Monday April 7th: “A” Day
Tuesday, April 8th: “B” Day
Agenda
Section 14.3: “Equilibrium Systems and Stress”
Le Châtelier’s principle, common-ion effect
Homework:
Section 14.3 review, pg. 518: #1-7
Le Châtelier’s principle worksheet
Concept Review: “Equilibrium Systems and
Stress”
Le Châtelier’s Principle
Stress is another word for something that causes
a change in a system at equilibrium.
Chemical equilibrium can be disturbed by a
stress, but the system soon reaches a new
equilibrium.
Le Châtelier’s principle: a system in equilibrium
will oppose a change in a way that helps
eliminate the change.
Le Châtelier’s Principle
Chemical equilibria respond to three kinds of
stress:
1. Changes in the concentrations of reactants or
products
2. Changes in temperature
3. Changes in pressure
When a stress is first applied to a system,
equilibrium is disturbed and the rates of the
forward and backward reactions are no longer
equal.
Le Châtelier’s Principle
The system responds to the stress by forming
more products or by forming more reactants.
A new chemical equilibrium is reached when
enough reactants or products form.
At this point, the rates of the forward and
backward reactions are equal again.
Increase in Reactant Concentration
If you increase a reactant’s concentration, the
system will respond to decrease the concentration
of the reactant by changing some of it into
product.
Therefore, the rate of the forward reaction must
be greater than the rate of the reverse reaction.
The equilibrium is said to shift right, and the
reactant concentration drops until the reaction
reaches equilibrium.
Increase in Reactant Concentration
Example
N2(g) + 3 H2(g)
2 NH3(g)
 If either N2 or H2 is increased, the equilibrium
will shift to the right to try to “get rid” of the
additional reactants.
Decrease in Reactant Concentration
If you decrease a reactant’s concentration, the
system will respond to increase the concentration
of the reactant by changing some of the product
into reactant.
Therefore, the rate of the reverse reaction must
be greater than the rate of the forward reaction.
The equilibrium is said to shift left, and the
product concentration drops until the reaction
reaches equilibrium.
Decrease in Reactant Concentration
Example
N2(g) + 3 H2(g)
2 NH3(g)
If either N2 or H2 is decreased, the
equilibrium will shift to the left to try to
replace the reactants that were removed.
Increase in Product Concentration
N2(g) + 3 H2(g)
2 NH3(g)
If NH3 is increased, the equilibrium will
shift to the left to try to get rid of the
additional NH3 added to the system.
Decrease in Product Concentration
N2(g) + 3 H2(g)
2 NH3(g)
If NH3 is decreased, the equilibrium will
shift to the right to try to replace the
product that was removed.
Changes in Temperature
Exothermic reactions have negative ΔH values,
which means they release energy.
Think of energy as a product in exothermic
reactions.
Endothermic reactions have positive ΔH
values, which means they absorb energy.
Think of energy as a reactant in
endothermic reactions.
Changes in Temperature
Exothermic Reaction Example
N2(g) + 3 H2(g)
2 NH3(g)
ΔH of forward reaction = -91.8 kJ
Think of it this way:
N2(g) + 3 H2(g)
2 NH3(g) + energy
Changes in Temperature
Exothermic Reaction Example
N2(g) + 3 H2(g)
2 NH3(g) + energy
If the temperature is raised during an
exothermic reaction, the equilibrium will
shift to the left to try to relieve the added
stress and additional reactants will be
made.
Changes in Temperature
N2(g) + 3 H2(g)
2 NH3(g) + energy
If the temperature is lowered during an
exothermic reaction, the equilibrium will shift
to the right to try to replace the energy that
was removed and more products will be
made.
Changes in Temperature
Endothermic Reaction Example
N2O4 (g)
2 NO2 (g)
ΔH of forward reaction= 55.3 kJ
Think of it this way:
N2O4 (g) + energy
2 NO2 (g)
Changes in Temperature
Endothermic Reaction Example
N2O4 (g) + energy
2 NO2 (g)
If the temperature is raised during an
endothermic reaction, the equilibrium
will shift to the right to relieve the added
stress and more products will be made.
Changes in Temperature
Endothermic Reaction Example
N2O4 (g) + energy
2 NO2 (g)
If the temperature is lowered during an
endothermic reaction, the equilibrium will
shift to the left to replace the energy that was
removed and more reactants will be made.
Changes in Pressure
Pressure has almost no effect on equilibrium
reactions that are in solution.
Gases in equilibrium may be affected by
changes in pressure.
Simply put, an increase in pressure shifts the
equilibrium to favor the reaction that
produces fewer gas molecules.
Changes in Pressure
Example
2 NOCl (g)
2 NO (g) + 1 Cl2 (g)
The left side of the equilibrium contains 2
moles of gas.
The right side of the equilibrium contains 3
moles of gas.
An increase in pressure will shift the
equilibrium to the left to produce fewer mole
of gas.
Changes in Pressure
Example
H2O (g) + CO (g)
H2 (g) + CO2 (g)
In this case, each side of the equilibrium
contains the same number of moles of gas, 2.
In such cases, a change in pressure will not
affect equilibrium.
Common-Ion Effect
The solubility of CuCl in pure water is 1.1  10−3 M.
The solubility of CuCl in sea water is 2.2  10−6 M.
CuCl is 500 times less soluble in sea water.
This dramatic reduction in solubility of CuCl
demonstrates the common-ion effect.
If you add chloride-rich ocean water to a saturated
solution of copper(I) chloride, the [Cl−] increases.
CuCl (s)
Cu+(aq) + Cl-(aq)
Common-Ion Effect
Remember, Ksp = [Cu+] [Cl-]
However, Ksp is a constant, so if the [Cl-] is increased,
[Cu+] must decrease.
This decrease can only occur by the precipitation
of the CuCl salt.
The ion Cl− is the common-ion in this case.
Common-ion effect: the phenomenon in which the
addition of an ion common to two solutes brings
about precipitation or reduces ionization.
You will see this when you add sodium citrate to a
saturated NaCl solution. The NaCl will precipitate out.
Practical Uses of Le Châtelier’s Principle
The chemical industry makes use of Le Châtelier’s
principle in the synthesis of ammonia by the
Haber Process.
High pressure is used to drive the following
equilibrium to the right.
The forward reaction converts 4 mol of gas into
2 mol of another gas, so it is favored at high
pressures.
Homework
Section 14.3 review, pg. 518: #1-7
Le Châtelier’s principle worksheet
Concept Review:”Equilibrium Systems and Stress”
– we’ve now finished the chapter…
Next time:
Section 4.3 Quiz
Chapter review work day
Chapter 14 Test/Concept Review Due:
Monday: April 21st: “A” Day
Tuesday, April 22nd: “B” Day