Gaseous Equilibria

A chemical reaction is typically drawn as starting materials followed by a right facing arrow and then products.

A B

In this type of reaction, A reacts to produce B. Such a reaction is considered to proceed from left to right.

In a chemical equilibrium, the reaction of A to produce B is reversible and is written using a double arrow. The double arrow shows that while A still reacts to produce B, B can also react to form A.

A B

If we start with pure A, then initially only the forward reaction can occur (A produces B). As B builds in concentration, the reverse reaction will also occur producing new A molecules.

As the forward reaction slows due to depletion of A, and the reverse reaction speeds up due to the formation of B, a point will come where the two rates of reaction are equal. At this point as much B is produced by the forward reaction as is destroyed by the reverse reaction and the concentrations of A and B will no longer change.

This point where the concentrations no longer change is called chemical equilibrium.

Even though there is no longer any outwardly visible change in the system, one must remember that the forward and reverse reactions continue unabated.

The Equilibrium Expression

For the reaction

A B

the concentrations of the reactants and products at chemical equilibrium can be described by the equilibrium expression.

K_eq = [B] / [A]

The equilibrium expression is the concentrations of the products (in this case B) divided by the concentrations of the reactants (in this case A).

The value K_eq is called the equilibrium constant. Whenever the reaction is at chemical equilibrium, the ratio of the concentrations as described by the equilibrium expression will be equal to K_eq.

In reactions where there are multiple reactants or products we take the mathematical product of all the product concentrations divided by the mathematical product of all of the reactant concentrations.

A + B C + D

K_eq = [C][D] / [A][B]

Finally, if the stoichiometry of the reaction is not 1 to 1, then in the equilibrium expression we raise the concentrations of the reactants and products to the power of their stoichiometric coefficients.

2A + B 3C

K_eq = [C]³ / [A]²[B]

Gaseous Equilibria - N₂O₄ 2NO₂

The compound N₂O₄ (nitrogen tetroxide) is a colourless gas. N₂O₄ undergoes a reversible reaction to produce two NO₂ molecules (nitrogen dioxide) which is a red coloured gas.

N₂O₄(g) 2NO₂(g)

The equilibrium expression for this reaction is

K_eq = [ NO₂ ]² / [N₂O₄]

To help us study the N₂O₄ - NO₂ system we will use some simple animations of a reaction vessel. The purpose of these animations is remind us of the composition of the reaction mixture and the types of reactions that are occurring. To accomplish this goal, we will employ a few simplications.

When studying the equilibrium reaction, we will be most interested in those molecules that react. Only those molecules will change the concentration of the N₂O₄ and NO₂.

If we have a vessel filled with colourless N₂O₄, rather than show the individual molecules we will shade the background a light gray to remind us of the presence of these molecules.

Similarly, if we have a vessel filled with reddish NO₂, we will shade the background a reddish colour to remind us of the presence of those molecules.

As we study cases where the N₂O₄ / NO₂ system is approaching equilibrium the colour of the background will give a good indication of the system's composition. Consider the example below where N₂O₄ is converted to NO₂. The background is initially gray. As more and more NO₂ is produced the background colour becomes more reddish.

Gaseous Equilibria - Experimental

Seen below is the stylized apparatus that we will use to conduct experiments involving N₂O₄ and NO₂. It consists of a 'reaction vessel' on the left in which we can visualize the relative amounts of each gas by the background colour as well as the reactions that are occurring. On the right we see a graph which will display and plot the amounts of each gas in moles per litre over the course of the experiment.

To keep calculations simple we will assume that the volume of the reaction vessel is exactly 1 litre. This way if we place 1 mole of a gas into the vessel, its concentration will also be exactly 1 mole per litre.

Also, because events can occur quickly we will occassionally 'freeze' time. When the small clock is not moving time is effectively frozen for our convenience.

For our first experiment, let's see what happens when we start with a 10 mole sample of pure N₂O₄.

In a real experiment this would be difficult to achieve since, of course, N₂O₄ and NO₂ exist in equilibrium and any gas cylinder would contain a mixture of the gases. For this 'thought' experiment we will imagine that the N₂O₄ sample was prepared by some method that produces only the desired gas.

Let's begin.

Let's pause for a moment and note what is happening. First, look at the plot on the right. Notice that the N₂O₄ concentration plotted as a gray line is decreasing, and the NO₂ concentration plotted in red is increasing. This change is reflected by a slight reddish tinge in the reaction vessel's background.

Also, you'll note that the main reaction so far has been N₂O₄ reacting to form NO₂ (signalled by a popping sound). As the NO₂ concentration builds the reverse reaction will become more frequent (signalled by a clicking sound).

Let's continue.

Note that the amounts of N₂O₄ and NO₂ are no longer changing. The system is at chemical equilibrium. It is important to note however that N₂O₄ and NO₂ continue to react as seen in the reaction vessel. So, although the amounts of each gas are static, the system as a whole is not.

To finish this part of the experiment, write down the amounts of N₂O₄ and NO₂ present at equilibrium and calculate K_eq for this reaction.

Let's repeat the experiment using a different starting amount of N₂O₄. Let's try 1 mole and see what happens.

Now that the system has reached chemical equilibrium, write down the amounts of each gas and recalculate the value of K_eq. Is it the same or different?