Stresses Applied to Equilibrium Systems
In this experiment you investigated Le Ch‚telier's
Principle, as applied to six equilibrium systems. Le Ch‚telier's principle states that, when a change is made to a
system that is already in equilibrium, the system reacts in such a way as to
return to equilibrium. For a
chemical reaction system, when the system "reacts so as to return to
equilibrium", this means that the system will either react in the forward direction
(so as to produce additional product) or in the reverse direction (so as to consume some
of the product).
Le Le Ch‚telier's principle is very important in the chemical and chemical engineering
industries. The goal in these industries is to get as much product as possible (which can
be then sold) for the given amounts of starting materials. Chemists and chemical engineers
designing new processes make great use of Le Le Ch‚telier's Principle in designing their
Results, Explanations, and Discussion
Since this was a "qualitative" experiment, there are obviously no
calculations to go through for your report..
Instead, let's try to understand each step of the procedure (what you did
and observed in lab) and try to relate this to Le Ch‚telier's Principle. Sometimes, given a long
procedure, it has been hard for students to appreciate exactly where Le Le Ch‚telier's
Principle enters into the discussion in some of the reactions. You should have ready both
your data pages (Pages 157-161) and also the procedure pages (Pages 149-155) for this.
Be careful! In your report, you are asked to tell what the stress
was that was applied to the equilibrium reaction. The stress is not
the chemical you added that caused a change. The stress can only
be a change in the concentration of one of the species that is written in the
equilibrium reaction equation. So, for example, if you add HCl to the system,
and it causes a change, the stress is not "adding HCl": the
stress must be interpreted in terms of the component species of the
A. The Chromate-Dichromate Equilibrium
Pages 149-150 and Page 157.
The equilibrium between chromate ion (CrO42-) and dichromate ion
(Cr2O72-) was studied. The indication for which way the
reaction shifts when changes were made in the system is based on the color of these
two ions in solution: chromate ion is bright yellow in solution, whereas dichromate ion is
bright orange. So, if the solution turned orange when you added something to it, the
equilibrium was shifted to the right; if the solution turned yellow, the equilibrium was
shifted to the left.
CrO42- (yellow) + 2H+ = Cr2O72-
(orange) + H2O
- In Step A1 on Page 150, you added sulfuric acid (H2SO4) to a
sample of potassium chromate, K2CrO4. What happened to the color of
the K2CrO4 when you did this? (see your observation on Page 157).
Based on the color change you saw, which way did the equilibrium shift, right or left?
Sulfuric acid is a source of H+: any source of H+ would have
caused the same shift in this equilibrium. The "stress" applied to
the system is not the addition of sulfuric acid! The
"stress" must be explained only in terms of the species
present in the equilibrium reaction equation: sulfuric acid is a source of hydrogen
ions and it is an increase in the hydrogen ion concentration that
causes the equilibrium to shift.
- This is an example of where students have trouble interpreting what they have done, and
how the equilibrium is affected by what has been done to the system. First of all, NaOH is
clearly not one of the species involved in the equilibrium written above. Yet, when
you added a few drops of NaOH to the mixture from Step A1, you should have seen a change.
Although NaOH is not part of the equilibrium above itself, NaOH reacts with one of the
components of the equilibrium, effectively removing that species from the equilibrium.
NaOH is a strong base and reacts with acids: NaOH reacts with the hydrogen
ion in the equilibrium system and removes it. If the H+ is removed from the
equilibrium system, the system will have to react in the direction that replaces some of
the H+ to restore equilibrium. Is this consistent with the color change you
H+ (from the equilibrium) + OH- (from the NaOH) H2O
The change in the system occurred when you added the NaOH, but
the "stress" to the system must be described in terms of the equation
for the equilibrium
CrO42- (yellow) + 2H+ = Cr2O72-
(orange) + H2O
So, since NaOH effectively removes hydrogen ion from the
system by converting it to water, the "stress" from the
equilibrium's point of view would be a decrease in the hydrogen ion
B. The Iron(III) - Thiocyanate Equilibrium
Pages 150-151 and Page 158
As you saw in the previous experiment, Fe3+ ion and SCN- react
with each other to form a red complex ion, [FeSCN]2+. In the previous
experiment, you used very dilute (0.00200 M) solutions of Fe3+ and SCN-
and so the color of the product was a faint reddish-orange. In this experiment, you use
much more concentrated solutions (0.1 M), and so the product solution is a deep red
(almost the color of blood).
Fe3+(colorless) + SCN- (colorless) = [FeSCN]2+
In this system, we can tell in which direction the equilibrium shifts when a change is
made by monitoring the intensity of color of the system. If the system gets darker
red in color, then the equilibrium must be shifting to the right (toward producing more of
the colored product). If the color of the system gets fainter (or disappears altogether),
the equilibrium must be shifting toward the left (toward the colorless components).
You had to be very observant on this part to see the change in
intensity of the red color when a drop of reagent was added: if you waited too
long, the localized change in color as the droplet enters the equilibrium
reaction would fade out.
- This step of the procedure is just setting up the equilibrium system: you mixed
some Fe(NO3)3 and some KSCN to generate the system in the equation
above. Ionic reactions are very fast, and the system came to equilibrium within a few
seconds, which was indicated by the appearance of the deep red color. Since this color is
so deep, you were told to divide the mixture into several wells, and to add water to it
until it had been diluted enough that you would be able to see color changes in the
- In this step, you added additional Fe(NO3)3 dropwise to one of the
wells containing the colored equilibrium mixture. If additional Fe3+ was added
to the system which was already in equilibrium, this would have been too much
Fe3+ present in the system. The system would have to react (either to the left
or to the right) to remove some of the additional Fe3+ from the mixture. In
which direction would the system react to remove Fe3+? If the system reacted to
remove the excess Fe3+, would the net result be more or less of the colored
product? Based on your observation (Page 158) as to whether the mixture turned darker or
lighter when the additional Fe3+ was added, which way did the equilibrium
shift? Was your color observation consistent with the idea that the system would have to
remove some of the additional Fe3+? You should realize that the change in the
system would have been taken with the first drop of additional Fe3+
added: if you were not observant, you might have missed the color change!
- In this step, you added additional KSCN dropwise to one of the wells containing the
colored equilibrium mixture. If additional SCN- were added to the system which
was already in equilibrium, this would have been too much SCN-
present in the system. The system would have to react to remove some of the additional SCN-
from the system. Based on this discussion, and your observations of the system from Page
158, which way did the equilibrium shift: toward the right (more product, more color) or
to the left (less product, less color).
- Here's a situation which is similar to that raised in section A2 above: NaOH is not
part of the equilibrium reaction given. However, the procedure (Page 151) has a special
note, reminding you that Fe(OH)3 is insoluble in water. If Fe(OH)3
is insoluble in water, then adding NaOH to the equilibrium system should cause the
reaction below to take place: adding NaOH precipitates Fe3+ as Fe(OH)3
from the equilibrium system. If Fe3+ is removed from the equilibrium
system, then the system will no longer be in equilibrium and will have to react in the
direction that restores some Fe3+. Based on your observations (Page 158) and
this discussion, which way would the equilibrium shift if Fe3+ is removed from
the system? Would the amount of colored complex increase or decrease?
Fe3+ (from the equilibrium) + 3OH- (from the NaOH)
Fe(OH)3 (solid precipitate)
C. The Cobalt(II)-Chloride Ion Equilibrium
We omitted this part of the experiment!
D. The Equilibrium in a Saturated NaCl Solution
Page 152 and Page 159
In Parts A and B above, we considered equilibrium systems that were based on true chemical
reactions. In this Part, we consider a physical equilibrium based on
When a crystalline ionic solid is dissolved in water, particles of solid enter the
solvent to form the solution. But this only happens up to a point. As the concentration of
ions in solution increases, the likelihood that ions will be attracted by the remaining
ions in the crystal increases, and ions which were dissolved begin to re-enter the
crystals of solid. Eventually, the rate at which ions re-enter the crystals becomes equal
to the rate at which ions leave the crystals, and a "steady state" is
established. Once this steady state is reached, the net concentrations of the ions in
solution no longer increases and the solution is said to be "saturated". Since
we have two opposing processes going on at the same rate, an equilibrium exists between
undissolved solute and the saturated solution.
Na+ (in solution) + Cl- (in solution) = NaCl
- This is just setting up the 24-well plate for the tests that follow. Perhaps you noticed
that there were some crystals at the bottom of the bottle of saturated NaCl solution as
you took your samples. These were to ensure that the solution would be saturated with NaCl
when you took some. Saturated NaCl contains Na+ and Cl- ions each at
a concentration of 5.4 M (moles/L).
- In this step, you added a few drops of concentrated hydrochloric acid (HCl) to one of
your saturated NaCl samples. Concentrated HCl contains H+ and Cl-
ions each at a concentration of 12.0 moles/L. In other words, when you added concentrated
HCl to the saturated NaCl solution you were adding a large amount of chloride ion, Cl-,
to the solution. The solubility equilibrium should have responded by shifting in the
direction which would remove some of the excess chloride ion from solution. What
did you observe (Page 159) in your solution that indicated that excess chloride ion was
being removed from the solution?
- In this part of the experiment, you added 3 M HCl to a second sample of saturated
NaCl. Notice that 3 M HCl contains Cl- ion at a lower
concentration than in the saturated NaCl solution (5.4 M) you added it to. When you added 3 M
HCl to the saturated NaCl solution, you probably did not observe any change (Page
159). Adding the 3 M HCl to the saturated sodium chloride solution
actually lowers the concentration of chloride ion in the solution.
E. The Equilibrium of Saturated Barium Chromate
This is another example of a solubility equilibrium, with a little "twist"
coming up in Step 2.
Ba2+ (in solution) + CrO42- (in
solution; yellow) = BaCrO4 (yellow solid)
- In this step, you just generate the equilibrium by mixing BaCl2 and K2CrO4
solutions. A yellow solid should have precipitated, with a clear yellow solution above it.
The clear yellow solution is a saturated BaCrO4 solution (see your observations
on Page 159).
- In this step, you added a few drops of concentrated hydrochloric acid, HCl, to the
mixture prepared in Step 1. What happened to the precipitate of BaCrO4 (see
your observations on Page 159) produced in Step 1 when you did this? On Page 159, you are
asked a question about why HCl affects the BaCrO4 equilibrium, even though HCl
does not appear as part of the equilibrium reaction above for BaCrO4. Look back
at Part A of this experiment and see if you can figure why an acid like HCl affects
the BaCrO4 equilibrium. What does addition of an acid do to CrO42-
ion in solution, and how would this affect the BaCrO4 equilibrium?
F. Equilibrium of Saturated Ferric Hydroxide
In this part you study really two equilibria involving Fe3+ ion and
hydroxide ion, OH-.
Fe3+ (in solution) + 3OH- (in solution) = Fe(OH)3
Fe3+ (in solution) + 4OH- (in solution) = Fe(OH)4-
which can also be written as
Fe(OH)3 (reddish solid) + OH- (in solution) =
Fe(OH)4- (in solution)
- In this step, you are merely generating the first equilibrium written above. When you
add sodium hydroxide solution to the iron(III) nitrate solution, a reddish precipitate of
Fe(OH)3 forms (see your observations on Page 160).
- When you add HCl (a strong acid) to the mixture from Step 1, the acid reacts with
hydroxide ion according to the reaction
H+ + OH- H2O.
- As the introduction to Part F states, an acid effectively removes OH- from
the equilibrium system. What happened to the precipitate from Step 1 when you added the
HCl (see your observations on Page 160) and how is this consistent with the equilibrium?
3. In this part you add additional NaOH to the second well containing Fe(OH)3
you generated in Step 1. Adding additional NaOH brings into play the second equilibrium.
How are your observations about what happened to the precipitate of Fe(OH)3
when additional NaOH added consistent with the second (and third) reactions above?
G. Equilibrium of Ammonia in Water
We omitted this part of the experiment!
H. Equiliibrium of Saturated AgCl
In this Part, you study two related equilibria involving silver chloride. You'll have
to pay close attention in the following discussion to keep straight which equilibrium
we're talking about, since the procedure is quite involved.
When silver ion (Ag+) and chloride ion (Cl-) are combined, a
precipitate of silver chloride forms, leaving a saturated solution of AgCl above the
solid. The silver ions and chloride ions remaining in solution are in equilibrium with the
AgCl ( solid) = Ag+ (in solution) + Cl- (in
Note that this solubility equilibrium has been written in the opposite sense to the
solubility equilibria in Parts D and E above: this doesn't really matter, since an
equilibrium reaction "goes in both directions". Just keep straight that your
explanations should be in terms of the way the reaction equation is given in this Part.
Silver ion also forms an equilibrium with ammonia
Ag+ (in solution) + 2NH3(in solution) = Ag(NH3)2+
- In this step, you are just generating the first equilibrium above (the solubility
equilibrium for AgCl). When you added HCl (a concentrated Cl- ion solution) to
AgNO3, a white precipitate of AgCl formed. The solution above the precipitate
was a saturated solution of Ag+ and Cl- ions.
- All you are doing in this step is removing most of the saturated solution from the
precipitate in Step 1. Realize though, that the precipitate in Step 1 is still wet with
saturated AgCl solution after you remove the bulk of the liquid from it. The solution you
remove is a saturated AgCl solution. The mixture left in the well from Step 1 is solid
AgCl still in equilibrium with the remaining coating of saturated AgCl solution.
- In this step, you play with both of the equilibria above. When you add additional
AgNO3 solution to the liquid removed in Step 2, this is an increase in the
concentration of Ag+ ion. What direction would the first equilibrium above
shift if additional Ag+ ion were added? How is this reflected in your
observation on Page 161? What is the precipitate that formed when you added AgNO3
to the liquid from Step 2? Then, you added concentrated ammonia to the mixture. What
happened to the precipitate that had just formed? This is the second equilibrium above:
when ammonia is added to a solution containing Ag+ ion, the ammonia forms a
complex with the Ag+ ion and removes it from solution. What happens to
the precipitate if Ag+ ion is removed from solution by ammonia?
- When you add ammonia to the mixture of solid AgCl and saturated AgCl solution, what
happens to the precipitate? As we saw in Step 3 above, ammonia removes Ag+
ion. What would have to happen to the precipitate if the Ag+ ion were removed
from the solution coating it? Then you added some nitric acid (HNO3). Nitric
acid reacts with ammonia (ammonia is a base): HNO3 + NH3 NH4NO3. The ammonia is thus
removed from the system. If ammonia is removed from the second equilibrium above, how does
that equilibrium shift? If the second equilibrium above shifts, how does that
affect the first equilibrium? This is getting really tricky, huh!!