Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107

Fundamentals of Chemistry

Fall 2009

Lecture Notes: 1 December

© R. Paselk 2005
 
     
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The Equilibrium and Mass Action Expressions

Recall last time we introduced a qualitative view of equilibria with Le Châtelier's Principle: If stress is applied to a system at equilibrium, the equilibrium will shift in such a way as to relieve the stress. We now want to go on with a quantitative point of view.

Quantitatively we can look at the relationship in a reaction at equilibrium represented by

A + B equilibrium double arrow C + D

At equilibrium this system consists of two reactions proceeding in opposite directions at the identical rate:

A + B right arrow C + D characterized by a constant, k1 to give a rate, r1 = k1[A] [B]

A + B left arrow C + D characterized by a constant, k2 to give a rate, r2 = k2[C] [D]

But if r1 = r2, then

k1[A] [B] = k2[C] [D], and gathering constants

k1/k2 = [C] [D]/[A] [B], the ratio of constants is given a new name, the equilibrium constant, Keq.

The Equilibrium Expression is then:

Keq = [C] [D]/[A] [B].

A similar expression is the Mass Action Expression:

Q = [C] [D]/[A] [B].

The mass action expression is algebraically identical to the equilibrium expression, but it applies to a more general case. That is, the equilibrium expression requires that the values in the expression give the equilibrium constant, whereas the mass action expression allows any set of values. Thus the mass action expression is used to describe a system which has not yet reached equilibrium, while the equilibrium expression is a special case of the mass action expression for a system at equilibrium.

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Let's look at a number of examples and see how to use the equilbirium expression to follow the equilibrium process.

Example: Consider the gas phase reaction:

PCl5 equilibrium double arrow PCl3 + Cl2

Keq = 5.0 x 10-2 @ 150 °C

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

 

Example: Consider the reaction of carbon monoxide and water to give carbon dioxide and hydrogen. Calculate the concentrations of all species at equilibrium if we start with 0.341 moles each of carbon monoxide and water in a 2.71 L container @ 600 K. Keq = 302.

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. Keq = 2.6 x 10-10 @ 100 °C. (Notice the very small value of Keq. This indicates that this reaction will not progress very far towards products, rather it will stay mostly in the form of the initial reactant.)

Heterogeneous Equilibria

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 particles 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:

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

Keq = 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!

Consider the equilibrium of calcium carbonate dissociating to calcum ion and carbonate ion:

CaCO3 (s) equilibrium arrow Ca2+ + CO32-

Find the solubility if K = 8.7 x 10-9

Note the solubility will be the amount of calcium or carbonate ion, since that much calcium carbonate must have dissolved to produce them!

K = [Ca2+] [CO32-] / [CaCO3 (s)] ; but recall that for a solid, M = 1, so rewriting

K = [Ca2+] [CO32-] = 8.7 x 10-9

and [Ca2+] = [CO32-] = (8.7 x 10-9 )1/2 = 9.3 x 10-5 M = Solubility


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Last modified 1 December 2009