Unit outcomes

By the end of this unit you will be able to:

• Define rates of reactions and identify factors effecting rate and refer to examples.
• State mechanism of reaction and of catalysis.
• Define chemical equilibrium and identify conditions for equilibrium.
• Identify factors affecting equilibrium apply and evaluate the effect on chemical reactions using Le Chatelier’s principle.
• Define and interpret equilibrium constant.

What you should know

Before you start this unit, make sure you can:

• Describe the kinetic-molecular theory. Refer to level 3 subject outcome 5.4 unit 1 if you need help with this.
• Identify, analyse, and apply energy changes in chemical reactions. Refer to level 3 subject outcome 6.1 unit 1 if you need help with this.
• Write chemical equations and balance them. Refer to level 3 subject outcome 6.2 unit 1 if you need help with this.
• Describe the concept of the mole and molar gas volume. Refer to level 3 subject outcome 6.2 unit 2 if you need help with this.
• Define the concentration of a substance. Refer to level 2 subject outcome 6.4 unit 1 if you need help with this.

Introduction

Parts of the text in this unit were sourced from Siyavula Physical Science Gr 21 Learner’s Book, Chapter 7, released under a CC-BY licence.

In this unit you will learn how to determine the rate of a reaction and the factors that can affect how fast a reaction happens. You will also learn that some reactions are reversible where the products can be converted back into the reactants and a situation called equilibrium is reached. This is not desirable in the chemical industry, so methods need to be used to manage this process to get maximum yield of the products.

Rate of a reaction

Some chemical reactions are fast, for example, burning a match. Other chemical reactions take a long time, for example, the rusting of an iron pipe. We will look at why reactions proceed at different rates (speeds) and how we can change the rate of the reaction.

In a chemical reaction, the substances that are undergoing the reaction are called the reactants, while the substances that form as a result of the reaction are called the products. The reaction rate describes how quickly or slowly the reaction takes place. So how do we know whether a reaction is slow or fast? One way of knowing is to look either at how quickly the reactants are used during the reaction or at how quickly the products form. For example, iron and sulfur react according to the following equation:

[latex]\scriptsize \text{ F}{{\text{e}}_{{(s)}}}+\text{ }{{\text{S}}_{{(s)}}}\text{ }\to \text{ Fe}{{\text{S}}_{{(s)}}}[/latex]

In this reaction, we can observe the speed of the reaction by measuring how long it takes before there is no iron or sulfur left in the reaction vessel. In other words, the reactants have been used up. Alternatively, one could see how quickly the iron sulfide (the product) forms. Since iron sulfide looks very different from either of its reactants, this is easy to do.
Here is another example:

[latex]\scriptsize \text{2M}{{\text{g}}_{{(s)}}}\text{ + }{{\text{O}}_{{2\text{ }(g)}}}\text{ }\to \text{ 2Mg}{{\text{O}}_{{(s)}}}[/latex]

In this case, the reaction rate depends on the speed at which the reactants (oxygen gas and solid magnesium) are used, or the speed at which the product (magnesium oxide) is formed.

Reaction rate is therefore defined as how quickly products are formed, or how quickly reactants are used up during a chemical reaction.

Reaction rates and collision theory

Collision theory is used to explain the rate of a reaction on a microscopic level. You know that all matter is made of particles which are continually moving.

For a reaction to occur, the particles that are reacting must collide with one another. Only a fraction of all the collisions that take place actually cause a chemical change.

These are called successful or effective collisions. Reactant particles must collide with sufficient energy (called the activation energy) and in the correct orientation at the moment of impact for the reactants to change into products. These successful collisions are necessary to break the existing bonds (in the reactants) and form new bonds (in the products).

Factors affecting reaction rate

Any change in conditions under which a reaction is taking place that will increase or decrease the number of successful collisions per unit time will affect the reaction rate. There are a number of factors that can affect the average rate of a reaction. It is important to know these factors so that reaction rates can be controlled. This is particularly important when it comes to industrial reactions, where greater productivity leads to greater profits for companies. The following are some of the factors that affect the average rate of a reaction:

• the nature of reactants (how reactive the substance is)
• the concentration of solutions or pressure of gases
• the surface area of solid reactants
• the temperature
• presence of catalysts.

Activity 1.1: Investigate the factors that affect the rate of chemical reactions

Time required: 10 minutes

What you need:

• internet connection

What to do:

Watch the video, Factors affecting the rate of a reaction (Demonstrations), on demonstrations of experiments to show how different factors affect the rate of a reaction.

Factors affecting the rate of a reaction (Demonstrations) (Duration: 08.35)

What did you find?

The nature of reactants: the more reactive the reactants, the faster the reaction rate.
The concentration of solutions: the higher the concentration of a solution, the faster the reaction rate.

The surface area of solid reactants: increasing the surface area of a solid (by grinding it into a powder) will increase the rate of a reaction.

The temperature: the higher the temperature, the faster the reaction rate.

Adding a catalyst: the presence of a catalyst will increase the rate of a reaction.

1. Concentration

An increase in the concentration implies that there are more particles of solute per unit volume of solution. When more particles are present in a given amount of space, a greater number of collisions will naturally occur between those particles. Since the rate of a reaction is dependent on the number of collisions occurring between reactants, the rate increases as the concentration increases.

2. Pressure of a gas

When the pressure of a gas is increased, its particles are forced closer together, decreasing the amount of empty space between the particles. Therefore, an increase in the pressure of a gas is also an increase in the concentration of the gas. For gaseous reactions, an increase in pressure increases the rate of reaction as a result of a greater number of collisions between reacting particles.

3. Surface area

Surface area is larger when a given amount of a solid is present as smaller pieces. A powdered reactant has a greater surface area than the same mass of reactant as a solid chunk. In order to increase the surface area of a substance, it may be broken into smaller pieces or ground into powder. Increasing the surface area allows more particles to be exposed and therefore increases the number of collisions that can take place per unit time.

Serious accidents have been caused by the failure to understand the relationship between surface area and reaction rate. An industry where this is important is flour mills. A grain of wheat is not very flammable. It takes significant effort to get a grain of wheat to burn. When the wheat is ground to make flour, it is ground into a fine powder and some of the powder gets scattered around in the air. A small spark is then sufficient to start a very rapid reaction which can destroy the entire flour mill. Efforts are now made in flour mills to have huge fans circulate the air in the mill through filters to remove the majority of the flour dust particles.

Another example is in the operation of coal mines. Coal in the presence of oxygen burns, but it takes an effort to get the coal to start burning; and once it is burning, it burns slowly because only the surface particles are available to collide with oxygen particles. The inner particles of coal have to wait until the outer surface of the coal lump burns off before they can collide with oxygen. In coal mines, huge blocks of coal must be broken up before the coal can be brought out of the mine. In the process of breaking up the huge blocks of coal, drills are used to drill into the walls of coal. This drilling produces fine coal dust that mixes in the air; a spark from a tool can then cause a massive explosion in the mine. Coal dust explosions have contributed to the death of many miners. In modern coal mines, sprinklers are used to spray water through the air in the mine and this reduces the coal dust in the air, and eliminates coal dust explosions.

4. Temperature

When reactant particles are heated, they move faster and faster. This results in a greater frequency of collisions. A more important effect of the temperature increase is that the collisions occur with a greater force and are thus more likely to surmount the activation energy barrier and go on to form products. Increasing the temperature of a reaction increases the number of effective collisions between reacting particles, so the reaction rate increases.

We use the effects of temperature on reaction rate every day. Food storage is a prime example of how the temperature affects reaction rate. Food is stored in freezers and refrigerators to slow down the processes that cause it to spoil. The decrease in temperature decreases the rate at which food will break down or be broken down by bacteria.

When milk, for example, is stored in the refrigerator, the molecules in the milk have less energy. This means that while molecules will still collide with other molecules, fewer of them will react (which means in this case ‘go sour’) because the molecules do not have sufficient energy to overcome the activation energy barrier. The molecules do have energy and are colliding, however, and so, over time, even in the refrigerator, the milk will spoil. Eventually the higher energy molecules will gain the energy needed to react and when enough of these reactions occur, the milk becomes ‘soured’.

However, if that same carton of milk was at room temperature, the milk would react much more quickly. Most of the molecules would have sufficient energy to overcome the activation energy barrier at room temperature, and many more collisions would occur. This allows for the milk to spoil in a much shorter amount of time.

This is also the reason why most fruits and vegetables ripen in the summer when the temperature is much warmer. When a banana ripens, numerous reactions occur that produce all the compounds that we expect to taste in a banana. But this can only happen if the temperature is high enough to allow the reactions that make those products.

Catalysis

A catalyst increases reaction rates in a slightly different way from other methods of increasing reaction rate. The function of a catalyst is to lower the activation energy so that a greater proportion of the particles have enough energy to react. A catalyst can lower the activation energy for a reaction by:

• orienting the reacting particles in such a way that successful collisions are more likely
• reacting with the reactants to form an intermediate state (called the activated complex) that requires lower energy to form the product.

Some metals, for example platinum, copper and iron, can act as catalysts in certain reactions. In our own bodies, we have enzymes that are catalysts, which help to speed up biological reactions.

Make sure you can illustrate the action of a catalyst using an energy profile diagram. These were studied in the section on energy changes in chemical reactions in level 3.

Much research is done by industrial companies and university research laboratories to find out how catalysts work and to improve their effectiveness. If catalytic activity can be improved, it may be possible to lower the temperature and/or the pressure at which the process occurs and thus save fuel which is one of the major costs in a large-scale chemical process. Further, it may be possible to reduce the amount of reactants that are wasted forming unwanted by-products.

An important application of catalysts is to speed up the hardening of fibreglass resin. Without the addition of a catalyst, the soft resin takes a very long time to harden which would be a disadvantage in the manufacturing industry that uses fibreglass. A similar application of catalysts is used to ensure the rapid hardening of cement.

Exercise 1.1

1. Explain the following terms:
   1. reaction rate
   2. activation energy
   3. concentration
   4. collision theory
   5. catalyst
2. List the six factors that can determine the rate of a reaction.
3. State which of the following will happen faster and explain your reasoning:
   1. lighting a fire using small twigs or large logs
   2. dissolving a lump of sugar in warm water or cold water
   3. taking a headache tablet or a headache powder of the same mass

The full solutions are at the end of the unit.

Chemical equilibrium

To define chemical equilibrium we need to ask some important questions about reactions:

• Does a reaction always run its full course so that all the reactants are used up?
  .
  When all the reactants in a reaction are used up the reaction is said to have gone to completion. However, in some reactions not all the reactants are used. Not all reactions go to completion.
• Does a reaction always proceed in the same direction, or can it be reversed? In other words, does a reaction always proceed reactants → products, or is it possible that a reaction will reverse and go products → reactants?
  .
  Reactions that go to completion are irreversible. However, in some reactions the reactants form products (in a forward reaction), and the products can change back into reactants (in a reverse reaction).
• Can a reaction reach a point where reactants are still present, but there does not seem to be any further change taking place in the reaction?
  .
  In a reversible reaction, the forward reaction starts at a fast rate but as the reactants get used up the rate will slow down. But as the amount of product increases the reverse reaction will start speeding up. Eventually the rate of the forward reaction (reactants → products) equals the rate of the reverse reaction (products → reactants).
  .
  At this point there are still reactants present but there does not appear to be any further change taking place. As fast as reactants are used up to make products, products are used up to replace reactants. The reaction is said to be in chemical equilibrium.
  .
  Chemical equilibrium is the state of a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. While a reaction is in equilibrium the concentration of the reactants and products remain constant.

Activity 1.2: Demonstrate equilibrium

Time required: 5 minutes

What you need:

• 2 glasses
• a saucer
• Vaseline
• water

What to do:

1. Half-fill two glasses with water and mark the level of the water in each case.
2. Place some Vaseline around the rim of one glass and place the saucer on top of the glass.
3. Leave the glasses and, over the course of a day or two, observe how the water levels in the two beakers change.

What did you find?

You should notice that in the beaker that is uncovered, the water level drops more quickly than in the covered beaker. This is because of evaporation. In the beaker that is covered, there is an initial drop in the water level, but after a while evaporation appears to stop and the water level in this beaker is higher than that in the one that is open.

In the open glass, liquid water becomes water vapour as a result of evaporation and the water level drops. A small amount of gas molecules will condense again, but because the gas molecules can escape from the system there is much less condensation than evaporation.

In the covered glass, evaporation also takes place. However, in this case, the vapour comes into contact with the surface of the saucer, and it cools and condenses to form liquid water again. This water is returned to the glass. Once condensation has begun, the rate at which the water level drops will start to decrease. At some point, the rate of evaporation will be equal to the rate of condensation, and there will be no change in the water level in the beaker. This can be represented as follows:

liquid ⇌ vapour

In this example, the reaction (in this case, a change in the phase of water) can proceed in either direction. In the forward direction there is a change in phase from liquid to gas (water vapour). The reverse reaction can also take place, when vapour condenses to form liquid again.

An open system is one in which matter or energy can flow into or out of the system. In the liquid-gas demonstration in activity 1.2, the uncovered glass was an example of an open system because the glass could be heated or cooled (a change in energy), and water vapour (the matter) could evaporate from the beaker. An equilibrium will not be established because the water vapour can escape from the glass.

A closed system is one in which energy can enter or leave, but matter cannot. The covered glass is an example of a closed system. The glass can still be heated or cooled, but water vapour cannot leave the system because the saucer is a barrier. Condensation changes the vapour to liquid and returns it to the beaker. In other words, there is no loss of matter from the system.

Take note!

There are two requirements for a chemical equilibrium to be established:

• the reaction must be reversible
• the reaction must take place in a closed system.

Le Chatelier’s principle

Once a reaction reaches equilibrium, the amount of product remains constant. This is not a desirable situation in an industrial process. It is therefore important to find ways to ‘shift’ the equilibrium. This means changing the conditions under which the reaction is happening to make one of the reactions happen faster than the other. In industry this would be to make the forward reaction happen faster than the reverse reaction so that the yield of the product can be increased.

Any factor that can affect the rate of either the forward or reverse reaction relative to the other can potentially affect the equilibrium position. The following factors can change the chemical equilibrium position of a reaction:

• concentration (for solutions and gases)
• temperature
• pressure (for gaseous reactants).

It is important to understand what effect a change in one of these factors will have on a system that is in chemical equilibrium. However, performing an experiment every time to find out would waste a lot of time. Towards the end of the 1800s the French chemist Henry Louis Le Chatelier came up with principle to predict those effects.

Le Chatelier’s principle helps to predict what effect a change in temperature, concentration or pressure will have on the position of the equilibrium in a chemical reaction. This is very important, particularly in industrial applications, where yields must be accurately predicted and maximised.

Le Chatielier’s principle states that, when a stress (change in pressure, temperature or concentration) is applied to a system in equilibrium, the equilibrium will change in such a way as to reduce the effect of the stress.

Take note!

When one of the above conditions is manipulated and causes the equilibrium position to change, we use the following phrases to describe the effect:

• the equilibrium shifts to the right/left or
• the forward/reverse reaction is favoured.

This indicates that one of the reactions (forward or reverse) will happen faster than the other.

The effect of concentration on equilibrium

If the concentration of a substance is changed, the equilibrium will shift to minimise the effect of that change.

If the concentration of a reactant is increased, the equilibrium will shift in the direction of the reaction that uses the reactants, so that the reactant concentration decreases. The forward reaction is favoured, or the equilibrium shifts to the right.

The forward reaction is also favoured if the concentration of the product is decreased, so that more product is formed.

If the concentration of a reactant is decreased, the equilibrium will shift in the direction of the reaction that produces the reactants, so that the reactant concentration increases. The reverse reaction is favoured, or the equilibrium shifts to the left.

The reverse reaction is also favoured if the concentration of the product is increased, so that product is used.

For example, look at the reaction between sulfur dioxide and oxygen to produce sulfur trioxide:

[latex]\scriptsize 2\text{S}{{\text{O}}_{{2\text{ (g)}}}}+\text{ }{{\text{O}}_{{2\text{ (g)}}}}\text{ }\rightleftharpoons \text{ 2S}{{\text{O}}_{{3\text{ (g)}}}}[/latex]

If the [latex]\scriptsize \text{S}{{\text{O}}_{2}}[/latex] or [latex]\scriptsize {{\text{O}}_{2}}[/latex] concentration was increased, Le Chatelier’s principle predicts that the equilibrium will shift to decrease the concentration of reactants. The forward reaction uses up [latex]\scriptsize \text{S}{{\text{O}}_{2}}[/latex] and [latex]\scriptsize {{\text{O}}_{2}}[/latex], so it will be favoured (happen faster than the reverse) and more [latex]\scriptsize \text{S}{{\text{O}}_{3}}[/latex] will be made.

If the [latex]\scriptsize \text{S}{{\text{O}}_{2}}[/latex] or [latex]\scriptsize {{\text{O}}_{2}}[/latex] concentration was decreased, Le Chatelier’s principle predicts that the equilibrium will shift to increase the concentration of reactants. The reverse reaction makes [latex]\scriptsize \text{S}{{\text{O}}_{2}}[/latex] and [latex]\scriptsize {{\text{O}}_{2}}[/latex], so it will be favoured and the amount of [latex]\scriptsize \text{S}{{\text{O}}_{3}}[/latex] will decrease.

If the concentration of [latex]\scriptsize \text{S}{{\text{O}}_{3}}[/latex] was decreased, Le Chatelier’s principle predicts that the equilibrium will shift to increase the concentration of products. The forward reaction makes [latex]\scriptsize \text{S}{{\text{O}}_{3}}[/latex], so it will be favoured.

If the concentration of [latex]\scriptsize \text{S}{{\text{O}}_{3}}[/latex] was increased, Le Chatelier’s principle predicts that the equilibrium will shift to decrease the concentration of products. The reverse reaction uses up [latex]\scriptsize \text{S}{{\text{O}}_{3}}[/latex], so it will be favoured.

The effect of temperature on equilibrium

If the temperature of a reaction mixture is changed, the equilibrium will shift to minimise that change.

If the temperature is increased the equilibrium will shift to favour the reaction which will reduce the temperature. The endothermic reaction is favoured.

If the temperature is decreased the equilibrium will shift to favour the reaction which will increase the temperature. The exothermic reaction is favoured.

Take note!

Remember that the energy change in a chemical reaction is given by the [latex]\scriptsize \Delta \text{H}[/latex] value. For reversible reactions this value applies to the forward reaction. If the forward reaction is exothermic, the reverse reaction will be endothermic and vice versa.

For example, in the reaction below:

[latex]\scriptsize {{\text{N}}_{{2\text{ (g)}}}}\text{ + 3}{{\text{H}}_{{2\text{ (g)}}}}\rightleftharpoons \text{ 2N}{{\text{H}}_{{3\text{ (g)}}}}[/latex] [latex]\scriptsize \Delta \text{ H = -92 kJ}\text{.mo}{{\text{l}}^{{-1}}}[/latex]

The forward reaction shown is exothermic (shown by the negative value for [latex]\scriptsize \Delta \text{H}[/latex]). This means that the forward reaction, where nitrogen and hydrogen react to form ammonia, gives off heat, increasing the temperature. In the reverse reaction, where ammonia decomposes into hydrogen and nitrogen gas, heat is taken in by the reaction, cooling the vessel (the reverse reaction is endothermic).

An increase in temperature would therefore favour the reverse reaction because it is endothermic and will take in energy (cooling the container). The yield of ammonia ([latex]\scriptsize \text{N}{{\text{H}}_{3}}[/latex]) will decrease.

A decrease in temperature will favour the forward reaction because it is exothermic and will release energy (warming the container). The yield of [latex]\scriptsize \text{N}{{\text{H}}_{3}}[/latex] will increase.

The effect of pressure on equilibrium

If the pressure of a gaseous reaction mixture is changed the equilibrium will shift to minimise that change.

If the pressure is increased, the equilibrium will shift to favour a decrease in pressure.

If the pressure is decreased, the equilibrium will shift to favour an increase in pressure.

When the volume of a system is increased (and the temperature is constant), the pressure will decrease. There are fewer collisions with the walls of the container and therefore a lower pressure.

The equilibrium will shift in a direction that increases the number of gas molecules so that the pressure is also increased. So, to predict in which direction the equilibrium will shift to change the pressure, you need to look at the ratio of the number of moles of gas in the reactants and number of moles of gas in the products, using the balanced equation for the reaction. (Remember that one mole of any substance contains Avogadro’s number of particles)

For example, the equation for the reaction between nitrogen and hydrogen is shown below:

[latex]\scriptsize {{\text{N}}_{{2\text{ (g)}}}}\text{ + 3}{{\text{H}}_{{2\text{ (g)}}}}\rightleftharpoons \text{ 2N}{{\text{H}}_{{3\text{ (g)}}}}[/latex]

The ratio in the balanced equation is [latex]\scriptsize 1:3:2[/latex]. That is, for every [latex]\scriptsize 1[/latex] mole of [latex]\scriptsize {{\text{N}}_{2}}[/latex] gas, there are [latex]\scriptsize 3[/latex] moles of [latex]\scriptsize {{\text{H}}_{2}}[/latex] gas and [latex]\scriptsize 2[/latex] moles of [latex]\scriptsize \text{N}{{\text{H}}_{3}}[/latex] gas. Therefore, the ratio is [latex]\scriptsize 4[/latex] moles of reactant gas to [latex]\scriptsize 2[/latex] moles of product gas.

An increase in pressure will favour the reaction that decreases the number of moles of gas. There are fewer moles of product gas than reactant gas, so the forward reaction is favoured and the yield of [latex]\scriptsize \text{N}{{\text{H}}_{3}}[/latex] will increase.

A decrease in pressure will favour the reaction that increases the number of moles of gas. There are more moles of reactant gas, so the reverse reaction is favoured and the yield of [latex]\scriptsize \text{N}{{\text{H}}_{3}}[/latex] will decrease.

The effect of a catalyst on equilibrium

If a catalyst is added to a reaction, both the forward and reverse reaction rates will be increased. If both rates increase, then the concentrations of the reactants and products will remain the same. This means that a catalyst has no effect on the equilibrium position.

However, a catalyst will affect how quickly equilibrium is reached. This is very important in industry where the longer a process takes, the more money it costs. So, if a catalyst reduces the amount of time it takes to form specific products, it also reduces the cost of production. Catalysts will therefore be used in conjunction with changing other conditions to favour the forward reaction.

Example 1.1

Look at this reaction at equilibrium:

IMAGE/EQUATION

State, and explain, how the position of equilibrium would change if:

1. the pressure of the system is increased.
2. the temperature of the system is decreased.
3. some nitrogen dioxide is removed from the system.

The following strategy should be used to answer the question:

Step 1: Identify the disturbance or stress on the system

Step 2: Use Le Chatelier’s principle to decide how the system will respond

Step 3: Look at the given equation and decide whether the rate of the forward reaction or the rate of the reverse reaction is increased to counteract the stress

Step 4: Explain how the chosen reaction counteracts the stress

Step 5: Where appropriate, link equilibrium shift to any observed change in the system, e.g. a colour change, effect on yield

Solutions

1. The pressure on the system is increased.
   According to Le Chatelier’s principle the system will respond to reduce the pressure.
   The reverse reaction will be favoured because it produces fewer moles of gas moles which will decrease the pressure.
   The colour will go lighter/the yield of [latex]\scriptsize \text{N}{{\text{O}}_{2}}[/latex] will decrease.
2. The temperature on the system is decreased.
   According to le Chatelier’s principle the system will respond to increase the temperature.
   The reverse reaction will be favoured as it is exothermic and will release energy to raise the temperature (positive [latex]\scriptsize \Delta \text{H}[/latex] value applies to forward reaction).
   The colour will go lighter/the yield of [latex]\scriptsize \text{N}{{\text{O}}_{2}}[/latex] will decrease.
3. The concentration of nitrogen dioxide is decreased.
   According to le Chatelier’s principle the system will respond to increase the concentration of nitrogen dioxide.
   The forward reaction will be favoured as it produced nitrogen dioxide.
   The colour will go darker/the yield of [latex]\scriptsize \text{N}{{\text{O}}_{2}}[/latex] will increase.

Exercise 1.2

1. Bromomethane can be produced using the following chemical reaction:
   [latex]\scriptsize \text{C}{{\text{H}}_{3}}\text{O}{{\text{H}}_{{(g)}}}+\text{ HB}{{\text{r}}_{{(g)}}}\rightleftharpoons \text{ C}{{\text{H}}_{3}}\text{B}{{\text{r}}_{{(g)}}}+\text{ }{{\text{H}}_{2}}{{\text{O}}_{{(g)}}}\text{ }\Delta \text{H= -37}\text{.2 kJ}\text{.mo}{{\text{l}}^{{^{{-1}}}}}[/latex]
   .
   For each of the changes given below, predict the effect that the change would have on the yield of bromomethane.
   1. An increase in pressure.
   2. A continuous removal of bromomethane.
   3. An increase in temperature.
2. For the reaction below, state five conditions which will increase the yield of ammonia ([latex]\scriptsize \text{ N}{{\text{H}}_{3}}[/latex]).
   [latex]\scriptsize {{\text{N}}_{{2\text{ (}g\text{)}}}}\text{ + 3}{{\text{H}}_{{2(g)}}}\rightleftharpoons \text{ 2N}{{\text{H}}_{{3(g)}}}\text{ }\Delta \text{H }[/latex]is negative

The full solutions are at the end of the unit.

The equilibrium constant

When a reaction is at equilibrium the concentrations (or amounts) of all substances remains constant because as fast as they are being made, they are being used up. It is useful to know how much of each substance is in the container – In particular the amount of products compared to the amount of reactants. A simple ratio from the balanced chemical equation gives us a number called the equilibrium constant ([latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex]).
[latex]\scriptsize {{\text{K}}_{\text{c}}}=\text{ }\displaystyle \frac{{[\text{products }\!\!]\!\!\text{ }}}{{\text{ }\!\![\!\!\text{ reactants }\!\!]\!\!\text{ }}}[/latex] where square brackets mean concentration.

The equilibrium constant is defined as the ratio between the concentration of the products and reactants in a chemical reaction.

When the concentration of the reactants is much larger than the concentration of the products, the [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] will be small (less than one). When the concentration of the reactants is less than that of the products, the [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] will be greater than one.

The formula for [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] has the concentration of the products in the numerator and the concentration of reactants in the denominator. A [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] greater than 1 means that there are more products than reactants in the container at equilibrium.

A [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] between 0 and 1 means that there are more reactants than products in the container at equilibrium.

Take note!

[latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] cannot be [latex]\scriptsize 0[/latex] or less.

Yield describes the quantity of product in the container.

A high [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] value means that:

• the concentration of products is high
• the reaction has a high yield.

A low [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] value (close to 0) means that:

• the concentration of reactants is high
• the reaction has a low yield.

Exercise 1.3

1. Define and give the symbol for the equilibrium constant.
2. Why is it useful to know the value of the equilibrium constant for a particular reaction?
3. A certain reaction has an equilibrium constant of [latex]\scriptsize 7.5\text{ x 1}{{\text{0}}^{{-2}}}[/latex]. What information does this give about the reaction?

The full solutions are at the end of the unit.

Summary

In this unit you have learnt the following:

• The average rate of a reaction describes how quickly reactants are used, or how quickly products form.
• A number of factors can affect the average rate of a reaction. These include the nature of the reactants, the concentration of solutions, the surface area of solids or pressure of gases, the temperature of the reaction and the presence or absence of a catalyst.
• Collision theory provides one way of explaining why each of the factors can affect the average rate of a reaction.
• In order for a reaction to happen the particles of the reactants need to collide with sufficient energy and in the correct orientation. These are called effective or successful collisions.
• A catalyst is used to lower the activation energy so that the reaction is more likely to take place. A catalyst does this by providing an alternative, lower energy pathway for the reaction.
• A catalyst therefore speeds up a reaction but remains unchanged after the reaction is complete.
• A reaction is reversible when reactants can react to form products, and products can react to form the reactants again.
• A reaction is in chemical equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.
• In an open system energy and matter can enter and leave the system. In a closed system energy can enter and leave the system, but matter cannot.
• Le Chatelier’s principle states that if an external stress (change in pressure, temperature, or concentration) is applied to a system in chemical equilibrium, the equilibrium will change in such a way as to reduce the effect of the stress.
• The principles of equilibrium are very important in industrial applications so that productivity can be maximised.
• The equilibrium constant ([latex]\scriptsize \displaystyle {{\text{K}}_{\text{c}}}[/latex]) gives information about the concentration of the reactants and products at equilibrium.
• A high [latex]\scriptsize \displaystyle {{\text{K}}_{\text{c}}}[/latex] value means that the concentration of products at equilibrium is high, and the reaction has a high yield of the products. A low [latex]\scriptsize \displaystyle {{\text{K}}_{\text{c}}}[/latex] value means that the concentration of products at equilibrium is low, and the reaction has a low yield of the products.

Unit 1: Assessment

Suggested time to complete: 30 minutes

1. Give two reasons why a collision between reactant particles may not result in the formation of a product.
2. Study the descriptions of the following reactions and explain the difference in the rates of reaction for Experiment X and Y in each case:
   1. Experiment X: a solid piece of magnesium, mass [latex]\scriptsize 0.05\text{ g}[/latex], is placed in [latex]\scriptsize 100\text{ ml}[/latex] of hydrochloric acid of concentration [latex]\scriptsize 0.01\text{ mol}\text{.d}{{\text{m}}^{{-3}}}[/latex] and vigorous bubbling occurs.
      Experiment Y: a solid piece of zinc, [latex]\scriptsize 0.05\text{ g}[/latex], is placed in [latex]\scriptsize 100\text{ ml}[/latex] of hydrochloric acid of concentration [latex]\scriptsize 0.01\text{ mol}\text{.d}{{\text{m}}^{{-3}}}[/latex] and bubbles form slowly.
   2. Experiment X: [latex]\scriptsize 50\text{ ml}[/latex] of potassium iodate solution of concentration [latex]\scriptsize 0.01\text{ mol}\text{.d}{{\text{m}}^{{-3}}}[/latex] is added to [latex]\scriptsize 100\text{ ml}[/latex] of a sodium sulphite and starch mixture. The product makes the solution go blue in [latex]\scriptsize 20\text{ s}[/latex].
      Experiment Y: [latex]\scriptsize 50\text{ ml}[/latex] of potassium iodate solution of concentration [latex]\scriptsize 0.001\text{ mol}\text{.d}{{\text{m}}^{{-3}}}[/latex] is added to [latex]\scriptsize 100\text{ ml}[/latex] of sodium sulphite and starch mixture. The product makes the solution go blue in [latex]\scriptsize \text{35 s}[/latex] .
   3. Experiment X: A sugar cube is dipped in cigarette ash and ignites as soon as a flame is placed near it.
      Experiment Y: An undipped sugar will not ignite even when placed in a flame for [latex]\scriptsize 30\text{ s}[/latex].
   4. Experiment X: Calcium carbonate pieces of total mass [latex]\scriptsize 0.1\text{ g}[/latex] are placed in [latex]\scriptsize 100\text{ ml}[/latex] of hydrochloric acid of concentration [latex]\scriptsize 0.01\text{ mol}\text{.d}{{\text{m}}^{{-3}}}[/latex] and bubbles form slowly.
      Experiment Y: Calcium carbonate powder of total mass [latex]\scriptsize 0.1\text{ g}[/latex] is placed in [latex]\scriptsize 100\text{ ml}[/latex] of hydrochloric acid of concentration [latex]\scriptsize 0.01\text{ mol}\text{.d}{{\text{m}}^{{-3}}}[/latex]and bubbles form vigorously.
3. Explain why a baker places bread dough in a warm place to rise, rather than in a fridge.
4. The following theoretical reaction is at equilibrium. State whether the statements about the reaction are TRUE or FALSE. Correct the statement if it is FALSE.
   [latex]\scriptsize \text{ A + B }\rightleftharpoons \text{ AB}[/latex]
   1. The rate of the forward reaction is equal to that of the reverse reaction.
   2. The amount of the reactants and products are equal.
5. Define Le Chatelier’s principle.
6. Water gas is a mixture of carbon monoxide and hydrogen. It is produced by passing steam over red-hot coals according to the following reaction:
   [latex]\scriptsize {{\text{C}}_{{(s)}}}+\text{ }{{\text{H}}_{2}}{{\text{O}}_{{(g)}}}\text{ }\rightleftharpoons \text{ }{{\text{H}}_{{2\text{ (g)}}}}\text{ + C}{{\text{O}}_{{(g)}}}\text{ }\Delta \text{H=131 kJ}\text{.mo}{{\text{l}}^{{-1}}}[/latex]
   How is the equilibrium affected by:
   1. continually pumping in steam?
   2. using red-hot coals?
   3. continually adding more coal?
   4. operating the system at a high pressure?
7. Given the following reaction at equilibrium:
   [latex]\scriptsize \text{C}{{\text{H}}_{{3\text{ }(g)}}}\text{ + 2}{{\text{O}}_{{2\text{ (g)}}}}\text{ }\rightleftharpoons \text{ C}{{\text{O}}_{{2\text{ (g)}}}}\text{ + 2}{{\text{H}}_{2}}{{\text{O}}_{{(l)}}}\text{ + heat}[/latex]
   use Le Chatelier’s principle to explain the effect if the following changes are made.
   1. more oxygen is added to the equilibrium mixture
   2. the temperature of the system is decreased
   3. the pressure of the system is increased.
8. In a closed bottle of fizzy cooldrink that is ice cold, the carbon dioxide in the drink is in equilibrium with water as follows:
   [latex]\scriptsize \text{C}{{\text{O}}_{{2\text{ (g)}}}}\text{ + }{{\text{H}}_{2}}{{\text{O}}_{{(l)}}}\text{ }\rightleftharpoons \text{ }{{\text{H}}_{2}}\text{C}{{\text{O}}_{{3\text{ (aq)}}}}\text{ }\Delta \text{H }[/latex]is negative
   Use Le Chatelier’s principle to explain the following:
   1. As soon as the bottle is opened the gas bubbles up and escapes.
   2. If a warm bottle is opened the gas bubbles up more quickly.
9. The following reaction is at equilibrium:
   IMAGE/EQUATION
   1. Explain what would happen to the colour of the solution if more [latex]\scriptsize \text{F}{{\text{e}}^{{3+}}}[/latex] was added.
   2. When the solution is heated, the red colour fades. Is the forward reaction exothermic or endothermic? Explain your answer.
10. At a certain temperature, the [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] for the following reaction is [latex]\scriptsize 0.18[/latex].
    [latex]\scriptsize \text{PC}{{\text{l}}_{{3\text{ (g)}}}}\text{ + C}{{\text{l}}_{{2\text{ (g)}}}}\text{ }\rightleftharpoons \text{ PC}{{\text{l}}_{{5\text{ (g)}}}}[/latex] [latex]\scriptsize \Delta \text{H}[/latex]is negative
    1. What information does the [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex] value indicate about the yield of [latex]\scriptsize \text{PC}{{\text{l}}_{5}}[/latex]?
    2. List five ways to increase the yield of [latex]\scriptsize \text{PC}{{\text{l}}_{5}}[/latex].

The full solutions are at the end of the unit.

Unit 1: Solutions

Exercise 1.1

1. .
   1. reaction rate: how quickly reactants are used, or how quickly products form.
   2. activation energy: the minimum amount of energy required for a reaction to take place.
   3. concentration: a measure of the number of particles in a given amount of space
   4. collision theory: the theory that explains that a chemical reaction will only take place if particles collide with sufficient energy and in the correct orientation.
   5. catalyst: a substance that can increase the rate of a reaction by lower the amount of activation energy required
2. nature of the substances
   surface area of a solid
   concentration of a solution
   pressure of a gas
   temperature
   presence of a catalyst
3. Lighting a fire using small twigs – greater surface area will increase the probability of successful collision per unit time, therefore increase the reaction rate.
   Dissolving a lump of sugar in warm water – a higher temperature means particles will collide with greater energy (force) and the number of successful collisions per unit time will increase, therefore the reaction rate will increase.
   Taking a headache powder – increased surface area will increase the probability of successful collisions per unit time and therefore increase the reaction rate.

Back to Exercise 1.1

Exercise 1.2

1. .
   1. No effect – the number of moles of gas in the reactants is the same as the number of moles of gas in the products.
   2. The yield of bromomethane would increase – if concentration is decreased, the reaction that makes that substance will be favoured. The forward reaction makes bromomethane, therefore the yield will increase.
   3. The yield of bromomethane would decrease – if temperature is increased, the endothermic reaction will be favoured to reduce the temperature. The reverse reaction is endothermic (negative [latex]\scriptsize \Delta \text{H}[/latex] value indicates forward reaction is exothermic).
2. To increase yield, the forward reaction needs to be favoured. Ways to do this will be:
   • increase concentration of [latex]\scriptsize {{\text{N}}_{2}}[/latex] (forward reaction uses up [latex]\scriptsize {{\text{N}}_{2}}[/latex])
   • increase concentration of [latex]\scriptsize {{\text{H}}_{2}}[/latex] (forward reaction uses up [latex]\scriptsize {{\text{H}}_{2}}[/latex])
   • decrease concentration of [latex]\scriptsize \text{N}{{\text{H}}_{3}}[/latex] (forward reaction makes [latex]\scriptsize \text{N}{{\text{H}}_{3}}[/latex])
   • decrease the temperature (forward reaction is exothermic so will release heat to increase temperature)
   • increase the pressure (the products contain more moles of gas compared to the reactants)

Back to Exercise 1.2

Exercise 1.3

1. The equilibrium constant is the ratio between the concentration of the products and reactants in a chemical reaction. The symbol is [latex]\scriptsize {{\text{K}}_{\text{c}}}[/latex].
2. The equilibrium constant indicates whether there are more products in the reaction vessel at equilibrium or more reactants.
3. The value for the equilibrium constant is less than1, therefore, when the reaction has reached equilibrium there are more reactants in the container than products. The equilibrium lies to the left and the yield of the product is low.

Back to Exercise 1.3

Unit 1: Assessment

1. The colliding particles may not have sufficient energy to overcome the activation energy requirement, or they may not be colliding in the correct orientation in order to break their bonds.
2. .
   1. Experiment X happened at a faster rate because magnesium is a more reactive metal than zinc, so the nature of the substance determined which reaction happened faster.
   2. Experiment X happened faster because the concentration of the potassium iodate solution was greater. The higher the concentration of a solution, the greater the probability of effective collisions per unit time, therefore the faster the reaction rate.
   3. Experiment X happened faster because the cigarette ash acted as a catalyst. Catalysts speed up reaction rate by lowering the activation energy requirement for the reaction.
   4. Experiment Y happened faster because the magnesium powder has a greater surface area therefore more effective collisions can occur per unit time and speed up the reaction rate.
3. Reaction rates are faster at higher temperatures. The reaction in the bread dough that causes it to rise will happen faster in a warm place than in a cold place.
4. .
   1. TRUE (definition of equilibrium).
   2. FALSE. The amount of the reactants and products remains constant.
5. Le Chatelier’s principle states that if an external stress (change in pressure, temperature, or concentration) is applied to a system in chemical equilibrium, the equilibrium will change in such a way as to reduce the effect of the stress.
6. .
   1. continually pumping in steam favours the forward reaction (the equilibrium shifts to the right), and the concentration of the products will increase.
   2. using red-hot coals favours the forward reaction because it is endothermic and will absorb the excess heat. The equilibrium shifts to the right and the concentration of the products will increase.
   3. continually adding more coal will not affect the equilibrium. It is not possible to change the concentration of a solid – only solutions and gases.
   4. high pressure will favour the reverse reaction because the reactants contain less moles of gas than the products.
7. .
   1. If more oxygen is added to the system the equilibrium will shift to use up oxygen. The forward reaction will therefore be favoured, and the concentration of the products will increase.
   2. If the temperature is decreased the equilibrium will shift to produce heat. The forward reaction produces heat so it will be favoured, and the concentration of the products will increase.
   3. If the pressure is increased the reaction that produces fewer moles of gas will be favoured in order to reduce the pressure. The forward reaction will be favoured because it produces only one mole of gas compared to the reverse reaction which produces three moles of gas.
8. .
   1. Opening the bottle reduces the pressure and this favours the reverse reaction in order to produce more moles of gas in order to increase the pressure, but the bottle is open so the gas escapes.
   2. Heating favours the reverse reaction as it is endothermic. The reverse reaction produces the carbon dioxide gas.
9. .
   1. The solution would become darker red because increasing the concentration of [latex]\scriptsize \text{F}{{\text{e}}^{{3+}}}[/latex] will favour the forward reaction in order to use up the extra [latex]\scriptsize \text{F}{{\text{e}}^{{3+}}}[/latex] and the forward reaction produces the red [latex]\scriptsize \text{FeCN}{{\text{S}}^{{2+}}}[/latex] ions.
   2. If the red colour fades, the reverse reaction is being favoured by heating. Therefore the reverse reaction must be endothermic. Thus the forward reaction must be exothermic.
10. .
    1. The yield of [latex]\scriptsize \text{PC}{{\text{l}}_{3}}[/latex] is low.
    2. The forward reaction must be favoured so the following changes can be made:
       • increase the concentration of [latex]\scriptsize \text{PC}{{\text{l}}_{3}}[/latex]
       • increase the concentration of [latex]\scriptsize \text{C}{{\text{l}}_{2}}[/latex]
       • decrease the concentration of [latex]\scriptsize \text{PC}{{\text{l}}_{5}}[/latex]
       • decrease the temperature (forward reaction is exothermic)
       • decrease the pressure (forward reaction produces less moles of gas).

Back to Unit 1: Assessment