+ + SCN- q e FeSCN2+
+
Fe(H2O)63+ + SCN- q e Fe(SCN)(H2O)52+ + H2O
Pink to blue
Co(H2O)6Cl2 → Co(H2O)4Cl2 + 2 H2O
Co(H2O)6Cl2 q e Co(H2O)4Cl2 + 2 H2O
Blue to pink
Co(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2
Chemical Equilibrium
qe
5
6
Examples of
Chemical Equilibria
Fe3+ + SCN- q e FeSCN2+
Phase changes such as
H2O(s) q e H2O(liq)
• After a period of time, the concentrations of
reactants and products are constant.
• The forward and reverse reactions continue
after equilibrium is attained.
Page 1
7
8
Chemical
Equilibria
Examples of
Chemical
Equilibria
CaCO3(s) + H2O(liq) + CO2(g) q e Ca2+(aq) + 2 HCO3-(aq)
• At a given T and P of CO2, [Ca2+] and [HCO3-] can
• Formation of
stalactites and stalagmites
be found from the EQUILIBRIUM
CONSTANT.
CaCO3(s) + H2O(liq) + CO2(g) q e Ca2+(aq) + 2 HCO3-(aq)
Example
9
10
• Remember we are looking at a net change.
• Reactions at equilibrium still react. It is just that the
forward and reverse reactions are equal.
H2O(g) + CO(g) q e H2(g) + CO2(g)
• Figure 13.3: (a) H2O and CO are mixed in equal
numbers and begin to react (b) to form CO2 and
H2. After time has passed, equilibrium is
reached (c) and the numbers of reactant and
product molecules then remain constant over
time (d).
• Figure 13.2: The changes in concentrations with time for
the reaction H2O(g) + CO(g) H2(g) + CO2(g) when equimolar
quantities of H2O(g) and CO(g)
are mixed.
Characteristics of Chemical
Equilibrium
11
• Figure 13.4: The changes with time in the rates of
forward and reverse reactions for H2O(g) + CO(g)
q e H2(g) + CO2(g) when equimolar quantities of
H2O(g) and CO(g) are mixed. The rates do not change
in the same way with time because the forward
reaction has a much larger rate constant than the
reverse reaction.
12
• The second reason applies.
N2(g) + 3H2(g) q e 2NH3(g)
• Remember
• When we mix N2, H2, and
– N2 triple bond = 941 kJ/mol
NH3 in a closed container at
– H2 single bond = 432 kJ/mol
25°C, we do no see a
change in the
• How can we speed up the
concentrations over time
reaction?
regardless of the original
– Catalyst
amounts of gas.
• Two possibilities exist
– The system is at chemical
equilibrium
– The forward and reverse
reactions are so slow that the
system moves towards
equilibrium at a rate that cannot
be detected.
Page 2
The Equilibrium Constant
13
The Law of Mass Action
• NOT TO BE CONFUED WITH THE RATE CONSTANT!!!!!
• Equilibrium Constant: The value obtained when
equilibrium concentrations of the chemical species are
substituted in the equilibrium expression.
• Equilibrium Expression: The expression (from the law
of mass action) obtained by multiplying the product
concentrations and dividing by the multiplied reactant
concentrations, with each concentration raised to a
power represented by the coefficient in the balanced
equation.
• Law of Mass Action: A general description of the
equilibrium condition; it defines the equilibrium
constant expression.
• WHAT THE *%#@!
• Two Norwegian chemist
– Cato Maximilian Guldberg (1836-1902)
– Peter Waage (1833-1900)
• Proposed in 1864 a general description of an
equilibrium condition.
– Based upon observations of many chemical reactions.
• For some reaction
jA + kB ↔ lC + mD
• The law of mass action is represented by the
equilibrium expression & K is the constant:
l
m
K=
Equilibrium Expression Example
15
•
•
•
•
•
6
NO2 H2O
K=
4
7
NH3 O2
Example
C D
A j Bk
Example
4NH3(g) + 7O2(g) q e 4NO2(g) + 6H2O(g)
4
14
K=
[NH3] = 3.1×10-2 M
[N2] = 8.5×10-2 M
[H2] = 3.1×10-3 M
N2(g) + 3H2(g) q e 2NH3(g)
What is the value of K?
[NH3 ]2
(3.1× 10 −2 )2
=
= 3.8 × 10 4
[N2 ][H2 ]3 (8.5 × 10 −1 )(3.1× 10 −3 )3
17
Example
•
•
•
•
•
•
•
•
•
•
•
[NH3] = 3.1×10-2 M
[N2] = 8.5×10-2 M
[H2] = 3.1×10-3 M
2NH3 (g)q e N2(g) + 3H2(g)
Just the reverse of the previous
example
• What is the value of K’?
[NH3] = 3.1×10-2 M
[N2] = 8.5×10-2 M
[H2] = 3.1×10-3 M
½ N2(g) + 3/2 H2(g) q e NH3(g)
Just ½ of the original example
What is the value of K’’?
K' ' = (K)1/2 = 1.9 × 102
[N ][H2 ]3 1
K' = 2
= = 2.6 × 10 −5
[NH3 ]2
K
Page 3
16
18
Equilibrium with Pressure
19
Notes on Equilibrium Expressions (EE)
• We have been talking about
equilibrium of gases in terms of
concentration.
• Why not pressure?
• You should know
The
Equilibrium Expression for a
reaction is the reciprocal of that
for the reaction written in reverse.
When the equation for a reaction is
multiplied by n, EEnew = (EEoriginal)n
Knew
PV = nRT & M =
= (Korginal)n
The
units for K depend on the
reaction being considered.
Customarily
P=
written without units.
Equilibrium with Pressure
P
RT
21
22
But
• There must be some sort of relationship
between K and KP.
• So we have the following reaction.
jA + kB q e lC + mD
2
∴
KP =
mols
Vol(L)
n
RT
V
P = CRT & C =
CNH3
[NH3 ]2
K=
=
= Kc
3
3
[N2 ][H2 ]
(CN2 )(CH2 )
PNH3
20
We KNOW
2
l
K=
3
(PN2 )(PH2 )
m
(P )(P )
[C]l [D]m
& KP = C j D k
j
k
[A] [B]
(PA )(PB )
P = CRT
23
24
Example
KP = K(RT)∆n
l
KP =
=
KP = 1.9×103 @ 25 °C
2NO(g) + Cl2(g) q e 2NOCl(g)
Calculate K from KP
KP = K(RT)∆n
1.9×103 = K(RT)∆n
∆n = 2 – (2+1) = -1
1.9×103 = K(RT)-1
1.9×103(RT) = K
1.9×103(0.082057)(298) = K = 4.6×104
m
(PC )(PD ) (CC × RT)l (CD × RT)m
=
j
k
j
k
(PA )(PB ) (C A × RT) (CB × RT)
(CC )l (CD )m (RT)l+m
×
= K(RT)(l+m)−(j+k)
(C A ) j (CB )k (RT) j+k
= K(RT) ∆n
Page 4
25
26
Writing and Manipulating K
Expressions
Heterogeneous Equilibria
• Heterogeneous Equilibrium: an equilibrium involving
reactants and/or products in more than one phase.
• Example: CaCO3(s) q e CaO(s) + CO2(g)
• You would think that the equilibrium constant would
be
K' =
Solids and liquids NEVER
appear in equilibrium
expressions.
[CO2 ][CaO]
[CaCO3 ]
S(s) + O2(g) q e SO2(g)
• But heterogenous equilibrium does not depend on the
amounts of pure solid or liquids present.
• Therefore
• K=[CO2]
K=
The Reaction Quotient, Q
27
Writing and Manipulating K
Expressions
are characterized by their REACTION
QUOTIENT, Q.
NH3(aq) + H2O(liq) q e
NH4+(aq) + OH-(aq)
• If Q = K, then system is at equilibrium.
• If Q > K, then conc. of products too large shifts
towards reactants.
• If Q < K, then conc. of reactants too large shifts
towards products.
[NH 4+ ][OH- ]
[NH3 ]
The Meaning of K
29
The Meaning of K
[NH3 ]2
[N2 ][H2 ]3
30
For AgCl(s) q e
Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 1.8 x 10-5
1.
Can tell if a reaction is productfavored or reactant-favored.
For N2(g) + 3 H2(g) q e 2 NH3(g)
Kc =
28
In general, all reacting chemical systems
Solids and liquids NEVER
appear in equilibrium
expressions.
K=
[SO2 ]
[O2 ]
= 3.5 x 10 8
Conc. of products is much
less than that of
reactants at equilibrium.
The reaction is strongly
Ag+(aq) + Cl-(aq)
q e AgCl(s)
reactant-favored.
Conc. of products is much greater
than that of reactants at equilibrium.
The reaction is strongly product-
is product-favored.
favored.
Page 5
31
The Meaning of K
Product- or Reactant Favored
32
K comes from thermodynamics.
See Chapter 19, page 812-813
∆G˚ < 0: reaction is product favored
∆G˚ > 0: reaction is reactant-favored
∆Go = -RT ln K
Product-favored
If K > 1, then ∆G˚ is negative
If K < 1, then ∆G˚ is positive
Reactant-favored
The Meaning of K
33
The Meaning of K
2. Can tell if a reaction is at equilibrium.
If not, which way it moves to approach
equilibrium.
n-butane
H H H H
H—C—C—C—C—H
H H H H
K =
[iso]
[n]
n-butane
H H H H
H—C—C—C—C—H
H H H H
iso-butane
H H H
H—C—C—C—H
H
H
H C H
H
K =
35
=
36
• The inherent tendency for a reaction to occur it
indicated by the magnitude of the equilibrium
constant.
• A value of K much larger than 1 mean?
– At equilibrium the reaction system will consist of mostly
products.
– Lies to the right.
If Q = K, then system is at equilibrium.
conc. of iso
conc. of n
= 2.5
The Extent of a Reaction
All reacting chemical systems are
characterized by their REACTION
QUOTIENT, Q.
product concentrations
Q =
reactant concentrations
Q =
[n]
iso-butane
H H H
H—C—C—C—H
H
H
H C H
H
If [iso] = 0.35 M and [n] = 0.15 M, are you
at equilibrium?
If not, which way does the reaction “shift”
to approach equilibrium?
= 2.5
The Meaning of K
[iso]
34
• A very small value of K
0.35
= 2.3
0.15
– At equilibrium the reaction system will consist of mostly
reactants.
– Lies to the left.
• Remember the size of K and the time required
to reach equilibrium are NOT directed related.
• The time is dependent on the Activation
Energy.
Q (2.33) < K (2.5)
Reaction is NOT at equilibrium, so [iso] must
become ________ and [n] must ____________.
Page 6
MANY MANY EXAMPLES
N2O4(g)q e 2NO2(g) 38
• N2O4 used as one of the
fuels on the lunar lander.
• Closed container of
N2O4(g) allowed to reach
equilibrium.
• KP = 0.133
• PN2O4 at eq = 2.71 atm
2
• PNO2 = ?
PNO2
37
• For the reaction
N2(g) + 3H2(g)q e 2NH3(g)
• K=6.0×10-2 @ 500 °C
• Predict the direction in which the system will shift
to reach equilibrium if
• [NH3] = 1.0×10-5 M
• [N2] = 1.0×10-5 M
• [H2] = 2.0×10-3 M
Q=
KP =
Apollo II lunar
landing module at
Tranquility Base,
1969.
[NH3 ]2
(1.0 × 10 −3 )2
=
= 1.3 × 107
3
[N2 ][H2 ]
(1.0 × 10 −3 )(2.0 × 10 −3 )3
Therefore, since Q>K the direction goes towards the reactants
PN2O4
= 0.133
∴
0.133 =
X2
2.71atm
X = 0.360 = 0.600
39
40
Determining K
2 NOCl(g) q e 2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At
equilibrium you find 0.66 mol/L of NO.
Calculate K.
Solution
Set of an “ICE” table of concentrations
[NOCl]
[NO]
[Cl2]
Initial
2.00
0
0
Change
Equilibrium
0.66
Determining K
41
42
Determining K
2 NOCl(g) q e 2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At
equilibrium you find 0.66 mol/L of NO.
Calculate K.
Solution
Set of a table of concentrations
[NOCl]
[NO]
[Cl2]
Initial
2.00
0
0
-0.66
+0.66
+0.33
Change
Equilibrium
1.34
0.66
0.33
2 NOCl(g) q e 2 NO(g) + Cl2(g)
[NOCl]
[NO]
[Cl2]
Initial
2.00
0
0
Change
-0.66
+0.66
+0.33
Equilibrium
1.34
0.66
0.33
K=
K=
Page 7
[NO]2 [Cl2 ]
[NOCl]2
[NO]2 [Cl2 ]
[NOCl]2
=
(0.66) 2 (0.33)
(1.34)2
= 0.080
43
H2(g) + I2(g) q e 2 HI(g)
Kc = 55.3
Typical Calculations
PROBLEM: Place 1.00 mol each of H2 and I2 in
a 1.00 L flask. Calc. equilibrium
concentrations.
44
Step 1. Set up ICE table to define
EQUILIBRIUM concentrations.
H2(g) + I2(g) q e 2 HI(g)
Initial
[H2]
[I2]
[HI]
1.00
1.00
0
Change
Equilib
[HI]2
Kc =
= 55.3
[H2 ][I2 ]
45
H2(g) + I2(g) q e 2 HI(g)
Kc = 55.3
H2(g) + I2(g) q e 2 HI(g)
Kc = 55.3
Step 1. Set up ICE table to define
EQUILIBRIUM concentrations.
[H2]
[I2]
[HI]
Initial
1.00
1.00
0
Change
-x
-x
+2x
Equilib
1.00-x
1.00-x
2x
46
Step 2. Put equilibrium concentrations
into Kc expression.
Kc =
[2x]2
= 55.3
[1.00 - x][1.00 - x]
where x is defined as am’t of H2 and I2
consumed on approaching equilibrium.
48
47
H2(g) + I2(g) q e 2 HI(g)
Kc = 55.3
Step 3. Solve Kc expression - take
square root of both sides.
Kc =
[2x]2
= 55.3
[1.00 - x][1.00 - x]
7.44 =
Nitrogen Dioxide
Equilibrium
2x
1.00 - x
N2O4(g) q e 2 NO2(g)
x = 0.79
Therefore, at equilibrium
e
[H2] = [I2] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M
Page 8
49
50
Nitrogen Dioxide Equilibrium
N2O4(g) q e 2 NO2(g)
Kc =
Nitrogen Dioxide Equilibrium
N2O4(g) q e 2 NO2(g)
[NO2 ]2
= 0.0059 at 298 K
[N2O4 ]
Kc =
If initial concentration of N2O4 is 0.50 M, what are
the equilibrium concentrations?
Step 1. Set up an ICE table
[N2O4]
[NO2]
Initial
0.50
0
Change
Equilib
Nitrogen Dioxide Equilibrium
N2O4(g) q e 2 NO2(g)
If initial concentration of N2O4 is 0.50 M, what are
the equilibrium concentrations?
Step 1. Set up an ICE table
[N2O4]
[NO2]
Initial
0.50
0
Change
-x
+2x
Equilib
0.50 - x
2x
51
Nitrogen Dioxide Equilibrium
N2O4(g) q e 2 NO2(g)
Step 2. Substitute into Kc expression and solve.
K c = 0.0059 =
Rearrange:
a = 4
0.0059 (0.50 - x) = 4x2
0.0029 - 0.0059x = 4x2
4x2 + 0.0059x - 0.0029 = 0
-0.0059 ±
x =
x = -0.00074 ±
-b ±
-0.0059 ±
c = -0.0029
2
b - 4ac
2a
(0.0059)2 - 4(4)(-0.0029)
2(4)
x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027
53
54
Solving Quadratic
Equations
(0.0059)2 - 4(4)(-0.0029)
2(4)
1/8(0.046)1/2
b = 0.0059
x =
This is a QUADRATIC EQUATION
ax2 + bx + c = 0
a = 4
b = 0.0059
c = -0.0029
x =
52
Solve the quadratic equation for x.
ax2 + bx + c = 0
[NO2 ]2
(2x) 2
=
[N2O 4 ]
(0.50 - x)
Nitrogen Dioxide Equilibrium
N2O4(g) q e 2 NO2(g)
[NO2 ]2
= 0.0059 at 298 K
[N2O4 ]
• Recommend you
solve the equation
exactly on a
calculator or use
the “method of
successive
approximations”
= -0.00074 ± 0.027
x = 0.026 or -0.028
But a negative value is not reasonable.
Conclusion: x = 0.026 M
[N2O4] = 0.050 - x = 0.47 M
[NO2] = 2x = 0.052 M
Page 9
Solving Equilibrium Problems
55
H2(g) + F2(g) q e HF(g)
• K = 1.15×102 @ some temperature.
• 3.00 mols of each component is added to a
1.50 L flask.
• Calculate the equilibrium concentrations of
all species.
1. Balance the equation.
2. Write the equilibrium expression.
3. List the initial concentrations.
4. Calculate Q and determine the shift to
equilibrium.
5. Define equilibrium concentrations.
6. Substitute equilibrium concentrations
into equilibrium expression and solve.
7. Check calculated concentrations by
calculating K.
I
I
2HF(g)
2.00 M
2.00 M
2.00 M
+ F2(g) q e
2HF(g)
2.00 M
2.00 M
2.00 M
E
Keep Going…
57
+ F2(g) q e
H2(g)
C
But which direction does the reaction go?
H2(g)
56
Example
58
H2(g)
+ F2(g) q e
2HF(g)
I
2.00 M
2.00 M
2.00 M
C
-x
-x
+2x
E
2.00 - x
2.00 -x
2.00 + 2x
C
E
Q=
K = 1.15 × 102 =
[HF]2
(2.000) 2
=
= 1.000
[H2 ][F2 ] (2.000)(2.000)
[HF]
(2.00 + 2x)2
=
[H2 ][F2 ] (2.00 − x)(2.00 − x)
2
∴
∴
Q < K, so the reaction go towards the products
1.15 × 10 =
2
2.00 + 2x
2.00 − x
∴
x = 1.528
59
Finally
•
•
•
•
•
60
Example
•
•
•
•
x = 1.528
So,
[H2] = 2.00 – 1.528 = 0.472 M
[F2] = 2.00 – 1.528 = 0.472 M
[HF] = 2.00 + 2(1.528) = 5.056 M
Page 10
H2(g) + F2(g) q e HF(g)
K = 1.15×102 @ some temperature.
3.00 mols of H2 in a 3.000 L flask
6.00 mols F2 in a 3.000 L flask
Calculate the equilibrium concentrations of
all species.
H2(g)
+ F2(g) q e
2HF(g)
I
1.00 M
2.00 M
0.00 M
C
-x
-x
+2x
E
1.00 - x
2.00 - x
2x
K = 1.15 × 10 2 =
[HF]2
(2x)2
=
[H2 ][F2 ] (1.00 − x)(2.00 − x)
61
Another #$%& Example!
62
H2(g) + I2(g) q e 2 HI(g)
• KP = 1.00×102
• The following gasses are mixed in
a 5.000 L Flask
∴
1.11× 10 2 (x 2 ) − 3.45 × 10 2 (x) + 2.30 × 10 2 = 0
∴
– PHI = 5.000×10-1 atm
– PH2 = 1.000×10-2 atm
– PI2 = 5.000×10-3 atm
− b ± b 2 − 4ac
2a
∴
x = 2.14 M or 0.968 M
x=
• Check Q
Cannot be 2.14 M
Why? (Look at [H2 ])
x = 0.968
63
Q=
(PHI °) 2
(5.00 × 10 −1 atm) 2
=
(PH 2 °)(PI 2 °) (1.000 × 10 − 2 atm)(5.000 × 10 − 3 atm)
64
KP =
Q = 5.000 × 10 3
(PHI ) 2
(5.00 × 10 −1 - 2x)2
=
(PH 2 )(PI 2 ) (1.000 × 10 − 2 + x)(5.000 × 10 − 3 + x)
Q>K, therefore the system will shift toward the reactants
Reduces to
H2(g)
I2(g)
+
2HI(g)
qe
(9.60 × 10 )x + 3.5x − (2.45 × 10 −1 ) = 0
1
I 1.00×10-2
5.00 ×10-3
5.00×10-1
C +x
+x
-2x
E 1.00×10-2 + x
5.00 ×10-3 + x 5.00×10-1 - 2x
x = 3.55 × 10- 2
ONE MORE
65
H2(g)
+
I2(g)
2
2HI(g)
qe
I
1.00×10-2 M
5.00 ×10-3 M
5.00×10-1 M
C
+x
+x
-2x
E
1.00×10-2 + x
5.00 ×10-3 + x 5.00×10-1 - 2x
66
2NOCl(g) q e 2NO(g) + Cl2(g)
• In an experiment in which 1.0 mol NOCl
is placed in a 2.0 L flask, what are the
equilibrium concentrations?
K=
x = 3.55×10-2 atm
• Therefore at eq.
[NO]2 [Cl 2 ]
= 1.6 × 10 − 5
[NOCl ]2
2NOCl(g)q e
– PHI = 4.29×10-1 atm
– PH2 = 4.55×10-2 atm
– PI2 = 4.05×10-3 atm
Page 11
2NO(g)
+
Cl2(g)
I
0.50
0
0
C
-2x
+2x
+x
E
0.5 - 2x
2x
x
67
K=
2
68
2
[NO] [Cl 2 ]
( 2x ) ( x )
= 1.6 × 10 − 5 =
2
[ NOCl ]
(0.50 − 2x) 2
• The problem is you end up with an
x3 component, which are not fun
to solve for
• But, since K is so small the
reaction will not proceed that far
to the right.
• Which means that x is a relatively
small number.
• So 2x does nothing to 0.50
K=
K=
2NO(g)
+
0.50
0
0
C
-2x
+2x
+x
E
0.5 - 2x
2x
x
x=
[NO] [Cl 2 ]
( 2x ) ( x )
= 1.6 × 10 − 5 =
2
[ NOCl ]
(0.50 − 2x) 2
[NO]2 [Cl 2 ]
( 2x ) 2 ( x )
4x 3
−5
=
1
.
6
×
10
=
=
[NOCl]2
(0.50) 2
(0.50) 2
EQUILIBRIUM AND
EXTERNAL EFFECTS
Cl2(g)
I
2
Therefore, x = 1.0×10-2
69
2NOCl(g)q e
2
70
• Temperature, catalysts, and changes in
concentration affect equilibria.
• The outcome is governed by LE
CHATELIER’S PRINCIPLE
1.0×10-2
• “...if a system at equilibrium is
disturbed, the system tends to shift its
equilibrium position to counter the
effect of the disturbance.”
• Therefore at eq.
– [NOCl]I = 0.50 M
– [NO] = 2.0×10-2 M
– [Cl] = 1.0×10-2 M
71
72
EQUILIBRIUM AND EXTERNAL EFFECTS
EQUILIBRIUM AND
EXTERNAL EFFECTS
• Temperature change --->
change in K
• Consider the fizz in a soft drink
CO2(aq) + HEAT q e CO2(g) + H2O(liq)
Henri Le Chatelier
1850-1936
Studied mining
engineering.
Interested in glass
and ceramics.
• K = P (CO2) / [CO2]
• Increase T. What happens to equilibrium
position? To value of K?
• K increases as T goes up because P(CO2)
increases and [CO2] decreases.
• Decrease T. Now what?
• Equilibrium shifts left and K decreases.
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73
74
EQUILIBRIUM AND EXTERNAL EFFECTS
Temperature Effects
on Equilibrium
• Add catalyst ---> no change in
• A catalyst only affects the RATE of
approach to equilibrium.
N2O4 (colorless) + heat
K
q e 2 NO2 (brown)
∆Ho
Kc =
= + 57.2 kJ
[NO2 ]2
[N2O 4 ]
Kc (273 K) = 0.00077
Kc (298 K) = 0.0059
Catalytic exhaust system
75
76
Haber-Bosch
Ammonia Synthesis
NH3
Production
Fritz Haber
1868-1934
Nobel Prize, 1918
N2(g) + 3 H2(g) q e 2 NH3(g) + heat
K = 3.5 x 108 at 298 K
Carl Bosch
1874-1940
Nobel Prize, 1931
77
78
EQUILIBRIUM AND EXTERNAL EFFECTS
Le Chatelier’s Principle
• Concentration changes
–no change in K
–only the position of equilibrium
changes.
Adding a “reactant” to a chemical
system.
CH13_slide78.mov
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79
80
Le Chatelier’s Principle
Le Chatelier’s Principle
Removing a “reactant” from a chemical
system.
Adding a “product” to a chemical
system.
CH13_slide79.mov
CH13_slide80.mov
81
82
Le Chatelier’s Principle
butane
ButaneIsobutane
Equilibrium
K =
[isobutane]
= 2.5
[butane]
isobutane
Removing a “product” from a chemical
system.
CH13_slide81.mov
Butane
Isobutane
butane
83
Butane
qe
Isobutane
Assume you are at equilibrium with [iso] = 1.25 M and [butane] =
0.50 M. Now add 1.50 M butane. When the system comes to
equilibrium again, what are [iso] and [butane]? K = 2.5
Solution
Calculate Q immediately after adding more
butane and compare with K.
isobutane
Q =
• At equilibrium with [iso] = 1.25 M and
[butane] = 0.50 M. K = 2.5.
• Add 1.50 M butane.
• When the system comes to equilibrium
again, what are [iso] and [butane]?
[isobutane]
1.25
=
= 0.63
[butane]
0.50 + 1.50
Q is LESS THAN K. Therefore, the
reaction will shift to the ____________.
CH13_slide83.mov
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84
85
Butane
qe
86
Isobutane
Butane
Solution
Q is less than K, so equilibrium shifts right —
away from butane and toward isobutane.
Set up ICE table
[butane]
[isobutane]
Initial
0.50 + 1.50
1.25
Change
-x
+x
Equilibrium
1.25 + x
2.00 - x
EQUILIBRIUM AND
EXTERNAL EFFECTS
e
Isobutane
You are at equilibrium with [iso] = 1.25 M and
[butane] = 0.50 M. Now add 1.50 M butane.
Solution
You are at equilibrium with [iso] = 1.25 M and [butane]
= 0.50 M. Now add 1.50 M butane.
K = 2.50 =
[isobutane]
1.25 + x
=
[butane]
2.00 - x
x = 1.07 M
At the new equilibrium position,
[butane] = 0.93 M and [isobutane] = 2.32 M.
Equilibrium has shifted toward isobutane.
87
88
Le Chatelier’s Principle
• Change T
• Temperature, catalysts, and changes in
concentration affect equilibria.
– change in K
– therefore change in P or concentrations at
equilibrium
• The outcome is governed by LE
• Use a catalyst: reaction comes more
quickly to equilibrium. K not changed.
• Add or take away reactant or product:
CHATELIER’S PRINCIPLE
• “...if a system at equilibrium is
disturbed, the system tends to shift its
equilibrium position to counter the
effect of the disturbance.”
– K does not change
– Reaction adjusts to new equilibrium “position”
89
90
Nitrogen Dioxide Equilibrium
N2O4(g) q e 2 NO2(g)
Nitrogen Dioxide Equilibrium
N2O4(g) q e 2 NO2(g)
Kc =
e
Kc =
[NO2 ]2
= 0.0059 at 298 K
[N2O4 ]
Increase P in the system by reducing the
volume.
In gaseous system the equilibrium will shift to
the side with fewer molecules (in order to
reduce the P).
[NO2 ]2
= 0.0059 at 298 K
[N2O4 ]
Increase P in the system
by reducing the volume
(at constant T).
Therefore, reaction shifts LEFT and P of NO2
decreases and P of N2O4 increases.
CH13_slide89.mov
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