The concept of dynamic equilibrium

Some reactions are able to go in two directions; forward and reverse. They are known as reversible reactions.
The forward and reverse reactions occur at the same time and never stop. As a result, they are called dynamic reactions.
Example: the reaction of iron(III) ions with thiocyanate ions

Process of Equilibrium

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When the rate of the forward reaction is equal to the rate of the reverse reaction the reaction is said to have reached equilibrium.

Be aware!
At equilibrium, the concentrations of the reactants and products are constant but are not necessarily equal.

When there are more products than reactants present the position of equilibrium lies to the right.
Example: in ester formation there are more products than reactants at equilibrium

When there are more reactants than products present the position of equilibrium lies to the left.
Example: in water, only a small proportion of the molecules have split to form ions at equilibrium:

It doesn't matter whether the reaction starts with 100% reactants or 100% products, the reaction will always reach the same equilibrium position.

Shifting the equilibrium position

Making changes to the concentration, pressure or temperature of a reaction can alter the position of the equilibrium.
The rule is that any change made to a reaction which is in equilibrium, will result in the equilibrium position moving to minimise the change made (Le Chatelier's Principle).
Changes to the concentration of a reactant or product
When chlorine gas dissolves in water, the following equilibrium is produced:

If potassium chloride (a source of chloride ions) is added to the equilibrium mixture, the equilibrium will shift to the left to remove the chloride ions added.
If potassium hydroxide is added, the hydroxide ions will react with the hydrogen ions and remove them from the mixture. The equilibrium will now move to the right to replace the lost hydrogen ions.
Changes to temperature or pressure
A good example of how changes to temperature and pressure alter the position of equilibrium can be seen in the Haber Process (the industrial manufacture of ammonia).
During the manufacture of ammonia the following equilibrium is present (which lies to the left):

Temperature
In the Haber Process, the forward reaction is an exothermic reaction. Although it is not a substance, in exothermic reactions heat can be imagined to be a product:

If the temperature is increased, then the equilibrium will shift to the left (the endothermic direction) to remove the extra heat added. This is why only a moderately high temperature (380 - 450°C) is used in the Haber Process.

Be aware!
Increasing the temperature always favours the endothermic reaction. Decreasing the temperature always favours the exothermic reaction.

Pressure
Reactions in which the equilibrium mixture is made up of only liquids and/or solids will not be affected by changes in pressure.
If there is at least one gas present in the equilibrium mixture then a change in the pressure may affect the position of the equilibrium.
Using the balanced equation, it can be seen that there are 4 volumes of reactants and only 2 volumes of product:

If the pressure is increased, the equilibrium will shift towards the right creating more product. This is because the volume of the product is smaller than the volume of the reactants and so the pressure will reduce to minimise the change.
If the pressure is reduced, the equilibrium will shift towards the left resulting in more reactants. This is because the volume of the reactants is greater than the volume of the product and so the pressure will be increased to minimise the change.
In the Haber Process the actual pressure used is around 200 atmospheres which favours more product:

·  High pressure always favours the side with the lowest volume of gases.

·  Low pressure always favours the side with the highest volume of gases.

Be aware!
Changes in pressure will not affect the position of equilibrium if there is an equal number of moles of gases on both sides of the equation. Remember liquids and solids contribute nothing to the volume of the equilibrium mixture.

Adding a catalyst
A catalyst reduces the time taken to reach equilibrium but does not change the position of the equilibrium. This is because the catalyst increases the rate of the forward and reverse reactions by the same amount.
In the Haber process, the catalyst used is iron.
The Haber Process is a continuous process. Ammonia is constantly being separated from the reaction mixture and the unreacted nitrogen and hydrogen is recycled back into the reaction vessel. As a result, equilibrium is never reached. Instead, the equilibrium is constantly shifting to the right to replace the ammonia which has been removed.

The concept of strong and weak

Strong acids

In aqueous solution, a strong acid is one that completely dissociates (splits up) into ions.
Example:
hydrochloric acid

This is not a reversible reaction and 100% of the HCl molecules dissociate into ions.
Other strong acids include nitric acid and sulphuric acid.

Weak acids

In aqueous solution, a weak acid is one that only partially dissociates into ions. As a result, a state of equilibrium is produced.
Example:
carbonic acid

Since the equilibrium lies very much to the left there are few hydrogen ions produced. This leads to the weakly acidic nature of carbonic acid.
In the case of both ethanoic and sulphurous acids, the equilibrium lies very much to the left:
sulphurous acid

ethanoic acid

Characteristics of strong and weak acids

Equimolar solutions of strong and weak acids differ in pH, conductivity and reaction rate but not in stoichiometry.

Property / 0.1M HCl / 0.1M CH3COOH
pH / 1 / 3
conductivity / high / low
reaction with magnesium / fast / slow
volume of acid required to neutralise 10cm3 of 0.1M sodium hydroxide (stoichiometry) / 10cm3 / 10cm3

Strong bases

In aqueous solution, a strong base is one that completely ionises.
Example:
sodium hydroxide

This is not a reversible reaction and 100% of the NaOH(s) splits into free ions.

Weak bases

In aqueous solution, a weak base is one that only partially ionises. Just as with weak acids, a state of equilibrium is produced.
Example:
ammonia solution

Since the equilibrium lies very much to the left, there are few hydroxide ions produced. This leads to the weakly basic nature of ammonia solution.

Characteristics of strong and weak bases

Equimolar solutions of strong and weak bases differ in pH and conductivity but not in stoichiometry.

Property / 0.1M NaOH / 0.1M NH3(aq)
pH / 13 / 11
conductivity / high / low
volume of base required to neutralise 10cm3 of 0.1M hydrochloric acid (stoichiometry) / 10cm3 / 10cm3

Be aware!
The terms 'strong' and 'weak' must not be confused with 'concentrated' and 'dilute'. 'Strong' and 'weak' refer to the ability of an acid or base to ionise. 'Concentrated' and 'dilute' refer to the quantity of acid or base present in the aqueous solution.

The pH of salt solutions

The pH of a salt depends on whether the acid and base from which it is formed are strong or weak.
The table below is a summary of strong and weak acids and bases. Decide whether the acids and bases below are strong or weak. Move the cursor over a box to reveal if you are correct.


In general, the salt of a strong acid and weak base will form an acidic solution when dissolved in water.
Example:
ammonium nitrate
When ammonium nitrate dissolves in water the following two equations are involved in producing the ions present in the solution:

This additional equilibrium removes hydroxide ions but not hydrogen ions. The water equilibrium shifts to the right to replace these lost hydroxide ions and, in so doing, also produces more hydrogen ions. There are now more hydrogen ions than hydroxide ions present, leading to a pH of less than 7.
The salt of a weak acid and strong base will form an alkaline solution when dissolved in water.
Example:
sodium ethanoate
When sodium ethanoate dissolves in water the following two equations are involved in producing the ions present in the solution:

This additional equilibrium removes hydrogen ions but not hydroxide ions. The water equilibrium shifts to the right to replace these lost hydrogen ions and, in so doing, also produces more hydroxide ions. There are now more hydroxide ions than hydrogen ions present, leading to a pH of more than 7.
Soaps are also salts of weak acids (such as stearic and oleic acid) and strong bases (such as sodium or potassium hydroxide). As a result, soaps are usually alkaline in nature.
The salt of a strong acid and strong base will form a neutral solution when dissolved in water
Example:
sodium chloride
When sodium chloride dissolves in water the following two equations are involved in producing the ions present in the solution:

The equal proportions of hydrogen and hydroxide ions in the solution are not disturbed by additional equilibria and so the pH of the solution is equal to 7.

Oxidising and reducing agents

Oxidation is the loss of electrons by a substance during a chemical reaction. The substance which accepts these electrons is called the oxidising agent.
Reduction is the gain of electrons by a substance during a chemical reaction. The substance which donates these electrons is called the reducing agent.
The substances being oxidised and reduced and the oxidizing and reducing agents can be identified in ionic equations.
Example:
Zinc metal can displace copper ions from a solution of copper(11) sulphate. Copper metal and a solution of zinc sulphate are produced.
Ionic equation: Zn(s) + Cu2+(aq) + SO42 -(aq) Cu(s) + Zn2+(aq) + SO42- (aq)
The equation shows that the zinc atoms (metal) have formed positively charged zinc ions - they have lost electrons. This means the zinc atoms have been oxidised.
The positively charged copper ions have formed copper atoms - they have gained the electrons which the zinc atoms have lost. This means the copper ions have been reduced.
(The sulphate ions have neither lost nor gained electrons - they have not taken part in the reaction. They are spectator ions)
The copper ion is acting as the oxidizing agent because it has accepted electrons from the zinc atom ie. caused the zinc atom to be oxidised.
The zinc atom is acting as the reducing agent because it has donated electrons to the copper ion ie. caused the copper ion to be reduced.
Example:
When chlorine reacts with iron(II) ions in solution, chloride ions and iron(III) ions are produced:
Cl2(g) + 2Fe2+(aq) 2Cl-(aq) + 2Fe3+(aq)
The equation shows that the Fe2+ ions have formed Fe3+ ions - each ion has lost an electron. This means the Fe2+ ions have been oxidised.
The Cl2 molecules have formed Cl- ions - they have gained the electrons which the Fe2+ ions have lost. This means the Cl2 molecules have been reduced.
The Cl2 molecule is acting as the oxidising agent because it has accepted electrons from the Fe2+ ions ie. caused the Fe2+ ions to be oxidised.
The Fe2+ ion is acting as the reducing agent because it has donated an electron to the Cl2 molecule ie. caused the Cl2 molecules to be reduced

Ion-electron equations from the data booklet

Although oxidation and reduction happen together they are often shown separately in ion-electron equations. Many of these equations are shown in the SQA data booklet, page 11. They are written as reductions but can be reversed to give the oxidation equation.
When copper ions are displaced from solution by zinc atoms, the copper ions are reduced to copper atoms so the ion-electron equation is written as it is in the data booklet:
Cu2+(aq) + 2e- Cu(s)
The zinc atoms are oxidised to zinc ions so the ion-electron equation in the data booklet has to be reversed:
Zn(s) Zn2+(aq) + 2e-
The reduction and oxidation equations can be added together to make the overall redox equation - the electrons must be balanced.
Cu2+(aq) + 2e- Cu(s) reduction
Zn(s) Zn2+(aq) + 2e- oxidation
Add: Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) redox
If the electrons in the two ion-electron equations don't balance then the equations have to be multiplied up to make them balance before the two equations are added together.
Example:
When iron(III) nitrate reacts with sodium iodide, the iron(III) ions are reduced to iron(II) ions and the iodide ions are oxidised to iodine molecules ( the nitrate and sodium ions are spectator ions).
From the data booklet, the reduction equation is:
Fe3+(aq) + e- Fe2+(aq)
The oxidation equation is:
2I-(aq) I2(s) + 2e-
The reduction equation has to be multiplied by two to balance the number of electrons in the oxidation equation.
2Fe3+(aq) + 2e- 2Fe2+(aq)
The balanced reduction and oxidation equations can now be added to give the redox equation
2I-(aq) I2(s) + 2e- oxidation
2Fe3+(aq) + 2e- 2Fe2+(aq) reduction
Add: 2I-(aq)+ 2Fe3+(aq) I2(s) + 2Fe2+(aq) redox

Working out ion electron equations not in the data booklet

A number of redox reactions will not take place unless there is acid present - the acid supplies hydrogen ions. During the reaction these hydrogen ions react to form water.
Example:
Iron(II) ions are oxidised to iron(III) ions by reaction with acidified potassium permanganate.
The reduction equation is in the data booklet:
MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l)
This could have been worked out in a number of steps, if the formulae of the main reactant and product in the reduction are known. In this example:
MnO4-(aq) Mn2+(aq)
Step 1 Balance all the elements except hydrogen and oxygen. In this case there are the same number of Mn on each side of the equation (one):
MnO4-(aq) Mn2+(aq)
Step 2 If oxygen is unbalanced add H2O to the side short of oxygen - in this case the right hand side (rhs). Multiply up until the number of oxygens on each side is the same.
MnO4-(aq) Mn2+(aq) + 4H2O(l)
Step 3 If hydrogen is unbalanced add H+(aq) to the side short of hydrogen - in this case the left hand side (lhs). Multiply up until the number of hydrogens on each side is the same.
MnO4-(aq) + 8H+(aq) Mn2+(aq) + 4H2O(l)
(the order of steps 2 and 3 can be swapped if need be - see the next example)
Step 4 The total charge must be the same on both sides of the equation. Only electrons can be added to one side or the other. Balance the charge by adding the appropriate number of electrons to the side which is short of negative charge - in this case the lhs.