Introducing Chemical Equilibrium

Some crystals of iodine, I2(s), are placed in an open gas jar (an open system) maintained at a constant room temperature. A purple vapour is seen to form above the crystals, the crystals gradually disappear, and eventually so does the purple iodine vapour.

The experiment is repeated but this time a lid is placed on the gas jar (a closed system). After a short time, a uniform purple vapour forms and some crystals of iodine remain. At the constant temperature the system remains like this indefinitely.

In this closed system an equilibrium has established. The equilibrium is between the iodine solid and the iodine vapour. Here we do not have a static condition but a dynamic condition. What is happening all of the time is that I2 molecules are leaving the solid and entering the vapour phase at the same rate as they are leaving the vapour and entering the solid state. The forward and reverse rates have become the same and the system is at equilibrium. However, this is not a chemical equilibrium but a physical equilibrium.

I2(s)      I2(g)

Incidentally, in the closed system above, the pressure of the iodine vapour is equal to its vapour pressure. This does depend on temperature, but is independent of the volume of the vessel and of the amount of iodine solid, provided some solid is always present.

A chemical reaction can also reach a state of equilibrium

At the start of an experiment you have two sealed flasks maintained at 720K, one containing a mixture of H2(g) and I2(g), and the other HI(g) only.

At intervals of time during the experiment the amount of HI(g) in each is measured (and expressed as the percent of all H and I atoms present as HI).

Starting with H2(g) and I2(g) only
Starting with HI(g) only 

Show both curves
H2(g) + I2(g)    ®    2HI(g)
2HI(g)    ®    H2(g) + I2(g)

Here's what's happening...

Starting with H2(g) and I2(g) only, the two react producing HI(g) in a forward reaction. The rate of this reaction decreases as it proceeds owing to the concentrations of H2(g) and I2(g) diminishing. As HI(g) is formed, its molecules begin to react to reform H2(g) and I2(g) in a reverse reaction. As the concentration of HI(g) increases as a result of the forward reaction, the rate of the reverse reaction increases. Eventually, the concentrations of all the chemical components will attain values such that the forward and reverse reactions are occuring at the same rate.

We now have a reversible reaction that is said to have attained chemical equilibrium.

In an equilibrium reaction, all chemical components are both reactants and products, though by tradition those shown on the left of the chemical equation are called reactants and those on the right products.

H2(g) + I2(g)        2HI(g)

Remember, although at equilibrium the forward and reverse reactions may appear to have stopped they are, in fact, taking place all of the time and at the same rate - equilibrium is a dynamic not a static condition.

What quantities of reactants and products are present at equilibrium?

Equilibrium refers to equality of forward and reverse reaction rates, not of quantities of reactants and products. In the above chemical equilibrium, 78% of the mixture is in the form of HI(g), so appreciable quantities of both reactants and products are present. Some reactions establish equilibrium after the formation of only minute amounts of products. In others, only minute amounts of reactants remain at equilibrium. The composition of the equilibrium mixture is sometimes referred to in terms of equilibrium position.

How long does it take for equilibrium to be reached?

In the case of HI(g), the mixture reaches equilibrium in less than 2 hours at 720 K. Other reactions may take only a tiny fraction of a second, and there are many more that would be nowhere close to equilibrium after a million years.

The Equilibrium Constant

Now consider we have a mixture of reactants and products that is not yet at equilibrium, or one that is at equilibrium.

Which way will the reaction go in order to reach equilibrium?

How far will the reaction go, that is, what will be the amounts of reactants and products at equilibrium?

If we make changes to the system, such as to its volume, or add more of a reactant, how will the equilibrium composition be affected?

Answer these questions...

To be able to answer these questions we have to know about the Equilibrium Constant. Now back to the equilibrium involving H2(g), I2(g) and HI(g) at 720K.

H2(g) + I2(g)        2HI(g)

In four separate experiments at 720 K, chemical equilibrium was allowed to establish. Two were started with different concentrations of H2(g) and I2(g) only in each, and two with differing initial concentrations of HI(g) only. In each of the four experiments, the equilibrium concentration of H2(g), I2(g) and HI(g) were measured. These are given in the table below:

Equilibrium concentrations (mol dm-3)
[H2(g)]eqm[I2(g)]eqm[HI(g)]eqm
1.14 x 10-20.12 x 10-22.52 x 10-2
0.92 x 10-20.20 x 10-22.96 x 10-2
0.34 x 10-20.34 x 10-22.35 x 10-2
0.86 x 10-20.86 x 10-25.86 x 10-2

If we substitute these equilbrium concentrations into the expression below:

then a constant value, within experimental error, is obtained. The square brackets represent concentrations in mol dm-3.

The value of this ratio of equilibrium concentrations (in mol dm-3) is known as the equilibrium constant, represented by the symbol Kc.

Many other chemical equilibrium reactions have been studied and in each case an equilibrium constant relating to the stoichiometric chemical equation has been calculated. This observation can be universally applied, in what is sometimes called the Law of Chemical Equilibrium, as illustrated by the general chemical equilibrium:

aA(g) + bB(g)        cC(g) + dD(g)

Can the equilbrium constant be expressed in other terms?

The Ideal Gas Equation shows that the pressure of a gas is proprtional to its concentration.

pV = nRT

where p is pressure of a particular gas (its partial pressure) in an equilibrium mixture, V is the total volume, n is the number of moles of the particular gas, R is the general gas constant, and T is the absolute temperature.

In the above equation where the temperature is also constant,

P a n / V

The equilibrium constant can therefore also be expressed in terms of partial pressures, and is denoted as Kp:

Does the equilbrium constant have units?

The equilibrium constant (Kc or Kp) may or may not have units; this depends precisely upon the equilibrium expression.

What values can the equilibrium constant have?

Equilibrium constants come in all sizes, from almost zero (like 10-50) to extremely large (like 1050). When K (Kc or Kp) is much greater than 1, the partial pressures or concentrations of the products are large in comparison to those of the reactants.

These are ways in which we describe this situation...

"...the reaction goes nearly to completion..." or " ...goes mostly to the right..." or "...the equilibrium favours the products...".

Does the equilibrium constant depend on temperature?

Yes. An equilibrium constant is constant only at a constant temperature. Changing the temperature of a chemical system at equilibrium will change the value of the equilibrium constant, and also the position of the equilibrium.

Are Kc and Kp related?

Yes. The relationship depends on the reaction and how it is written. Specifically, it depends on the number of moles of gaseous reactants and products.

The equation is:

Kp = Kc(RT)Dn

where Dn is the number of moles of gaseous products minus the number of moles of gaseous reactants, R is the gas constant, and T is the absolute temperature at which the equilibrium exists.

Now let's make some use of the equilibrium constant...

Return to: The Equilibrium Constant

Which way will the reaction go in order to reach equilibrium?

We have a mixture of SO2(g), O2(g), and SO3(g) at 1000K which has still to reach equilibrium. The partial pressures are pSO2 = 0.48 atm, pO2 = 0.18 atm, and pSO3 = 0.72 atm. Kp = 3.40 atm-1. The reaction is:

2SO2(g) + O2(g)      2SO3(g)

Which way must the reaction go to reach equiilibrium?

See the answer...

How far will the reaction go, that is, what will be the amounts of reactants and products at equilibrium?

A sample of air is compressed so that [N2(g)] = 0.80 mol dm-3 and [O2(g)] = 0.20 mol dm-3. (It might help to imagine the container has a volume of 1 dm3.) It is then heated to 1500 K and allowed to reach equilibrium.

N2(g) + O2(g)      2NO(g)

How much of each reactant and product will be present at equilibrium? Kc = 1.0 x 10-5 at 1500K.

See the answer...

If we make changes to the system, such as to its volume, or add more of a reactant, how will the equilibrium composition be affected?

See the answer...

Le Chatelier's Principle

It is of great practical importance to be able to predict, and to control, the equilibrium composition of a chemical reaction. In an industrial process it may be desirable to convert as much as possible of reactants to prodcuts. At equilibrium the product yield may not be satisfactory. But, the composition of the equilibrium mixture (equilibrium position) is dependent on certain conditions, such as temperature and pressure, and it might be possible to change these to achieve a better yield. Le Chatelier's Principle helps us deal with this.

Now consider another chemical reaction in a state of equilibrium at a constant temperature.

N2(g)   +   3H2(g)        2NH3(g)    DH°f, 298 = - 46.0 kJ mol-1

In a closed system and at a pressure of 250 atmospheres in the presence of a finely divided iron catalyst at 500 °C, about 15 percent of the gases are converted into ammonia at equilibrium.

We can impose the following changes [both an increase and decrease] upon this chemical system at equilibrium:

  1. Amount of a substance / concentration of a reactant or product.
  2. Pressure / Volume (pressure and volume are inversely related).
  3. Temperature.

In general terms, this is what Le Chatelier's Principle says...

If a change is imposed upon a chemical system at equilibrium then it will respond in such a way as to undo, in part, the effect of the change imposed upon it.

Here are some examples:

Example 1

The reaction is at equilibrium in a closed vessel of fixed volume. The temperature is constant. Suddenly some nitrogen gas is added. The concentration of nitrogen has increased. The partial pressure of nitrogen is raised. The system is no longer at equilibrium. Le Chatelier's Principle says that the equilibrium will adjust so as to reduce the concentration of the nitrogen. This can only be achieved if the nitrogen reacts. The nitrogen can only react with hydrogen forming ammonia. The equilibrium therefore shifts from left to right. A new equilibrium position is established. Since the system responds 'to undo, in part, the effect of the change', at the new equilibrium position there will be slightly more nitrogen, less hydrogen, and more ammonia present.

Example 2

The reaction is at equilibrium in a closed vessel. The temperature is constant. The volume of the vessel is suddenly reduced. This is equivalent to increasing the total pressure of the system. The system is no longer at equilibrium. Le Chatelier's Principle says that the equilibrium will adjust so as to reduce the pressure. It can only achieve this by shifting in the direction of fewer moles of gas (fewer gaseous molecules). There are 2 moles of gas on the right of the equilibrium and 4 on the left. The equilibrium therefore shifts from left to right. A new equilibrium position is established. The yield of ammonia is increased.

Example 3

The reaction is at equilibrium in a closed vessel of fixed volume. The temperature of the system is suddenly reduced. The system is no longer at equilibrium. Le Chatelier's Principle says that the equilibrium will adjust so as to bring about a rise in temperature of the system. A reaction needs to occur so as to produce heat for this to happen. The equilibrium must shift in the exothermic direction, that is, from left to right. A new equilibrium position is established. The yield of ammonia is increased.

Getting the best yield of ammonia...

It seems that by raising the pressure of the system and at the same time reducing its temperature we can maximise the yield of ammonia at equilibrium. This is not without its problems. Maintaining a higher pressure is expensive; more robust plant (thicker pipes and stronger joints, for example) is required, and more electricity would have to be used to drive pumps to maintain this pressure. Lowering the temperature slows the rate at which chemical reactions take place resulting in a longer wait for equilibrium to be achieved. A 'compromise' has to be settled upon. The conditions selected approximate to those given above.

Does a catalyst affect the equilibrium position?

No. The addition or removal of a catalyst does not cause a shift in equilibrium. A catalyst is a substance that affects the rate of a reaction, but is not consumed in the reaction. It does this by providing an alternative mechanism for the reaction but of lower activation enthalpy. It therefore changes the rate of approach of equilibrium and so affects the forward and reverse reaction rates in the same way. The composition of the equilibrium mixture is unchanged.

Heterogeneous Equilibrium

So far we have considered only chemical equilibria involving all gases, that is, homogeneous equilibria. An equilibrium involving more than one phase - gas and solid, for example, or liquid and solid - is said to be heterogeneous. For example:

H2(g) + I2(s)        2HI(g)

In the above example, the equilibrium expressions for this reaction are:

Notice that I2 is missing from the equilibrium expressions. This is because, at room temperature, iodine is a solid. The composition of the equilibrium mixture of H2(g) and HI(g) is independent of the amount of solid iodine present, as long as some of it is always present. [The partial pressure of I2(g) (which is its vapour pressure) in equilibrium with I2(s) is constant at a constant temperature, and is independent of the volume of the container and of the quantity of solid.]

Equilibrium in Solution

Here we will consider homogeneous and heterogeneous chemical equilibria in which the solvent is much more abundant than all the other components put together. The solvent may also be a reactant or product itself.

First of all, let's look at an equilibrium that is not in solution:

CH3COOH(l) + CH3CH2OH(l)        CH3COOCH2CH3(l) + H2O(l)

In this case, water is not a solvent but a product only.

Now consider the same equilibrium in very dilute aqueous solution. The chemical equation is:

CH3COOH(aq) + CH3CH2OH(aq)        CH3COOCH2CH3(aq) + H2O(l)

And the equilibrium expression is:

Water is now both a product and a solvent. In reaching a state of equilibrium the concentration of the water changes so little that it is effectively constant. For this reason its value is taken along with the equilibrium constant, Kc.

Here are some more examples of chemical equilibria in aqueous solution:

CH3COOH(aq) + H2O(l)        CH3COO-(aq) + H3O+(aq)

Ka is called the acid dissociation constant. Again, the concentration of water changes negligibly on reaching equilibrium and so its value is taken along with the equilibrium constant (Kc) to form the 'modified' equilibrium constant, Ka.

NH3(aq) + H2O(l)        NH4+(aq) + OH-(aq)

Kb is called the base dissociation constant. Again, the concentration of water changes negligibly on reaching equilibrium and so its value is taken along with the equilibrium constant (Kc) to form the 'modified' equilibrium constant, Kb.

AgCl(s)        Ag+(aq) + Cl-(aq)

Ks is the Solubility Product. It is the equilibrium constant for the equilibrium that exists between a slightly soluble salt and its ions in saturated solution. The solid, AgCl, is omitted from the equilibrium expression. As long as there is some solid present the concentration of the saturated solution of ions remains constant at a given temperature.