Mr. Lou's Chemistry Website

Daniel McIntyre Collegiate

Lesson 1 - Introduction to Chemical Equilibrium

Typically when we think of what happens during a chemical reaction we think of the reactants getting totally used up so that none are left and ending up with only products. Also, we generally consider chemical reactions as one-way events. You may well have learned during earlier science classes that this is one way to distinguish chemical change from physical changes - physical changes (such as the melting and freezing of ice) are easily reversed, but chemical changes cannot be reversed (pretty tough to un-fry an egg).

Physical Equilibrium

Just remember how does the water cycle work? Water is recycled over and over. When liquid in an open container is left to sit, it will eventually evaporate to dryness. The particles on the surface of the liquid vaporize and leave the container. If the container is sealed, the vapour is trapped and cannot leave the container. Eventually, the space above the liquid becomes saturated with vapour and for every molecule that evaporates, another condenses.

Liquid in a closed container, at a constant temperature, reaches equilibrium with its vapour. At equilibrium,

rate of evaporation = rate of condensation

The pressure created by the vapour at equilibrium is known as the vapour pressure, abbreviated Pvap.

It is called equilibrium dynamic because even though the amount of vapour and liquid, for example, do not change there is constant conversion between the two states. The amounts remain constant because the rate of the forward process (evaporation) is equal to the rate of the reverse process (condensation). As one liquid molecule vapourizes, another vapour molecule condenses. This occurs simultaneously.

Reversible Reactions

Typically when we think of what happens during a chemical reaction we think of the reactants getting totally used up so that none are left and ending up with only products. Also, we generally consider chemical reactions as one-way events. You may well have learned during earlier science classes that this is one way to distinguish chemical change from physical changes - physical changes (such as the melting and freezing of ice) are easily reversed, but chemical changes cannot be reversed (pretty tough to un-fry an egg).

In this unit we will see that this isn't always the case. We will see that many chemical reactions are, in fact, reversible under the right conditions. And because many reactions can be reversed, our idea of a reaction ending with no reactants left, only products, will need to be modified.

Here are some examples of reactions that can be reversed:

1.

Nitrogen dioxide, NO2, a reddish-brown gas, reacts to form colourless dinitrogen tetroxide, N2O4:

2 NO2 (g) → N2O4 (g)

But the reaction can also go the other way - dinitrogen tetroxide also readily breaks down to form nitrogen dioxide:

N2O4 (g) → 2 NO2 (g)

We typically write a reaction that can go in both directions by using a double arrow (which will sometimes appear as ↔ in these online notes):

2 NO2 (g) N2O4 (g)

Because the reaction continues in both directions at the same time, we never run out of either NO2 or N2O4. NO2 is continually being used up to form N2O4, but at the same time N2O4 is forming more NO2

2.

When hydrogen gas is passed over heated iron oxide, iron and steam are produced:

(1)

Fe3O4 (s) + 4 H2 (g) → 3 Fe(s) + 4 H2O(g)

The reverse reaction can occur when steam is passed over red-hot iron:

(2)

3 Fe(s) + 4 H2O(g) → Fe3O4 (s) + 4 H2 (g)

We can write these two equations together as:

(3)

Fe3O4 (s) + 4 H2 (g) 3 Fe(s) + 4 H2O(g)

When we have a reversible reaction written in this way, we need to be able to distinguish between which way the reaction is headed. As written above in Reaction (3), we would say that in the forward reaction iron oxide and hydrogen gas, the reactants, produce the products iron and steam.

During the reverse reaction, iron reacts with steam to produce the products iron oxide and hydrogen gas.

It is important to understand the terminology, and to use the terms correctly.

Does it matter which way we write our reversible reaction? Could we write it as

3 Fe(s) + 4 H2O(g) Fe3O4 (s) + 4 H2 (g)

Yes, we could. Now iron and steam are reactants of the forward direction, and iron oxide and hydrogen gas would be the reactants of the reverse direction.

Equilibrium

Here's another example of a reversible reaction - dissolving salt in a beaker of water, described by the following reaction:

NaCl(s) at equilibrium NaCl(aq)

If you keep adding more and more solid salt, eventually you'll reach the point where no more salt dissolves, and the excess sits at the bottom of the beaker. At this point we have a saturated solution. Has the dissolving reaction stopped? It would appear so, but that's not the case (wouldn't that be too easy?).

What happens in our saturated solution, which has reached the point of equilibrium, is that both the forward

NaCl(s) → NaCl(aq)

and reverse

NaCl(aq) → NaCl(s)

reactions are still going on, but at the same rate. This in effect cancels out any observable, or measurable, changes in our system. At the same rate that solid NaCl produces aqueous NaCl (dissolved salt), the dissolved salt is recrystallizing to form more solid NaCl.

Equilibrium is the state at which the rate of the forward reaction equals the rate of the reverse reaction.

At the point of equilibrium, no more measurable or observable changes in the system can be noted.

It is important for you to understand that equilibrium means the rates of the forward and reverse reactions are equal; it does not mean that there are equal amount of reactants and products present at equilibrium.

For example, the following reaction was allowed to come to the point of equilibrium, and concentrations of all reaction participants were measured at that time:

H2 (g) + I2 (g) at equilibrium with  2 HI(g)

At equilibrium: [H2] = 0.022 M

[I2] = 0.022 M

[HI] = 0.156 M

For this particular reversible reaction, there is more HI at equilibrium (0.156 M) than there is of H2 and I2 (both at 0.022M).

We say the product side of the reaction is favoured.

Equilibrium does not mean equal amounts at equilibrium!

Here's something to help you understand about how equilibrium works. Imagine yourself on a escalator that is going down. You start at the top (reactants) and end up at the bottom (products). But when you are partway down you start walking up the escalator as it continues going down. If you match your rate of walking up to the same rate that the escalator is going down, you make no progress and appear to be at a standstill. To an observer it would look as if you and the escalator had come to a stop, but actually both upward and downward movements continue.

Equilibrium is dynamic - both forward and reverse reactions continue, even though the reaction appears to have stopped. And this equilibrium does not need to occur right in the middle of two floors - you could be near the bottom, near the top, or anywhere in between when you carry out your reverse process.

In order for a reversible reaction to reach the point of equilibrium, the reaction must be carried out in a closed system - no additional reactants can be added or products removed. If, in our last example, the product HI was removed as it formed, the reaction would never reach the point of equilibrium - instead H2 and I2 would continue to react to produce HI until one or both of the reactants was used up.

If reactants are constantly being added, and products removed as they form, the system would appear to be at equilibrium because to an outside observer it would appear that the reaction has stopped, but that would not be the case. This situation - with new material constantly being added as products are removed - is called a steady state system. A factory with an assembly line is a steady state system - new raw materials are constantly being added; finished products are removed. A campfire with wood being added to the fire is another steady state system. Be careful not to confuse steady state with equilibrium.

How do the rates of the forward and reverse reactions as the reaction heads towards equilibrium (before it reaches equilibrium?)

If we start our above reaction with H2 and I2, and with no HI, the two gases will react at a certain rate. But remember that the rate of a reaction slows down over time, as the reactants get used up (and lower their concentration). Eventually, however, the amount of the product HI increases, and it will begin producing H2 and I2. Thus the rate of the reverse reaction starts out slowly (there is no HI present), but will speed up as the concentration of HI increases. Eventually both rates will level off (not always to the same level as shown by this example, however):

The Equilibrium Constant, Kc

Chemists have found that there is a mathematical relationship that exists between the concentration of the reactants and products once equilibrium has reached that is independent of the initial concentration of the participants.

For any general reaction:

aA + bB at equilibrium with  cD + dD

an equilibrium constant expression can be written as:

Kc

=

[C]c × [D]d


[A]a × [B]b

This mathematical relationship exists for all equilibrium systems, and produces a constant ratio called the equilibrium constant, Kc.

This equation is sometimes called the mass-action expression.

This relationship will be very important to us for the next few units, so it is important that you understand how to set this relationship up and what it tells us about an equilibrium system.

The products of the reaction (C and D) are placed in the numerator, and their concentrations are raised to the power of the coefficients from the balanced equation. The reactants (A and B) are placed in the denominator, with their concentrations raised to the power of their coefficients.

For the reaction between hydrogen and iodine gas to produce hydrogen iodide:

H2 (g) + I2 (g) at equilibrium with  2 HI(g)

the equilibrium constant expression will be:

Kc

=

[HI]2


[H2] × [I2 ]

Using the example we examined in our last section, equilibrium concentrations for each substance were measured at equilibrium and found to be:

At equilibrium: [H2] = 0.022 M

[I2] = 0.022 M

[HI] = 0.156 M

We substitute these values into our equilibrium expression and solve for Keq:

Kc

=

[HI]2


[H2] × [I2 ]

=

(0.156)2


(0.022)(0.022)

 

=

50.3

The value of Keq, which has no units, is a constant for any particular reaction, and its value does not change unless the temperature of the system is changed. It does not depend on the initial concentrations used to reach the point of equilibrium.

Writing the Equilibrium Law

Homogeneous equilibria are those in which the reactants and products are all in the same phase, gases (g) or aqueous (aq).

Example 1. Write the equilibrium law for the following equation:

N2(g) + 3 H2(g) 2 NH3(g)

Solution:

We write the product concentrations over the reactant concentrations, then write in the exponents. The coefficient for N2 is 1, so it is written as [N2] in the mass action expression. The coefficient for H2 is 3, so it is written as [H2]3 and NH3 is written as [NH3]2.

Since all reactants and products are gaseous, the equilibrium law would be

Heterogeneous equilibria involve reactants and products in more than one state. When writing the mass action expression or equilibrium law, substances which are solids (s) or liquids (l) are omitted.

Solids and liquids rarely change in concentration, therefore are not included. For example, the concentration of liquid water will not change significantly over the course of a reaction.

Example 2. Write the equilibrium law for the following equation:

C(s) + H2O(g) CO(g) + H2(g)

Solution:

Since C is in the solid state, it is omitted from the equilibrium law.

Equilibrium Constant

The equilibrium constant, abbreviated KC, KP or just K, is the ratio of product concentrations/pressures to reactant concentrations/pressures. Since the equilibrium constant is based on the specific rate constant in rate laws, the only factor affecting K is temperature.

The equilibrium constant can indicate whether products or reactants are favoured at equilibrium. That is, whether there are more products or reactants at equilibrium.

If K = 1, [products] = [reactants].

Neither reactants nor products are favoured.

If K>1, the amount of product is greater than reactant.

We say “the products are favoured” or “the equilibrium position lies to the right”.

For example, if the reaction A + B C + D has a KC = 1×105 mixing A and B results in almost a complete conversion to products C and D.

If K<1, the product concentrations are less than reactant concentrations.

We say “the reactants are favoured” or “the position of equilibrium lies to the left”.

For example, if the reaction E + F G + H has a KC = 1×10–5 mixing E and F results in almost no conversion to products.