CfE Higher Chemistry – Unit 3 – Chemistry in Society
Chemical Industry
1. Because of the vast range of materials that the chemical industry makes the
quality of our lives is greatly improved.
2. The chemical industry creates a great number of exports and this benefits the
national economy.
3. Stages in the manufacture of a new product can include research, pilot
study, scaling up, production then review.
4. A feedstock is a reactant from which other chemicals can be extracted or
produced.
5. The raw materials in the chemical industry are fossil fuels, metal ores and
minerals, air and water.
6. Chemical manufacture may be organised as a batch or a continuous
process.
7. Two important issues in a chemical plant are (i) safety for the operators and
local residents, and (ii) environmental issues e.g. pollution of local rivers.
8. The high cost of energy means that it must be not wasted.
9. Manufacturing costs include capital costs, fixed costs and variable costs.
10. The chemical industry in the United Kingdom is largely capital intensive rather
than labour intensive.
11. The location of a chemical industry can be affected by geographical transport
and social factors as well as by practical factors (e.g. the local availability of a
raw material).
12. Factors influencing the choice of a particular synthetic route for a chemical
process include cost, availability and suitability of a reactant, yield of
product, opportunity for recycling unreacted reactant and
removal/separation of by products.
Mole Calculations
1. A balanced equation tells us the mole ratio between different reactants
2. The number of moles can be calculated by dividing the mass by the GFM for
solids, multiplying the concentration by the volume for solutions and dividing
the volume by the molar volume for gases.
3. The reaction is limited by the reactant not in excess – i.e the reactant of which
there is not enough moles.
4. Molar volume is the volume that one mole of any gas occupies.
5. Percentage yield is calculated by actual yield divide by theoretical yield
multiplied by 100.
6. Atom economy is mass of what you are interested in divide by mass of
reactants multiplied by 100.
Equilibiria
1. In many chemical reactions the products can change back into the reactants.
Reactions like this are called reversible reactions.
2. In a reversible reaction when the rate of reverse reaction is equal to the rate
of the forward reaction a dynamic equilibrium exists.
3. At equilibrium the concentrations of all the reactants and products remain
constant although not necessarily (and not usually) stationery/equal.
4. Catalysts do not alter the position of equilibrium (since they speed up both
the forward and reverse reactions equally). Catalysts decrease the time
taken for the reaction mixture to reach equilibrium.
5. For a system at equilibrium a change in the temperature of the system or a
change in the concentrations of the chemicals will alter the position of
equilibrium. A change in the pressure can alter equilibrium in some reactions
involving reactions.
6. In the Haber Process: N2(g) + 3H2(g) ↔ 2NH3(g) ∆H= -92kJmol‾¹
 Removing some NH3 as it is formed will cause more NH3 to be
formed.
 Increasing the pressure will cause the equilibrium to move so that
more ammonia will be made.
 Increasing the temperature will shift the equilibrium to make less
ammonia.
7. In the Haber Process the continuous removal of ammonia prevents an
equilibrium being attained and the recycling of unreacted gases means that
no chemicals are wasted.
Chemical Energy
1. If ∆H for a reaction a + b → c+d equals -65kJ mol ‾¹ then ∆H for the reaction
c+d → a+b equals +65kJ mol‾¹.
2. Hess’s law states that:”the enthalpy change for a chemical reaction is
independent of the route taken.
3. Arranging suitable chemical reactions in energy cycles allows us to apply
Hess’s law. Hess’s law states that in oxidisng carbon to carbon dioxide the
enthalpy change will be the same by route A as by route B. In this energy
cycle ∆H1 and ∆H3 can be measured by experiment. ∆H2 could be calculated
from these values since, according to Hess’s law ∆H1= ∆H2+∆H3.
4. Hess’s law can also be determine from bond enthalpies
5. Energy can be calculated by E=cmΔT (Divide by number of moles) and then
used to calculate enthalpy of solution, neutralization or combustion.
Redox
1. Since oxidizing agent (oxidizes/reduces) the other reactant, i.e. causes it to
(gain/lose) electrons, that oxidizing agent must (gain/lose) electrons.
2. It follows that reducing agents lose electrons.
3. Ion-electron equations can involve atoms and ions, or molecules and ions e.g.
Cu(s) → Cu²+ + 2e‾ and Cl2(s) + 2e‾ → 2Cl‾(aq)
4. The first of these is a(n) oxidization reaction and the second is a reduction
reaction.
5. If the reactant and product species are known then ion-electron equations
which involve H+(aq) and H20(ℓcan be worked out e.g.
6.
7.
8.
9.
Cr2o7²‾(aq) + 14H+ + 6e‾ →2Cr³+(aq) +7H2o
Ion-electron equations can be combined to produce a balanced equation for
the overall redox reaction. Before combining, the ion-electron equations must
be multiplied by suitable number so that the numbers of electrons lost or
gained are the same.
the numbers of moles reacting can be seen in the balanced equation for a
redox reaction.
By carrying out a redox titration the volume of one reactant that reacts
completely with a known volume of the other reactant is found.
After titration, if the concentration of either reactant is known, the
concentration of the other reactant can be calculated using
where
n= number of moles
c=concentration in mol l‾
v=volume
in
l
Chromatography
1. Explain how chromatography (TLC, GLC and column) works
2. Calculate Rf values
3. Be able to use chromatograms