Download Module 3 -- Lesson 4

RETSD Adult Ed Program
CHEM 40S
Module 3, Lesson 4
LESSON 4 -- LE CHATELIER’S PRINCIPLE
Definition: Le Chatelier’s Principle: If a system is in equilibrium and a condition is
changed, then the equilibrium will shift toward restoring the original conditions. The
conditions affecting equilibrium are temperature, pressure, and concentration of either
reactants or products.
If stress is put on a reversible reaction at equilibrium, the equilibrium will shift in such a
way as to relieve the stress.
Example:
The reaction for the preparation of ammonia by the Haber process is:
N2(g) + 3H2(g)  2NH3(g) + heat
Consider the effects of concentration, pressure, and temperature on the equilibrium.
If the concentration of either of the reactants is increased, the number of collisions
between reactant particles will increase. The result is an increase in the reaction rate
toward the right, or in other words, an increase in the rate of the forward reaction.
Then, as the amount of NH3 increases, the rate of the reverse reaction also increases.
However, the net result to the system as a whole is to shift the equilibrium toward the
right. That is, more product is produced. For example, if the [H2] is increased, the [N2] is
decreased and the [NH3] increases. The value of the equilibrium constant remains the
same.
The removal of a product will have the effect of driving the equilibrium to the right to
replace the lost product. Consider the example of an organic acid reacting with an
alcohol to produce an ester and water. An ester is a compound with a pleasant odor that
can be synthesized in the laboratory by reacting an alcohol and an organic acid
(chemistry 30S students do this as part of their course at the CLC). Esters account for the
distinctive odors of many fruits such as bananas, pineapples, and oranges.
Organic acid + alcohol  ester + water
The reactions are very slow; therefore the reactants must be heated, and a catalyst must
be added in order to make the reaction occur with any speed. In esterification reactions,
the products are in chemical equilibrium with the reactants. They will seldom go to
completion unless the water produced in the reaction is removed as it forms. The catalyst
added is concentrated sulfuric acid, which also acts as a dehydrating agent (it removes the
water from the equilibrium). The removal of water helps to force the equilibrium to the
right, producing more product (ester).
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CHEM 40S
Module 3, Lesson 4
If a reactant or a product is added to a system at equilibrium, the system will
shift away from the added component. If a reactant or product is removed, the
system will shift toward the removed component.
Returning to the Haber process as an example, if the pressure is increased, more product
is formed. What if the pressure was doubled? Doubling the pressure exerted on a gas will
decrease its volume to half the original volume. Therefore, the concentrations of
nitrogen, hydrogen, and ammonia are all doubled.
N2(g) + 3H2(g)  2NH3(g)
Kr = [NH3]2
The reverse reaction must then speed up by a factor of 4, as indicated by the rate law for
the reverse reaction. On the other hand, the concentration of hydrogen is cubed. Further,
it is multiplied by the concentration of nitrogen:
Kf = [N2] [H2]3
Doubling the pressure, then, should increase the rate of the forward reaction by 23 x 2, or
16 times. The net result is clearly an increase in product.
Increased pressure on a reaction system with a gas phase has the same effect as increased
concentration. Pressure increases concentration, not Keq. Temperature changes cause a
change in the value Keq.
An increase in pressure on a system at equilibrium favors the side with fewer
gas molecules -- the equilibrium will shift to that side.
Using the Haber process as an example, consider the application of pressure on the whole
system at equilibrium. By shifting to produce more product, NH3, the system reduces the
stress because it reduces the number of gas molecules and thus the pressure in the
container is reduced. For every four reactant molecules, two product molecules are
produced. To reduce the added pressure, the system will shift to the side with fewer gas
molecules.
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CHEM 40S
Module 3, Lesson 4
When the volume of the container holding a gaseous system is reduced, the
system responds by reducing its own volume. This is done by decreasing the
total number of gaseous molecules in the system.
Example:
In the reaction H2(g) + Cl2(g)  2HCl(g), all substances are gases. Pressure would not
shift the equilibrium, as the rate in each direction would be affected the same way. The
number of gas molecules is the same on both sides. Pressure, of course, has an effect
only on the gases in the reaction. A reaction that takes place in solution would be
unaffected by pressure.
TEMPERATURE
It is important to realize that although the changes we have just discussed may alter the
equilibrium position, they do not alter the equilibrium constant, Keq. For example, the
addition of a reactant shifts the equilibrium position to the right but has no effect on the
value of the equilibrium constant; the new equilibrium concentrations still work to
calculate the same value of Keq in the equilibrium expression. The effect of temperature
on equilibrium is different, however, because the value of Keq changes with temperature.
Both the forward and reverse reactions of any system at equilibrium are sped up by an
increase in temperature. However, their rates are increased by different amounts. Also,
the value of the equilibrium constant itself is changed by a change in temperature.
One easy way to predict the shift in an equilibrium subjected to a temperature change is
to consider heat as a reactant or product. In the Haber process:
N2(g) + 3H2(g)  2NH3(g) + heat
heat is produced when hydrogen and nitrogen react. If heat is considered to be a product,
the addition of heat (a product) will increase the concentration of the product. The
equilibrium will shift to the left. In the Haber process, the reverse reaction (the
decomposition of ammonia) is favored by the addition of heat.
Optimum conditions are those which produce the highest yield of product.
In industry, a desired chemical can sometimes be obtained by a reversible reaction.
However, equilibrium is often attained before enough product is produced to make the
process economical. In such circumstances, the equilibrium can be shifted to get a higher
yield of product. The chemist determines what conditions will tend to produce the highest
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CHEM 40S
Module 3, Lesson 4
yield. The conditions which produce the highest yield are called the optimum
conditions.
For example, the optimum conditions for the Haber process are:





High concentration of H2 and N2 must be maintained
removal of NH3 as it is formed
precise temperature control - high enough to maintain a reasonable rate but low
enough not to favor the reverse reaction
use of a contact catalyst to lower the required activation energy
a high pressure should be maintained through the reaction
Each condition increases the yield by shifting the equilibrium to favor the product.
LESSON 4, ASSIGNMENT 1
Answer the following questions. For each situation, predict the shift in equilibrium using
Le Chatelier’s Principle and explain the shift using collision - rate theory.
1. When a few drops of a 0.20 mol/L Fe(NO3)3(aq) solution are added to 50 mL of a 1.0
mol/L KSCN(aq) solution, the following equilibrium is established:
Fe3+(aq) +
pale yellow
SCN(aq)  FeSCN2+(aq)
colorless
red
a. A crystal of KSCN(s) is added to the equilibrium mixture. Hint: KSCN is an
ionic compound – when added to the solution, it will dissociate into its
component ions.
b. A few drops of dilute NaOH(aq) are added to the equilibrium mixture. NaOH is
also an ionic compound which will dissociate into its ions when added.
2. For each of the following systems at equilibrium, predict how they will shift as
temperature is increased:
a. N2(g) + O2(g) + energy  2NO (g)
b. 2SO2(g) + O2(g)  2SO3(g) (forward reaction is exothermic)
3. For the exothermic reaction
2SO2(g) + O2(g)  2SO3(g)
predict the equilibrium shift caused by each of the following changes:
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RETSD Adult Ed Program
CHEM 40S
a.
b.
c.
d.
Module 3, Lesson 4
SO2 is added
SO3 is removed
The volume is decreased
The temperature is decreased
4. Complete the following table:
Reaction: Energy + N2O4(g)  2NO2(g)
Change
Addition of N2O4(g)
Addition of NO2(g)
Removal of N2O4(g)
Removal of NO2(g)
Decrease in container volume
Increase in container volume
Increase in temperature
Decrease in temperature
Shift (left/right)
Reading Assignment
Chemistry: The Study of Matter
Sections 18-4 to 18-8: Pages 518 to 532
Problem Assignment 2
Chemistry: The Study of Matter
Practice Problems No. 9 to 11: Page 523
Practice Problems No. 14 & 15: Page 529
End of Module 3 – Read over and do all questions a second time in order to be well
prepared for the test. Following is a list of important terms you should know:
Reaction rate
collision theory
activated complex
activation energy
REACTION MECHANISM
rate determining step
Concentration
rate law expression
specific rate constant
heterogeneous reactions
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CHEM 40S
Module 3, Lesson 4
activated complex
catalyst
homogeneous catalyst
coefficient
equilibrium
solubility
solvent
solute
saturated
spontaneous reaction
nonspontaneous reaction
CHEMICAL EQUILIBRIUM
equilibrium constant
Keq
law of mass action
incomplete reaction
complete reaction
homogeneous equilibria
heterogeneous equilibria
Le Chatelier’s Principle
the Haber process
factors necessary for equilibrium to exist
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