Lesson

Consider this for a moment. On any given school day, there is an equilibrium between the people inside the building and outside the building.

Generally the rates of people entering and leaving the building are the same.

This equilibrium can be disturbed when a fire drill takes place. The sound of the fire alarm prompts a sharp increase in the number of people leaving the school.

When everything settles down again, the only people going into or out of the school are the firefighters. A new equilibrium exists.

The old equilibrium has been disturbed and a new one established. The position of the new equilibrium is different from the position of the old one.

This analogy introduces you to two important aspects of equilibrium:

  • an equilibrium exists when opposing rates of change are equal
  • a disturbance can result in a change in the position of an equilibrium

The Position of an Equilibrium

In a system at equilibrium, the concentrations of the reactants and products do not have to be equal; in fact, they are unequal in almost all cases. The concentrations of reactants are either greater than or less than those of the products. The only equal thing in an equilibrium is the opposing rates of change.

The position of an equilibrium is a function of the relative concentrations of reactants and products:

  • if the concentration of the reactants is greater than the concentration of the products, then we say that the position of the equilibrium lies to the left. 
  • if the concentration of the products is greater than the concentration of the reactants, then we say the position of the equilibrium lies to the right.

This is a simplistic view which we will refine in the next section, but for now it will help you to work through the rest of this lesson.

Changing the Position of an Equilibrium

A system at equilibrium will stay at equilibrium as long as:

  1. the system remains closed - nothing is added or removed, and
  2. temperature and pressure are kept constant.

If a system is disturbed, then the position of the equilibrium may shift. You will explore three important types of disturbance in this lesson:

  • changing the concentration of a species.
  • changing the volume of a system.
  • changing the temperature. 

The position of an equilibrium can be altered by imposing one or more changes on a system. We'll look at the types of change one at a time.

You will also be introduced to Le Chatelier's Principle and have a look at disturbances that do not have an effect on the position of an equilibrium.

Changing the Concentration of a Species

You can either increase or decrease the concentration of a species in a system. In order to describe the effect of changing the concentration of a species, it is first necessary to describe a system. In the example that follows, the concentration of one species will be increased.

The System

An equilibrium is established by mixing aqueous solutions of iron(III) nitrate and potassium thiocyanate. Since only certain ions in these solutions undergo a change in bonding, we represent the system using a net ionic equation:

A net ionic equation shows only those species that undergo a change in bonding. Potassium and nitrate ions are omitted because they are not changed during the reaction.

The system is at equilibrium when all species in the system have stable molar concentrations. This point is labelled teq on the graph below.

Two of the species in this system have distinctive colours as aqueous ions. Iron(III) (Fe3+) ions are yellow-brown, and iron thiocyanate (FeSCN2+) ions are brick red. Thiocyanate (SCN-) ions are colourless.

The colour of the system at equilibrium allows you to assess the relative concentrations each species:

  • a yellow-brown colour means [Fe3+] > [FeSCN2+].
  • a brick red colour means [Fe3+] is less than< [FeSCN2+].
  • a yellowish-orange colour means [Fe3+] ~ [FeSCN2+].

At time teq, a test tube containing this system at equilibrium might look like this.

Disturbing the System

Suppose you add a crystal of KSCN to the system. The crystal will dissolve and dissociate to produce aqueous K+ and SCN- ions. The result is an increase in [SCN-].

According to the collision theory, an increase in the concentration of a reactant should cause an increase in the reaction rate. In this case, an increase in [SCN-] should result in an increase in the rate of formation of FeSCN2+.

Initially, the rate of FeSCN2+ is high, but it tapers off as Fe3+ is consumed. Since FeSCN2+ has a deep red colour, the result is a darker colour in the test tube. Compare before and after:

          

In the new equilibrium, [FeSCN2+] is greater and [Fe3+] is lower. [SCN-] will be higher too, but not as high as it would be if you just added the existing amount to the amount added. Some of the SCN- had to be consumed to make FeSCN2+. [Fe3+] is lower because it was consumed by the added SCN- to make FeSCN2+.

When the concentrations restabilize, a new equilibrium is established. We say that the position of the equilibrium has shifted to the right.

We say shifted to the right because the concentration of the species on the right side of the equation has increased while the concentration of a species on the left decreased.

Summary of Example
  • Mixing iron(III) nitrate and potassium thiocyanate solutions to form iron thiocyanate, results in an equilibrium.
  • The addition of a KSCN crystal to the system at equilibrium represents a disturbance causing an increase in the forward reaction rate.
  • The forward reaction rate then slows down again as concentrations of reactants decrease and it eventually matches the reverse reaction rate again.
  • Since the concentration of a reactant species has decreased and the concentration of the product has increased, the position of the equilibrium has shifted to the right.

The equilibrium shifted in response to counteract the effect of the disturbance.

Can you think of other ways to disturb this system?

The Effect of Decreasing the Concentration of a Species

Let's reconsider the iron thiocyanate system.

A net ionic equation shows only those species that undergo a change in bonding. Potassium and nitrate ions are omitted because they are not changed during the reaction.

One way to decrease the concentration of an aqueous species is to precipitate it out of solution. For example, you could decrease [Fe3+] by adding hydroxide ions (OH-) to form Fe(OH)3(s).

A decrease in [Fe3+] causes a decrease in the forward reaction rate.

In other words, the forward rate would be slower compared to the reverse rate.

For a short time, the rate of FeSCN2+ decomposition would exceed its formation, so its concentration will decrease. Meanwhile, the [SCN-] increases.

The position of the equilibrium shifts to the left.

The colour in the test tube might look like this:

Concentrations of Gases

It is important to point out that a change in concentration only applies to dissolved species and gases.

When dealing with systems involving gases, concentration can be decreased by removing a gas from the system altogether. For example, in the system:

the equilibrium can be disturbed by removal of CO2 as it forms.

Alternatively, condensing a species to the liquid state also decreases its concentration. This technique is used in the production of ammonia, NH3.

If some ammonia is removed by condensation, the reverse reaction rate becomes slower relative to the forward rate, and the position of the equilibrium shifts to the right.

Changing the Volume of a System

If a system contains one or more gases, a change in volume causes a change in the pressure of the gas.

  • increasing the volume decreases gas pressure.
  • decreasing the volume increases gas pressure.

Pressure and volume are inversely proportional for fixed amounts of gas at constant temperature.

Pressure changes can be equated with changes in concentration. For example, each gas in this cylinder has its own molar concentration.

If the piston is pushed in,

     

the volume of the cylinder is reduced, and the concentrations of the gases increase because there are more moles of each gas per unit of volume. (Can you see how the molecules are squeezed together at lower volume?)

When a system is at equilibrium, a change in volume changes the pressure/concentration of all gas species in the system. According to the theory of chemical reactions, an increase in concentration should result in an increase in reaction rate. The question that arises from this is, if a system is at equilibrium, are both the forward and reverse reaction rates increased? The answer is yes.

However, if you ask the question: are both reactions equally affected? The answer is it depends.

Consider this system:

Let's increase the pressure by decreasing volume of the system at equilibrium

    

and graph the effect of the change:

If both rates were equally affected by the volume decrease (pressure increase), then the amount of each gas would not change at tdisturbance. However, there appears to be a decrease in the amounts of H2 and N2 and an increase in the amount of NH3 (count the molecules in each piston).

The forward rate must be favoured by the pressure increase. The position of the equilibrium shifts to the right.

Increasing the volume of a chemical system involving gases has the opposite effect.

The amount of NH3 decreases, while the amounts of N2 and H2 increase. The reverse reaction is favoured in this case. The position of the equilibrium shifts to the left.

Compare the number of molecules in each cylinder before and after the pressure changes. What do you notice? What do you notice about the relative amounts of space between the molecules before and after?

A change in the volume of a system causes a shift in the position of the equilibrium if the total number of gas moles on one side of the chemical equation is different from the total number of gas moles on the other side of the equation.

For any system at equilibrium involving gases:

  • if volume is decreased, then pressure increases and the reaction which forms fewer gas moles is favoured thereby reducing the total number of gas molecules in the system.
  • if volume is increased, then pressure decreases and the reaction which forms more gas moles is favoured thereby increasing the total number of gas molecules in the system.

Does a change in gas volume always result in a shift in the position of an equilibrium? The answer is no. It only has an effect on the position of an equilibrium if the number of gas moles on the left differs from the number of gas moles on the right.

For example, increasing the pressure on this system increases the rates of reaction equally and does not affect the position of the equilibrium.

The relative concentrations of the species involved do not change. Thus,

  • a volume change on a system in which there are equal numbers of reactant gas moles and product gas moles has no effect on the position of the equilibrium.

Changing the Temperature

So far in this discussion of equilibrium position, we have completely ignored the role of energy. The assumption has been that when a system is disturbed by changing either the concentration of a species or the volume of the system, temperature has been kept constant.

Recall that increasing the temperature of a system causes more frequent, but more importantly, more intense collisions between particles, and therefore faster reaction rates.

When a system is at equilibrium, an increase in temperature has an effect on both the forward and reverse reaction rates because all particles collide with greater intensity. However, one of the two reactions is more profoundly affected by the temperature increase.

Compare the activation energy of the forward and reverse reactions. Are they the same? Roll your mouse over the graph to see the activation energy for the reverse reaction.

For any reversible reaction, the activation energy of the endothermic reaction is greater than that of the exothermic reaction. Increasing the temperature of the system, or in other words, adding heat, favours the endothermic reaction.

In the case of the SO2, O2, SO3 system, the forward rate will be faster than the reverse rate for a time and this will result in an increase in the amounts of SO2, and O2, and a decrease in the amount of SO3. Eventually, the build-up of SO2 and O2 coupled with the decrease in SO3 will result in equal forward and reverse rates. You would say that the equilibrium position has shifted to the right.

Another way of stating it is

  • when heat is added to a system, the position of an equilibrium shifts away from the location of the energy term.

Conversely,

  • when heat is removed from a system, the position of an equilibrium shifts towards the location of the energy term.

Le Châtelier's Principle

Let's take a moment to look back at the effect of each disturbance:

  • increase concentration of a species, system responds by using up that species and anything else it reacts with to increase the amount of the species it produces
  • decrease concentration of a species, system responds by replacing that species
  • increase the pressure on a gas system by decreasing volume, system responds by reducing the number of gas molecules to relieve the effect of increased pressure
  • decrease the pressure on a gas system by increasing volume, system responds by increasing the number of gas molecules to counteract the effect of decreased pressure
  • increase the temperature of a system, system responds by favouring the endothermic reaction
  • decrease the temperature of a system, system responds by favouring the exothermic reaction

Can you come up with a general statement about what happens to an equilibrium when it is disturbed in some way?

Henri Le Châtelier did. He summarized the effect of a disturbance on a system at equilibrium in this way:

when disturbed, a system at equilibrium will respond in some way to counteract the effect of the disturbance.

It is important to note here that Le Châtelier's principle does not explain why a change has a particular effect. His principle is used to make predictions only. That's what makes Le Châtelier's Principle different from a theory. A theory has the power to explain and predict observations.

You use the collision theory to explain the effect of a change on the rates of forward and reverse reactions. You apply Le Châtelier's principle to make predictions about the effect of a change on the position of an equilibrium.

Changes That Do Not Affect Equilibrium Position

Are there disturbances that do not cause a shift in the position of an equilibrium?

When Gas Moles are Equal

As mentioned above, a change in the pressure of a system by changing volume does not always result in the shift in the position of an equilibrium. Consider this system:

If the volume of this system is halved, the pressure on the system doubles. The rates of both the forward and reverse reactions increase due to the higher concentrations of the species on the right and left sides of the system; however, both rates are equally affected because there are two moles of gas on each side of the system.

The total number of gas molecules at the higher pressure won't be any different than the total number of gas molecules before the pressure change, so the position of the equilibrium is the same.

When a Catalyst is Added

A catalyst is a species that speeds up a reaction by providing a reaction mechanism with lower activation energy. Take a second look at the effect by rolling your mouse over the graph.

You should notice that the reaction pathway for the catalyzed reaction has lower activation energy, but more importantly, you should also see that the activation energies of the forward and reverse reactions are lower by the same amount.

The net result of the addition of a catalyst is that the forward and reverse rates are equally affected and there is no change in the position of the equilibrium; however, there is a reduction in the time taken to reach the equilibrium.

Recall that there are three steps in the establishment of any equilibrium from starting materials:

  1. forward reaction proceeds quickly and slows progressively.
  2. reverse reaction starts slowly and progressively speeds up.
  3. both rates become equal.

If a catalyst speeds up the forward reaction, then the equilibrium point should be reached faster.

When Surface Area is Increased

A way to speed up a reaction involving a solid is to increase its surface area. This allows more collisions per unit of time and a faster reaction rate.

The question arising from this is, does increasing the surface area of a species affect the position of an equilibrium? The answer is no it doesn't.

Consider this system:

If you powder chunks of Fe3O4 , you will increase the rate of the forward reaction; however, since the reactants are produced faster, the reverse reaction will also be faster.

The net effect of increasing the surface area is reduced time to establish the equilibrium. There is no change in the position of the equilibrium.

Activity

Textbook Readings
  • none
Textbook Practice Items
  • page 535: items 4, 5, 6, and 7
  • pages 538-541: items 15, 16, 17, 38 and 43
Practice Items

When you look at a bottle of pop like this one, you see a colourless liquid and a small amount of a gas over the liquid.

The liquid is actually a mixture of water, sugar, and dissolved carbon dioxide (carbonic acid, H2CO3). Carbonic acid is the product of a reaction between dissolved carbon dioxide and water. The reaction is reversible.

In a carbonated drink, the position of the CO2 - H2CO3 equilibrium lies far to the left.

Typically, less than 1% of the dissolved carbon dioxide molecules react with water to become carbonic acid.

  1. Describe the effect of increasing [CO2] on the position of the equilibrium.
  2. Describe the effect of heating on the position of the equilibrium.

The dissolved carbon dioxide comes from the gas beneath the bottle cap. This is an example of a physical equilibrium.

The gas over the solution is under pressure. The gas pressure is about 400 kPa or about four times normal atmospheric pressure (100 kPa). That's why you here a phitz sound when you remove the cap of a pop bottle - its the sound of escaping gas as the pressure is released. Opening the system decreases the concentration of CO2 gas. The fizzing that follows is a result of dissolved CO2 becoming CO2 gas.

  1. Describe the effect of reducing gas pressure on the position of the equilibrium.

Test Yourself

There is no self test for this lesson.