CHEMISTRY 1220/1250 - Lab 19: Explanation of H2O Effect on Equilibrium
CHEMISTRY 1220/1250 - Lab 19: Explanation of H2O Effect on Equilibrium
Last Modified: Sunday, 16-Dec-2018 15:15:20 EST
How does the Addition of H2O, a Pure Liquid, Shift Equilibrium?
Consider the reaction on page 59 (p. 129 for Chem 1250):
SbCl3 (aq) + H2O (l) = SbOCl (s) + 2 H+ (aq) + 2 Cl- (aq)
The equilibrium constant, K, for this reaction is:
[H+]2[Cl-]2
K = ------------
[SbCl3]
Where these are the equilibrium concentrations.
Note the omission of H2O and SbOCl. Concentrations of pure liquids
and pure solids do not appear in an equilibrium expression. This is due to the
fact that the molar concentration of a pure liquid or solid is constant.
However, if you add H2O to the reaction mixture the reaction shifts
to the right. Why does this happen if it is a pure liquid and it does not
appear in the expression for K?
This shift occurs because by adding H2O you are changing the
concentrations of the aqueous substances and because of the form of the
expression for K (the fact there is a squared term in the numerator but not in
the denominator). Let's see why by doing some substitutions into the above
expression.
Before any additional H2O is added and the reaction is at
equilibrium, we have the following:
[H+]2eq1[Cl-]2eq1
K = -----------------
[SbCl3]eq1
Now let's add enough H2O to cut the concentrations of each of these substances
in half (double the volume of solution by adding enough H2O).
[H+] = 1/2 * [H+]eq1
[Cl-] = 1/2 * [Cl-]eq1
[SbCl3] = 1/2 * [SbCl3]eq1.
This means the reaction is no longer at equilibrium and we have to calculate the
reaction quotient, Q, and compare it to K.
[H+]2[Cl-]2 {1/2*[H+]eq1}2{1/2*[Cl-]eq1}2 1/4{[H+]2eq1}*1/4{[Cl-]2eq1}
Q = ----------- = ----------------------------- = ----------------------------
[SbCl3] 1/2*[SbCl3]eq1 1/2[SbCl3]eq1
Thus,
Q = 1/8 * K.
Since Q < K the reaction is not at equilibrium and will have to shift to
reestablish a new equilibrium. When Q < K this means the numerator is too small
(denominator too large), which means there is not enough product and too
much reactant to be at equilibrium. The reaction must SHIFT to the
RIGHT to use up some reactant and make more product until equilibrium
is reestablished.
This is all due to the fact that the concentrations of the aqueous substances
changed when H2O was added and not directly due to the addition of
the H2O.
You can think about this in the same way you deal with pressure changes and
equilibrium. When you increase the pressure on a reaction the reaction shifts
in the direction of fewer moles of gas. Increasing the pressure (decreasing
the volume) increases the concentrations of the substances.
If you decrease the pressure the reaction shifts in the direction of more moles
of gas. Decreasing the pressure (increasing the volume) decreases the
concentration of the substances.
Pressure changes have no effect if there is no change in the number of moles of
gas in the reaction (the numerator and denominator in K have the same form,
i.e., overall powers).
In a reaction involving aqueous substances in solution, if you add H2O
you decrease the concentrations of everything. As long as there is a change in
moles of aqueous substances, the reaction will shift in the direction of more
moles of aqueous substances. If you remove some H2O you increase the
concentrations of everything and the reaction shifts in the direction of fewer
moles of aqueous substances. If there is no change in moles of the aqueous
substances adding H2O will have no effect (no shift occurs).
In the example above for exp. 19 there are more moles of aqueous
substances on the right hand side of the equation (product side). There are
4 moles of aqueous product compared to 1 mole of aqueous reactant. If you
decrease the concentration of these substances by adding H2O you
shift to the right, towards more moles of aqueous substances.