[3 Mark Question Answer]

Question 13 Marks

(i) Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C.
(ii) In the contact process for the manufacture of sulphuric acid, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two-step procedure is used. Write the equations for the two steps involved in D.
(iii) What type of substance will liberate sulphur dioxide from sodium sulphite in step E?
(iv) Write the equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F ?

Answer

(i) C = Vanadium pentoxide is the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide.
(ii) D involves the following two steps:
$a) SO_3 + H_2SO_4 \rightarrow H_2S_2O_7$
$b) H_2SO_7 + H_2O \rightarrow 2H_2SO_4$
(iii) In step E, Dilute sulphuric acid will help in the liberation of sulphur dioxide from sulphites.
(iv) This is the reaction by which sulphur dioxide is converted to sodium sulphite in step F.
$2NaOH + SO_2 \rightarrow Na_2SO_3 + H_2O$

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Question 23 Marks

Sulphuric acid is manufactured by contact process
Give the conditions for the oxidation of $SO_2$

Answer

(b) The conditions for the oxidation of $SO_2$ are:
(i) The temperature should be as low as possible. The yield has been found to be maximum at about $410^0C-450^oC$
(ii) High pressure (2 atm) is favoured because the product formed has less volume than reactant.
(iii) Excess of oxygen increases the production of sulphur trioxide.
(iv) Vanadium pentoxide or platinised asbestos is used as catalyst.

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Question 33 Marks

Copy and complete the following table:

Column $1$	Column $2$	Column $3$
Substance reacted with acid	Dilute or concentrated acid	Gas
		Hydrogen
		Carbon dioxide
		Only chlorine

Answer

Column $1$	Column $2$	Column $3$
Substance reacted with acid	Dilute or concentrated acid	Gas
Zinc	Dilute sulphuric acid	Hydrogen
Calcium carbonate	Concentrated sulphuric acid	Carbon dioxide
Bleaching power $CaOCl_2$	Dilute sulphuric acid	only chlorine

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Question 43 Marks

Sulphuric acid used in the preparation of $HCI$ and $HNO_3$? Give equations in both cases.

Answer

Concentrated sulphuric acid has a high boiling point ($356^\circ C$). So, it is considered to be non-volatile. Hence, it is used for preparing volatile acids like Hydrochloric acid and Nitric acids from their salts by double decomposition.
$NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl$
$NaNO_3 + H_2SO_4 \rightarrow NaHSO_4 + HNO_3$

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Question 53 Marks

Give a chemical test to distinguish between : dilute sulphuric acid and conc. Sulphuric acid.

Answer

1. Dilute sulphuric acid treated with zinc gives Hydrogen gas which bums with pop sound. Concentrated $H _2 SO _4$ gives $SO _2$ gas with zinc and the gas turns Acidified potassium dichromate paper green.
2. Barium chloride solution gives white ppt. with dilute $H _2 SO _4$, This white ppt. is insoluble in all acids. Concentrated $H _2 SO _4$ and NaCl mixture when heated gives dense white fumes if glass rod dipped in Ammonia solution is brought near it.

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Question 63 Marks

Give two balanced reactions of each type to show the following properties of sulphuric acid:
Non-volatile nature

Answer

Balanced reactions are:
Non-volatile nature:
It has a high boiling point (356°C) so it is considered to be non-volatile. Therefore, it is used for preparing volatile acids like hydrochloric acid, nitric acid from their salts by a double decomposition reaction.
$NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl$
$KCl + H_2SO_4 \rightarrow KHSO_4 + HCl$

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Question 73 Marks

Give two balanced reactions of each type to show the following properties of sulphuric acid:
Oxidising agent

Answer

Balanced reactions are:
Oxidising agent:
$H_2SO_4 \rightarrow H_2O + SO_2 + [O]$
Nascent oxygen oxidizes non-metals, metals and inorganic compounds.
For example,
Carbon to carbon dioxide
$C+H_2SO_4 \rightarrow CO_2 +H_2O +2SO_2$
Sulphur to sulphur dioxide
$S +H_2SO_4 \rightarrow 3SO_2 +2H_2O$

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Question 83 Marks

Give two balanced reactions of each type to show the following properties of sulphuric acid:
Acidic nature.

Answer

Balanced reactions are:
Acidic nature:
(i) Dilute $H_2SO_4$ reacts with basic oxides to form sulphate and water.
$2 NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$
$(ii) CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O$
(iii) It reacts with carbonate to produce $CO_2.$
$Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 ↑$

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Question 93 Marks

Some properties of sulphuric acid are listed below. Choose the property $A, B, C$ or $D$ which is responsible for the reactions (i) to (v). Some properties may be repeated:

Acid
Dehydrating agent
Non-volatile acid
Oxidising agent

$(i) C_{12}H_{22}O_{11} +nH_2SO_4 \rightarrow 12C + 11H_2O + nH_2SO_4$
$(ii) S + 2H_2SO_4\rightarrow 3SO_2 +2H_2O$
$(iii) NaCl +H_2SO_4 \rightarrow NaHSO_4+ HCl$
$(iv) CuO + H_2SO_4 \rightarrow CuSO_4 +H_2O$
$(v) Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O +CO_2$

Answer

(i) $C _{12} H _{22} O _{11}+ nH _2 SO _4 \rightarrow 12 C +11 H _2 O + nH _2 SO _4-$ Dehydrating agent
(ii) $S +2 H _2 SO _4 \rightarrow 3 SO _2+2 H _2 O$ - Oxidising agent
(iii) $NaCl + H _2 SO _4 \rightarrow NaHSO _4+ HCl$ - Non-volatile acid
(iv) $CuO + H _2 SO _4 \rightarrow CuSO _4+ H _2 O -$ Acid
(v) $Na _2 CO _3+ H _2 SO _4 \rightarrow Na _2 SO _4+ H _2 O + CO _2$ - Acid

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Question 103 Marks

Concentrated sulphuric acid is both an oxidizing agent and a non-volatile acid. Write one equation. Each illustrates the above-mentioned properties of sulphuric acid.

Answer

Concentrated sulphuric acid as
(i) Oxidising agent : The oxidising property of conc. sulphuric acid its due to the fact that on thermal decomposition, it yeilds nacent oxygen [O].
$H_2SO_4\rightarrow H_2O + SO_2+ [O]$
(ii) Non-volatile acid:
conc. sulphuric acid has a high boiling point $\left(338^{\circ} C \right)$ that why it is said to be a non-volitile compound, therefore it is used for preparing volatile acids like hydrochloric acids, nitric acids from there salt by double decomposition
$H_2SO_4 + NaCl \rightarrow NaHSO_4 + HCl$

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Chemistry [3 Mark Question Answer]

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