LAW OF CHEMICAL EQUILIBRIUM:
LE CHATELIER'S PRINCIPLE
In this experiment, you will study chemical equilibria and investigate the effects
that changes in concentrations and changes in temperature have on reactions that are
at equilibrium.
Background:
Two important questions are asked about every chemical reaction: (a) How
much product is produced and (b) how fast is it produced? The first question involves
chemical equilibrium and the second question belongs to the domain of chemical
kinetics. Some reactions are irreversible and they go to completion (100% yield). An
example would be when you ignite methane gas (the major component in natural gas)
in the presence of air. The methane will burn completely and irreversibly with the
oxygen in the air and form carbon dioxide and water:
CH4 + 2O2  CO2 + 2H2O
Other reactions do not go to completion and are therefor reversible. In such
cases the reaction can go in either the forward direction or the reverse direction. For
example, the following reaction between iron (III) and thiocyanate (which takes place in
an aqueous solution):
Fe+3+ SCN-  FeSCN+2
is often used to illustrate reversible reactions.
When one mixes a solution containing the yellow iron (III) with another solution
containing the colorless thiocyanate (where there is an equal number of moles of both
ions), a reaction forming the deep red FeSCN+2 ion occurs. This is the forward
reaction. At the beginning when we mix iron (III) and thiocyanate ion, the rate of the
forward reaction is at a maximum. As time goes on, this rate decreases because we
have less and less iron (III) and thiocyanate to react.
Analysis shows, however, that there is still iron (III) and thiocyanate ions present.
This is because at the same time the forward reaction occurs, some of the red FeSCN+2
ion formed also decomposes and forms the yellow iron (III) ion and thiocyanate ion.
This is the backward (reverse) reaction. At the beginning when we mix iron (III) and
thiocyanate ion, the rate of the reverse reaction (which began at zero) gradually
increases. Eventually the forward and reverse reaction rates become equal. When this
point is reached, we call the process a dynamic equilibrium or just an equilibrium.
When in equilibrium at a particular temperature, a reaction mixture obeys the Law of
Chemical Equilibrium. This Law is a result of a condition on the concentration of
reactants and products expressed in the equilibrium constant, K. For the above
reaction between iron (III) and thiocyanate ions, the equilibrium constant can be written
in the form of an equilibrium expression:
-
K = [FeSCN+2] / [Fe+3][SCN ]
or in general:
K = [products] / [reactants]
The brackets, [ ], indicate concentration in moles/L at equilibrium. As the name implies
the equilibrium constant is a constant at a set temperature for a particular reaction.
Its magnitude tells if a reaction goes to completion or if it is far from completion
(reversible reaction). A number much smaller than 1 for K usually indicates that at
equilibrium only a small amount of product(s) are formed and the mixture consists
mainly of reactants. We say that the equilibrium lies far to the left. A number much
larger than 1 for K usually indicates that at equilibrium a large amount of product(s) are
formed and the mixture consists mainly of products. We say that the equilibrium lies far
to the right. With an irreversible reaction (such as the combustion of methane), the
equilibrium constant would be infinite. The reaction between iron (III) and thiocyanate
has an equilibrium constant of about 200 at room temperature, indicating that the
equilibrium lies to the right but does not go to completion. Thus at equilibrium, both
reactants and product are present, although the amount of product molecules far outnumbers the reactant molecules.
The Law of Chemical Equilibrium is based on the constancy of the equilibrium
constant. This means that if one disturbs the equilibrium, for example, by adding more
reactant molecules, there will be an increase in the number of product molecules in
order to maintain the product/reactant ratio thus preserving the numerical value of the
equilibrium constant. Le Chatelier's Principle expresses this as follows: If an external
stress is applied to a system in equilibrium, the system reacts in such a way as to
partially relieve the stress. In our present experiment we demonstrate Le Chatelier's
Principle in two manners: (a) disturbing the equilibrium by changing the concentration of
a product or reactant and (b) changing the temperature .
CHANGES IN CONCENTRATION:
A
In the first experiment we add ammonia to a pale blue copper (II) sulfate solution.
The ionic reaction is:
Cu+2 + 4NH3  Cu(NH3)4+2
A change in the color indicates the formation of the copper-ammonia complex ion.
Adding a strong acid such as HCl (which provides H3O+ ions) to this equilibrium causes
the ammonia to react with the acid:
NH3 + H3O+  NH4+ + H2O
Thus we removed some reactant ammonia molecules from the equilibrium mixture in
the first reaction equation. As a result we expect the equilibrium to shift to the left
reforming hydrated copper(II) ions and that there should be a color change.
B
In the second reaction we demonstrate the common ion effect. As an example,
when we have a mixture of H2PO4- with HPO4-2 in solution, the following equilibrium
exists:
H2PO4- + H2O  H3O+ + HPO4-2
If we add a few drops of aqueous HCl (a strong acid) to the solution, we've added an
ion, H3O+, that is common to the equilibrium above. We expect, on the basis of the Le
Chatelier Principle that the equilibrium will shift to the left. Thus, the solution will not
become that acidic.
C
For the final demonstration of effects in changes in concentration, we will use the
iron (III) - thiocyanate reaction mentioned in the background section of this procedure
sheet. When a solution of iron (III) chloride and potassium thiocyanate are mixed, the
ionic reaction is:
Fe
+3
-
+ SCN
+2
 FeSCN
where chloride and potassium ions are spectator ions. Nevertheless, their
concentrations may also influence the equilibrium. For example, when chloride ions are
+3
in excess, the yellow color of the Fe will disappear with the formation of a colorless
FeCl4 complex ion.
Fe+3 + 4Cl-  FeCl4This also removes some of the iron (III) ions from the iron (III) - thiocyanate reaction and
therefor causes a shift in that equilibrium and color change.
CHANGES IN TEMPERATURE:
Most reactions are accompanied by some energy changes. Frequently the
energy is in the form of heat. We talk of endothermic reactions if heat is consumed
during the reaction and therefor we can consider heat as one of the reactants.
Conversely, in an exothermic reaction heat is evolved, and we can consider heat as
one of the products. Therefore, if we heat an equilibrium mixture of an endothermic
reaction, it will behave as if we added one of its reactants (heat) and the equilibrium will
shift to the right. Heating the equilibrium mixture of an exothermic reaction results in
the equilibrium shifting to the left since we are now adding one of the products. We will
demonstrate the effect of temperature on the reaction between cobalt ion and chloride
ion forming the cobalt-chloride complex ion:
Co+2 + 4Cl-  CoCl4-2
You will observe a change in the color depending if the equilibrium was established at
room temperature or at 100°C in boiling water. From the color change you should be
able to tell if the reaction was endothermic or exothermic.
EXPERIMENTAL PROCEDURES:
Some general instructions for this lab:
1) When using pipets (or medicine droppers), remember that the bulb of the
pipet must always be at a higher elevation relative to the tip of the pipet.
2) As an approximation, 1 mL is equal to about 20 drops from a pipet.
CHANGES IN CONCENTRATION:
Your test tubes for parts A-C must be clean, but they need not be dry. After
washing test tubes, just vigorously shake out any excess water.
A) Place 20 drops of 0.1 M CuSO4 solution into a 20 x 150 mm or 16 x 150 mm test
tube. Keeping track of the number of drops, add 1 M NH3 solution (also labeled as
NH4OH and called ammonium hydroxide) mixing the contents after each drop. Initially,
the solution will turn milky, but continue to add the ammonia solution until the milky
precipitate dissolves and the color changes. Note the new color and the number of
drops needed. To this new equilibrium mixture add dropwise (keeping track of the
number of drops) 1 M HCl solution until the color changes back to pale blue and the
solution is clear. Dispose these solutions in the correctly labeled copper waste
container in the hood.
B) Place 2 mL of H2PO4- / HPO4-2 (also called phosphate buffer) solution into a 13 x 100
mm test tube. Using a pH paper, test the pH of this solution (to within 0.5 pH units) and
determine if it is acidic or basic. To do this, dip a clean glass stirring rod into the
solution and then touch it to a small strip of pH paper that is about 2 - 3 cm long. The
color that the spot becomes can be correlated to a color chart on the pH paper
dispenser and thus, the pH determined. An acidic solution has a pH less than 7, and a
basic solution has a pH greater than 7. An increase in pH means a decrease in [H 3O+]
and a decrease in pH means an increase in [H3O+]. In a similar manner, using a new
strip of pH paper, determine the pH (to within 0.5 pH units) of 2 mL of distilled water.
Do not touch your glass stirring rod with the distilled water to the same spot that you
had touched with the H2PO4- / HPO4-2 solution. Add a drop of 1 M HCl to both the
H2PO4- / HPO4-2 solution and to the distilled water, mix, and then determine the pH of
each solution to within 0.5 pH units. Dispose these solutions down the drain.
C) Prepare a stock solution by adding 0.25 mL of 0.1 M iron (III) nitrate, Fe(NO3)3, and
0.25 mL 0.1 M potassium thiocyanate (pronounced thigh-o-sigh-a-nate), KSCN, to 12
mL of distilled water in a beaker. Set up four 20 x 150 mm or 16 x 150 mm test tubes
and label them. To each test tube add about 2 mL of the stock equilibrium mixture you
just prepared. Use the first test tube as a standard to which you can compare the color
of the other solutions. Add 10 drops of distilled water to the first test tube and mix
thoroughly. To the second test tube add 10 drops of 0.1 M iron (III) chloride solution
and mix thoroughly, to the third add 10 drops of 0.1 M NaSCN solution and mix
thoroughly and to the fourth add 10 drops of saturated NaCl solution and mix
thoroughly. Observe the color in each test tube and record your observations.
Dispose these solutions in the correctly labeled iron waste container in the hood.
CHANGES IN TEMPERATURE:
Note: It is important that the test tubes used in this part of the experiment be dry.
Select three 25 x 200 mm test tubes and perform a visual inspection on each for
cleanliness. If the test tube is clean, use it without washing. If it is not clean,
select another test tube.
To a dry test tube, place 5 drops of 1 M Co(NO3)2. Label it #1. Note the color of this
solution.
To a second dry test tube (labeled #2); place 5 drops of 1 M Co(NO 3)2. Add
concentrated HCl, 5 drops at a time with mixing, noting the color with each 5-drop
increment. Do not add more than 25 drops. CAUTION: Do not allow skin contact and
do not inhale the HCl vapors. Compare the color of this new solution to test tube #1.
To a third dry test tube (labeled #3); place 5 drops of 1 M CoCl2. Immerse the test tube
into a boiling water bath for about 1 minute. Compare the color of the solution in test
tube #3 (while in the boiling water) to test tube #1. There should be a slight color
change in test tube #3 that may be more obvious with residual solution that coats the
side of the test tube. Dispose these solutions in the correctly labeled cobalt waste
container in the hood.
REPORT SHEET: LE CHATELIER'S PRINCIPLE
Name:
Lab Partner:
Section:
CHANGES IN CONCENTRATION:
A:
1) What is the color of the copper sulfate solution
a) before the addition of ammonia solution?
b) after the addition of ammonia solution?
2) What is the color of copper (II) ions?
3) What is the color of the copper-ammonia complex ion?
4) How many drops of 1 M ammonia did you add to cause a color change?
5) How many drops of 1 M HCl did you add to cause a color change?
6) How many drops of 1 M HCl did you expect to cause a color change? Why?
B:
1) What is the pH of the phosphate buffer
a) before adding a drop of 1 M HCl?
b) after adding a drop of 1 M HCl?
2) What is the pH of distilled water
a) before adding a drop of 1 M HCl?
b) after adding a drop of 1 M HCl?
3) Using Le Chatelier's Principle, explain why the phosphate buffer should have gone
through a smaller pH change compared to distilled water. Be specific, using
appropriate chemical reaction equations in your explanation.
C:
1) Compare the colors of test tubes #2, #3, and #4 with test tube #1:
#1
#2
#3
#4
2) For test tubes #2, #3, and #4; in which direction did the equilibrium shift for each test
tube? Explain why it shifted in that direction for each test tube, using Le Chatelier's
Principle and appropriate chemical reaction equations in your explanation. Be specific.
CHANGES IN TEMPERATURE:
1) What is the color of the Co(NO3)2 solution
a) before the addition of concentrated HCl solution?
b) after the addition of 5 drops of concentrated HCl solution?
c) after the addition of 10 drops of concentrated HCl solution?
d) after the addition of 15 drops of concentrated HCl solution?
e) after the addition of 20 drops of concentrated HCl solution?
f) after the addition of 25 drops of concentrated HCl solution?
2) What is the color of cobalt (II) ions?
3) What is the color of the cobalt-chloride complex ion after the addition of 25 drops of
concentrated HCl solution?
4) In which direction did the equilibrium shift after you added the concentrated HCl?
Explain why it shifted in that direction, using Le Chatelier's Principle and appropriate
chemical reaction equations in your explanation. Be specific.
5) What is the color of CoCl2 solution at room temperature?
6) What is the color of CoCl2 solution at boiling water temperature?
7) In which direction did the equilibrium shift upon heating? Explain how arrived at your
answer from your observations in this part of the lab.
8) Is this reaction exothermic or endothermic? Explain how you arrived at your answer.
Be specific.
(over)
POST LAB QUESTIONS
1) Adding HCl to the H2PO4- / HPO4-2 mixture is an example of the common ion effect.
Of the three other equilibrium reactions in this experiment, identify one that involved a
shift in equilibrium that involved the common ion effect. Explain.
2) Silver ion (Ag+) forms a water insoluble precipitate with chloride (Cl-), and effectively
removes it by forming AgCl. In the following equilibrium, Co+2 + 4Cl-  CoCl4-2,
would you expect to observe a color change, and if so what type of change, if you
added a large quantity of silver ions? Explain.
3) Write the reaction equation where the numerical value of the equilibrium constant
changed. Did it get larger or smaller? Explain.
4) The manufacture of ammonia from nitrogen and hydrogen is an exothermic reaction.
Which temperature would give a greater yield of ammonia, room temperature or
100oC? Explain.
POST LAB QUESTIONS
1) Adding HCl to the H2PO4- / HPO4-2 mixture is an example of the common ion effect.
Of the three other equilibrium reactions in this experiment, identify one that involved a
shift in equilibrium that involved the common ion effect. Explain.
2) Silver ion (Ag+) forms a water insoluble precipitate with thiocyanate (SCN-), and
effectively removes it by forming AgSCN. In the following equilibrium,
+3
+2
Fe
+ SCN  FeSCN , would you expect to observe a color change, and if so
what type of change, if you added a large quantity of silver ions? Explain.
3) Write the reaction equation where the numerical value of the equilibrium constant
changed. Did it get larger or smaller? Explain.
4) The manufacture of ammonia from nitrogen and hydrogen is an exothermic reaction.
Which temperature would give a greater yield of ammonia, room temperature or
100oC? Explain.