Cfe Higher Chemistry Unit 3 Chemistry in Society

CfE Higher Chemistry – Unit 3 – Chemistry in Society

Chemical Industry

Because of the vast range of materials that the chemical industry makes the quality of our lives is greatly improved.
The chemical industry creates a great number of exports and this benefits the national economy.
Stages in the manufacture of a new product can include research, pilotstudy, scaling up, production then review.
A feedstock is a reactant from which other chemicals can be extracted or produced.
The raw materials in the chemical industry are fossil fuels, metal ores and minerals, air and water.
Chemical manufacture may be organised as a batch or a continuous process.
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.
The high cost of energy means that it must be not wasted.
Manufacturing costs include capital costs, fixed costs and variable costs.
The chemical industry in the United Kingdom is largely capital intensive rather than labour intensive.
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).
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

A balanced equation tells us the mole ratio between different reactants
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.
The reaction is limited by the reactant not in excess – i.e the reactant of which there is not enough moles.
Molar volume is the volume that one mole of any gas occupies.
Percentage yield is calculated by actual yield divide by theoretical yield multiplied by 100.
Atom economy is mass of what you are interested in divide by mass of reactants multiplied by 100.

Equilibiria

In many chemical reactions the products can change back into the reactants.Reactions like this are calledreversible reactions.
In a reversible reaction when the rate of reverse reaction is equal to the rate of the forward reaction a dynamic equilibrium exists.
At equilibrium the concentrations of all the reactants and products remain constant although not necessarily (and not usually) stationery/equal.
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.
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.
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.

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

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.
Hess’s law states that:”the enthalpy change for a chemical reaction is independent of the route taken.
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.
Hess’s law can also be determine from bond enthalpies
Energy can be calculated by E=cmΔT (Divide by number of moles) and then used to calculate enthalpy of solution, neutralization or combustion.

Redox

Since oxidizing agent (oxidizes/reduces) the other reactant, i.e. causes it to (gain/lose) electrons, that oxidizing agent must (gain/lose) electrons.
It follows that reducing agents lose electrons.
Ion-electron equations can involve atoms and ions, or molecules and ions e.g. Cu(s) → Cu²+ + 2e‾ and Cl2(s) + 2e‾ → 2Cl‾(aq)
The first of these is a(n) oxidization reaction and the second is a reduction reaction.
If the reactant and product species are known then ion-electron equations which involve H+(aq) and H20(ℓcan be worked out e.g. 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