Study of Compounds — Sulphuric Acid

Question 1

State why sulphuric acid was called – 'oil of vitriol'.

Answer

Sulphuric acid was initially called 'oil of vitriol' because it was prepared by distilling green vitriol [FeSO4.7H₂O] and hence the name – 'oil of vitriol' was given.

2FeSO₄.7H₂O ⟶ Fe₂O₃ + SO₂ + SO₃ + 14H₂O

SO₃ + 14H₂O ⟶ H₂SO₄ + 13H₂O

Question 2

State how you would convert (i) sulphur (ii) chlorine (iii) sulphur dioxide — to sulphuric acid.

Answer

(i) By action of heat on sulphur and nitric acid:
S + 6HNO₃ [conc.] ⟶ 6NO₂ + 2H₂O + H₂SO₄

(ii) By passage of chlorine through an aqueous solution of sulphur dioxide:
Cl₂ + SO₂ + 2H₂O ⟶ 2HCl + H₂SO₄

(iii) By oxidation of an aq. soln. of sulphur dioxide:
2SO₂ + 2H₂O + O₂ ⟶ 2H₂SO₄

Question 3

State the purpose of the 'Contact process'.

Answer

The purpose of the 'Contact process' is to manufacture sulphuric acid.

Question 4

In the Contact process

(i) State how you would convert (a) sulphur (b) iron pyrites to sulphur dioxide in the first step of the Contact process.

(ii) State the conditions i.e. catalyst, promoter, temperature and pressure in the catalytic oxidation of sulphur dioxide to sulphur trioxide in the Contact tower. Give a balanced equation for the same.

(iii) State why the above catalytic oxidation reaction supplies energy.

(iv) Give a reason why – vanadium pentoxide is preferred to platinum during the catalytic oxidation of sulphur dioxide.

(v) Give a reason why the catalyst mass is heated electrically – only initially.

(vi) State why sulphur trioxide vapours are absorbed in concentrated sulphuric acid and not in water to obtain sulphuric acid.

Answer

(i) Sulphur dioxide is produced by burning sulphur [S] or iron pyrites [FeS₂]

S + O₂ ⟶ SO₂

4FeS₂ + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂

(ii) Catalyst : Vanadium pentoxide [V₂O₅] or platinum [Pt.]

Temperature : 450 - 500°C , Pressure : 1 to 2 atmosphere

$2\text{SO}_2 + \text{O}_2 \xrightleftharpoons[450-500\text{\degree C} /1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + \Delta$

Conversion ratio: 98% of sulphur dioxide converted to sulphur trioxide.

(iii) The catalytic oxidation of SO₂ to SO₃ is an exothermic reaction, hence, supplies energy.

(iv) Vanadium pentoxide is preferred to platinized asbestos as a catalyst since it is comparatively cheaper and less easily poisoned or susceptible to impurities.

(v) The catalyst mass is only initially heated electrically, since the catalytic oxidation of sulphur dioxide is an exothermic reaction and the heat produced maintains the temperature at 450 – 500°C.

(vi) Even though sulphur trioxide is an acid anhydride of sulphuric acid it is not directly absorbed in water to give sulphuric acid.

The reaction is highly exothermic resulting in production of a dense fog of sulphuric acid particles which do not condense easily.

Hence sulphur trioxide vapours are dissolved in conc. sulphuric acid to give oleum which on dilution with the requisite amount of soft water in the dilution tank gives sulphuric acid of the desired concentration [about 98%].

Question 5

Give a reason why concentrated sulphuric acid is kept in air tight bottles.

Answer

As, concentrated sulphuric acid is hygroscopic and has a great affinity for water hence, it absorbs moisture from the atmosphere. Therefore, it is kept in air tight bottles.

Question 6

State the basic steps with reasons, involved in diluting a beaker of conc. H₂SO₄.

Answer

For dilution of conc. H₂SO₄, the conc. acid is added to water and not the water to the acid even though heat is evolved in both cases.

1. If water is added to the acid : There is a sudden increase in the temperature and the acid being in bulk tends to spurt out.
2. If acid is added to water : The water is in bulk and the acid being heavier settles down. The heat evolved is dissipated in the water itself and hence the spurting of the acid is minimized.

Question 7

Give reasons why dilute sulphuric acid:

(i) behaves as an acid when dilute.

(ii) is dibasic in nature.

Answer

(i) Acidic properties of sulphuric acid are due to the presence of hydronium ions [H₃O⁺] formed when H₂SO₄ dissociates in aqueous solution. Hence, it behaves as an acid when dilute.

(ii) Sulphuric acid dissociates in aqueous solution giving two hydrogen ions (H⁺) ions per molecule of the acid. Hence, it is dibasic in nature.

H₂SO₄ ⇌ 2H⁺ + SO₄^2-

H₂SO₄ + 2H₂O ⇌ 2H₃O⁺ + SO₄^2-

Question 8

Convert dil. H₂SO₄ to –

(i) Hydrogen

(ii) Carbon dioxide

(iii) Sulphur dioxide

(iv) Hydrogen sulphide

(v) An acid salt

(vi) A normal salt.

Answer

(i) Zn + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂ (g)

(ii) 2KHCO₃ + H₂SO₄ ⟶ K₂SO₄ + 2H₂O + 2CO₂

(iii) S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

(iv) FeS + H₂SO₄ (dil.) ⟶ FeSO₄ + H₂S

(v) NaOH + H₂SO₄ (dil.)⟶ NaHSO₄ + H₂O

(vi) 2NaOH + H₂SO₄ (dil.)⟶ Na₂SO₄ + 2H₂O

Question 9

Give equations for formation of two different acids from conc. H₂SO₄. State the property of sulphuric acid involved in the above formation.

Answer

$\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_3$

$\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}$

The property of being a non-volatile acid is used in the above reactions.

Question 10

Give equations for oxidation of conc. H₂SO₄ giving the oxidised products –

(i) Carbon dioxide

(ii) Sulphur dioxide

(iii) Phosphoric acid

(iv) Copper (II) sulphate

(v) Iodine

(vi) Sulphur

respectively

Answer

(i) C + 2H₂SO₄ (conc.) ⟶ CO₂ + 2SO₂ + 2H₂O

(ii) S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

(iii) 2P + 5H₂SO₄ (conc.) ⟶ 2H₃PO₄ + 5SO₂ + 2H₂O

(iv) Cu + 2H₂SO₄ (conc.) ⟶ CuSO₄ + SO₂ + 2H₂O

(v) 2HI + H₂SO₄ (conc.) ⟶ I₂ + SO₂ + 2H₂O

(vi) H₂S + H₂SO₄ (conc.) ⟶ S + SO₂ + 2H₂O

Question 11

Give a reason why concentrated and not dil H₂SO₄ behaves as an oxidising and dehydrating agent.

Answer

Oxidising property of concentrated sulphuric acid is due to the fact that on thermal decomposition, it yields nascent oxygen [O] which helps in oxidation.

H₂SO₄ ⟶ H₂O + SO₂ + [O]

Nascent oxygen oxidizes non-metals, metals and inorganic compounds.

On the other hand, dil H₂SO₄ on heating does not decompose to give nascent oxygen and as such cannot behave as an Oxidising agent.

Question 12

Give the equation for the reaction of conc. sulphuric acid with –

(i) glucose

(ii) sucrose

(iii) cellulose

(iv) an organic acid containing one carbon atom and two hydrogen atoms

(v) an organic acid containing two carbon and two hydrogen atoms

(vi) an alcohol

(vii) hydrated copper [II] sulphate

Answer

(i) Reaction of conc. sulphuric acid with glucose:

$\text{C}_{6}\text{H}_{12}\text{O}_{6}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{6C} + \text{6H}_2\text{O}$

(ii) Reaction of conc. sulphuric acid with sucrose:

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}$

(iii) Reaction of conc. sulphuric acid with cellulose:

$[\text{C}_{6}\text{H}_{10}\text{O}_{5}]\text{n}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{6[C]}_\text{n} + \text{5[H}_2\text{O}]_\text{n}$

(iv) Reaction of conc. sulphuric acid with formic acid [H.COOH]:

$\text{H.COOH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CO} + \text{H}_2\text{O}$

(v) Reaction of conc. sulphuric acid with oxalic acid [H₂C₂O₄]:

$\text{H}_{2}\text{C}_{2}\text{O}_{4} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CO} + \text{CO}_{2} + \text{H}_2\text{O}$

(vi) Reaction of conc. sulphuric acid with ethyl alcohol:

$\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{C}_{2}\text{H}_{4} \small{\space[ethene-ethylene]} + \text{H}_2\text{O}$

(vii) Reaction of conc. sulphuric acid with hydrated copper sulphate:

$\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

Question 13

State the observation seen when conc. H₂SO₄ is added to –

(i) sucrose

(ii) hydrated copper [II] sulphate.

Answer

(i) Conc. H₂SO₄ dehydrates sucrose to a black spongy charged mass of carbon sugar charcoal.

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}$

(ii) The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

$\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

Question 14

State how addition of –

(i) copper

(ii) NaCl

to hot conc. H₂SO₄ serves as a test for the latter.

Answer

(i) When Cu is added to hot conc. H₂SO₄, colourless suffocating (SO₂) gas is evolved. The gas turns orange coloured acidified potassium dichromate solution to green colour.

Cu + 2H₂SO₄ (conc.) $\xrightarrow{\Delta}$ CuSO₄ + SO₂ + 2H₂O

(ii) When NaCl is heated with conc. H₂SO₄, colourless pungent smelling HCl gas is evolved.

$\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}] \xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}$

Dense white fumes are produced when a rod dipped in NH₃ soln. is brought near the test tube containing HCl gas.

Question 15

Give two tests for dilute sulphuric acid with balanced equations. State why

(i) BaCl₂

(ii) Pb(NO₃)₂

are used for the above tests

Answer

(i) BaCl₂ solution on treating with dilute sulphuric acid forms white ppt. of barium sulphate, which is insoluble in dilute sulphuric acid.

BaCl₂ + H₂SO₄ (dil) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

(ii) Pb(NO₃)₂ on treating with dilute sulphuric acid gives white ppt of lead sulphate, which is insoluble in dil. H₂SO₄.

Pb(NO₃)₂ + H₂SO₄ (dil) ⟶ 2HNO₃ + PbSO₄ ↓ [white ppt. formed]

BaCl₂ and Pb(NO₃)₂ are used for the above tests as they react with H₂SO₄ and form BaSO₄ and PbSO₄ respectively, which are the only compounds insoluble in dil. sulphuric acid.

Question 16

Give a test to distinguish dilute sulphuric acid from dilute HCl and dilute HNO₃.

Answer

Barium chloride (BaCl₂) solution reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO₄) but with dil. HCl and dil. HNO₃ no white ppt. is produced since BaCl₂ and Ba(NO₃)₂ is soluble in dil. H₂SO₄.

H₂SO₄ (dil.) + BaCl₂ (aq.) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

HNO₃ (dil.) + BaCl₂ (aq.) ⟶ No ppt.

HCl (dil.) + BaCl₂ (aq.) ⟶ No ppt.

Question 17

State three different chemical compounds other than acids manufactured industrially from sulphuric acid.

Answer

1. Ammonium sulphate [(NH₄)₂SO₄] used as a fertilizer
2. Superphosphate of lime [Ca(H₂PO₄)₂ + (CaSO₄)] used as a fertilizer
3. Tri-nitro toluene (T.N.T.) used as an explosive.

Question 1

Sulphuric Acid — Manufacture-Contact process

Answer

Question 2

Chemical properties of sulphuric acid

Answer

Question 1(2002)

State the substance/s reacted with dilute or concentrated sulphuric acid to form the following gases:

(i) Hydrogen

(ii) Carbon dioxide.

State whether the acid used in each case is dilute or concentrated.

Answer

(i) Zinc, Iron or any active metal reacts with dilute Sulphuric Acid (H₂SO₄) to form Hydrogen. The reactions are shown below:

Zn + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂ [g]

Fe + H₂SO₄ (dil.) ⟶ FeSO₄ + H₂ [g]

(ii) Carbonates and bicarbonates react with dilute Sulphuric Acid (H₂SO₄) to give Carbon dioxide (CO₂). The reactions are shown below:

Na₂CO₃ + H₂SO₄ (dil.) ⟶ Na₂SO₄ + H₂O + CO₂

2KHCO₃ + H₂SO₄ (dil.) ⟶ K₂SO₄ + 2H₂O + 2CO₂

Question 2(2002)

Write the equations for the lab. preparation of:

(i) Na₂SO₄ using dil. H₂SO₄,

(ii) PbSO₄ using dil. H₂SO₄

Answer

(i) 2NaOH + H₂SO₄ (dil) ⟶ Na₂SO₄ + 2H₂O

(ii) Pb(NO₃)₂ + H₂SO₄ (dil) ⟶ PbSO₄ ↓ + 2HNO₃

Question 1(2003)

State the name of the process by which H₂SO₄ is manufactured. Name the catalyst used.

Answer

Contact process is the name of the process by which H₂SO₄ is manufactured.

Catalyst — Vanadium pentoxide [V₂O₅] or platinum [Pt.]

Question 2(2003)

Concentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is ............... [less volatile / stronger] in comparison to these two acids.

Answer

Concentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is less volatile in comparison to these two acids.

Question 3(2003)

Write the equations for the laboratory preparation of the following salts using sulphuric acid:

(i) Copper sulphate from copper

(ii) Lead sulphate from lead nitrate

Answer

Cu + 2H₂SO₄ (conc.) ⟶ CuSO₄ + SO₂ + 2H₂O

Pb(NO₃)₂ + H₂SO₄ (dil.) ⟶ PbSO₄ ↓ + 2HNO₃

Question 1(2004)

Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide.

Answer

Vanadium pentoxide [V₂O₅] or platinum [Pt.]

Question 2(2004)

In the Contact process, 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.

Answer

The equations for the two steps involved are:

SO₃ + H₂SO₄ ⟶ H₂S₂O₇

H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

Question 1(2005)

Write balanced equations for the following:

(i) Potassium hydrogen carbonate and dilute sulphuric acid.

(ii) Sodium nitrate and concentrated sulphuric acid.

Answer

2KHCO₃ + H₂SO₄ ⟶ K₂SO₄ + 2H₂O + 2CO₂

$\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_\bold{3}$

Question 2(2005)

Choose the property of sulphuric acid (A, B, C or D), which is relevant to each of the preparations (i) to (iii) :

A: Dil. acid (typical acid properties)

B: Non-volatile acid

C: Oxidizing agent

D: Dehydrating agent

Preparation of :

(i) HCl

(ii) ethene from ethanol

(iii) copper sulphate from copper oxide.

Answer

(i) In preparation of HCl — B: Non-volatile acid

$\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}$

(ii) In preparation of ethene from ethanol — D: Dehydrating agent

C₂H₅OH + H₂SO₄ (conc.) ⟶ C₂H₄ + H₂O

(iii) In preparation of copper sulphate from copper oxide — A: Dil. acid (typical acid properties)

CuO + H₂SO₄ (dil.) ⟶ CuSO₄ + H₂O

Question 1(2006)

Name the process used for the large scale manufacture of sulphuric acid.

Answer

Contact process.

Question 2(2006)

Which property of sulphuric acid accounts for it's use as a dehydrating agent.

Answer

Sulphuric acid removes water of crystallization from hydrated salts.

Question 3(2006)

Conc. H₂SO₄ is an oxidizing agent and a non volatile acid. Write an equation for each property.

Answer

Sulphuric acid as an Oxidizing agent —

C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2SO₂ + 2H₂O

Sulphuric acid as an non-volatile acid —

$\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_3$

Question 1(2007)

Write balanced equation for the following reactions:

(i) Lead sulphate from lead nitrate solution and dilute sulphuric acid.

(ii) Copper sulphate from copper and conc. sulphuric acid.

Answer

(i) Pb(NO₃)₂ + H₂SO₄ [dil] ⟶ PbSO₄ + 2HNO₃

(ii) Cu + 2H₂SO₄ [conc.] ⟶ CuSO₄ + 2H₂O + SO₂

Question 2(2007)

Properties of H₂SO₄ are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v).

A : Acid
B: Dehydrating agent
C: Non-volatile acid
D: Oxidizing agent

(i) C₁₂H₂₂O₁₁+ nH₂SO₄ ⟶ 12C + 11H₂O + nH₂SO₄

(ii) S + 2H₂SO₄ ⟶ 3SO₂ + 2H₂O

(iii) NaCl + H₂SO₄ ⟶ NaHSO₄ + HCl

(iv) CuO + H₂SO₄ ⟶ CuSO₄ + H₂O

(v) Na₂CO₃ + H₂SO₄ ⟶Na₂SO₄ + H₂O + CO₂

(Some properties may be repeated)

Answer

(i) B: Dehydrating agent

(ii) D: Oxidizing agent

(iii) C: Non-volatile acid

(iv) A : Acid

(v) A : Acid

Question 3(2007)

Dilute hydrochloric acid and dilute sulphuric acid are both colourless solutions. How will the addition of barium chloride solution to each help to distinguish between the two.

Answer

Barium chloride soln. reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO₄) but with dil. HCl no white ppt. is produced.

H₂SO₄ (dil.) + BaCl₂ (aq.) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

HCl (dil.) + BaCl₂ (aq.) ⟶ no reaction

Question 4(2007)

From HCl, HNO₃, H₂SO₄ state which has the highest boiling point and which has the lowest.

Answer

H₂SO₄ — highest boiling point (338°C).

HCl — lowest boiling point (-83°C).

Question 1(2008)

Dilute H₂SO₄ will produce a white ppt. when added to a solution of:

A. Copper nitrate
B. Zinc nitrate
C. Lead nitrate
D. Sodium nitrate

Answer

Lead nitrate

Pb(NO₃)₂ + H₂SO₄ (dil) ⟶ 2HNO₃ + PbSO₄ ↓ [white ppt. formed]

White ppt. of PbSO₄ is visible.

Question 2(2008)

Identify the following substance : Liquid E can be dehydrated to product ethene.

Answer

Liquid E is Ethanol [C₂H₅OH]

C₂H₅OH + H₂SO₄ (conc.) ⟶ C₂H₄ + H₂O

Question 3(2008)

Copy and complete the following table relating to an important industrial process and it's final output.

Answer

Question 4(2008)

Making use only of substances given: dil. sulphuric acid, Sodium carbonate, Zinc, Sodium Sulphite, Lead, Calcium carbonate:

Give equations for the reactions by which you could obtain:

(i) hydrogen

(ii) sulphur dioxide

(iii) carbon dioxide

(iv) zinc carbonate (2 steps)

Answer

(i) Zn + H₂SO₄ [dil.] ⟶ ZnSO₄ + H₂

(ii) Na₂SO₃ + H₂SO₄ [dil.] ⟶ Na₂SO₄ + H₂O + SO₂

(iii) Na₂CO₃ + H₂SO₄ [dil.] ⟶ Na₂SO₄ + H₂O + CO₂

(iv) Zn + H₂SO₄ [dil.] ⟶ ZnSO₄ + H₂
ZnSO₄ + Na₂CO₃ [dil.] ⟶ ZnCO₃ + Na₂SO₄

Question 5(2008)

What property of conc. H₂SO₄ is used in the action when sugar turns black in it's presence.

Answer

Dehydrating agent.

C₁₂H₂₂O₁₁+ nH₂SO₄ ⟶ 12C + 11H₂O

Question 6(2008)

Write the equations for:

(1) dil. H₂SO₄ and barium chloride.

(2) dil. H₂SO₄ and sodium sulphide.

Answer

(i) BaCl₂ + H₂SO₄ (dil) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

(ii) Na₂S + H₂SO₄ ⟶ Na₂SO₄ + H₂S

Question 7(2008)

Which property of conc. H₂SO₄ allows it to be used in the preparation of HCl and HNO₃ acids.

Answer

Conc. H₂SO₄ is a non volatile acid.

Question 1(2009)

Name the gas evolved [formula is not acceptable] – The gas that can be oxidized to sulphur.

Answer

Hydrogen sulphide gas (H₂S)

H₂S + H₂SO₄ [conc.] ⟶ S + SO₂ + 2H₂O

Question 1(2010)

Give the equation for:

(i) Heat on sulphur with conc. H₂SO₄

(ii) Reaction of - sugar with conc. H₂SO₄

Answer

(i) S + 2H₂SO₄ [conc.] ⟶ 3SO₂ + 2H₂O

(ii) C₁₂H₂₂O₁₁+ nH₂SO₄ ⟶ 12C + 11H₂O

Question 2(2010)

Give a balanced equation for the conversion of zinc oxide to zinc sulphate.

Answer

ZnO + H₂SO₄ [dil.] ⟶ ZnSO₄ + H₂O

Question 3(2010)

Select from A, B, C –

A: Sodium hydroxide solution.

B: A weak acid.

C: Dilute sulphuric acid.

The solution which liberates sulphur dioxide gas, from sodium sulphite.

Answer

Dilute sulphuric acid

Na₂SO₃ + H₂SO₄ [dil.] ⟶ Na₂SO₄ + H₂O + SO₂

Question 1(2011)

State your observation when – Sugar crystals are added to conc. sulphuric acid.

Answer

Steam is evolved from the test tube and it becomes very hot. Black spongy mass of carbon is formed.

C₁₂H₂₂O₁₁+ nH₂SO₄ ⟶ 12C + 11H₂O

Question 2(2011)

Choose the correct answer from the choices:

The gas evolved when dil. sulphuric acid reacts with iron sulphide.

A: Hydrogen sulphide
B: Sulphur dioxide
C: Sulphur trioxide
D: Vapour of sulphuric acid.

Answer

Hydrogen sulphide

FeS + H₂SO₄ [dil.] ⟶ FeSO₄ + H₂S

Question 3(2011)

Give a balanced equation for – Dilute sulphuric acid is poured over sodium sulphite

Answer

Na₂SO₃ + H₂SO₄ (dil.) ⟶ Na₂SO₄ + H₂O + SO₂

Question 4(2011)

Give balanced equations for the manufacture of sulphuric acid by the Contact process.

Answer

Contact process:

Sulphur or Pyrite Burner:
S + O₂ ⟶ SO₂
4FeS₂ + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂

Contact Tower :

$2\text{SO}_2 + \text{O}_2 \xrightleftharpoons[450-500\text{\degree C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + 45 \text{ K cals}$

Absorption Tower :
SO₃ + H₂SO₄ ⟶ H₂S₂O₇

Dilution Tank:
H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

Question 5(2011)

State the property of sulphuric acid shown by the reaction of conc. sulphuric acid when heated with

(a) Potassium nitrate

(b) Carbon

Answer

(a) Sulphuric acid behaves as a non volatile acid.

$\text{KNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{KHSO}_\text{4} + \text{HNO}_\bold{3}$

(b) Sulphuric acid behaves as an oxidizing agent

C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂

Question 1(2012)

Name the gas produced on reaction of dilute sulphuric acid with a metallic sulphide.

Answer

Hydrogen sulphide (H₂S)

Question 2(2012)

Some properties of sulphuric acid are listed below. Choose the role played by sulphuric acid – A, B, C, or D which is responsible for the reactions (i) to (v).

Some role/s may be repeated.

A : Dilute acid
B : Dehydrating agent
C : Non-volatile acid
D : Oxidising agent

(1) $\text{CuSO}_4.5\text{H}_2\text{O} \xrightarrow{\text{Conc. H}_2 \text{SO}_4} \text{CuSO}_4 + 5\text{H}_2\text{O}$

(2) S + 2H₂SO₄ [conc.] ⟶3SO₂ + 2H₂O

(3) $\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_\bold{3}$

(4) MgO + H₂SO₄ ⟶ MgSO₄ + H₂O

(5) Zn + 2H₂SO₄ [conc.] ⟶ ZnSO₄ + SO₂ + 2H₂O

Answer

(1) B: Dehydrating agent

(2) D: Oxidizing agent

(3) C: Non-volatile acid

(4) A: Dilute acid

(5) D: Oxidising agent

Question 3(2012)

Give balanced equation for the reaction : Zinc sulphide and dilute sulphuric acid.

Answer

ZnS + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂S

Question 1(2013)

State one appropriate observation for : conc. H₂SO₄ is added to a crystal of hydrated copper sulphate.

Answer

The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

$\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

Question 2(2013)

In the given equation :

S + 2H₂SO₄ ⟶ 3SO₂ + 2H₂O

Identify the role played by conc. H₂SO₄ i.e.

(A) Non-volatile acid

(B) Oxidising agent

(C) Dehydrating agent

(D) None of the above.

Answer

Oxidizing agent

Question 3(2013)

Give a balanced equation for : Dehydration of concen­trated sulphuric acid with sugar crystals.

Answer

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}$

Question 4(2013)

Identify the substance underlined: A dilute mineral acid which forms a white precipitate when treated with barium chloride solution.

Answer

Dilute sulphuric acid.

BaCl₂ + H₂SO₄ (dil) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

Question 1(2014)

Write balanced equations for the following: Action of concentrated sulphuric acid on carbon.

Answer

C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2H₂O + 2SO₂

Question 2(2014)

Distinguish between the following pairs of compounds using the test given within brackets: Dilute sulphuric acid and dilute hydrochloric acid [using barium chloride solution]

Answer

Barium chloride soln. reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO₄) but with dilute hydrochloric acid no white ppt. is produced since barium chloride is soluble in dil. sulphuric acid.

H₂SO₄ (dil.) + BaCl₂ (aq.) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

HCl (dil.) + BaCl₂ (aq.) ⟶ No reaction

Question 3(2014)

State – Any two conditions for the conversion of sulphur dioxide to sulphur trioxide.

Answer

Two conditions for the conversion of sulphur dioxide to sulphur trioxide are:

1. Temperature — 450 to 500°C
2. Pressure — 1 to 2 atmosphere

Question 4(2014)

Give one equation each to show the following properties of sulphuric acid:

(i) Dehydrating property

(ii) Acidic nature

(iii) As a non-volatile acid

Answer

(i) $\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

(ii) Na₂SO₃ + H₂SO₄ (dil.) ⟶ Na₂SO₄ + H₂O + SO₂

(iii) $\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_\bold{3}$

Question 1(2015)

Identify the acid in each case:

(i) The acid which is used in the preparation of a non­-volatile acid.

(ii) The acid which produces sugar charcoal from sugar.

(iii) The acid on mixing with lead nitrate soln. produces a white ppt. which is insoluble even on heating.

Answer

(i) Conc. nitric acid

(ii) Conc. sulphuric acid

(iii) Dilute sulphuric acid

Question 2(2015)

Give equations for the action of sulphuric acid on —

(a) Potassium hydrogen carbonate.

(b) Sulphur

Answer

(a) 2KHCO₃ + H₂SO₄ ⟶ K₂SO₄ + 2H₂O + 2CO₂

(b) S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

Question 3(2015)

In the manufacture of sulphuric acid by the Contact process, give the equations for the conversion of sulphur trioxide to sulphuric acid.

Answer

SO₃ + H₂SO₄ ⟶ H₂S₂O₇

H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

Question 1(2016)

Write balanced chemical equation for: Action of dilute sulphuric acid on Sodium Sulphite.

Answer

Na₂SO₃ + H₂SO₄(dil) ⟶ Na₂SO₄ + H₂O + SO₂

Question 2(2016)

State your observations when:

(i) Barium chloride soln. is mixed with sodium sulphate soln.

(ii) Concentrated sulphuric acid is added to sugar crystals.

Answer

(i) White coloured precipitate of barium sulphate is formed when sodium sulphate is mixed with barium chloride.

Na₂SO₄ (dil.) + BaCl₂ (aq.) ⟶ 2NaCl + BaSO₄ ↓ [white ppt. formed]

(ii) Black spongy charred mass of carbon is formed. Steam is seen to be evolved.

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}$

Question 3(2016)

A, B, C and D summarize the properties of sulphuric acid depending on whether it is dilute or concentrated.

A: Typical acid property

B: Non-volatile acid

C: Oxidizing agent

D: Dehydrating agent

Choose the property [A, B, C or D] depending on which is relevant to each of the following:

(i) Preparation of hydrogen chloride gas.

(ii) Preparation of copper sulphate from copper oxide.

(iii) Action of conc. sulphuric acid on sulphur.

Answer

(i) B: Non-volatile acid

$\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}$

(ii) A: Typical acid property

CuO + H₂SO₄ (conc.) ⟶ CuSO₄ + H₂O

(iii) C: Oxidizing agent

S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

Question 1(2017)

Write the balanced chemical equation for – Action of concentrated sulphuric acid on sulphur.

Answer

S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

Question 2(2017)

State one relevant observation for – Action of conc. sulphuric acid on hydrated copper sulphate.

Answer

The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

$\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

Question 3(2017)

State – How will you distinguish between dilute hydrochloric acid and dilute sulphuric acid using lead nitrate solution.

Answer

Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.

H₂SO₄ (dil.) + Pb(NO₃)₂ ⟶ 2HNO₃ + PbSO₄ ↓ [white ppt. formed which does not dissolve on warming the mixture].

Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution.

2HCl (dil.) + Pb(NO₃)₂ ⟶ 2HNO₃ + PbCl₂ ↓ [white ppt. formed which dissolves on warming the mixture].

Hence, dilute hydrochloric acid and dilute sulphuric acid can be distinguished using lead nitrate solution.

Question 4(2017)

Write balanced chemical equations to show –

(i) The oxidizing action of conc. sulphuric acid on carbon.

(ii) The behaviour of H₂SO₄ as an acid when it reacts with magnesium.

(iii) The dehydrating property of conc. sulphuric acid with sugar.

(iv) The conversion of SO₃ to sulphuric acid in the Contact process.

Answer

(i) C + 2H₂SO₄ (conc.) ⟶ CO₂ + 2SO₂ + 2H₂O

(ii) Mg + H₂SO₄ (dil.) ⟶ MgSO₄ + H₂ (g)

(iii) $\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}$

(iv) SO₃ + H₂SO₄ (conc.) ⟶ H₂S₂O₇ (oleum)
H₂S₂O₇ + H₂O ⟶ 2H₂SO₄ (conc.)

Question 1(2018)

Choose the correct answer from the options given:

The catalyst used in the Contact Process is:

1. Copper
2. Iron
3. Vanadium pentoxide
4. Manganese dioxide

Answer

Vanadium pentoxide

Question 2(2018)

Write a balanced chemical equation for - Action of concentrated sulphuric acid on carbon.

Answer

C + 2H₂SO₄ (conc.) ⟶ CO₂ + 2SO₂ + 2H₂O

Question 3(2018)

State which property of sulphuric acid is shown by the reaction of concentrated sulphuric acid with:

(i) Ethanol

(ii) Carbon

Answer

(i) Dehydrating property

$\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{C}_{2}\text{H}_{4} \small{\space[ethene-ethylene]} + \text{H}_2\text{O}$

(ii) Oxidizing property

C + 2H₂SO₄ (conc.) ⟶ CO₂ + 2SO₂ + 2H₂O

Question 1(2019)

Write balanced chemical equations for the following reactions:

Action of dilute sulphuric acid on —

(i) Sodium hydroxide

(ii) Zinc sulphide

Answer

(i) 2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

(ii) ZnS + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂S

Question 2(2019)

Distinguish between the following pairs of compounds using the reagent given in the bracket:

Dilute hydrochloric acid and dilute sulphuric acid [using lead nitrate soln.]

Answer

Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.

H₂SO₄ (dil.) + Pb(NO₃)₂ ⟶ 2HNO₃ + PbSO₄ ↓ [white ppt. formed which does not dissolve on warming the mixture].

Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution.

2HCl (dil.) + Pb(NO₃)₂ ⟶ 2HNO₃ + PbCl₂ ↓ [white ppt. formed which dissolves on warming the mixture].

Hence, dilute hydrochloric acid and dilute sulphuric acid can be distinguished using lead nitrate solution.

Question 3(2019)

Complete the following equation:

C + conc. H₂SO₄ ⟶

Answer

C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2SO₂ + 2H₂O

Question 4(2019)

Copy and complete the following table which refers to the industrial method for preparation of sulphuric acid.

Answer

Question 1(2020)

Choose the correct answer from the options:

The acid which can produce carbon from cane sugar, is —

1. Concentrated Hydrochloric acid
2. Concentrated Nitric acid
3. Concentrated Sulphuric acid
4. Concentrated Acetic acid

Answer

Concentrated Sulphuric acid

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}$

Question 2(2020)

Identify the substance underlined in the following :

The acid that is a dehydrating as well as a drying agent.

Answer

Sulphuric acid

Question 1

Select the correct answer from the choice in brackets.

1. The oxidised product obtained when sulphur reacts with conc. H₂SO₄. [H₂S/SO₂/H₂SO₃].
2. The dehydrated product obtained when cane sugar reacts with conc. H₂SO₄. [CO/C/CO₂]
3. The type of salt formed when excess of caustic soda reacts with sulphuric acid. [acid salt / normal salt]
4. The reduced product obtained when hydrogen sulphide reacts with conc. H₂SO₄. [SO₂/S/H₂O]
5. The salt which reacts with dil. H₂SO₄ acid to give an insoluble ppt. [Cu(NO₃)₂/Zn(NO₃)₂/Pb(NO₃)₂]

Answer

1. SO₂
2. C
3. Normal salt
4. SO₂
5. Pb(NO₃)₂

Question 2

$\text{Iron pyrites} \xrightarrow{\text{A}} \text{Colourless acidic gas} \xrightarrow{\text{B}} \text{Sulphur trioxide} \xrightarrow{\text{C}} \text{Oleum} \xrightarrow{\text{D}} \text{Sulphuric acid}$

1. Give a balanced equation for the conversion 'A'.
2. The gaseous mixture of the product of conversion 'A' and air contains dust particles as an impurity. Name another impurity in the same mixture.
3. Is the conversion 'B' an exothermic or an endothermic reaction. Would lowering the temperature favour or retard the forward reaction.
4. If the product of conversion 'B' is an acid anhydride of H₂SO₄, the anhydride of conversion 'A' is ............... .
5. State why water is added for the conversion 'D' and not for the conversion 'C'

Answer

1. 4FeS₂ + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂
2. Arsenious oxide (As₄O₆)
3. The catalytic oxidation of SO₂ to SO₃ is an exothermic reaction. According to Le Chatelier's principle, higher yield of the product is obtained by lowering the temperature.
4. acid anhydride of H₂SO₃ (Sulphurous acid)
5. Sulphur trioxide is not directly absorbed in water to give sulphuric acid as the reaction is highly exothermic resulting in production of a dense fog of sulphuric acid particles which do not condense easily.
   Hence, sulphur trioxide vapours are dissolved in conc. sulphuric acid to give oleum which on dilution with the requisite amount of soft water in the dilution tank gives sulphuric acid of the desired concentration.

Question 3

Give balanced equations for the following reactions using sulphuric acid.

1. Formation of a black mark on a piece of wood on addition of conc. H₂SO₄ to it.
2. Oxidation of a foul smelling acidic gas, heavier than air and fairly soluble in H₂O by conc. H₂SO₄.
3. Formation of an acid salt from sulphuric acid and (a) an alkali (b) a sodium salt.
4. Formation of a hydrocarbon from an organic compound
5. Formation of sulphur dioxide using a metal below hydrogen in the activity series.

Answer

1. $[\text{C}_{6}\text{H}_{10}\text{O}_{5}]\text{n}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{6[C]}_\text{n} + \text{5[H}_2\text{O}]_\text{n}$

2. H₂S + H₂SO₄ (conc.) ⟶ S + 2H₂O + SO₂

3. (a) Formation of an acid salt from sulphuric acid and an alkali (NaOH):
   NaOH [insufficient] + H₂SO₄ (dil.)⟶ NaHSO₄ + H₂O
   
   (b) Formation of an acid salt from sulphuric acid and a sodium salt (NaCl):
   $\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}] \xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}$

4. $\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{C}_{2}\text{H}_{4} \small{\space[ethene-ethylene]} + \text{H}_2\text{O}$

5. Cu + 2H₂SO₄ (conc.) ⟶ CuSO₄ + 2H₂O +SO₂

Question 4

Match the conversions in column 'X' using sulphuric acid, with the type of chemical property of sulphuric acid A to E it represents in column 'Y'

Answer

Question 5

Select the correct substance from the substances A to J which react with the sulphuric acid to give the product 1 to 10. [State whether the acid used in each case is dilute or concentrated].

A : Iron

B : Sodium carbonate

C : Sodium chloride

D : Formic acid

E : Sodium nitrate

F : Sodium sulphite

G : Ethyl alcohol

H : Sodium sulphide

I : Sodium hydroxide (excess)

J : Hydrogen sulphide

1. Product — Sulphur dioxide
2. Product — Sulphur
3. Product — Hydrogen
4. Product — Hydrochloric acid
5. Product — Sodium sulphate
6. Product — Carbon dioxide
7. Product — Carbon monoxide
8. Product — Nitric acid
9. Product — Hydrogen sulphide
10. Product — Ethene

Answer

1. F: Sodium sulphite and dil. sulphuric acid
2. J: Hydrogen sulphide and conc. sulphuric acid
3. A: Iron and dil. sulphuric acid
4. C: Sodium chloride and conc. sulphuric acid
5. I: Sodium hydroxide and dil. sulphuric acid
6. B: Sodium carbonate and dil. sulphuric acid
7. D: Formic acid and conc. sulphuric acid
8. E: Sodium nitrate and conc. sulphuric acid
9. H: Sodium sulphide and dil. sulphuric acid
10. G: Ethyl alcohol and conc. sulphuric acid

Question 6.1

Give reasons for the following:

Sulphuric acid forms two types of salts with an alkali.

Answer

Sulphuric acid forms two types of salts i.e., acid salt and normal salt with an alkali since it's basicity is two.

NaOH [insufficient] + H₂SO₄ ⟶ H₂O + NaHSO₄ [acid salt]

2NaOH [excess] + H₂SO₄ ⟶ 2H₂O + Na₂SO₄ [normal salt]

Question 6.2

Give reasons for the following:

Conc. sulphuric acid is used as a laboratory reagent in the preparation of iodine from hydrogen iodide.

Answer

As conc. H₂SO₄ oxidises hydrogen iodide to iodine, hence, conc. sulphuric acid is used as a laboratory reagent in the preparation of iodine from hydrogen iodide.

Question 6.3

Give reasons for the following:

Barium chloride solution can be used to distinguish between dil. H₂SO₄ and dil HNO₃.

Answer

Barium chloride soln. reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO₄) but with dil. HNO₃ no white ppt. is produced since Ba(NO₃)₂ is soluble in dil. H₂SO₄.

H₂SO₄ (dil.) + BaCl₂ (aq.) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

HNO₃ (dil.) + BaCl₂ (aq.) ⟶ No ppt.

Question 6.4

Give reasons for the following:

The gaseous product obtained differs when zinc reacts with dilute and with conc. H₂SO₄ respectively.

Answer

Concentrated Sulphuric acid is a strong oxidising agent because on thermal decomposition it yields nascent oxygen which helps in oxidation. Hence, when zinc reacts with conc. H₂SO₄, Zn is oxidized to ZnSO₄ and sulphur dioxide gas is evolved along with hydrogen gas.

Zn + 2H₂SO₄ (conc.) ⟶ ZnSO₄ + 2H₂ + 2SO₂

Dilute Sulphuric acid on the other hand behaves as a typical acid. Hence, when zinc reacts with dil. H₂SO₄, displacement reaction takes place, ZnSO₄ is formed and hydrogen gas is evolved.

Zn + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂

Question 6.5

Give reasons for the following:

Ethanol can be converted to ethene using sulphuric acid.

Answer

As H₂SO₄ is a strong dehydrating agent, so it removes elements of water i.e. hydrogen and oxygen in the ratio 2:1 from carbohydrates, organic compounds etc. Hence, ethanol can be converted to ethene using sulphuric acid.

$\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{C}_{2}\text{H}_{4} \small{\space[ethene-ethylene]} + \text{H}_2\text{O}$