Unfortunately, Wonko's answers to parts C and F of your question are
incorrect, and his answers to parts A and E are only correct under
special circumstances.
This is actually a very tricky problem because of the nature of the
reaction chosen by the authors of the question. If this question is
for an introductory chemistry class, it was a very poor choice of a
reaction to use. On the other hand, if this is for a more advanced
chemistry class, then it's a very good question because of the
subtleties it introduces.
Le Chatelier's principle basically says that if one perturbs a system
at equilibrium, the system readjusts to counteract or reduce that
perturbation. If one cools down a system, the system will (if it can)
try to heat back up. If one heats a system, it will try to cool down.
If one dilutes the concentration of one component in the system, the
system will adjust to increase that concentration. etc.
In the case at hand, we have a system consisting of solid ferric iron
hydroxide in equilibrium with an aqueous solution containing cyanide
(CN-) ions, and we are asked to predict how the system will respond to
various perturbations. Because the system is initially at equilibrium
at a fixed temperature, each of the dissolved species (CN-,
Fe(CN)6^3-, and OH-) have specific concentrations that are related by
the equilibrium constant for this reaction:
K_eq = [OH-]^3 * [Fe(CN)6^3-] / [CN-]^6
where the brackets indicate concentrations (really chemical
activities, but that is something we'll ignore here). If we change
the concentration of any or all of these aqueous species from their
initial value(s) at equilibrium (keeping the temperature constant),
the right hand side of the above equation (which is usually referred
to as the "ion activity product" or IAP) will no longer be equal to
the value of the equilibrium constant, and the system will readjust
the concentrations so that the IAP is once again equal to the
equilibrium constant. Any change in concentrations that makes the IAP
larger than the equilibrium constant (i.e., increases the
concentration of products) will cause the reaction to run to the left.
Similarly, any change that makes the IAP less than the equilibrium
constant will cause the reaction to run to the right.
Let's look at each of the perturbations in turn.
----------------------------------------------------
a. Add KCN (aq)
Wonko's answer is the simple answer, and the one that your text or
teacher is probably looking for. Adding KCN will cause the reaction
to run to the right (R), but this is only always true if the KCN is
added as a solid! In this case, the only effect on the equilibrium
is to increase the concentration of CN- ions. The system will respond
by trying to counteract that change, and the reaction will run to the
right, consuming both CN- and Fe(OH)3. However, the question states
that the KCN is added in the form of an aqueous solution. In this
case, the effect depends on both the concentration of KCN in the
solution and the amount of solution added, so the question cannot be
answered with the information given! To see why this is so, let's
calculate the IAP for the perturbed solution and compare it to the
equilibrium constant. For simplicity, we'll assume that there is 1
liter of the initial solution and we'll add X liters of aqueous KCN
solution that has a concentration of equal to [KCN]* (= [CN-]* = [K+]*
because KCN dissociates completely). We'll also assume that this
solution is at room temperature and has a neutral pH and so it has a
concentration of hydroxide ions [OH-] = 10^-7 Molar.
The IAP for the system after we've added the KCN solution and before
it has had a chance to reequilibrate is given by:
IAP = {(([OH-] + 10^-7 * X)/(1+X))^3 * [Fe(CN)6^3-]/(1+X)}/{([CN-]
+ X*[CN-]*)/(1+X)}^6
where the "unprimed" concentrations in brackets are the concentrations
of the species in the initial, unperturbed system.
If we divide the IAP by the equilibrium constant, we can see if the
result is great than 1 (meaning the IAP is larger than K and the
reaction will run to the left), less than 1 (meaning the IAP is less
than K and the reaction will run to the right) or equal to 1 (which
means that the IAP didn't change, and there will be no effect).
IAP/K_eq= {{(([OH-] + 10^-7 * X)/(1+X))^3 *
[Fe(CN)6^3-]/(1+X)}/{([CN-] + X*[CN-]*)/(1+X)}^6}/{[OH-}^3 *
[Fe(CN)6^3-]/[CN-}^6}
Simplifying this expression yields:
IAP/K_eq = (1+x)^2 * {([OH-] + 10^-7*X)/[OH-]}^3 * {[CN-]/([CN-] + X*[CN-]*)}^6
This is a nasty-looking expression, but, in general, it will not be
equal to one. The first two factors will always be greater than one,
and the third factor will always be less than one. The net effect
will depend on which factor dominates. If the added solution is very
dilute in KCN (i.e., [CN-]* << [CN-]) then the last factor is almost
equal to one, and the first two factors are both greater than one, so
the IAP > K_eq, and the reaction will run to the left. If, on the
other hand, the added solution is very concentrated, and you don't add
much of it (i.e., X*[CN-]* >> [CN-]) then the first two factors will
be almost equal to one, but the last factor will be << 1, so the IAP
is less than K_eq, and the reaction will run to the left.
----------------------------------------------------
b. add more Fe(OH)3
Wonko's answer is correct. Adding more solid to the system when the
system contains a solid at equilibrium with the solution will have no
effect on the equilibrium.
----------------------------------------------------
c. add liquid water
Wonko's answer is incorrect. Oddly enough, adding water to this
system will cause the reaction to run to the left, inducing
precipitation of the solid! The reason for this is that the solid in
this case is what is called a "hydrous phase" - it can react to
release water. (Consider the reaction that forms Fe(OH)3 from simple
ferric iron oxide: Fe2O3 + 3H20 = 2Fe(OH)3). If we add water to the
sytem, we will actually drive the reaction towards the side that
includes the hydrous phase. In addition, any aqueous reaction that
includes an H+ or OH- ion on one side of the reaction can be rewritten
so that it explicitly involves water because one can add or subtract
the dissociation reaction for water (H2O = H+ + OH-) from the original
reaction to reexpress the reaction in terms of OH- and H2O (if the
original reaction involved just H+) and vice versa. Let's do that for
this reaction because we're going to need that rewritten reaction to
answer part F of the question anyway.
Add together the reactions:
Fe(OH)3 + 6CN- <--> Fe(CN)6^3- + 3OH-
3H+ + 3OH- <--> 3H20
and we get:
Fe(OH)3 + 6CN- (aq) + 3H+ + 3OH- <--> Fe(CN)6^3- + 3OH- + 3H2O
cancelling the 3OH-'s on each side gives:
Fe(OH)3 + 6CN- (aq) + 3H+ <--> Fe(CN)6^3- + 3H2O
Adding water to the system will drive the reaction to the left. In
essence, what is happening in this reaction is that the Fe(OH)3 solid
is dissolving, releasing Fe3+ ions and OH- ions to solution. The Fe3+
ions react with the CN- ions to form dissolved ferric cyanide ions,
and the ON- ions that are released will react with any available H+
ions to form water! If we add more water, we drive the reaction
backwards.
One can also see this from consideration of the effect on the IAP. As
an exercise, try writing down the expression for the IAP of the
original reaction (which does not explicitly include H2O as a reactant
or product) if you add X liters of pure water water with [OH-] = 10^-7
M to an initial system containing 1 liter of solution. You should
find that the ratio of the IAP to K_eq is equal to:
(1+X)^2 * (1 + (10^-7*X)/[OH-])^3
This quantity is always > 1, so the reaction will run to the left.
----------------------------------------------------
d. increase the temperature
Wonko's answer is correct. Increasing the temperature will drive an
exothermic reaction to the left.
----------------------------------------------------
e. add KOH (aq)
The full answer to this part of the question parallels that for part
A. If the KOH is added as a solid, then the answer is simple and the
reaction runs to the left. If the KOH is added as an aqueous
solution, then one also has to consider the effects of dilution, but
in this case, it turns out that the effect will *always* be to drive
the reaction to the left because both dilution of the original
solution and addition of OH drive the reaction to the left. (In part
A, addition of CN- to the solution tends to drive the reaction to the
right, but dilution drives the reaction to the left -- see answer to
part C --, so the effects compete with one another.)
----------------------------------------------------
f. add HCL (aq)
Wonko's answer is incorrect, even for the simplest case in which one
bubbles HCl gas through the original solution. If the HCl is added as
a dilute aqueous solution we also get the complications due to
dilution that we've encountered in parts A, C, and E.
To see why Wonko is wrong at the simplest level, consider the way we
rewrote the reaction in part C (by adding the dissociation constant
for water to the original reaction):
Fe(OH)3 + 6CN- (aq) + 3H+ <--> Fe(CN)6^3- + 3H2O
If we add concentrated HCl (i.e., hydrochloric acid, which completely
dissociates to H+ and Cl- ions) to the system (or bubble HCl gas
through the solution) we will not dilute the system, but we will
increase the concentration of H+ ions. This will drive the reaction
to the right. If the HCl solution is very dilute, however, the
dilution effect could dominate, and the reaction could be forced to
run to the left. Consider the effect on the IAP as done in my answer
to part A. As written (the HCl is added as an aqueous solution), you
cannot answer the question because insufficient information is given;
you need to know how much of the HCl solution is added, and what its
concentration is.
----------------------------------------------------
g. bubble HCN gas through the solution
Wonko's answer is correct. Bubling hydrogen cyanide gas through the
system will increase the concentrations of both H+ and CN- ions in the
system, but not have the complicating effects due to dilution. As we
have seen above, increasing either the H+ or CN- ion concentrations
will drive the reaction to the right. |
