The idea of excess
In a balanced equation like the one below, it is often assumed that all
of the reacting chemicals change into products.
In that case, at the end of the reaction, no CaCO3 or HCl will be left
behind. However, if there is a shortage of, say, CaCO3 then the
reaction will stop when the CaCO3 runs out. Some HCl will be left
over, unable to react as there is no more CaCO3. The HCl is said to
be in excess.
Example:
Calculate which reactant is in excess when 10 g of calcium carbonate
reacts with 50 cm3 of 2 mol l-1 hydrochloric acid.
Step 1: Calculate the number of moles of each reactant present:
Step 2: From the balanced equation, work out the mole ratio for the
reactants:
Step 3: Using the number of moles in step 1, choose one reactant
and work out the number of moles of the other reactant needed to
react with it:
There is not enough HCl available to react with 0.1 moles of CaCO3.
This means that some of the CaCO3 will be left over when the HCl is
used up.
Answer: CaCO3 is the reactant which is in excess.
Potential energy
In a reaction mixture the reactants and products contain potential
energy. This potential energy is also known as enthalpy (symbol H).
During a chemical reaction the enthalpy of the reactants changes as
the reactants form new products.
The enthalpy change (symbol H) for a reaction can be calculated
from a potential energy diagram.
H = H(products) - H(reactants)
H is measured in kilojoules per mole (kJ mol
Exothermic reactions
In exothermic reactions some of the reactant's potential energy or
enthalpy is released into the surroundings, usually in the form of heat.
As a result, the enthalpy of the products is less than the enthalpy of
the reactants.
Be aware!
The enthalpy change for an exothermic reaction is always negative.
Example: A potential energy (enthalpy) diagram for an exothermic
reaction is shown below:
The activation energy (Ea) for the forward reaction is shown by (a):
Ea (forward) = H (activated complex) - H (reactants) = 200 - 150 = 50
kJ mol-1
The activation energy (Ea) for the reverse reaction is shown by (b):
Ea (reverse) = H (activated complex) - H (products) = 200 - 50 = 150
kJ mol-1
The enthalpy change for the reaction is shown by (c):
H = H (products) - H (reactants) = 50 - 150 = -100 kJ mol-1
Endothermic reactions
In endothermic reactions heat energy is taken in from the
surroundings and turned into potential energy in the products. As a
result, the enthalpy of the products is greater than the enthalpy of the
reactants.
Be aware!
The enthalpy change for an endothermic reaction is always positive.
Example: the potential energy (enthalpy) diagram for an endothermic
reaction is shown below
The activation energy (Ea) for the forward reaction is shown by (a):
Ea (forward) = H (activated complex) - H (reactants) = 400 - 100 = 300
kJ mol-1
The activation energy (Ea) for the reverse reaction is shown by (b):
Ea (reverse) = H (activated complex) - H (products) = 400 - 300 = 100
kJ mol-1
The enthalpy change for the reaction is shown by (c):
H = H (products) - H (reactants) = 300 - 100 = +200 kJ mol-1
Enthalpy changes
Enthalpy changes can also be calculated by experiment using the
equation:
H = cm
T
Where,
c = specific heat capacity of water (4.18 kJ kg-1 °C-1)
m = mass of water heated (kg)
T = change in temperature of the water (°C)
Be aware!
If there is a rise in temperature, the reaction is exothermic and
involves a negative enthalpy change.
If there is a drop in temperature the reaction is endothermic and
involves a positive enthalpy change.
Example:
When 90cm3 of methane is burned, the heat produced is used to
raise the temperature of 100cm3 of water from 20°C to 29.1°C.
Calculate the enthalpy change for the reaction (remember to convert
the volume of water into mass of water in kilograms. 1cm3 of water
has a mass of 0.001kg).
H = cm T
= 4.18 x 0.1 x 9.1
= 3.80 kJ
Be aware!
Combustion reactions are always exothermic so the enthalpy change
will always be negative. So, the enthalpy change for this reaction is =
-3.80 kJ.
Example:
When 1.5g of ammonium nitrate is dissolved in water, there is a drop
in temperature. The enthalpy change for the reaction is +0.335 kJ.
Calculate the enthalpy of solution for ammonium nitrate. The enthalpy
of solution is the enthalpy change when one mole of a substance
dissolves completely in water.
H = +0.335 kJ for 1.5g of ammonium nitrate
Gram formula mass (1 mole) of ammonium nitrate = 80g
The enthalpy of solution for ammonium nitrate is positive because it
involves a drop in temperature.
Example:
When 150cm3 of dilute nitric acid (concentration 1 mol l-1) is
neutralised by 150cm3 of sodium hydroxide there is a temperature
rise of 6.81°C. Calculate the enthalpy of neutralisation of nitric acid.
Be aware!
The enthalpy of neutralisation is the enthalpy change when one mole
of water is produced in the reaction of an acid with an alkali.
Step 1: Work out the enthalpy change for this reaction.
(there is a temperature rise so the enthalpy change is negative)
Step 2: Calculate the number of moles of water produced.
From the balanced equation it can be seen that the number of moles
of water is equal to the number of moles of nitric acid:
Step 3 : Calculate the enthalpy change for one mole of water.
H = -8.54 kJ for 0.15 moles of water
The enthalpy of neutralisation is roughly -57 kJ mol-1 for any acid and
alkali. This is because, omitting spectator ions, the reaction occurring
in neutralisation is always just:
Example:
In an experiment to determine the enthalpy of combustion of ethanol,
the following results were obtained:
Initial mass of ethanol and burner
Final Mass of ethanol and burner
Mass of ethanol burned
Initial temperature
Final temperature
Temperature rise
Mass of water
= 199.8g
= 199.6g
= 0.2g
= 22°C
= 33°C
= 11°C
= 0.1kg
Using these results, calculate the enthalpy of combustion of ethanol.
The enthalpy of combustion is the enthalpy change when one mole of
a substance burns completely in oxygen.
Step 1: Calculate the enthalpy change for the 0.2g of ethanol burned
(combustion reactions always have a negative enthalpy)
Step 2: Calculate the enthalpy change for one mole of ethanol
H = -4.60kJ for 0.2g of ethanol
Gram formula mass (1 mole) of ethanol = 46g
This value is much less than that quoted in the data booklet (-1367 kJ
mol-1).
Possible reasons for the lower experimental value include:



heat is lost during the experiment to the surrounding air and the
water container.
there may have been some incomplete combustion going on.
some of the ethanol may have escaped through evaporation.
Be aware!
The enthalpies of solution, combustion and neutralisation are
measured in kJ mol-1. Enthalpy changes not involving one mole of a
substance should be quoted in kJ only.
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Avogadro constant
Mole and number of formula units
One mole of a substance is its gram formula mass (GFM).
One mole of a substance contains 6.02 x 1023 formula units.
This number is known as the Avogadro Constant.
The Avogadro Constant can be found in the SQA data booklet and is
expressed as 6.02 x 1023 mol-1.
The term formula unit is a general term that relates to the type of
particles that make up a substance. In general, it refers to the formula
normally used for the substance.
In diamond, one formula unit is a carbon atom (C).
In oxygen, one formula unit is an oxygen molecule (O2).
In sodium chloride, one formula unit is one sodium ion and one
chloride ion (Na+Cl-).
In silicon dioxide, one formula unit is one silicon atom and two oxygen
atoms (SiO2).
Equimolar quantities of substances contain equal numbers of formula
units.
The idea of the mole links the mass of a substance to the number of
formula units it contains.
The calculations below involve calculating number of formula units
given number of moles and vice versa.
The link here is:
Example: How many molecules are contained in 0.65 moles of
sulphur dioxide, SO2?
Answer: 3.91 x 1023 molecules are in 0.65 moles of SO2.
Example: How many moles of hydrogen contain 5.5 x 1024 molecules
of H2?
Answer: 9.14 moles contain 5.5 x 1024 molecules of H2.
Mass and number of formula units
The examples involve calculating number of formula units from a
given mass and vice versa.
The link here is:
As the mass of moles is involved, it is necessary to calculate the
GFM of the substance.
Example: How many molecules are in 150g of water, H2O?
Answer: 5.02 x 1024 molecules are in 18g of water, H2O.
To work out the number of atoms, multiply the number of molecules
by three, which is the number of atoms in one molecule.
Example: What mass of carbon disulphide (CS2) contains 2.04 x 1022
molecules of CS2?
Answer: 2.58g contains 2.04 x 1022 molecules of CS2.
Example: What mass of sodium chloride (NaCl) would contain 9.03 x
1021 ions?
Note that one formula unit of NaCl contains two ions.
Answer: 0.44g of NaCl contains 9.03 x 1021 ions.
Calculating molar volume
The molar volume is the volume occupied by one mole of a gas.
The units used for the molar volume are l mol-1 (litres per mole).
Calculating molar volume from mass
Example:
976 cm3 of oxygen was found to have a mass of 1.3g
Calculate the molar volume of oxygen under these conditions:
Remember that the molar volume is the volume occupied by one
mole of oxygen, O2.
GFM of oxygen, O2, is 32.0g
Answer: 23.8 l is the volume of 32.0g of O2. So the molar volume
of oxygen is 24.0 l mol-1 under these conditions.
The above can be carried out for other gases under the same
conditions of temperature and pressure.
The molar volume is the same for all gases at the same temperature
and pressure.
Calculating molar volume from density
The molar volume of a gaseous element can be calculated from its
density.
The SQA data booklet shows the densities of selected elements.
Density figures are quoted using the units g cm-3 and are measured at
standard temperature and pressure (0°C and one atmosphere).
Example: Calculate the molar volume of argon at standard
temperature and pressure (s.t.p) using the density value quoted on
page 3 of the SQA data booklet.
Answer: 22.2 litres is the volume of 40.0g of argon. So the molar
volume of argon is 22.2 l mol-1 under these conditions.
Be aware!
The actual value for the molar volume of a gas depends on the
conditions of temperature and pressure used when making the
measurement.
In the second example above, the molar volume of argon was
calculated as 22.2 l mol-1 under conditions of standard temperature
and pressure (0°C and one atmosphere pressure). If measurements
had been made under different conditions the actual value for the
molar volume of argon calculated would be different.
Calculations involving molar volume
Volume to moles
Example: Under certain conditions the molar volume of methane,
CH4, is 23 l mol-1. How many moles of methane are present in 250
cm3 of the gas?
Answer: 0.011 moles are in 250 cm3 of CH4 gas.
Moles to volume
Example: What is the volume of 0.025 moles of hydrogen when the
molar volume is 24.0 l mol-1? Give your answer in cm3.
Answer: 600 cm3 is the volume of 0.025 moles of hydrogen.
Reacting volumes
Under the same conditions of temperature and pressure equal
number of moles of gas occupy the same volume.
This means that the volume of a gaseous reactant or product can be
calculated from the balanced equation for the reaction.
Volume of one gas from another gas
The examples below involve the calculation of the volume of a
product gas from the volume of a reactant gas and vice versa.
Example: What volume of propane, C3H8, must be burned in oxygen
to give 30 cm3 of carbon dioxide gas as a product (under the same
conditions of temperature and pressure)?
Answer: 10 cm3 of propane must be burned.
Example: What volume of carbon dioxide is made when 50 cm3 of
butene, C4H8, is completely burned in oxygen (under the same
conditions of temperature and pressure)?
Answer: 200 cm3 of CO2 is produced.
Reactant gas in excess
When a reaction takes place almost always one of the reactants will
be in excess unless exact quantities have been used.
The example that follows shows how to work out which reactant is in
excess when gases are involved and by how much.
Example: Which gas is in excess and by what volume, if 50 cm3 of
methane reacts with 125 cm3 of oxygen?
Select one of the reactants and calculate the volume of this reactant
which would be needed to react with the given volume of the other
reactant.
The volume available is greater than the volume needed so O2 is in
excess.
This means that the reaction quantities are determined by methane,
CH4.
Answer: O2 is in excess by 25 cm3.
The final total gas volume and composition for a reaction depends on
the volumes of any gases produced and the unreacted volume of the
excess gas reactant.
Be aware!
Note that the volumes of liquids and solids can be ignored in such
calculations. This is because the volumes of solids and liquids are
negligible compared to equimolar volumes of gases.
In the above example, the volume and composition of the resulting
gas mixture is :
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Using Hess's Law
Hess's law states that the enthalpy change in converting reactants
into products is the same regardless of the route taken.
So the enthalpy change for two routes from the same reactants to the
same products will be equal.
Example:
Hess's Law states that:
Hx =
H1 +
H2
If all but one of the above enthalpy changes are known then the
remaining enthalpy change can be calculated.
Example:
The formation of ethane can be represented by the following
equation:
Calculate the enthalpy change for the reaction using the enthalpies of
combustion of carbon, hydrogen and ethane given in the SQA data
book.
H1 = 2 × enthalpy of combustion of carbon (2 × -394).
H2 = 3 × enthalpy of combustion of hydrogen (3 × -286).
H3 = reverse of enthalpy of combustion of ethane. The enthalpy of
combustion of ethane has a negative value so the reverse will have a
positive value (+1560).
Using Hess's Law:
Answer: The enthalpy change for the reaction is -86 kJ mol-1
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Calculations involving pH
The pH of a solution depends on the concentration of H+(aq) ions
present.
A shorthand method of denoting concentration of hydrogen ions is
[H+(aq)]. The square brackets surrounding the ion symbol mean
'concentration in moles per litre'.
When the concentration of hydrogen ions is expressed using powers
of 10 the pH is equal to the negative of the power number.
Example:
The exact pH of a solution can be calculated using:
pH = -log[H+(aq)]
For most solutions the pH lies in the range 0 to 14.
Example: In a solution of pH 2 what is the concentration of H+(aq)?
Example: A salt solution has a hydrogen ion concentration of 10-9 mol
l-1. What is its pH?
Calculations Involving the ionic product
In pure water the concentration of hydrogen ions is equal to the
concentration of hydroxide ions.
The pH of pure water is 7.
This means that the concentrations of both the hydrogen and
hydroxide ions are 10-7 mol l-1.
Written, [H+(aq)] = [OH-(aq)] =10-7 mol l-1
Multiplying the concentration of hydrogen ions and the concentration
of hydroxide ions gives the ionic product for water:
Remember that the ionic product for water is given on the last page of
the Higher data booklet.
No matter what the solution is, the ionic product is always 10-14 mol2 l-2
pH [H+(aq)]
[OH-(aq)] [H+(aq)] × [OH-(aq)]
6
10-6 mol l-1 10-8 mol l-1 10-14 mol2 l-2
7
10-7 mol l-1 10-7 mol l-1 10-14 mol2 l-2
8 10-8 mol l-1 10-6 mol l-1 10-14 mol2 l-2
Example: In a solution of pH 4 what is the concentration of OH- (aq)
ions?
Example: If [OH-(aq)] is 0.001 mol l-1 what is the pH value?
Working out concentration
The concentration of a reactant can be calculated from the results of
a redox titration.
In order to calculate concentration we have to work out the number of
moles reacting and multiply that by the volume. The number of moles
of electrons involved must also be taken into account.
This can be done using the following relationship:
Example:
Calculate the concentration of an iron(II) sulphate solution given that
20 cm3 of the solution reacts completely with 23.5 cm3 of 0.02 mol l-1
potassium permanganate solution, in the presence of acid.
The oxidation equation is:
The reduction equation is:
Answer
Where, x refers to Fe2+(aq) and y refers to MnO4 -(aq)
Answer: The concentration of the iron(II) sulphate solution is
0.12 mol l-1
Example:
In an experiment to find the concentration of sodium sulphite solution,
25 cm3 samples of the solution were titrated with 0.01 mol l-1 acidified
potassium permanganate solution. The results were as follows:
Titre Volume of permanganate (cm3)
1
15.5
2
15.3
3
15.2
The redox equation for the reaction is:
Calculate the concentration of the sulphite solution.
Answer
Step 1
In order to work out n x and n y the oxidation and reduction equation
have to be written - they are found on page 11 of the SQA data
booklet.
From this n x = 2 and n y = 5
Step 2
To find V y the average titre must be calculated. The first volume is
always omitted. The other two titres are then averaged:
So, V y = 15.25 cm3
Step 3
From the question, V x = 25 cm3 and C y = 0.01 mol l-1
Step 4
C x can be calculated from:
Answer: The concentration of sodium sulphite solution is 0.015
mol l-1
Working out mass
In the Prescribed Practical Activity (PPA) the mass of a vitamin C
tablet has to be calculated by titrating a solution of the tablet with
iodine solution.
The number of moles of vitamin C has to first be calculated. This is
done indirectly by first working out the number of moles of iodine
required and then using the redox equation to work out the number of
moles of vitamin C present. Moles can then be converted to mass.
Example:
A 250 cm3 solution of vitamin C (C6H8O6) was prepared by dissolving
a tablet in de-ionised water. A 25 cm3 sample of the solution was
titrated against 0.015 mol l-1 iodine solution, using starch indicator.
The average titre was 17.4 cm3.
Calculate the mass of vitamin C in the original tablet, given the gram
formula mass of vitamin C is 176g and the redox equation:
Answer
Step 1 Calculate the number of moles of iodine (n):
n = C × V(litres) = 0.015 × 0.0174 = 2.61 ×10-4 moles
Step 2 Calculate the number of moles of vitamin C:
From the redox equation, 1 mole of iodine (I2) reacts with 1 mole of
vitamin C (C6H8O6) so, the number of moles of vitamin C in 25 cm3 is
2.61 × 10-4 moles.
Step 3 Calculate the number of moles of vitamin C in 250 cm3 (the
whole tablet):
Moles of vitamin C in 25 cm3 = 2.61 × 10-4 moles
Moles of vitamin C in 250 cm3 = 2.61 × 10-3 moles
Step 4 Calculate mass from number of moles:
GFM of vitamin C = 176g
So, mass of vitamin C in tablet = 176 × 2.61 × 10-3 = 0.46 g
Answer: Mass of vitamin C in one tablet is 0.46g
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Calculating mass and volume
In order to do these calculations you must know that:


the charge associated with 1 mole of electrons is 96,500 C.
the production of one mole of an element from its ion always
requires n x 96,500 C, where n is the number of moles of
electrons in the ion-electron equation.
Example
Calculate the mass of aluminium produced when molten aluminium
oxide is electrolysed using a current (I) of 20,000 A for 5 hours, 21
min and 40s.
Step 1: Work out the quantity of charge (Q) passed in the time (t)
given.
Step 2:From the ion-electron equation, work out the relationship
between the number of moles of electrons and number of moles of
product.
Step 3:Use the information in steps 1 and 2 to calculate the mass of
aluminium produced.
Answer: Mass of aluminium produced is 36 kg
The volume of gas produced during electrolysis can be calculated if
the molar volume is known.
Example
In the industrial production of chlorine gas, a current of 50,000 A was
passed through a salt solution for 1 hour. Calculate the volume of gas
which would be produced. (Take the molar volume to be 25.0 l mol-1)
Step 1: Work out the quantity of charge (Q) passed in the time given:
Step 2:From the ion-electron equation, work out the relationship
between the number of moles of electrons and number of moles of
product gas:
Step 3:Use the information from steps 1 and 2 and the molar volume
to calculate the volume of gas produced:
Answer: The volume of chlorine produced is 2.33 x 104 litres.
Calculating quantity of electricity
Example
Calculate the charge which would be required to produce 200 g of
magnesium each minute, when molten magnesium chloride is
electrolysed.
Step 1: From the ion-electron equation, work out the relationship
between the number of moles of electrons and number of moles of
product:
Step 2:Use the information from the question and step 1 to calculate
the charge:
Answer: The quantity of charge needed is 1.59 x 106 C
Example
Calculate the charge which would be required to produce 2000 litres
of fluorine gas in 1 hour by electrolysis. Take the molar volume to be
26.5 l mol-1.
Step 1: From the ion-electron equation work out the relationship
between the number of moles of electrons and number of moles of
product gas:
Step 2:Use the information in the question and step 1 to calculate the
charge:
Answer: The quantity of charge needed is 1.46 x 107 C.