TOPIC: CHEMICAL SYSTEMS - FERTILISER INDUSTRY …accelerated-learning.co.za/exam/Subjects...

GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME

PHYSICAL SCIENCES GRADE 12 SESSION 17 (LEARNER NOTES)

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TOPIC: CHEMICAL SYSTEMS - FERTILISER INDUSTRY AND BATTERIES

SECTION A: TYPICAL EXAM QUESTIONS

QUESTION 1: 25 minutes

(Taken from the DoE Physical Sciences Exemplar Paper 2 2008)

The flow diagram below shows the main steps in the industrial preparation of two important solid fertilisers.



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1.1 Write down the balanced chemical equation for the formation of the brown gas.(3)

1.2 Write down the name of process Y. (2)

1.3 Write down the chemical formula of liquid E. (2)

1.4 Write down the chemical formulae of fertilisers C and D respectively. (4)

The following extract comes from an article on fertilisers:

A world without food for its people –

A world with an environment poisoned through the actions of man –

Are two contributing factors towards a disaster scenario.

1.5 Write down THREE ways in which the use of fertilisers poisons the environment. (6)

[17]


(Taken from the DoE Physical Sciences Exemplar Paper 2 2008)

A dry cell, as shown in the diagram below, does not contain a liquid electrolyte. The electrolyte in a typical zinc-carbon cell is a moist paste of ammonium chloride and zinc chloride.

The paste of ammonium chloride reacts according to the following half-reaction:

2NH

4 (aq) + 2e- → 2NH3(g) + H2(g) … (i)

Manganese(IV) oxide is included in the cell to remove the hydrogen produced during half-reaction (i), according to the following reaction:

2MnO2(s) + H2(g) ⇌ Mn2O3(s) + H2O(ℓ) ... (ii)

Metal cap (+)

Carbon rod

Manganese(IV) oxide

Moist paste of ammonium chloride

and zinc chloride

Metal bottom (-)



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The combined result of these two half-reactions can be represented by the following

half-reaction:

2NH

4 (aq) + 2MnO2(s) + 2e- ⇌ Mn2O3(s) + 2NH3(g) + H2O(ℓ) … (iii)

2.1 Explain why it is important that the hydrogen produced in half-reaction (i) is

removed by the manganese(IV) oxide. (2)

In a zinc-carbon cell, such as the one above, half-reaction (iii) and the half-reaction

that takes place in the Zn/Zn2+ half-cell, produce an emf of 1,5 V under standard

conditions.

2.2 Write down the half-reaction occurring at the anode. (2)

2.3 Write down the net ionic equation occurring in the zinc-carbon cell. (2)

2.4 Calculate the reduction potential for the cathode half-reaction. (4)

2.5 When in use the zinc casing of the dry cell becomes thinner, because it is

oxidised. When not in use, it still corrodes. Give a reason for the latter

observation. (2)

2.6 Dry cells are generally discarded when 'flat'. Why is the carbon rod the most

useful part of the cell, even when the cell is flat? (2) [14]


(Taken from the DoE Physical Sciences Additional Exemplar Paper 2 2008)

3.1 A group of learners set up an electrochemical cell using lead and copper half cells.

3.1.1 Which ONE of copper or lead will be the negative electrode? Give a reason for your answer. (2)

3.1.2 Use the Table of Standard Reduction Potentials and write down the reduction half-reaction that will take place in this cell. (2)

3.1.3 A 2V bulb is connected to the cell instead of the bulb. Will the bulb light up? Justify your answer with a calculation. (5)

3.1.4 A voltmeter is now connected to the cell instead of the bulb. It is observed that after a while the reading on the voltmeter drops to zero. We say the cell is “flat” or “dead”.

Explain this observation in terms of the concentrations of the solutions in the cell. (3)



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3.2 A cell such as the one described above is not much useful. However, the principle is used in batteries for cars, torches, computers, etc. These batteries are called secondary cells.

One such battery is the mercury cell. The half reactions occurring in this cell are shown below.

Zn (s) + 2OH- (aq) → ZnO (s) + H2O + 2e- ………….. (1)

HgO (s) + H2O + 2e- → Hg (l) + 2OH- (aq)………… (2)

3.2.1 Write down the overall cell reaction. (2)

3.2.2 Why does the use of this battery pose an environmental hazard? (1) [15]

SECTION B: ADDITIONAL CONTENT NOTES

FERTILISERS

In order to make fertiliser, we need to manufacture nitric acid, ammonia and sulphuric acid. All of these industrial processes make use of our understanding of equilibrium and rates. Let us further investigate these industrial processes.

SULPHURIC ACID

Uses of sulphuric acid

S acts as a drying agent – used by chemical engineers to mop up water.

S used as an electrolyte in car batteries (lead accumulator batteries).

S used in the recovery of metals such as uranium and copper.

S used to clean metal surfaces.

S used to produce fertilisers.

THE CONTACT PROCESS

S (s) + O2 (g) SO2 (g)

SO2 (g) + O2 (g) 2 SO3 (g) (V2O5 catalyst)

SO3 (g) + H2SO4 H2S2O7

H2S2O7 + H2O 2 H2SO4

S Sulphur is burnt in oxygen.

S Sulphur dioxide is further reacted with oxygen in contact with a vanadium pentoxide catalyst to form sulphur trioxide. This is a reaction that is governed by equilibrium principles.

S Sulphur trioxide is then added to sulphuric acid (note only 1 molecule) to form pyrosulphuric acid/fuming sulphuric acid/oleum.

S Pyrosulphuric acid is mixed with water and sulphuric acid is formed (note 2 molecules). One molecule of sulphuric acid is „invested‟ to make two molecules.



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Looking specifically at step 2:

2S02(g) + O2(g) 2S03(g) ΔH = -95 kJ.mol"1

Vanadium pentoxide is used as a catalyst

Temperature: Exothermic reaction. Lowering the temperature shifts the equilibrium to the right. However, lower temperature also slows the reaction down so an optimum temperature must be found (723 K)

Pressure: 3 molecules on the left and 2 molecules on the right, therefore, high pressure favours the forward reaction. (1 atm is good enough without unnecessary high cost for high pressure reaction vessels)

Concentration: Sulphur(VI)oxide is continuously removed, to reduce concentration and shift the equilibrium to the right. The sulphur(VI)oxide is dissolved in water to form sulphuric acid.

S Sulphate salts are often added to fertilisers especially in the form of ammonium sulphate and potassium sulphate.

AMMONIA

A Ammonia is produced by means of the Haber process. This process is also governed by equilibrium principles.

HABER PROCESS

N2 (g) + 3 H2 (g) 2 NH3 (g) (iron oxide catalyst) ΔH = -46 kJ.mol-1

Temperature: Exothermic reaction. Lowering the temperature shifts the equilibrium to the right. However, lower temperature also slows the reaction down so an optimum temperature must be found (600 K - 800 K)

Pressure: 4 molecules on the left and 2 molecules on the right; therefore, high pressure favours the forward reaction. (350 atm)

Concentration: Ammonia is continuously removed, to reduce ammonia concentration and shift the equilibrium to the right.

Uses of ammonia

A as a cleaning agent when dissolved in water.

A in the manufacture of fertilisers. Ammonium salts are highly soluble in water so they can be dissolved and absorbed by plants, thus making nitrogen available to the plant.

A in the manufacture of ammonium carbonate used in the textile industry.

A in the manufacture of nitric acid.



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NITRIC ACID

N Nitric acid is produced by the Ostwald process.

THE OSTWALD PROCESS

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O

2 NO (g)+ O2 (g) 2 NO2 (g)

4 NO2 + 2 H2O + O2 4 HNO3

N Ammonia is oxidised to nitrogen monoxide and water N Nitrogen monoxide is further oxidised to nitrogen dioxide N Nitrogen dioxide is then added to water from the first step and oxygen to make nitric acid. N Nitrate salts are highly soluble in water and, therefore, assist in allowing nitrogen to be

absorbed by plants

CHALLENGES FOR THE CHEMICAL INDUSTRY

safety issues regarding the transport of chemicals. pollution control. waste disposal. energy consumption.

THE FERTILISER INDUSTRY

Commercial fertilisers are described in terms of their relative proportions of nitrogen (N), phosphorus (P) and potassium (K). NPK numbers are printed on fertiliser bags. E.g 5:13:5 means the content of the bag is

5% nitrogen

13% phosphorus

5% potassium

Sources of potassium: Salts containing potassium are referred to as POTASH when used as fertilisers.

Mined potassium salts, such as KNO3 and K2SO4 are used. Sources of phosphorus

bonemeal that was acidified to yield phosphoric acid H3PO4 mined calcium phosphate Ca3(PO4)2 Calcium phosphate is insoluble in water and cannot be absorbed by plants.

Calcium phosphate is treated with sulphuric acid to yield water soluble superphosphate that can be absorbed.

Ca3(PO4)2 + H2SO4 → 2CaSO4 + Ca(H2PO4)2

EUTROPHICATION

r Extensive utilisation of fertilisers leads to pollution of surface and ground water and eutrophication of lakes, dams and rivers. This is mainly due to fertiliser wash off

r Eutrophication is caused by excess phosphates and nitrates in rivers, lakes and dams. r Phosphates promote the growth of algae which cover the surface of the water and kill

aquatic plants below. r Bacteria decompose the dead aquatic life causing oxygen levels in the water drop to such

a low level that aquatic animals suffocate and die.



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BATTERIES

Batteries are GALVANIC (VOLTAIC) cells. They convert chemical energy into electrical energy through a spontaneous Redox reaction Cell Capacity - the ability of a fully charged battery to deliver a specific quantity of current

over a specific

period of time (minutes or hours)

Q = I∆t

where Q is measured in Amp. hours (A·h)

Cell emf - gives the voltage that the battery can provide

- the voltage and charge together gives the amount of energy (work done)

the battery can provide

W = VQ or W = VI∆t

Where W is the work done (energy transferred) and measured in Joules (J)

Example of cell capacity and emf :

Two 12 V batteries are rated at 95 A·h and 36 A·h respectively and can supply current

continuously for 20 hours.

(a) What continuous current can be drawn from these batteries in 20 hours?

The batteries can also be rated to produce the following currents:

1. The 95 A·h battery is rated at 460 A for 60s

2. The 36 A·h battery is rated at 200 A for 60s

(b) What is the emf value for these batteries? What does this value indicate?

(c) Calculate the amount of energy provided by each battery during the 60s period.

Question adapted from Physical Science 12 – M. Mann



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Answers

(a) For the 95 A·h battery

Q = 95 Ah Q = I∆t

∆t = 20 hours 95 = I x 20

I = 95

20

I = 4,75 A

For the 36 A·h battery

Q = 36 Ah Q = I∆t

∆t = 20 hours 36 = I x 20

I = 36/20

I = 1,8 A

It provides 12 Joules of energy per coulomb of charge

(c) For 95 A·h battery

I = 460 A Q = I∆t

∆t = 60 s = 460 x 60

Q = 27 600 C

V = 12 V Now W = VQ

= 12 x 27 600

W = 331 200 J



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For 36 A·h battery

I = 200 A Q = I∆t

∆t = 60 s = 200 x 60

Q = 12 000 C

V = 12 V thus W = VQ

= 12 x 12 000

W = 144 000 J

Primary and Secondary cells

Primary Cells - these are cells that can be discarded after use.

- they become “flat” and thus can no longer produce electrical energy required

to make an appliance work

At the anode : Zn Zn2+ + 2e

At the cathode : 2NH4+ + 2MnO2 + 2e Mn2O3 + H2O + 2NH3

Secondary cells - these are cells that can be recharged by driving a current through them

- this converts electrical energy into chemical energy by reversing the

chemical reaction, i.e. the products of the reaction during discharge are

reconverted into the starting materials

The Leclanche cell

The very familiar “torch battery” is an example of the Leclanche

cell and is a primary cell.

Cathode - is a carbon rod running through the centre of the cell

surrounded by a paste of manganese dioxide (MnO2)

Anode - the zinc metal casing of the cell acts as the anode

Electrolyte - a paste of ammonium chloride (NH4Cl ) and zinc

chloride (ZnCl2)



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At the anode : Pb + SO42- PbSO4 + 2e

At the cathode : PbO2 + SO42- + 4H+ + 2e PbSO4 + 2H2O

Overall :

Pb + PbO2 + 2H2SO4 2PbSO4 + 2H2O

SECTION C: HOMEWORK


1.1. Every bag of fertiliser is marked with an NPK number. If a bag is marked 3:10:5 it means the bag contains:

A. 5% nitrogen

B. 10% phosphorus

C. 10% potassium

D. 5% phosphorus

1.2. Sulphuric acid is manufactured via the Contact Process. The purpose of the vanadium pentoxide catalyst in this process is to:

A. Speed up the reaction

B. Make more sulphuric acid

C. Allow for contact between all the chemicals

D. Prevent the process from over reacting.

The Lead Acid cell (the car battery)

Anode - lead alloy grids (Pb)

Cathode - grids packed with lead dioxide

(PbO2)

Electrolyte - dilute sulphuric acid (H2SO4)



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1.3. Eutrophication is a type of pollution. Which element or compound is primarily responsible for this problem?

A. Sulfur dioxide

B. Carbon dioxide

C. Cloro flouro carbons

D. Nitrogen

1.4. Which element which does occur in fertilisers, is not utilised from the fertiliser by the plant?

A. Nitrogen

B. Oxygen

C. Phosphorous

D. Potassium

1.5. Which property, common to both ammonia and nitrate ions, is useful in their application as fertilisers?

A. Their pungent smell

B. Their acidic properties

C. Their basic properties

D. Their solubility in water

1.6. Nitric acid is manufactured via the:

A. Ostwald Process

B. Haber Process

C. Contact Process

D. Castner Process

1.7. The Contact Process

2SO2 (g) + O2 (g) 2 SO3 (g) ; (V2O5 catalyst) , ΔH < 0

can give the highest yield, under the following conditions of temperature and pressure:

Temperature Pressure

A High Low

B High High

C Low Low

D Low High

(7 x 2) = [14]



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The following half-reactions take place in a fuel cell:

2H2 + 4OH- ↔ 4H2O + 2e- EΘ = -0,83 V

O2 + 2H2O + 4e- ↔ 4OH- EΘ = 0,40 V

2.1 What type of reaction takes place in a fuel cell? (2)

2.2 What is a fuel cell? (2)

2.3 Which electrode is the cathode? (2)

2.4 Give the net overall ionic equation. (4) [10]

SECTION D: SOLUTIONS AND HINTS TO SECTION A

QUESTION 1

1.1 2NO + O2 → 2NO2 √√√ (3)

1.2 Catalytic oxidation of ammonia √√ (2)

1.3 HNO3 √√ (2)

1.4 D – (NH4)2SO4 √√ E - NH4NO3 √√ (4)

1.5 ANY THREE:

Nitrogen saturated soil allows other nutrients to be washed away. √√

The top soil becomes acidic because of the washing. √√

The build up of nitrates in the rivers causes algae bloom and this then leads to the depletion of the oxygen in the water and thus the aquatic life dies. √√

Alien plants can start to grow.

The groundwater can become acidic.

The build up of nitrates in the water can cause death in infants due to lack of oxygen in their blood known as blue baby syndrome because of haemoglobin deficiency. (6)

[17]

QUESTION 2

2.1 H2 (g) builds up in a cell and can result in the cell bursting or exploding when it is

ignited (2)

2.2 Zn Zn2+ + 2 e- (2)



The SSIP is supported by

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2.3 Zn (s) Zn2+ (aq) + 2 e-

2 NH4+ (aq) + 2 MnO2 (s) + 2 e- Zn2+ (aq) + 2 NH3 (g) + H2O (ℓ)

Zn (s) + 2 NH4+ (aq) + 2 MnO2 Zn2+ (aq) + Mn2O3 (s) + 2 NH3 (g) + H2O (ℓ)

(2)

2.4 Ecell = Ecathode - Eanode

1,5 = Ekatode - (-0,76)

Ecathode = +0,74 V (4)

2.5 NH4+ is an acid and reacts with the zinc casing. (2)

2.6 Can be reused as an electrode in other situations. (2)

[14]

QUESTION 3

3.1.1 Lead √ because it is a stronger reducing agent OR is preferably oxidised √ (2)

3.1.2 Cu2+ (aq) + 2e- → Cu (s) √√ (2)

3.1.3 Eθ

cell = Eθ

cathode - Eθ

anode √

= 0,34 √ – (-0,13)√

Eθ

cell = 0,47 V √

Bulb will not light up, √ energy from cell not sufficient OR emf of cell is much less than 2V needed for the bulb. (5)

3.1.4 While the cell is in operation, the concentration of the reactants decreases.√ At the same time the concentration of the product increases. √ The result is a

gradual decrease in the cell potential until there is no further change in concentration and equilibrium is reached where the cell potential will be zero.√ (3)

3.2.1 Zn (s) + HgO (s) → ZnO (s) + Hg (l) (2)

3.2.2 Mercury is poisonous. √ (1)

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TOPIC: CHEMICAL SYSTEMS - FERTILISER INDUSTRY …accelerated-learning.co.za/exam/Subjects...

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