Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2013

Lecture Notes 35: 29 April


Equilibrium Expression Problems

I want to look at solving equilibrium problems, so we'll look at a number of examples and see how to use the equilibrium expression to follow the equilibrium process.

Example: Consider the gas phase reaction at equilibrium :

PCl5 equilibrium arrow PCl3 + Cl2

K = 5.0 x 10-2 @ 150 °C

(What will happen to this reaction if the volume is increased?

Shift to right to bring the total concentration of particles back up.)

Find [Cl2] if [PCl5] = 0.40 M and [PCl3] = 0.20 M.

Know that K = [PCl3] [Cl2] / [PCl5]

Substituting, K = (0.20)[Cl2] / (.040) = 5.0 x 10-2

Rearranging, [Cl2] = (5.0 x 10-2)(0.40)/(0.20) = 1.0 x 10-1 M

[Cl2] = 0.10 M

A very common type of reaction is one in which a dissociation takes place.

Example: Consider the gas phase dissociation of carbonyl chloride to carbon monoxide and chlorine @ 100 °C.

If 0.20 moles of carbonyl chloride (COCl2) is placed in a 2.5 L container at 100 °C calculate the concentrations of all species at equilibrium. K = 2.6 x 10-10 @ 100 °C.

  COCl2  equilibrium arrow CO + Cl2
Before reaction 0.20/2.5 = 0.080   0 0
@ Equilibrium 0.080 - x   x   x

K = [CO] [Cl2] / [COCl2]

K = (x)(x)/(0.080 -x) = 2.6 x 10-10

Notice that this will give a quadratic equation - yuk! This is to be avoided if possible, if for no other reason than more steps means more opportunities to make a mistake! So we will take advantage of the fact that we are working with experimental work - there is always an error, and if x is smaller than our error, then we can simplify.

The trick is to assume x<< 0.080 and if we look at K, its value (2.6 x 10-10) is small enough that it is likely that our assumption is correct. Note that we MUST write out any assumptions so we understand WHY, in this case, x disappears from the COCl2. So:

K = x2/ 0.080 = 2.6 x 10-10

x2 = (0.080)(2.6 x 10-10) = 2.08x 10-11

and x = 4.6 x 10-6

Note that x = 4.6 x 10-6<< 0.080, so our assumption is correct (x is much less than the significant figure error)!

therefore [CO] = [Cl2] = 4.6 x 10-6 M

[COCl2] = 0.080 M

Heterogeneous Equilibrium Systems

So far our discussion has dealt only with homogeneous systems, that is all of the components are in the same phase. What about heterogeneous systems where the components occupy different phases. For example look at the gas/solid system below:

CaO(s) + CO2 (g) equilibrium arrow CaCO3 (s)

We can write the equilibrium expression for this reaction as normal:

K = [CaCO3 (s)] / [CaO(s)][CO2 (g)]

The problem is, what is the concentration of the solids? In a sense each is dissolved in itself and does not change during the reaction (the lumps of stuff can get larger or smaller, but the concentrations remain constant). It turns out, for theoretical reasons we won't go into, the activity or "behavioral concentration" in the pure state is 1. Thus we can put in the concentration of 1 for each solid:

K = [1] / [1][CO2 (g)]

K = 1/[CO2 (g)]

So the equilibrium expression depends only on the concentration of the gas phase, in this case carbon dioxide, and the amounts of solid reactants and products is inconsequential! (Note that in a salt system such as this, the salts set the vapor pressure of the carbon dioxide, giving us a way of creating systems of defined vapor pressure.)

Solution examples

Iron hydroxides in Humboldt County water. Two iron equilibria are involved, ferrous and ferric hydroxides:

Fe(OH)2(s) equilibrium arrow Fe2+ + 2OH- ; Keq = 1.8x10-15

Fe(OH)3(s) equilibrium arrow Fe3+ + 3OH- ; Keq = 4x10-38

Bacterial metabolism in the soil results in acidic (low pH) with a very low oxygen content (very low pO2). When this water reachs ferrous hydroxide in "blue clay" deposits the ferrous hydroxide dissolves and percolates through geological formations until it comes up as springs. Contact with the air, with a high oxygen tension, pO2= 0.2 atm, converts the ferrous iron to ferric iron with a much lower solubility and we see yellow-brown "rust" deposits around the springs.

Carbonates, carbon dioxide and acidity. Two equilibrium reactions have very important impacts on a variety of geological (e.g. formation of limestone caves and cave formations) and biological processes (e.g. formation of carbonate based mollusc shells):

CaCO3 (s) + H+equilibrium arrow Ca2+ + HCO3-

HCO3- + H+equilibrium arrow CO2+ H2O

Discussion of cave formation and effects of changes of atmospheric carbon dioxide/ocean acidity and shell formation.


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© R A Paselk

Last modified 26 April 2013