Chapter 7 - Compounds of Non-metals - Ammonia

In the previous section, we completed the discussion on acids, alkalies and salts. In this section we will see Compounds of Non-metals.

Ammonia
We know that, nitrogen is an important element required for the growth of plants. Details here. We have also seen that plants do not get the required quantities of nitrogen by natural means alone. We have to supply nitrogen through fertilisers. So we have to produce large quantities of such fertilisers. In other words, we have to produce such fertilisers industrially. For the industrial production of nitrogenous fertilisers, we have to produce ammonia first.

Preparation of ammonia in the laboratory

Fig.7.1 shows the diagram for the preparation of ammonia in the laboratory.

Fig.7.1

1. Ammonium chloride (NH₄Cl) and Calcium hydroxide (Ca(OH)₂) are mixed well and heated.
Let us write the equation:

Reactants:

    ♦ Ammonium chloride. One molecule is NH₄Cl. 

    ♦ Calcium hydroxide. One molecule is Ca(OH)₂.

Products:

    ♦ Calcium chloride. One molecule is CaCl₂.

    ♦ Water. One molecule is H₂O
    ♦ Ammonia. One molecule is NH₃

• So skeletal equation is:

NH₄Cl + Ca(OH)₂ → CaCl₂ + H₂O + NH₃. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: NH₄Cl + Ca(OH)₂ → CaCl₂ + H₂O + NH₃
Step 2: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + H₂O + NH₃
Step 3: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + H₂O + 2NH₃
Step 4: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + NH₃ 

	Reactants						Products
	N	H	Cl	Ca	O		N	H	Cl	Ca	O
Step 1	1	6	1	1	2		1	5	2	1	1
Step 2	2	10	2	1	2		1	5	2	1	1
Step 3	2	10	2	1	2		2	8	2	1	1
Step 4	2	10	2	1	2		2	10	2	1	2

So the balanced equation is: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + NH₃ 

2. From the equation, we can see that water is also formed as a product. So we have to prevent the newly formed ammonia (NH₃) from dissolving in the newly formed water. 
3. For that, the test tube is kept in a slanting position. So that, the newly formed water will not collect at the bottom of the test tube. 
4. Still, there will be water in the vapour form. This will move out through the delivery tube, along with the NH₃. That means, the gas which comes out of the delivery tube will be a mixture of water vapour and ammonia. 
5. We have to remove the water vapour. For that, the delivery tube enters a drying tower. The upper chamber of the drying tower contains quick lime (CaO). This CaO, which is a drying agent, will absorb the water vapour. So the delivery tube that comes out of the drying tower will contain NH₃ only. 
6. This delivery tube enters a gas jar. Thus the ammonia gas is collected in the gas jar. We can see that the gas jar is kept in an inverted position. This is because, ammonia gas is lighter than air. That is., the density of ammonia is lesser than the density of air. So it will rise up. This rising ammonia will displace the air present in the jar, and will occupy it’s top position in the inverted jar.

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Drying agent

Drying agents are substances capable of absorbing moisture from substances. After the absorbtion, ths substances will become ‘dry’. In the above experiment, CaO is used as the drying agent. CaO is alkaline. The ammonia is also alkaline. The two will not react with each other, and so, the ammonia comes out of the drying tower.

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Ammonia is highly soluble in water. This can be proved using the 'fountain experiment'. The arrangement is shown in the fig.7.2 below: 

Fig.7.2

• The flask which is placed in inverted position is filled with ammonia gas. A jet tube enters the flask from the bottom. The bottom end of the jet tube is dipped in water contained in a beaker. To this water, some phenolphthalein is already added.

• A few drops of water is added into the flask using the syringe. We can see a fountain of pink water in the flask. How is this fountain formed? Why is it pink coloured? Let us analyse:

1. When a few drops of water is added into the flask using the syringe, Some of the ammonia gas dissolves in that water. This will create some vacuum in the flask. So the out side atmospheric pressure will be greater than the pressure inside the flask. This atmospheric pressure will push down on the water in the beaker. So the water rushes up through the jet tube. 
2. When more water enters the flask in this way, more ammonia gas will dissolve in that water. Thus new vacuum is created. Because of this additional vacuum, the atmospheric pressure will again push down on the water in the beaker. 
3. This continues as a cyclic process. Thus a fountain will be formed. The cycle will continue until all the ammonia is dissolved. When all the ammonia is dissolved, there will not be any formation of 'new vacuum'.

■ Now we will see the reason for the pink colour: The ammonia gas present in the flask dissolves in the water which rushes into the flask. So the water becomes a solution. A ‘solution of ammonia in water’. This solution is alkaline in nature. So the phenolphthalein that is already present, will turn pink. Thus we get a pink fountain.
A video showing the demonstration can be seen here.

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This experiment shows that ammonia gas is readily soluble in water. The equation is:

NH₃ + H₂O → NH₄OH. This is a balanced equation.

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A highly concentrated solution of ammonia in water is called liquor ammonia.

Ammonia gas can be easily liquefied by applying pressure. Liquefied ammonia is called liquid ammonia.

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We know that, ammonia is alkaline in nature. So let us see it’s reaction with an acid:

■ Introduce the tip of a glass rod which is dipped in concentrated hydrochloric acid, into a glass jar filled with ammonia. We can see the formation of thick white fumes. This is due to the formation of ammonium chloride (NH₄Cl). Let us write the equation:
NH₃ + HCl → NH₄Cl
This is a balanced equation
• The white fumes are caused by small white particles of NH₄Cl

■ In this way, ammonia reacts with acids yielding ammonium salts, which are used as chemical fertilisers

Properties of ammonia

• It has no colour

• It has a pungent smell

• It is alkaline in nature

• It is highly soluble in water

• It is lighter than air

Identification of ammonia gas

We can use the following tests:

1. The basic test is by the smell. Ammonia has a characteristic pungent smell.

2. Ammonia turns wet red litmus paper into blue colour, showing it’s alkaline property

3. When the tip of a glass rod which is dipped in concentrated hydrochloric acid is introduced into a glass jar filled with ammonia, thick white fumes are formed.

Identification of ammonium salts

1. Prepare a solution of the given ammonium salt. Take 5 ml of nesslers reagent in a test tube. 
2. Add a few drops of the prepared salt solution into this. 
3. If a brownish orange precipitate is formed, then the given salt is a salt of ammonia.

Industrial preparation of Ammonia

1. Nitrogen and hydrogen are taken in the ratio 1:3. This ratio is obvious because, in the final product which is ammonia (NH₃), There are 3 atoms of hydrogen for every one atom of nitrogen.
2. They are made to react with each other. Very high temperature and pressure is required for the reaction to take place. This is because, nitrogen does not take part in reactions very easily, due to it's triple bond.
3. Spongy iron is used as a catalyst.
4. This is known as the Haber process. The equation of the reaction is:
N₂ + 3H₂ → 2NH₃. This is a balanced equation.

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In the next section, we will see sulphuric acid. 

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Chapter 5 - Non-Metals

In the previous section, we completed the discussion on the Periodic table. In this section we will discuss about Non-metals. We will discuss about non-metals like oxygen, hydrogen, nitrogen, chlorine etc., For each of these non-metals, we will see:

• Uses • Importance • Occurrence in nature • Preparation in the lab.

First we will see oxygen:

Oxygen is a gas. It is present in the atmosphere. But in the atmosphere, there are lots of other gases also. If we take a sample of air from the atmosphere,

• 78.08% of that sample will be nitrogen • 20.95% will be oxygen • 0.9% will be argon • 0.038% will be carbon dioxide

• The remaining portion, that is [100 -(78.08 +20.95 +0.9 +0.038)] 0.032% will consist of other gases.

So we see that oxygen has a significant presence. It is the breath of life. It is most essential for the existence of life. We know that oxygen used up by living beings is replenished by plants.

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Oxygen was discovered by a British scientist Joseph Priestly. It was identified as an element by a French scientist Antoine Lavoisier. The word 'oxygen' comes from a Greek word that means 'acid former'.

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Preparation of oxygen in the lab:

• In the lab, oxygen is prepared by heating crystals of potassium permanganate in a dry ‘boiling tube’. 

• Note that the ‘boiling tube’ must be dry. That is., there must not be any presence of water or other substances.

The diagram of the process is shown in fig.5.1 below. [It may be noted that, all experiments should be carried out under the supervision of teachers. All safety precautions should be taken]

Fig.5.1

■ So what happens when the crystals are heated?

• The crystals break down into three new substances:

(i) Potassium manganate: K₂MnO₄

(ii) Manganese dioxide MnO₂ and 

(iii) Oxygen: O₂

• So we see that oxygen is one of the ‘three new substances’. It is in the gaseous form, and so will rise towards the top of the tube.

• If we introduce a glowing splinter into the tube, it will burst into flame. This will tell us about the presence of oxygen. A video can be seen here.

■ So we tested and confirmed the presence of oxygen using the glowing splinter. How do we know the presence of the other two: K₂MnO₄ and MnO₂ ?

• There are various tests and procedures to determine the contents in the tube after the reaction. Such tests and procedures are used both in labs and in industries. We will learn about them in higher classes. For the present discussion, it is sufficient to know which are the 'three new substances'.

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So the crystals of potassium permanganate changed into three different substances. It is a decomposition reaction. As heat is required for the decomposition to take place, it is called a thermal decomposition reaction.

1. Let us write the equation for the reaction:

KMnO₄ + heat → K₂MnO₄ + MnO₂ + O₂

2. Let us analyse the above equation. When we take some crystals of potassium permanganate in the tube, we are taking some molecules of it.

3.Any one molecule of potassium permanganate is KMnO₄. So this molecule is placed on the left side of the equation in (1).

4. When heat is applied, we get three ‘new substances’.

• One molecule of a new substance is K₂MnO₄. So it is placed on the right side of the equation in (1)

• One molecule of another new substance is MnO₂. So it is also placed on the right side of the equation in (1)

• One molecule of the third new substance is O₂. So it is also placed on the right side of the equation in (1)

5. The equation seems to be complete. But it is not. Writing such an equation will not give us the ‘number of molecules’ of each substance involved in the reaction. [Equation in (1) is called the 'skeletal equation']. So we must write a balanced equation. Details here.

6. The steps for writing the balanced equation is shown below:

7. So the balanced equation is: 2KMnO₄ + heat → K₂MnO₄ + MnO₂ + O₂

From the balanced equation, we can see that, 2 molecules of potassium permanganate will be required to produce one molecule of oxygen.
8. Also, from balanced equations, we can calculate accurate quantities of substances taking part in a reaction. We will learn more about calculation of quantities in higher classes.

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Another method for the preparation of oxygen:

1. During electrolysis of water, we get hydrogen and oxygen. So it can also be used for the preparation of oxygen. Let us analyse the equation:

2. The original substance that we take is water. So we will write one molecule of it on the left side.

3. The new substances that we get are hydrogen and oxygen. So we will write one molecule of each on the right side. So the skeletal equation is:

H₂O → H₂ + O₂

4. Now we must balance it. The steps are shown below:

The balanced equation is: 2H₂O → 2H₂ + O₂

5. So we see that oxygen can be prepared by the electrolysis of water also.

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Chemical reactions in which oxygen is one of the reactants

Reaction of oxygen with other non-metals:
Take a small quantity of sulphur in a spatula and burn it. We will get the smell of sulphur dioxide.

Sulphur dioxide is a gas. It is formed when sulphur is burnt. During the burning, a chemical reaction takes place. The reaction is between sulphur and oxygen. 
■ A process of burning a substance in oxygen is called combustion.

So we see that sulphur and oxygen reacts together to form sulphur dioxide. Let us write the equation:
• One molecule of sulphur is S. One molecule of oxygen is O₂. We will write them on the left side.
• One molecule of sulphur dioxide is SO₂. We will write it on the right side.
• So the skeletal equation is: S + O₂ → SO₂ . This equation is already balanced.

■ Sulphur is a non-metal. So this is an example of the reaction of oxygen with a non-metal. There are more examples in which oxygen reacts with other non-metals:
■ Reaction with carbon: Carbon burns in oxygen to form carbon dioxide. When the burning takes place, the carbon is reacting with oxygen. 
• We will write one molecule each of carbon and oxygen on the left side
• We will write one molecule of the product, which is carbon dioxide, on the right side
• So the skeletal equation is: C + O₂ → CO₂ . This skeletal equation is already balanced.
■ Reaction with hydrogen: Hydrogen and oxygen reacts together to form water. But the reaction can be carried out only in special labs with special equipments. It is an explosive reaction, and so must be carried out only under authorised technical supervision.
• We will write one molecule each of hydrogen and oxygen on the left side
• We will write one molecule of the product, which is water, on the right side
• So the skeletal equation is: H₂ + O₂ → H₂O.
• This is not a balanced equation. The steps for writing the balanced equation is shown below:

• So the balanced equation is: 2H₂ + O₂ → 2H₂O. 
• From the balanced equation, we can see that 2 molecules of hydrogen will be required to react with one molecule of oxygen to produce water. Also note that, the reaction will give two molecules of water as the product.

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In the next section, we will discuss the reaction of oxygen with metals. 

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