Download Topic 7.2 Equilibrium The Position of Equilibrium

7.2 The Position of Equilibrium.
Assessment Statements
7.2.1 Deduce the equilibrium constant expression
(Kc) from the equation for a homogeneous
reaction.
7.2.2 Deduce the extent of a reaction from the
magnitude of the equilibrium constant.
7.2.3 Apply Le Chatelier’s principle to predict
qualitative effects of changes of
temperature, pressure and concentration on
the position of equilibrium and on the value of
the equilibrium constant.
Assessment Statements
 7.2.4 State and explain the effect of a
catalyst on an equilibrium reaction.
 7.2.5 Apply the concepts of kinetics and
equilibrium to industrial process
 Suitable examples include the Haber and
Contact processes
Assessment Statements
7.2.1 Deduce the equilibrium constant expression
(Kc) from the equation for a homogeneous
reaction.
The Equilibrium Constant
 Changing the temperature affects the rates of
forward and reverse reactions differently
because they have different activation energies
 Kc = expressed in terms of molar concentration
(mol/L)
Homogeneous Equilibria
 Haber process to manufacture ammonia
3H2(g) + N2(g) < -- > 2NH3(g)
 Kc = [NH3]2
[H2]3[N2]
 Contact process to manufacture sulfuric acid
2SO2(g) + O2(g) < -- > 2SO3(g)
 Kc = [SO3]2
[SO2]2 [O2]
Calculating Kc
N2 + 3Cl2 <--> 2NCl3
Temp remains constant in 5 L flask.
Equilbrium was reached and the following was found:
0.007 mol N2 , 0.0022 mol Cl2 and 0.95 mol of NCl3.
Calculate the Kc for this reaction.


Find molar concentration of everything (mol/L)
Use equilibrium expression based on chemical
equation
Answer
1.
2.
[N2] = 0.007 mol / 5 L = 0.0014 M
[Cl2] = 0.0022 mol / 5 L = .00044 M
[NCl3] = 0.95 mol / 5 L = 0.19 M
Kc = [NCl3]2 = [0.19]2
[N2][Cl2]3
[0.0014][0.00044]3
= 3.0 x 10 11
Question
 CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
1. Write an expression for Keq
2. Calculate K at a given temperature if [CH4] =
0.020 M, [O2] = 0.042 M, [CO2] = 0.012 M, and
[H2O] = 0.030 M at equilibrium. (include units)
Assessment Statements
7.2.2 Deduce the extent of a reaction from the
magnitude of the equilibrium constant.
The value of K
 The value or magnitude of K tells us the extent to
which reactants have been converted into
products.
 Remember in the ratio for K, the concentration of
products divided by the concentration of
reactants.
 A small value for K means that very little of
the reactants were converted into products
before equilibrium was reached.
 This is stated as “reactants are favoured”.
 A large value of K means that most of the
reactants were converted into products
before equilibrium was reached.
 This is stated as “products are favoured”.
The Magnitude of Equilibrium Constants
If K >1, then products dominate at equilibrium and
equilibrium lies to the right.
If K <1, then reactants dominate at equilibrium and the
equilibrium lies to the left.
 If K = 1 neither reactants nor products are
favoured.
 The value of K does not indicate how long
it takes for equilibrium to be reached.
 The value of K varies with temperature and
that’s why its usually mentioned with K
Calculating
Equilibrium Constants
Nitrogen dioxide decomposes at high temperatures
according to this equation:
2 NO2 (g)  2 NO (g) + O2 (g)
If the equilibrium concentrations are as follows:
[NO2]= 1.20 M, [NO] = 0.160, and [O2] = 0.080 M;
calculate the equilibrium constant.
Keq =
[NO]2 [O2]
[NO2] 2
(0.160)2 (0.080)
So Keq =
(1.20)2
= 0.00142
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Calculating
Equilibrium Constants
The equilibrium equation for the oxidation of sulfur
dioxide is as follows:
2SO2 (g) +
O2 (g)  2 SO3 (g)
If the equilibrium concentrations are as follows:
[SO2 ]= 0.44 M, [O2] = 0.22, and [SO3] = 0.78 M ,
Calculate the equilibrium constant
Keq=
[SO3]2
[SO2]2 [O2]
(0.78)2
So
Keq =
(0.44)2
(0.22)
= 14.3
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Practice Problem 1
The equilibrium equation for the carbon monoxide with
steam to produce hydrogen gas is as follows:
CO2 (g) + H 2 (g)  CO (g) + H2O (g)
If the equilibrium concentrations are as follows:
[CO] = 1.00 M, [H2O] = 0.025, [CO2] = 0.075 M and
[H2] = 0.060 M, calculate the equilibrium constant.
Keq =
So Keq =
[CO] [H2O]
[CO2] [H2]
(1.00) (0.025)
(0.075) (0.060)
= 0.139
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Equilibrium Calculations –
Using I.C.E.Models
Equilibrium constants and concentrations can
often be deduced by carefully examining
data about the initial and equilibrium
concentrations
Initial Change Equilbrium
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Equilibrium Calculations
ICE Model problem 1
Hydrogen and iodine are in equilibrium with
Hydrogen iodide to this reaction:
H2 + I2  2HI
Suppose that 1.5 mole of H2 and 1.2 mole of I2
are placed in a 1.0 dm3 container. At equilibrium it
was found that there were 0.4 mole of HI.
Calculate the equilibrium concentrations of [H2] and
[I2] and the equilibrium constant.
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Equilibrium Calculations
ICE Model Problem 1 -- Solution
Hydrogen and iodine are in equilibrium with Hydrogen
iodide to this reaction:
H2 + I2  2HI
Suppose that 1.5 mole of H2 and 1.2 mole of I2 are
placed in a 1.0 dm3 container. At equilibrium it was
found that there were 0.4 mole of HI. Calculate the
equilibrium concentrations of [H2] and [I2] and the
equilibrium constant.
[H2 ]
[ I2 ]
[HI ]
Keq =
I
1.5
1.2
0
C
- x
- x
+2x
[HI]2
[H2 ] [ I2 ]
E
1.5- x
1.2 –x
0.4
=
Since 2x = 0.4, x = 0.2
[H2 ] = 1.5 – 0.2 = 1.3
[I2 ] = 1.2 – 0.2 = 1.0
(0.4)2
(1.3) (1.0)
= 0.123
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Equilibrium Calculations
ICE Model Problem 2
Sulfur dioxide reacts with oxygen to produce sulfur
trioxide according to this reaction:
2 SO2 + O2  2SO3
Suppose that 1.4 mole of SO2 and 0.8 mole of O2
are placed in a 1.0 dm3 container. At equilibrium it
was found that there were 0.6 mole of SO3.
Calculate the equilibrium concentrations of [SO2] and
[O2] and the equilibrium constant.
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Equilibrium Calculations
ICE Model Problem 2 - Solution
Sulfur dioxide reacts with oxygen to produce sulfur
trioxide according to this reaction:
2 SO2 + O2  2SO3
Suppose that 1.4 mole of SO2 and 0.8 mole of O2 are
placed in a 1.0 dm3 container. At equilibrium it was
found that there were 0.6 mole of SO3. Calculate the
equilibrium concentrations of [SO2] and [O2] and the
equilibrium constant.
[SO2 ]
[O2 ]
[SO3]
Keq =
I
1.4
0.8
0
C
-2x
-x
+2x
[SO3]2
[SO2 ]2 [O2 ]
E
1.4-2x
0.8 –x
0.6
=
Since 2x = 0.6, x = 0.3
[SO2 ] =1.4 – 2( 0.3) = 0.8
[O2 ] = 0.8 – 0.3 = 0.5
(0.6)2
= 0.281
(0.8)2(0.5)
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Calculating Equilibrium
Concentrations from Keq
What is the concentration for each substance at equilibrium for the
following gaseous reaction
C2H6  C2H4 + H2 KC = 1.01
if the initial concentration of ethene, C2H4 and that of hydrogen are both
0.300 M?
[C2H4] [H2]
[C2H6]
Keq =
Initial
Change Equilibrium
C 2H 4
0.300
-x
0.300-x
H2
0.300
-x
0.300-x
C 2H 6
0
+x
x
[0.300-x][0.300-x]
[x]
1.01x = 0.09 - 0.6x +x2
x2 - 1.61x +0.09 = 0
1.01 =
x = 1.61+ (1.61)2-4(1)(0.09)
2(1)
x = 0.0580 and 1.552. The
second root is extraneous.
so [C2H6] = 0.0580
and [H2 ] = [C2H4] = 0.242
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2NO2 (g) <--> N2O4 (g)
 Four experiments were
performed.
 The initial
concentrations of the
two chemicals were
different in each
experiment and the
concentration of each
gas was measured once
the system reached
equilibrium.
At Equ ilibrium
Exp #
[NO2]
[N2O4]
1
0.104
1.19
2
0.048
0.254
3
0.136
2.04
4
0.202
4.48
Calculate and Describe
 Using the data on the previous page, calculate the
K for each experiment.
 What did you find?
Exp #
1
2
At Equ
[NO2]
0.104
0.048
ilibrium
[N2O4]
1.19
0.254
3
4
0.136
0.202
2.04
4.48
Starter
 CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
1. Write an expression for Keq
2. Calculate K at a given temperature if [CH4] = 0.020
M, [O2] = 0.042 M, [CO2] = 0.012 M, and [H2O] = 0.030
M at equilibrium. (include units)
Objectives
 7.2.3 Apply Le Chatelier’s principle to
predict qualitative effects of changes of
temperature, pressure and concentration
on the position of equilibrium and on the
value of the equilibrium constant.
Le Chatelier’s Principle
When a system at equilibrium is placed under
stress, the system will undergo a change in
such a way as to relieve that stress.
Henri-Louis Le Chatelier
(1850-1936)
French industrial chemist
Le Chatelier Translated:
When you take something away from a system at
equilibrium, the system shifts in such a way as to
replace what you’ve taken away.
When you add something to a system at
equilibrium, the system shifts in such a way as to
use up what you’ve added.
Le Chatelier Example #1
A closed container of ice and water at
equilibrium. The temperature is raised.
Ice + Energy <-- > Water
The equilibrium of the system shifts to
right
the _______
to use up the added energy.
Le Chatelier Example #2
A closed container of N2O4 and NO2 at
equilibrium. NO2 is added to the container.
N2O4 (g) + Energy < - - > 2 NO2 (g)
The equilibrium of the system shifts to
left
the _______
to use up the added NO2.
Le Chatelier Example #3
A closed container of water and its vapor at
equilibrium. Vapor is removed from the system.
water + Energy  vapor
The equilibrium of the system shifts to
right to replace the vapor.
the _______
Le Chatelier Example #4
A closed container of N2O4 and NO2 at
equilibrium. The pressure is increased.
N2O4 (g) + Energy < - - > 2 NO2 (g)
The equilibrium of the system shifts to
left
the _______
to lower the pressure,
because there are fewer moles of gas
on that side of the equation.
Pressure Changes to system
If the volume decreases, the concentration
increases, and there will be a shift to the side
with the less amount of moles.
If the volume increases, the concentration
decreases, and there will be a shift to the side
with the more amount of moles.
7.2.3 Apply Le Chatelier’s principle to predict the qualitative
effects of changes of temperature, pressure and concentration on the
position of equilibrium and on the value of the equilibrium constant.
N2(g) + 3H2(g)
2NH3(g) ΔHo = -92 kJ
What effect does increasing the pressure have on the
equilibrium?
•Reaction will shift toward the side with the
fewest gas molecules.
•The left side has 4 gas molecules (1N2 & 3H2). The
right side has 2 gas molecules. Reaction will shift
to the right.
•N2 & H2 will react and more NH3 will be produced.
•Kc does not change
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7.2.3 Apply Le Chatelier’s principle to predict the qualitative
effects of changes of temperature, pressure and concentration on
the position of equilibrium and on the value of the equilibrium
constant.
N2(g) + 3H2(g)
2NH3(g) ΔHo = -92 kJ
What effect does decreasing the pressure have on the
equilibrium?
If a reaction has equal numbers of gas molecules on
the left and on the right, changing the pressure has no
effect.
2HI(g)
H2(g) + I2(g)
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Example;
If I increase the pressure, where is the shift?
(right)
If I decrease the pressure, where is the shift?
(left)
2SO2 + O2 <--> 2SO3
(3moles)
(2moles)
Effect of Concentration
1.
2.
3.
4.
If you add more reactant, it shifts to the right
increasing the formation of product, using up
the reactants.
If you add product, it shifts to the left
If you remove product, it shifts to the right,
increasing the formation of product.
If you remove reactant, it shifts to the left
Le Chatelier’s Principle –
The Concentration Effect
N2 (g) +3 H2 (g)
Keq =


2NH3 (g)
DH = - 92 kJ
[NH3]2
[N2] [H2]3
An increase in the concentration of N2 results in
a decrease H2 and an increase in NH3 in such a way to
keep the equilibrium constant the same
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Le Chatelier’s Principle –
The Concentration Effect
N2 (g) +3 H2 (g)
Keq =


2NH3 (g)
DH = - 92 kJ
[NH3]2
[N2] [H2]3
Likewise an increase in the concentration of H2 results
in a decrease in N2 and an increase in NH3 in such a way
to keep the equilibrium constant the same
44
Le Chatelier’s Principle –
The Concentration Effect
N2 (g) +3 H2 (g)
Keq =
 2NH (g) DH = - 92 kJ
3

[NH3]2
[N2] [H2]3
An increase in the concentration of NH3 results in a increase in N2 and an
increase in H2 in such a way to keep the equilibrium constant the same.
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Effect of temperature
 Energy is treated as a reactant if endothermic
equation, and as a product if exothermic equation.
 If cooling a system, then it shifts so more heat is
produced.
 If heating a system, then it shifts so extra heat
is used up.
Example for temp. changes for
Endothermic Reaction
Heating the below reaction causes the system
to shift to the right = more products, because
you treat energy like a reactant.
2NaCl +H2SO4 + energy < -- > 2HCl + Na2SO4
Cooling the above reaction causes the system to
shift to the left = less reactants, so need to
make up more
7.2.3 Apply Le Chatelier’s principle to predict the qualitative effects
of changes of temperature, pressure and concentration on the position
of equilibrium and on the value of the equilibrium constant.
N2(g) + 3H2(g)
2NH3(g) ΔHo = -92 kJ
Effects of Temperature Change
 Increasing temperature causes the equilibrium position to
shift in the direction that absorbs heat (endothermic).
 Decreasing temperature causes the equilibrium position
to shift in the direction that produces heat (exothermic).
 The value of Kc will change with a change in temp. If the
reaction shifts right, the value of Kc increases. If the
reaction shifts left, the Kc value decreases.
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48
7.2.3 Apply Le Chatelier’s principle to predict the qualitative
effects of changes of temperature, pressure and concentration on
the position of equilibrium and on the value of the equilibrium
constant.
N2(g) + 3H2(g)
2NH3(g) ΔHo = -92 kJ
What effect does increasing the temperature have on
the equilibrium?
•Increasing the temperature causes the reaction to shift
to use up some of the added heat (endothermic rx).
•The reaction as written is exothermic so the endothermic
rx is from right to left. The rx will shift left.
•[N2] increases, [H2] increases, [NH3] decreases.
•Kc value will decrease
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Effect of temp change on exothermic
reactions
 Heating the below reaction causes the system to
shift to the left, to use up the extra heat.
2SO2 + O2 <--> 2SO3 + energy
 Cooling the above reaction causes the system to
shift to the right, to make up for the lost heat.
7.2.3 Apply Le Chatelier’s principle to predict the qualitative
effects of changes of temperature, pressure and concentration on
the position of equilibrium and on the value of the equilibrium
constant.
N2(g) + 3H2(g)
2NH3(g) ΔHo = -92 kJ
What effect does decreasing the temperature have on the
equilibrium?
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7.2.3 Apply Le Chatelier’s principle to predict the qualitative
effects of changes of temperature, pressure and concentration on the
position of equilibrium and on the value of the equilibrium constant.
N2O4(g)
2NO2(g)
ΔHo = 58.0 kJ
Both gases are present in a flask at equilibrium. N2O4 is
a colorless gas while NO2 is brown.
1) What color will the contents of the flask be if the
pressure is increased? Explain.
2) State and explain three (3) ways the amount of NO2
production can be increased.
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Assessment Statements
 7.2.4 State and explain the effect of a
catalyst on an equilibrium reaction.
The effect of a catalyst on equilibrium
 Adding a catalyst speeds up a reaction by
providing an alternative mechanism with a
lower activation energy, thus speeding up both
the forward and backward reaction rate.
 It shortens the time needed to attain
equilibrium concentrations
 It has no effect on the position of
equilibrium, however equilibrium will be
attained more quickly.
7.2.4 State and explain the effect of a catalyst on an
equilibrium reaction.
1) A catalyst lowers the
activation energy barrier for
both the forward and the
reverse reactions.
2) Therefore a catalyst increase
the rates of both reactions by
the same factor.
3) A catalyst increases the rate
at which equilibrium is
achieved, but does not
change the final composition
of the substances.
4) The Kc value is not affected
by the presence of a catalyst.
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Heterogeneous Equilibria
 In a heterogeneous equilibrium, the components are in
two different phases.
 A common form of a heterogeneous equilibrium is the
equilibrium which exists between a solid and an
aqueous (water) solution.
 When a substance dissolves in water there is an
equilibrium established between the solid an its
dissolved ions
Example:
AgCl (s)  Ag+ (aq) + Cl- (aq)
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Heterogeneous Equilibria
 AgCl (s)
Ag+ (aq) + Cl- (aq)
 In this example the concentration of the solid phase
is essentially 1. The equilibrium constant then takes
the form
 K = [Ag+ ][Cl-].
 This form is known as a solubility product and is
usually designated as Ksp


57
Different designations of
equilibrium constants
 Keq General designation for an equilibrium constant
 Kc Equilibrium constant based on concentration
 Kp Equilibrium constant based on pressure (gases)
 Ksp Solubility product
 Ka Acid equilibrium constant
 Kb Base equilibrium constant
 Kw Ion product of water
58
Assessment Statements
 7.2.5 Apply the concepts of kinetics and
equilibrium to industrial process
 Suitable examples include the Haber and
Contact processes
7.2.5 Apply the concepts of kinetics and equilibrium to
industrial processes.
N2(g) + 3H2(g)
2NH3(g) ΔHo = -92 kJ
The Haber process to manufacture ammonia
 Ammonia is an important starting point for the
production of fertilizers, nitric acid, explosives and
polymers (nylon).
 Under what conditions can will an industrial chemist
run this reaction to increase the yield of ammonia?
 An optimum temperature must be found
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Haber Process
 N2(g) + 3H2(g) < - - > 2NH3(g)
kJ/mol
ΔH= -92
 Mixture’s volume is compressed and passed over a
heated iron catalyst.
 Conditions for his equilibrium is critical.


High pressure is favourable due to 4 moles on left and 2
moles on right. Increased pressure causes a shift to the
left, favouring product formation.
This is expensive to due and most production plants will
resist compressing gases in terms of operating costs.
Compromise will be met.
Compromise
 This is an exothermic reaction, so low
temperatures would be favourable to produce
product.
 Low temps mean low reaction rates, so we may get a
higher yield but it will take a long time to get it.
Not good for business.
 A compromise temp, as well as the use of a catalyst
will aid in speeding up the reaction to a more
acceptable standard.
Typical conditions
 Pressure between 20-100 MPa (200-1000 atm)
 Temperatures around 700 K
 The reaction is not allowed to reach
equilibrium, because reaction rate decreases
as we approach equilibrium, and typically only
20% of N2 and H2 is converted.
 The gases are cooled and NH3 is condensed
and removed, leaving unused N2 and H2
available for further production.
http://www.absorblearning.com/media/item.
action?quick=128#
Animation of Haber process
Ammonia’s Uses
 Manufacture of fertilizers (ammonia salts and
urea)
 Manufacturing nitrogen used in polymers for the
fabrication of nylon
 Used in the production of explosives (TNT,
dynamite)
Contact Process
Production of sulfuric acid by the oxidation
of sulfur.
Sulfur is burnt in air to form sulfur dioxide

1.
•
S(s) + O2(g) < - - > SO2(g)
•
2SO2(g) + O2(g) < - - > 2SO3(g) ΔH= -196 kJ/mol
•
SO3(g) + H2O(l)  H2SO4(l)
2. Sulfur dioxide is mixed with air and passed
over vanadium(V)oxide catalyst to produce
sulfur trioxide.
3. Sulfur trioxide is reacted with water to
produce sulfuric acid.
7.2.5 Apply the concepts of kinetics and equilibrium to
industrial processes.
2SO2(g) + O2(g)
2SO3(g) ΔHo = -192 kJ
The contact process to manufacture sulfuric acid
SO3(g) + H2O(l)
H2SO4(l)
 Sulfuric acid is used in many chemical processes
 Under what conditions can an industrial chemist run
this reaction to increase the yield of sulfur trioxide?
 An optimum temperature must be found.
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More money, more SO3…
 High pressure would favour the formation of
SO3 in the 2nd step, however its too expensive.
 Reactants are compressed to 2 atm to achieve
the desired flow rate in the reactor.
 Pure O2 would drive the equilibrium to the
right, however its an unnecessary expense.
 Low temperatures, because its exothermic,
would be best, but it slows the rate too much.
Compromised conditions
 Temp between 700-800 K (fast initial reaction
rate)
 The use of a finely divided V2O5 catalyst
 Oxidation is done in converters at lower
temperatures (slows reaction rate)
 Overall conversion is 90% to SO3
http://www.absorblearning.com/media/item.action?q
uick=12b
Contact process animation
Uses of H2SO4
 Fertilizers (converting insoluble phosphate
rock into soluble phosphates)
 Polymers
 Detergents
 Paints
 Pigments
 Petrochemical industry
 Processing of metals
 Electrolyte in car batteries
 Le Chatelier’s principle is a memory aid, it doesn’t
explain why these changes occur.
 Listen carefully and read over text pages to help
you develop further understanding of explanation.
 http://www.mhhe.com/physsci/chemistry/essenti
alchemistry/flash/lechv17.swf
 Haber process notes
 http://www.chemguide.co.uk/physical/equilibria/h
aber.html