Industrial Chemistry

Question 1

What is a natural product?

Answer 1

A natural product is a substance produced directly from something found in nature, with little or no modification.

Question 2

Why have industrial products replaced natural products?

Answer 2

The world’s natural resources are decreasing in supply due to the overwhelming demand from the rapidly increase world’s population.
As a result, natural products have become heavily depleted or no longer economically viable for use.

Also, industrial products are modified, possessing better properties than natural products.

Question 3

What is the chemical name for natural rubber?

Answer 3

polyisoprene

Question 4

Why is natural rubber not useful?

Answer 4

It is not very useful as it is inelastic, becomes soft when warmed and brittle when cooled, and is reactive due to its double bond.

Question 5

What property makes natural rubber useful?

Answer 5

Its double bond along polymer chains allows vulcanisation (cross-links) which rectifies these issues and allows for the development of growth of automotive industries and its needs for tyres, thus increasing the demand for rubber.

Question 6

What is the chemical name of synthetic rubber?

Answer 6

Styrene-butadiene rubber (SBR) is a synthetic rubber made from by-products of petroleum refining. It is not as reactive as natural rubber since it has fewer double bonds.

Question 7

Why was synthetic rubber developed and what are its disadvantages?

Answer 7

o It was developed due to the demand for rubber to make tyres and its raw materials being readily available.
o However, its disadvantages include its reliance on a non-renewable resource, lower biodegradability and the use of petrochemical starting materials (pollution).

Question 8

What is the procedure in the practical: modelling an equilibrium reaction?

Answer 8

1. Fill one 50 mL measuring cylinder, leaving the other cylinder empty.
2. Place the straw in the water-filled (first) measuring cylinder so that it touches the bottom.
3. Place a finger above the top of the straw and transfer all of the water from the straw into the empty (second) cylinder by release the finger off the top once the straw is inside the measuring cylinder.
4. Place the straw back into the second measuring cylinder so that it touches the bottom and transfer all of the water from the straw into the first measuring cylinder.
5. Record the amount of water displaced in each cylinder.
6. Repeat steps 2, 3, 4 and 5 until the amount of water in each cylinder remains relatively the same after each cycle.

Question 9

What are the results of the practical: modelling an equilibrium reaction?

Answer 9

An equilibrium was achieved when both of the water levels in each cylinder was 25 mL as it was the amount was relatively constant. The change was relatively quick at first but eventually slowed down to a point where the water levels does not change considerably.

Question 10

In the practical analysing a chemical equilibrium, what is the procedure?

Answer 10

1. Mix 10 mL of iron (III) nitrate and 10 mL of potassium thiocyanate (KSCN) solution in 150 mL beaker.
2. Add sufficient water to the beaker until the mixture is orange-red. Divide this solution into 3 test tubes.
3. Add a few crystals of KSCN into the 1st test tube.
4. Stopper the tube and agitate it.
5. Add a few crystals of Fe(NO3 )3 into the 2nd test tube.
6. Stopper the tube and agitate it.
7. Add a few drops of 1 molL^(-1) NaOH into the 3rd test tube.
8. Stopper the tube and agitate it.

Question 11

What is the equation in the practical: analysing a chemical equilibrium?

Answer 11

Fe^(3+)(aq) + SCN^-(aq) ⇌ FeSCN^(2+) (aq)

Fe^(3+) is yellow brown, SCN^- is colourless and FeSCN^(2+) is deep blood red.

Question 12

In the practical: analysing a chemical equilibrium, the observation’ blood red colour of the solution intensified in the first test tube’ was made. What is the explanation behind this?

Answer 12

The equilibrium has shifted to the right to use up some of the added SCN^- as predicted by Le Chatelier’s principle.

Question 13

In the practical: analysing a chemical equilibrium, the observation’ blood red colour of the solution intensified in the second test tube’ was made. What is the explanation behind this?

Answer 13

The equilibrium has shifted to the right to use up some of the added Fe^(3+)as predicted by Le Chatelier’s principle.

Question 14

In the practical:analysing a chemical equilibrium, the observation “A brown precipitate formed and the blood red colour of the solution faded.” was made. What is the explanation behind this?

Answer 14

When NaOH was added, Fe(OH)3 (s) formed as a precipitate.

Fe^(3+) (aq) + 3OH^- (aq)→Fe(OH)3 (s)

This effectively removes Fe^(3+) from the system so Le Chatelier’s principle predicts that the equilibrium will shift to the left in an attempt to produce more Fe^(3+).

Question 15

The magnitude of K can give an indication of the extent of the equilibrium, if K is very large (> 10^3), where does the equilibrium lie and what is favoured?

Answer 15

the equilibrium lies well to the right; the products are strongly favoured.

Question 16

The magnitude of K can give an indication of the extent of the equilibrium, if K is very large (< 10^-3), where does the equilibrium lie and what is favoured?

Answer 16

the equilibrium lies well to the left; the reactants are strongly favoured.

Question 17

o When a reaction is not at equilibrium, Q replaces K in the same expression.
Explain.

Answer 17

If Q = K, then the reaction is at equilibrium.
If Q ≠ K, then the reaction is NOT at equilibrium.
If Q < K, then the reaction will proceed from left to right towards the products until Q = K.
If Q > K, then the reaction will proceed from right to left towards the reactants until Q = K.

Question 18

What is the factor(s) that influence the value of K?

Answer 18

temperature

Question 19

What are some uses of sulfuric acid?

Answer 19

o Sulfuric acid is also used for the “pickling” of steel to remove any surface rust, grease and dirt prior to galvanising. It is used for this purpose because it is very corrosive and readily removes these coatings.
o Finally, sulfuric acid is also used as an electrolyte in lead-acid batteries.

Question 20

Explain the exothermic nature in the ionisation of sulfuric acid.

Answer 20

o Forming bond releases energy. Breaking bond absorbs energy.
o Ionisation is exothermic as hydrogen is ionised from an acid, absorbing energy. The hydrogen ion will react will form a hydronium ion which hence releases energy.
o However the energy released with a hydrogen reacting with water as a hydrogen ionising from an acid is greater, thus accounting for the exothermic nature of the ionisation of an acid.

Question 21

Why is sulfuric acid dangerous to humans?

Answer 21

o Since human bodies are comprised mainly of water, sulfuric acid will dehydrate our cells, causing it to die as it is a power dehydrating agent. It also causes severe burns as it involves in an exothermic reaction.
o Inhalation damages the mucous membranes and upper respiratory tract. Symptoms include irritation of nose and throat.
o Skin contact results in redness, pain severe burns which may be followed by circulatory collapse and death.

Question 22

Why is adding acid to water safer as opposed to adding water to acid?

Answer 22

o Sulfuric acid is a powerful dehydrating agent and oxidising agent as it has a strong affinity to water and will remove water from carbohydrates.
o It is important to add acid to water and never water to acid. Acid in water is an exothermic reaction and will produce large quantities of heat.
o Adding water to acid will cause water to oil and acid to splash. However adding acid to cold water while stirring dissipates the heat, but more important minimise the chance of boiling and splashing as water has a high specific heat capacity, minimising the amount of heat produced.

Question 23

What are some safety precautions in handling sulfuric acid?

Answer 23

 Use of googles, gloves and lab coat to prevent any skin contact with the acid.
 Having cold running supply of water nearby to flush out any acid that comes into any contact with the person.
 If concentrated sulfuric acid comes into contact with a person, wipe it off first and then run it under cold running water.

Question 24

What are the safety precautions in transporting and storing sulfuric acid?

Answer 24

o Concentrated sulfuric acid is stable but reacts violently with water and organic materials. It is a strong dehydrating agent and when water is absorbed, the process is highly exothermic.
o Thus reinforced air tight stainless steel is used as there are no ions to attack the metal. The container should be labelled, outline the risks and precautions.

Concentrated sulfuric acid is virtually all molecules and does not contain much free water so it can be stored in steel containers.
o In general, regardless of the method of transport, sulfuric acid should be split into many small containers, minimising the danger on the off-chance that a container breaks or there is a contamination of concentrated sulfuric acid with water.

For dilute sulfuric acid, it is stored in glass, followed by plastic. It cannot be stored in metal as it will react away the metal.

Question 25

The Frasch Process (Froth Flotation) Outline the industrial process.

Answer 25

1. A hole is drilled into the deposit and a set of three concentric pipes is inserted. Superheated water (about 170oC) is pumped down the outer concentric pipe to melt the sulfur. 2. Hot compressed air is then sent through the narrow innermost pipe and a low density liquid/ air emulsion of sulfur is formed. The hydrostatic pressure and the pressure of the compressed air forces the emulsion to the surface through the middle pipe. 3. The liquid sulfur quickly cools. As it solidifies, it separates from water.

Question 26

What are the properties of sulfur that allows The Frasch Process (Froth Flotation) to take place?

Answer 26

o The three properties of sulfur which allows the process to proceed include its low melting point (115OC), low density liquid (1.8 g cm-3) and its insolubility in water. o The low melting point of sulfur allows the addition of super-heated water to melt into a mobile liquid. o The low density enables the liquid sulfur emulsion to rise to the surface when formed by compressed air. Being insoluble as it solidifies from the surface will allow it to separate from water, making it easy to collect The product formed is very pure (~99.5%)

Question 27

What are the environmental issues of the frasch process?

Answer 27

o Land subsidence: there is a possibility of land collapse when a large amount of solid material is removed from underground. This results in habitat loss, thus disrupting the ecosystem and potentially the food chain. o CO2 Production: The energy required to produce superheated steam from water is obtained from the burning of fossil fuels. This greatly increases the concentration of CO2 in the atmosphere, contributing to global warming. o The sulfur brought to the surface can be readily oxidised to form sulfur This can have severe consequences even when in small concentrations, and as such the water used to flush the sulfur from the deposit must be reused, or recycled before it can be disposed of. o In addition, should the sulfur dioxide find its way into the atmosphere, large concentrations will result in the formation of acid rain, which can have serious repercussions upon manmade infrastructure as well as flora and fauna, particularly marine ecosystems.

Question 28

What is the first step of the contact process?

Answer 28

Step 1: Sulfur dioxide is obtained by burning sulfur in dry air. (30-50% excess oxygen is present). o This occurs in the furnace. S (s) + O2 (g) → SO2 (g)

Question 29

What is the second step of the contact process?

Answer 29

Step 2: Sulfur dioxide is oxidised to sulfur trioxide catalytically. This involves an equilibrium reaction, so the conditions must be carefully controlled to maximise the yield. o The catalyst is porous pellets of vanadium (V) oxide- V2O5. o The catalytic conversion occurs over multiple catalyst beds to maximise the conversion and speech with which the equilibrium is attained. The reaction takes place at 550o C and 1-2 atm. o The gases are cooled after each successive catalyst bed. o A catalyst bed is used instead of raising the temperature as raising the temperature consumes more energy and decreases the yield. o Multiple catalyst beds are used to provide a large surface area to maximise the contact and intermediate cooling of the reagents. o This occurs in the catalytic reactor. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) With catalyst V2O5(s) above reversible arrow.

Question 30

What is the third step of the contact process?

Answer 30

Step 3: In theory, sulfur trioxide can be added directly with water, but the reaction is so exothermic that a mist of sulfuric acid is produced which is difficult to manage. o Sulfur trioxide is dissolved in concentrated sulfuric acid in an absorption tower. The resulting liquid is called oleum (H2S2O7) o Oleum is a useful form of sulfuric acid for transportation purposes as it is less corrosive to metals and can be diluted upon arrival. This occurs in the absorber. SO3 (g) + H2SO4 (l) → H2S2O7 (l) Step 4: Oleum is dissolved in water to form ~98% pure sulfuric acid. This occurs in the diluter. H2S2O7 (l) + H2O (l) → 2H2SO4 (l)

Question 31

What is the third step of the contact process?

Answer 31

Step 3: In theory, sulfur trioxide can be added directly with water, but the reaction is so exothermic that a mist of sulfuric acid is produced which is difficult to manage. o Sulfur trioxide is dissolved in concentrated sulfuric acid in an absorption tower. The resulting liquid is called oleum (H2S2O7) o Oleum is a useful form of sulfuric acid for transportation purposes as it is less corrosive to metals and can be diluted upon arrival. This occurs in the absorber. SO3 (g) + H2SO4 (l) → H2S2O7 (l)

Question 32

What is the fourth step of the contact process?

Answer 32

Step 4: Oleum is dissolved in water to form ~98% pure sulfuric acid. This occurs in the diluter. H2S2O7 (l) + H2O (l) → 2H2SO4 (l)

Question 33

In the practical: To observe the reactions of sulfuric acid acting as an oxidant and a dehydrating agent What are the safety issues

Answer 33

Concentrated sulfuric acid is very corrosive so wear safety goggles, protective gloves and a lab coat must be worn. The experiment must be done in a fume cupboard since SO2 gas may be produced.

Question 34

In the practical: To observe the reactions of sulfuric acid acting as an oxidant and a dehydrating agent What is the method?

Answer 34

1. Add 50g of brown sugar into a beaker. | 2. Add 50 mL of concentrated sulfuric acid into the beaker and record all observations.

Question 35

In the practical: To observe the reactions of sulfuric acid acting as an oxidant and a dehydrating agent What are the results

Answer 35

A porous, smoking, black column expanded out from the beaker; the glucose has been dehydrated into carbon. C6H12O6(s) → 6C(s)+6H2O(l) Some SO2 is also produced as carbon may be oxidised by sulfuric acid. C(s)+2H2SO4 (l))→CO2(g)+SO2(g)+2H2O(l)

Question 36

For sulfuric acid Reactions as an oxidising Agent, | state some reactions with dilute and concentrated

Answer 36

In dilute sulfuric acid, the oxidant is the H^+ ion. Some reactive metals are oxidised in dilute sulfuric acid, releasing H2. Mg(s)+2H^+(aq)→Mg^(2+)(aq) + H2(g) In concentrated sulfuric acid, the SO4(2-) ion is the oxidant. Less reactive metals can be oxidised by concentrated sulfuric acid, releasing SO2. Cu(s)→Cu^(2+)(aq)+2e^- H2SO4(l)+2H^+(aq)+2e^-→ SO2(g) + 2H2O(l) The net reaction is: Cu(s) + H2SO4(l) + 2H^+(aq) →Cu^(2+)(aq) +SO2(g) +2H2O(l) Adding SO4^(2-) to both sides: Cu(s)+ 2H 2SO4(l)→CuSO4(aq)+ SO2(g)+ 2H2O(l)

Question 37

For Reactions as a Dehydrating Agent , write an equation.

Answer 37

o Concentrated sulfuric acid is an excellent dehydrating agent because of its strong affinity to water. o We do not use H2SO4 (aq) for dehydration reactions as when it is diluted, there is already water in excess such as aqueous sulfuric acid will not react. Example: Dehydration of ethanol. C2H5OH(g) ------> C2H4 (g) + H2O (g)