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EQUILIBRIUM
SCH4U
Learning Goals:
• Explain the concept of dynamic equilibrium, using
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•
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examples of physical and chemical equilibrium systems.
Explain the concept of chemical equilibrium and how it
applies to the concentration of reactants and products in a
chemical reaction at equilibrium.
Create and complete an ICE table for an equilibrium
system.
Draw graphs of c vs. t to illustrate a chemical system
approaching equilibrium.
Use appropriate terminology such as dynamic equilibrium,
reversible reaction, closed system, equilibrium
concentrations, phase equilibrium, solubility equilibrium,
chemical equilibrium, equilibrium position
Quantitative vs. Reversible
• So far we have studied quantitative reactions,
reactions which proceed until the limiting reactant is
consumed
• However, many reactions are not quantitative, they
are reversible meaning they can proceed in both
forward and reverse directions
•A + B
C+D
N 2( g )  3H 2( g )  2 NH 3( g )
H 2O(l )  H 2O( g )
Dynamic Equilibrium
• When the rate of the forward reaction and the rate of
How reaction
do the
the reverse
are equal, the system is in
following images
dynamic
equilibrium
represent
dynamic
equilibrium?
• This does
NOT mean that nothing is happening!
Consider the following…
Over time, what
happens to an open
beaker of water?
Over time, what
happens to the water
level if we cover the
beaker?
We can study equilibrium
under closed systems!
Write an equation to
describe the equilibrium
system.
Properties of Equilibrium Systems
• 1. Macroscopic properties (e.g. colour,
pressure, concentration, pH) are constant
• I.e. It appears as if nothing is changing
• 2. Can only be reached in a closed system
• 3. Forward rate = Reverse rate
• 4. Can be established from either direction
At equilibrium, the concentrations of all reactants
and products will remain constant
3 types of Equilibrium
1. Solubility Equilibrium
2. Phase Equilibrium
3. Chemical
Equilibrium
NaCl(s)
Na+ (aq) + Cl- (aq)
Types of Equilibrium
• Solubility Equilibrium: A dynamic
equilibrium between a solute and solvent
in a saturated solution in a closed system
• Phase Equilibrium: a dynamic equilibrium between
different physical states of a pure substance in a closed
system
• Ex: water/liquid in a sealed container
• Chemical Reaction equilibrium: a dynamic equilibrium
between reactants and products of a chemical reaction in
a closed system
Practice
Sketch a graph of Concentration of Reactants and products
vs. time for Experiment 1 and 2
More Practice
1. Consider the reaction:
• Hydrogen gas and hydrogen iodide gas are both clear
colourless, while iodine gas is purple.
• Explain why when hydrogen gas and iodine gas are
added to a closed flask and allowed to react, the colour
never fades to colourless.
2. Sketch a graph to illustrate the changes in concentration
of each component over time as equilibrium is reached for
Experiment 1
• 3. If experiment 1 did not reach
equilibrium, what would the
theoretical yield of HI be?
• 4. What is the actual yield of HI
in experiment 1?
Solving Equilibrium Problems
• Use ICE Tables
• I: Initial Concentration (mol/L)
• C: Change in concentration
• +x if increasing in concentration
• -x if decreasing in concentration
• Multiply x by the # of moles
• E: Equilibrium Concentration (mol/L)
Practice #1
• In a 4.00L container, 2.50 mol of carbon dioxide gas is
decomposed. At equilibrium, [CO2(g)]eq = 0.125 mol/L. Use
an ICE table to find [O2(g)]eq and [CO(g)]eq
Practice #2
• 0.500 mol of NOCl(g) is decomposed in a closed 2.00 L
container. [NO(g)]eq = 0.040 mol/L. Find [Cl2(g)]eq and
[NOCl(g)]eq.
Practice #3
• When 2.00 mol of ethene gas and 1.50 mol of bromine
vapour come to equilibrium in a closed 1.00 L container,
the equilibrium concentration of bromine vapour is
measured at 0.150 mol/L. Find [C2H4(g)]eq and
[C2H4Br2(g)]eq.
Practice #3 Continued
• Graph the equilibrium reaction for the addition of bromine
to ethene.
At equilibrium, the rate of the forward and reverse
reactions are equal
Equilibrium
The Equilibrium Constant, Keq
For the reaction:
aA + bB  cC + dD
At equilibrium:
rfwd = rrev
kfwd[A]a[B]b = krev[C]c[D]d
Rearrange
Sub in Keq for
kfwd/krev
Sub in rate
law equation
kfwd = [C]c[D]d
krev
[A]a[B]b
Keq = [C]c[D]d
[A]a[B]b
Equilibrium constant
Keq Conditions
• Note: the equilibrium constant expression only works with
reactions that occur in a single step
• Keq will remain the same as long as the temperature is
kept constant (changing the temperature changes the
forward and reverse reactions by different amounts and
therefore change the Keq)
Examples
Write the equilibrium constant expression for the
following two reactions
a
a))2
2 NO
NO22(( gg )) 
N
N 22O
O44(( gg ))
b
b)) N
N 22(( gg )) 
3
3H
H 22(( gg )) 
2
2 NH
NH 33(( gg ))
Heterogeneous Equilibria
• Equilibrium systems can involve all states of matter
• However, the concentration of a pure solid or liquid
cannot change
• Therefore, equilibrium constant expressions will not
include solids and liquids
H 2O
O(l ) 
H
H 2O
O( g )
aa))H
2 (l )
2 (g)
NH 4Cl
Cl( s ) 
 NH
NH 3( g ) HCl
HCl( g )
bb))NH
4
(s)
3( g )
(g)
Magnitude of Keq
Keq = [C]c[D]d
[A]a[B]b
Keq >> 1
At equilibrium there is more products
than reactants. The reaction is
product favoured
Keq = 1
At equilibrium there is an equal
amount of products and reactants
Keq << 1
At equilibrium there is more
reactants than products. The
reaction is reactant favoured
Example
Calculate the value of Keq for the following system
CO2( g )  H 2( g )  CO( g )  H 2O( g )
At Equilibrium:
[CO2] = 0.0954 mol/L
[H2] = 0.0454 mol/L
[CO] = [H2O] = 0.00460 mol/L
Practice
For each reaction and their respective equilibrium constant,
predict whether reactants or products are favoured.
• N2(g) + O2(g)  2NO(g)
Keq = 4.7 x 10-31
• NO(g) + CO(g) 2N2(g) + CO2(g)
Keq = 2.2 x 1059
Le Chatelier’s Principle
• When a chemical system at equilibrium is
disturbed by a stress, the system adjusts
(shifts) to oppose the change
• Stresses include:
• Change in concentration
• Change in pressure (or volume)
• Change in temperature
Change in Concentration
A(g) + 3B(g)  2C(g) + heat
• Increasing the concentration of the reactants OR
• Decreasing the concentration of the products
• Will favour the forward reaction, causing the equilibrium to
shift to the RIGHT
• Decreasing the concentration of the reactants OR
• Increasing the concentration of the products
• Will favour the reverse reaction, causing the equilibrium to
shift to the LEFT
• RECALL: Addition or removal of solid or liquids does not change the
concentration. Therefore does not cause a shift. I.e. only applies to gases
and aqueous solutions.
Change in Concentration
N2(g) + 3H2(g)  2NH3
Change in Pressure
A(g) + 3B(g)  2C(g) + heat
 volume  pressure
 volume  pressure
• Increasing the volume of the container OR Decreasing the
pressure
• Will cause a shift to the side with MORE gas molecules
• In our example, it will shift left (4 molreactants > 2 molproducts)
• Decreasing the volume of the container OR Increasing the
pressure
• Will cause a shift to the side with LESS gas molecules
• In our example, it will shift right (4 molreactants > 2 molproducts)
Change in Temperature
In an exothermic reaction:
• Increasing the
temperature will cause a
shift to the LEFT
• Decreasing the
temperature will cause a
shift to the RIGHT
Change in Temperature
In an endothermic reaction:
• Increasing the temperature
will cause a shift to the
RIGHT
• Decreasing the
temperature will cause a
shift to the LEFT
Change in Temperature
Recall: Keq is temperature dependent.
Therefore, changes in temperature will also
affect Keq
Shift right =  products,  Keq
Shift left =  reactants, Keq
DEMONSTRATION
Variables that do NOT Affect Equilibrium
• Catalysts
• Increases reaction rate by lowering activation energy (of
BOTH the forward and the reverse reactions equally)
• Decreases the time required to reach equilibrium but does
not affect the final position of equilibrium
• Inert Gases
• Increases the pressure, which will increase reaction rate
• Increases the probability of successful collisions for BOTH
products and reactants equally
• Decreases the time required to reach equilibrium but does
not affect the final position of equilibrium
Practice
The Reaction Quotient (Q)
• If a chemical system begins with reactants only, it is
obvious that the reaction will shift right (to form products).
• However, if BOTH reactants and products are present
initially, how can we tell which direction the reaction will
proceed?
• Use a trial value called the reaction quotient, Q
• When a reaction is NOT at equilibrium
• Q=Keq  the system is at equilibrium
• Q > Keq  the system shifts towards reactants to reach equilibrium
• Q < Keq the system shifts towards products to reach equilibrium
Practice #1
(p. 464) In a container at 450°C, N2 and H2 react to produce NH3. K = 0.064.
When the system is analysed, [N2] = 4.0 mol/L, [H2] = 2.0 X 10-2 mol/L, and
[NH3] = 2.2 X 10-4 mol/L. Is the system at equilibrium, if not, predict the
direction in which the reaction will proceed.
N2( g )  3H 2( g )  2 NH3( g )
Q
[ NH 3( g ) ]
2
[ N 2 ( g ) ][ H 2 ( g ) ]3
4 2
(2.2 10 )

2 3
(4.0)( 2.0 10 )
 1.5 10 3
0.0015  0.064
QK
 The system will shift right
Practice #2
In a container, carbon monoxide and water vapour are
producing carbon dioxide and hydrogen at 900oC.
CO(g) + H2O(g)  H2(g)
+ CO2(g)
Keq = 4.00 at 900oC
If the concentrations at one point in the reaction are: [CO(g)]
= 4.00 mol/L, [H2O(g)] = 2.00 mol/L, [CO2(g)] = 4.00 mol/L,
and [H2(g)] = 2.00 mol/L. Determine whether the reaction
has reached equilibrium, and, if not, in which direction it will
proceed to establish equilibrium.
Practice #2 Answer
products
Practice #3
Calculating Equilibrium Concentrations from Initial Concentrations
• Carbon monoxide reacts with water vapour to produce
carbon dioxide and hydrogen. At 900oC, Keq is 4.200.
Calculate the concentrations of all entities at equilibrium if
4.000 mol of each entity are initially place in a 1.00L
closed container.
Practice #4
Calculating Equilibrium Concentrations Involving a Quadratic Equation
• If 0.50 mol of N2O4 is placed in a 1.0L closed container at
150oC, what will be the concentrations of N2O4 and NO2
at equilibrium? (Keq = 4.50)
Practice #5
Simplifying Assumption: 100 rule (for small K values)
If: [initial reactant] > 100, you can simplify the Keq expression
K
Ex: 2CO2(g)   2CO(g) + O2 (g)
If K = 6.40 x 10-7, determine the concentrations of all
substances at equilibrium if it starts with [CO2] = 0.250 mol/L
Solubility Equilibrium
• Not all ionic compounds are equally soluble
• Ionic compounds dissociate into individual ions in aqueous
solutions
• This can be a reversible system
• Example: CaCl2(s)
Ca2+(aq) + 2Cl-(aq)
• Equilibrium can be reached between the solid substance and
its dissolved ions (saturation point)
• The solution is saturated at equilibrium (no more ions can dissociate)
Solubility Product Constant (Ksp)
• An equilibrium equation can be written for solubility
reactions
• Ex: AgCl (s)
Ag+ (aq) +
Cl- (aq)
Recall: Since AgCl is a solid, the concentration is not
changing, so it is “built in” to the K value:
• The new constant is the solubility product constant (Ksp)
Example
• Eg: Lead (II) chloride has a molar solubility of 1.62x102
mol/L at 25oC. What is the Ksp of this salt?
PbCl2

Pb2+ + 2Cl-
Ksp = [Pb2+][Cl-]2
[Pb2+] = [PbCl2] = 1.62x10-2mol/L
[Cl-] = 2[PbCl2] = 2(1.62x10-2mol/L) = 3.24 x 10-2mol/L
Ksp = [1.62x10-2mol/L][3.24 x 10-2mol/L]2
= 1.7x10-5
Example 2
• The Ksp of silver chloride at 25oC is 1.8x10-10. What is the
molar solubility of AgCl?
AgCl   Ag+ + ClAgCl

Ag+
Cl0
Initial
-
0
Change
-
+X
+X
Equilibrium
-
X
X
Ksp = [Ag+][Cl-]
1.8x10-10 = [X][X]
1.8x10-10 = X2
X = 1.34x10-5M
• The size of Ksp depends on the solubility of the salt.
• Large Ksp: [ions] at equilibrium is high, salt is very soluble
• Small Ksp: [ions] at equilibrium is low, salt has low solubility
• To determine whether a precipitate will form during a
reaction, a trial solubility product constant can be determine
which is denoted by the symbol Qsp.
Qsp < Ksp : Shifts right to equilibrium – all solid dissolving
Qsp > Ksp : Shifts left to equilibrium – precipitate forms
Qsp = Ksp : Equilibrium (saturated) – no precipitate
Example 1
• Will a ppt form if a solution of CaSO4 is made such that
[Ca2+] = 0.0104 mol/L and [SO42-]=0.0082 mol/L?
• Ksp =7.1x10-5
• (Ans: Q = 8.5 X 10-5, ppt forms)
Example 2
• Will a precipitate form when a solution is made by
dissolving PbCl2 into water.
• [Pb2+]=0.0001mol/L. [Cl-]=0.05mol/L. Ksp = 1.2x10-5