EQUILIBRIUM
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
the reverse reaction are equal, the system is in
dynamic equilibrium
• This does NOT mean that nothing is happening!
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
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
is decomposed in a closed 2.00 L
container. [NO(g)]eq = 0.040 mol/L. Find [Cl2(g)]eq and
[NOCl(g)]eq.
(g)
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
 HH 2OO( g )
aa))HH22OO(l(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
Practice 2
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 #3
Simplifying Assumption: 100 rule (for small K values)
If: [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