In the previous section, we discussed about ammonia. In this section we will a few more non-metallic compounds.
Concentrated H2SO4 has great affinity towards water. It has the ability to absorb moisture from the substances that come in contact with it. It will absorb water even if the water is chemically combined with the substance.
Let us see an example:
In the next section, we will see Hydrogen chloride.
Sulphuric acid
Sulphuric acid is a chemical
of utmost importance in industry. It's various industrial uses are
listed below:
•Manufacture of fertilisers •Manufacture of explosives •Manufacture of paints •Manufacture of fibres •Manufacture of other chemicals •Used for dehydration
■ Sulphuric acid is called the King of
chemicals.
Industrial preparation of Sulphuric acid
Sulphuric acid is industrially
prepared by Contact process. Let us see the various stages in the
process:
Stage 1: Sulphur is burnt in
air to produce sulphurdioxide (SO2). The equation is:
S + O2 → SO2. This is a balanced equation.
S + O2 → SO2. This is a balanced equation.
Stage 2: This SO2 is allowed
to combine with more oxygen. High temperature is required for this
process. Also vanadium pentoxide (V2O5) is used as a catalyst. The
product is sulphur trioxide (SO3).
Let us write the equation:
Let us write the equation:
Reactants:
♦ Sulphur dioxide. One molecule is SO2
♦ Oxygen. One molecule is O2
Products:
♦ Sulphur trioxide. One molecule is SO3
• So skeletal equation is:
SO2 + O2 → SO3. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: SO2 + O2 → SO3
Step 2: 2SO2 + O2 → 2SO3
So the balanced equation is: 2SO2 + O2 → 2SO3
Step 1: SO2 + O2 → SO3
Step 2: 2SO2 + O2 → 2SO3
| Reactants | Products | ||||
|---|---|---|---|---|---|
| S | O | S | O | ||
| Step 1 | 1 | 4 | 1 | 3 | |
| Step 2 | 2 | 6 | 2 | 6 | |
Stage 3: SO3 is dissolved in
concentrated sulphuric acid. The product is H2S2O7. It is called
oleum. The equation is: SO3 + H2SO4 → H2S2O7
This is a balanced equation.
Stage 4: This Oleum is mixed with water to produce sulphuric acid of the required concentration. Let us write the equation:
So the balanced equation is: H2S2O7 + H2O → 2H2SO4.
Stage 4: This Oleum is mixed with water to produce sulphuric acid of the required concentration. Let us write the equation:
Reactants:
♦ Oleum. One molecule is H2S2O7.
♦ Water. One molecule is H2O
Products:
♦ Sulphuric acid. One molecule is H2SO4.
• So skeletal equation is:
H2S2O7 + H2O → H2SO4. This is not a balanced equation. The steps for writing the balanced equation are shown below:
Step 1: H2S2O7 + H2O → H2SO4
Step 2: H2S2O7 + H2O → 2H2SO4
Step 1: H2S2O7 + H2O → H2SO4
Step 2: H2S2O7 + H2O → 2H2SO4
| Reactants | Products | ||||||
|---|---|---|---|---|---|---|---|
| H | S | O | H | S | O | ||
| Step 1 | 4 | 2 | 8 | 2 | 1 | 4 | |
| Step 2 | 4 | 2 | 8 | 4 | 2 | 8 | |
• Theoretically, there is no
need for stages 3 and 4. Let us see why this is so:
(i) At the end of stage 2, we get
sulphur trioxide (SO3). When this is dissolved in water, we should get
sulphuric acid. The equation is: SO3 + H2O → H2SO4. This is a balanced equation.
(ii) But there are practical
difficulties in carrying out this reaction. The dissolution of SO3 in
water is an exothermic process.
(iii) The H2SO4 which is initially formed, becomes fine smog like particles. These particles hinder further dissolution. So the complete dissolution according to the equation does not take place.
(iv) So, in the industrial production, we go through all the four stages.
The stages can be presented in the form of a flow chart given below:
(iii) The H2SO4 which is initially formed, becomes fine smog like particles. These particles hinder further dissolution. So the complete dissolution according to the equation does not take place.
(iv) So, in the industrial production, we go through all the four stages.
The stages can be presented in the form of a flow chart given below:
Physical properties of sulphuric acid
The important physical
properties of H2SO4 are: •Colourless •viscous in nature •Highly corrosive •Denser than water •Water soluble
Chemical properties of sulphuric acid
Affinity towards water:
• Take 5 ml of water in a test tube and slowly add concentrated sulphuric acid to it.
• If we touch the bottom of the test tube, we can feel the heat.
• This is because, the reaction between water and H2SO4 is highly exothermic.
• Take 5 ml of water in a test tube and slowly add concentrated sulphuric acid to it.
• If we touch the bottom of the test tube, we can feel the heat.
• This is because, the reaction between water and H2SO4 is highly exothermic.
■ While diluting sulphuric acid,
the acid should be added to water in very small quantities, while
stirring it. We should not do the other way round. That is., we
should not add water to the acid. If we do it, there will be spurting
and may cause burns to our body.
Concentrated H2SO4 has great affinity towards water. It has the ability to absorb moisture from the substances that come in contact with it. It will absorb water even if the water is chemically combined with the substance.
Let us see an example:
1. Take a little sugar in a watch
glass. Add a few drops of concentrated sulphuric acid to it. A
reaction takes place and a black residue is left in the watch glass.
Let us write the equation:
C12H22O11 → 12C + 11H2O. This is a balanced equation.
2. The 12 carbon atoms present in the left side as 'C12' is present as '12C' on the right side.
3. Now look at the other elements: Oxygen and hydrogen:
♦ In the left side, there are 22 Hydrogen atoms and 11 Oxygen atoms.
♦ So we can say: In a sugar molecule, the number of H atoms and number of O atoms are in the ratio 22:11 which is same as 2:1.
♦ This is the same 'ratio of hydrogen and oxygen' in water.
♦ On the right side, we see that 11 molecules of water are formed. Those 11 water molecules contain the same 22 H and 11 O on the left side
4. So, when sulphuric acid reacts with a substance, the H and O atoms get separated from that substance. Those separated atoms combine together to form water.
5. The residue that remains, after the removal of water is carbon.
■ After the separation, we say that the substance is dehydrated. Thus, H2SO4 has the ability to dehydrate substances.
C12H22O11 → 12C + 11H2O. This is a balanced equation.
2. The 12 carbon atoms present in the left side as 'C12' is present as '12C' on the right side.
3. Now look at the other elements: Oxygen and hydrogen:
♦ In the left side, there are 22 Hydrogen atoms and 11 Oxygen atoms.
♦ So we can say: In a sugar molecule, the number of H atoms and number of O atoms are in the ratio 22:11 which is same as 2:1.
♦ This is the same 'ratio of hydrogen and oxygen' in water.
♦ On the right side, we see that 11 molecules of water are formed. Those 11 water molecules contain the same 22 H and 11 O on the left side
4. So, when sulphuric acid reacts with a substance, the H and O atoms get separated from that substance. Those separated atoms combine together to form water.
5. The residue that remains, after the removal of water is carbon.
■ After the separation, we say that the substance is dehydrated. Thus, H2SO4 has the ability to dehydrate substances.
• Similar to sugar, glucose (C6H12O6) and
fructose (C6H12O6) also reacts with H2SO4 and get dehydrated. [Note that, glucose and fructose have the same chemical formula. But the arrangement of atoms is different]
• Number of H atoms : Number of O atoms ratio is the same 2:1 in glucose and fructose also.
The examples of sugar, glucose and fructose that we saw above are those in which, water is present in a chemically combined form.
Another case is there, in which water is not chemically combined. But it is present in crystalline form. That is., molecules of water are attached to the molecules of the substance by physical bonding.
• Copper sulphate is an example.
♦ It contains water in crystalline form. The chemical formula is: CuSO4.5H2O.
♦ It is called hydrous copper sulphate. It is blue in colour
♦ The H2SO4 reacts with CuSO4.5H2O and causes dehydration.
♦ After dehydration, it is called 'anhydrous copper sulphate' (CuSO4). It is white in colour.
• Number of H atoms : Number of O atoms ratio is the same 2:1 in glucose and fructose also.
The examples of sugar, glucose and fructose that we saw above are those in which, water is present in a chemically combined form.
Another case is there, in which water is not chemically combined. But it is present in crystalline form. That is., molecules of water are attached to the molecules of the substance by physical bonding.
• Copper sulphate is an example.
♦ It contains water in crystalline form. The chemical formula is: CuSO4.5H2O.
♦ It is called hydrous copper sulphate. It is blue in colour
♦ The H2SO4 reacts with CuSO4.5H2O and causes dehydration.
♦ After dehydration, it is called 'anhydrous copper sulphate' (CuSO4). It is white in colour.
In the previous section, we
saw the application of quick lime (CaO) as a drying agent in the
preparation of ammonia. Now we see that H2SO4 also can absorb water.
So it can also be used as a drying agent. In fact, it is used as a
drying agent in the manufacture of chlorine (Cl2), Sulphur dioxide (SO2) and Hydrogen chloride (HCl). We saw the case
of Cl2 in a previous section (see fig.5.4).
But it cannot be used as a drying agent in the manufacture of ammonia because of the following reason: H2SO4 is an acid, while ammonia is alkaline. The two will react together.
But it cannot be used as a drying agent in the manufacture of ammonia because of the following reason: H2SO4 is an acid, while ammonia is alkaline. The two will react together.
Reaction of sulphuric acid with salts
• We have seen that salts are
formed when acids and alkalies react together.
• Now, if we take such a salt, and allow it to react with H2SO4, we will get back the acid. Let us see an example:
• Now, if we take such a salt, and allow it to react with H2SO4, we will get back the acid. Let us see an example:
• We know that when the acid HCl
react with the alkali NaOH, we get the salt NaCl.
• Now, this NaCl is allowed to react with H2SO4. We will get back the acid HCl. The reaction is as shown below:
NaCl + H2SO4 → NaHSO4 + HCl. This is a balanced equation.
• Now, this NaCl is allowed to react with H2SO4. We will get back the acid HCl. The reaction is as shown below:
NaCl + H2SO4 → NaHSO4 + HCl. This is a balanced equation.
• Another example:
Sulphuric acid reacts with potassium nitrate (KNO3). Nitric acid is obtained as a product. The equation is:
KNO3 + H2SO4 → KHSO4 + HNO3. This is a balanced equation.
■So we can say: Concentrated
sulphuric acid can displace acids from their salts. This
method is employed in the manufacture of hydrochloric acid, nitric
acid etc., Sulphuric acid reacts with potassium nitrate (KNO3). Nitric acid is obtained as a product. The equation is:
KNO3 + H2SO4 → KHSO4 + HNO3. This is a balanced equation.
Oxidising nature of sulphuric acid
We have seen oxidation and
reduction in a previous chapter. Details here. Based on that
discussion, we can write the 'oxidation number' of any element in a
compound.
Consider the following
reaction:
C + H2SO4 → CO2 + 2H2O + 2SO2.
This is the reaction between
carbon and sulphuric acid. The reaction is carried out as follows:
• Concentrated sulphuric acid is
added to a small quantity of carbon in a test tube and heated.
From
the equation, we can see that, the products are CO2, H2O and SO2. Let us write the
oxidation number of each element on both sides of the equation:
C0 + H2+1S+6O4-2 → C+4O2-2 + 2H2+1O-2 + 2S+4O2-2.
From this we can see that, the
oxidation state of carbon increased from 0 to +4. That means, the
carbon is oxidised. The sulphuric acid acts as the oxidising agent.
Another example:
Cu + 2H2SO4 → CuSO4 + 2H2O + SO2.
Let us write the oxidation number of each element on both sides of the equation:
Cu0 + 2H2+1S+6O4-2 → Cu+2S+6O4-2 + 2H2+1O-2 + S+4O2-2.
From this we can see that, the oxidation state of copper increased from 0 to +2. That means, the copper is oxidised. The sulphuric acid acts as the oxidising agent.
In the above examples, sulphuric acid oxidised carbon
(a non-metal) and copper (a metal). So we get a common property of sulphuric acid:
■ Sulphuric acid oxidises some metals and non-metals
Identification of sulphate salts
• We know that, when
hydrochloric acid reacts with alkalies, we get chlorides.Example:
HCl + NaOH → NaCl + H2O.
• HCl also reacts with metals to
form chlorides. Example:
Zn + 2HCl → ZnCl2 + H2.
■ We can write many examples
where hydrochloric acid takes part in reactions to give chlorides.
■ In
the same way, sulphuric acid also takes part in chemical reactions to
give sulphates.
Let us see some examples:
• Sulphuric acid reacts with the
alkali sodium hydroxide (NaOH) to give sodium sulphate. The equation
is:
2NaOH + H2SO4 → Na2SO4 + 2H2O.
• Sulphuric acid reacts with the
metal zinc to give zinc sulphate. The equation is:
Zn + H2SO4 → ZnSO4 + H2O.
• Sulphuric acid reacts with calcium carbonate to give calcium sulphate. The equation is:
CaCO3 + H2SO4 → CaSO4 + H2O + CO2.
So we saw some methods by
which sulphates are formed. Let us now see how these sulphate salts
can be identified:
1. Prepare an aqueous solution of
the given salt. We want to check whether this salt is sulphate salt
or not.
2. Add some barium chloride
solution to it.
3. If it is a sulphate salt,
white precipitate of barium sulphate will be formed. The equation is:
BaCl2 + Na2SO4 → BaSO4↓ + 2NaCl.
4. Can we confirm that it is a
sulphate salt? That is., will any other salt give a white precipitate
when barium chloride is added?
In fact, like
sulphate salts, carbonate salts also give white precipitate. The
equation is:
BaCl2 + Na2CO3 → BaCO3↓ + 2NaCl.
BaCl2 + Na2CO3 → BaCO3↓ + 2NaCl.
5. So, when we test a given salt
by adding barium chloride, and if we get a white precipitate, the
given salt can either be sulphate or carbonate. We want to know which
one.
6. For that, add concentrated HCl to the precipitate.
7. If the precipitate dissolves in HCl and bubbles of CO2 are formed, it is a carbonate.
8. If the precipitate does not dissolve, it is a sulphate.
6. For that, add concentrated HCl to the precipitate.
7. If the precipitate dissolves in HCl and bubbles of CO2 are formed, it is a carbonate.
8. If the precipitate does not dissolve, it is a sulphate.
In the next section, we will see Hydrogen chloride.
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