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C h a pt e r
13
Chemical Equilibrium
Chemistry 4th Edition
McMurry/Fay
Dr. Paul Charlesworth
Michigan Technological University
The Equilibrium State
•
01
Chemical Equilibrium: A state achieved when the
rates of the forward and reverse reactions are
equal and the concentrations of the reactants and
products remain constant.
•
Equilibrium between phases is known as physical
equilibrium.
Prentice Hall ©2004
Chapter13
The Equilibrium State
Prentice Hall ©2004
Chapter13
Slide 2
02
Slide 3
1
The Equilibrium State
•
Chemists are interested in these reversible
reactions. One example is the following:
Prentice Hall ©2004
Chapter13
The Equilibrium State
Prentice Hall ©2004
Chapter13
The Equilibrium State
•
02
Slide 4
03
Slide 5
03
Graphs of reactant and product concentrations
change with time as shown below.
Prentice Hall ©2004
Chapter13
Slide 6
2
The Equilibrium State
Prentice Hall ©2004
04
Chapter13
Slide 7
Equilibrium Constant
•
The equilibrium expression compares reactant and
product concentrations.
[Pr oducts] = [B ] = [NO2 ] = 4.63 × 10−3
r
a
[Re ac tants ] [A] [N 2O4 ]
p
K=
•
01
b
2
From this we obtain a constant (K) for the reaction,
which is independent of concentration changes, but
dependent on the temperature.
Prentice Hall ©2004
Chapter13
Slide 8
Equilibrium Constant
•
02
Homogeneous Equilibrium:When all reacting
species are in the same phase, all reactants and
products are included in the expression.
•
Amounts of components are given as molarity or
partial pressure of a gas.
[NO ] 2
Kc = [ 2 ]
N2O4
Prentice Hall ©2004
Chapter13
2
PNO
2
Kp =
PN2O4
Slide 9
3
Equilibrium Constant
•
03
We can convert between Kc and Kp using an
equation derived from PV = nRT:
For
aA
æ
bB
Kp = Kc (0.0821T) ?n
?n = moles gas products – moles of gas reactants
?n = b – a
Prentice Hall ©2004
Chapter13
Equilibrium Constant
•
Slide 10
04
The following pictures rerepresent mixtures of A molecules
(red) and B molecules (blue), which interconvert according
to the equation A æ B. If Mixture (1) is at equilibrium, which
of the other mixtures is also at equilibrium?
Prentice Hall ©2004
Chapter13
Equilibrium Constant
•
Slide 11
05
Write the Kp and Kc expressions for:
2 N 2O5(g) æ 4 NO2(g) + O2(g)
•
The equilibrium concentrations for the reaction
between CO and Cl2 to form carbonyl chloride
(phosgene gas) CO(g) + Cl2(g) æ COCl2(g) at 74°C
are: [CO] = 1.2 x 10–2 M, [Cl2] = 0.054 M, and
[COCl2] = 0.14 M. Calculate Kc and Kp.
Prentice Hall ©2004
Chapter13
Slide 12
4
Equilibrium Constant
•
06
Methane (CH4) reacts with hydrogen sulfide to yield H2 and
carbon disulfide, a solvent used in manufacturing. What is
the value of Kp at 1000 K if the partial pressures in an
equilibrium mixture at 1000 K are 0.20 atm of CH 4, 0.25 atm
of H 2S, 0.52 atm of CS2, and 0.10 atm of H2?
Prentice Hall ©2004
Chapter13
Equilibrium Constant
Slide 13
07
•
Heterogeneous Equilibrium:When reacting
species are in different phases, solid and liquid
phases are excluded from the expression because
their concentrations “do not change.”
•
For CaCO3(s) æ CaO(s) + CO2(g)
Kc = [CO2] because CaCO 3 and CaO are solids.
Prentice Hall ©2004
Chapter13
Equilibrium Constant
Prentice Hall ©2004
Chapter13
Slide 14
08
Slide 15
5
Equilibrium Constant
•
09
Write the equilibrium equation for each of the
following reactions:
(a) CO2(g) + C(s) æ 2 CO(g)
(b) Hg(l) + Hg 2+(aq) æ Hg 22+(aq)
(c) 2 Fe(s) + 3 H2 O(g) æ Fe2O3(s) + 3 H2(g)
(d) 2 H2 O(l) æ 2 H2(g) + O2(g)
Prentice Hall ©2004
Chapter13
Using Equilibrium Constants
•
Slide 16
01
We can make the following generalizations
concerning the composition of equilibrium mixtures:
If K c > 103, products predominate over reactants. If Kc is
very large, the reaction is said to proceed to completion.
If K c is in the range 10–3 to 103, appreciable
concentrations of both reactants and products are present.
If K c < 10–3, reactants predominate over products. If Kc is
very small, the reaction proceeds hardly at all.
Prentice Hall ©2004
Chapter13
Using Equilibrium Constants
•
Slide 17
02
The reaction quotient (Q c) is obtained by
substituting initial concentrations into the
equilibrium constant. Predicts reaction direction.
Q c > Kc
System proceeds to form reactants.
Q c = Kc
System is at equilibrium.
Q c < Kc
System proceeds to form products.
Prentice Hall ©2004
Chapter13
Slide 18
6
Using Equilibrium Constants
•
Predicting the direction of a reaction.
Prentice Hall ©2004
Chapter13
Using Equilibrium Constants
•
03
Slide 19
04
The equilibrium constant (Kc ) for the formation of
nitrosyl chloride, from nitric oxide and chlorine gas:
2 NO(g) + Cl 2(g) æ 2 NOCl(g) is 6.5 x 104 at 35°C.
In an experiment, 2.0 x 10–2 moles of NO, 8.3 x 10–3
moles of Cl2, and 6.8 moles of NOCl are mixed in a
2.0-L flask. In which direction will the system
proceed to reach equilibrium?
Prentice Hall ©2004
Chapter13
Using Equilibrium Constants
Slide 20
05
Knowing K allows us to calculate equilibrium
concentrations from initial concentrations.
• We use the Initial Change Equilibrium method.
•
Initial (M)
Change (M)
Equil ibrium (M)
•
cis-stilbene æ trans-stilbene
0.850
0
–x
+x
(0.850–x)
x
Use Kc =24 to determine equilibrium concentrations.
Prentice Hall ©2004
Chapter13
Slide 21
7
Using Equilibrium Constants
•
06
A mixture of 0.500 mol H2 and 0.500 mol I2 was
placed in a 1.00-L stainless steel flask at 700°C.
The equilibrium constant Kc for the reaction
H2(g) + I2(g) æ 2 HI(g) is 57 at this temperature.
Calculate the equilibrium concentrations.
If the starting concentration of HI was 0.040 M, calculate
the new equilibrium concentrations.
If the initial concentrations are [H2] = 0.100 M and
[I 2] = 0.200 M, calculate the equilibrium concentrations.
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
•
01
Le Châtelier’s principle:
If an external stress
is applied to a system
at equilibrium, the
system adjusts in
such a way that the
stress is partially
offset.
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
•
Slide 22
Slide 23
02
Concentration Changes:
The concentration stress of an added reactant or
product is relieved by reaction in the direction
that consumes the added substance.
The concentration stress of a removed reactant or
product is relieved by reaction in the direction
that replenishes the removed substance.
Prentice Hall ©2004
Chapter13
Slide 24
8
Le Châtelier’s Principle
•
03
Haber process for synthesis of ammonia.
N2(g) + 3 H2(g) æ 2 NH3(g)
Kc = 0.291 at 700 K
•
Given an equilibrium mixture of 0.50 M N2, 3.00 M
H2, and 1.98 M NH 3 at 700 K, what happens when
the concentration of N2 is increased to 1.50 M?
•
Le Châtelier’s principle tells us the reaction will
relieve the stress by converting the N2 to NH 3.
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
Prentice Hall ©2004
Chapter13
Slide 25
04
Slide 26
05
Slide 27
9
Le Châtelier’s Principle
06
•
The reaction of iron(III) oxide with carbon
monoxide occurs in a blast furnace when iron ore
is reduced to iron metal:
•
Use Le Châtelier’s principle to predict the direction
of reaction when an equilibrium mixture is
disturbed by:
Fe2O3(s) + 3 CO(g) æ 2 Fe(l) + 3 CO2(g)
•(a)
Adding Fe2O 3
Prentice Hall ©2004
(b) Removing CO2 (c) Removing CO
Chapter13
Slide 28
Le Châtelier’s Principle
•
07
Volume and Pressure Changes: Only reactions
containing gases are affected by changes in
volume and pressure.
Increasing pressure = Decreasing volume
•
PV = nRT tells us that increasing pressure or
decreasing volume increases concentration.
Prentice Hall ©2004
Chapter13
Slide 29
Le Châtelier’s Principle
•
N2(g) + 3 H2(g) æ 2 NH3(g)
Prentice Hall ©2004
Chapter13
08
Kc = 0.291 at 700 K
Slide 30
10
Le Châtelier’s Principle
•
09
Consider the reaction: N2O4(g) æ 2 NO2(g), taking place
in a cylinder with a volume = 1 unit.
[NO2 ] = 2 mol/1 = 2
[N 2O4 ] = 1 mol/1 = 1
K=
Prentice Hall ©2004
[NO2 ]2
[N 2 O4 ]
=4
Chapter13
Slide 31
Le Châtelier’s Principle
•
10
The Volume is then halved, which is equivalent to
doubling the pressure.
[NO2 ] = 2 mol/0.5 = 4
[N2 O4 ] = 1 mol/0.5 = 2
Q=
•
[NO2 ]2
=8
[N2 O4 ]
Since Q > K, the [product] is too high and the
reaction progresses in the reverse direction.
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
•
Slide 32
11
Does the number of moles of reaction products
increase, decrease, or remain the same when
each of the following equilibria is subjected to a
decrease in pressure by increasing the volume.
1. PCl 5(g)
æ PCl 3(g) + Cl 2(g)
2. CaO(s)
+ CO2(g) æ CaCO3(s)
3. 3
Prentice Hall ©2004
Fe(s) + 4 H2O(g) æ Fe3O4(s) + 4 H2(g)
Chapter13
Slide 33
11
Le Châtelier’s Principle
•
12
The following picture
represents the equilibrium
mixture for the gas -phase
reaction A2 æ 2A.
•
Draw a picture that shows
how the concentrations
change when the pressure is
increased by decreasing the
volume.
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
Slide 34
13
•
Temperature Changes: Changes in temperature
can change the equilibrium constant.
•
Endothermic processes
are favored when
temperature increases.
•
Exothermic processes
are favored when
temperature decreases.
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
•
Slide 35
14
Consider the reaction N2(g) + 3 H2(g) æ 2 NH3(g)
which is exothermic by 92.2 kJ.
Prentice Hall ©2004
Chapter13
Slide 36
12
Le Châtelier’s Principle
•
15
In the first step of the Ostwald process for synthesis
of nitric acid, ammonia is oxidized to nitric oxide by
the reaction:
4 NH 3(g) + 5 O2(g) æ 4 NO(g) + 6 H2O(g) ? H° = –905.6 kJ
•
How does the equilibrium amount vary with an
increase in temperature?
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
Slide 37
16
•
The following pictures represent the composition of
the equilibrium mixture at 400 K and 500 K for the
reaction A(g) + B(g) æ AB(g).
•
Is the reaction
endothermic or
exothermic?
Prentice Hall ©2004
Chapter13
Le Châtelier’s Principle
•
Slide 38
17
Catalysis: No effect.
Prentice Hall ©2004
Chapter13
Slide 39
13
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