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 10cm3
0.1M sodium hydroxide (stoichiometry)
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)
SO42- (aq)
Cu(s) + Zn2+(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) + 2eZn(s)
Cu(s) reduction
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.
MnO4-(aq)
(one
negative)
overall
charge
+ 8H+(aq)
(eight
positive)
= two
positive
+ 5e(five
negative)
Mn2+(aq)
(two
positive)
overall
charge
+ 4H2O(l)
(no
charge)
= two
positive
Example:
This example is not in the data book.
Br2(l)
BrO- (aq)
Step 1 Br2(l)
2 BrO- (aq)
Step 2 (this time add H2O(l) to the lhs first to balance the oxygens on
the rhs)
Br2(l) + 2H2O(l)
2 BrO- (aq)
Step 3 (now add H+(aq) to the rhs to balance the hydrogens on the
lhs)
Br2(l) + 2H2O(l)
2 BrO- (aq) + 4H+(aq)
Step 4 ( two electrons have to be added to the rhs to make the
overall charge zero - the same as the lhs)
Br2(l) + 2H2O(l)
2 BrO- (aq) + 4H+(aq) + 2e-