Frank Brothers Solutions for ICSE Class 10 Chemistry Chapter 10 Study Of Sulphur Compound Sulphuric Acid

Study Of Sulphur Compound - Sulphuric Acid

Solution 1: Sulphuric acid is found in some of the hot springs.

📝 Teacher's Note: Use this natural occurrence to connect chemistry with geography. Students often remember better when they know real-world examples like Yellowstone's hot springs.

🎯 Exam Tip: Always mention "hot springs" as the natural source of sulphuric acid - it's a common one-mark question.

Solution 2: The theory of manufacture of sulphuric acid by contact process involves following steps:
i. Production of \( SO_2 \): \( SO_2 \) is produced by burning sulphur or roasting iron pyrites.
\( S + O_2 \rightarrow SO_2 \)
or
\( 4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2 \)
ii. Catalytic oxidation of \( SO_2 \) by air to give sulphur trioxide:
\( 2SO_2 + O_2 \rightleftharpoons 2SO_3 + Heat \)
iii. Absorption of sulphur trioxide in 98% sulphuric acid to form oleum:
\( SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)
iv. Dilution of oleum to get sulphuric acid of desired concentration:
\( H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4 \)

📝 Teacher's Note: Emphasize why SO₃ is not directly dissolved in water - it causes violent reaction. The 98% acid step prevents this dangerous situation.

🎯 Exam Tip: Remember the catalyst used is Vanadium pentoxide (V₂O₅) at 450°C - often asked in reactions questions.

Solution 3:
(i) Constant boiling mixture: Mixture which boils without any change in composition is known as constant boiling mixture.
(ii) Hygroscopic substance: The compound which absorbs water vapour from the atmosphere.
(iii) Oleum: Oleum is called pyrosulphuric acid or orthosulphuric acid. Formula is \( H_2S_2O_7 \).
(iv) Dehydrating agent: The compound which has more affinity for water. It removes atoms of hydrogen and oxygen in the form of water from the composition of a substance.

📝 Teacher's Note: Use everyday examples - silica gel packets (hygroscopic) and concentrated H₂SO₄ drying wet gases (dehydrating agent) to make concepts relatable.

🎯 Exam Tip: For definitions, include the formula where applicable (like H₂S₂O₇ for oleum) - examiners appreciate complete answers.

Solution 4:
(a) Products obtained by dissolving sulphur dioxide and chlorine in water are sulphuric acid and hydrochloric acid.
\( SO_2 + 2H_2O + Cl_2 \rightarrow H_2SO_4 + 2HCl \)
(b) Vanadium pentoxide is used as catalyst in contact process.
(c) When \( SO_3 \) is dissolved in 98% sulphuric acid it forms oleum.
\( SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)
(d) Zinc is treated with Sulphuric acid it forms Zinc sulphate and hydrogen.
\( Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2 \)
(e) When ferrous sulphide is treated with sulphuric acid it forms ferrous sulphate and hydrogen sulphide.
\( FeS + H_2SO_4 \rightarrow FeSO_4 + H_2S \)
(f) The precipitate of \( PbSO_4 \) is formed when lead nitrate is treated with dilute sulphuric acid.
\( Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4 \downarrow + 2HNO_3 \)
(g) When \( BaCl_2 \) is treated with sulphuric acid, precipitate of \( BaSO_4 \) is formed.
\( BaCl_2 + H_2SO_4 \rightarrow BaSO_4 \downarrow + 2HCl \)
(h) When carbon is treated with hot conc. Sulphuric acid then Water, sulphur dioxide and carbon dioxide are formed.
\( C + 2H_2SO_4 \rightarrow 2H_2O + 2SO_2 + CO_2 \)
(i) The property used to prepare HCl and \( HNO_3 \) from \( H_2SO_4 \) is that Suphuric acid is non-volatile acid. So when treated with salts of more volatile acids and heated, concentrated sulphuric acid displaces the more volatile acids.
\( KCl + H_2SO_4 \xrightarrow{conc} KHSO_4 + HCl \)
\( KNO_3 + H_2SO_4 \xrightarrow{conc} KHSO_4 + HNO_3 \)

📝 Teacher's Note: Group these reactions by type - displacement, precipitation, oxidation - to help students recognize patterns in sulphuric acid behavior.

🎯 Exam Tip: Always write balanced equations with correct states (↓ for precipitates) and mention conditions like "conc" or "hot" where applicable.

Solution 5:
(i) When water is dissolved in large scale then sulphuric acid is formed.
\( SO_3 + H_2O \rightarrow H_2SO_4 \)
(ii) When concentrated sulphuric acid is added to equal volume of cold water it limit down the heat which is released.
(iii) When 100 ml of 98% sulphuric acid is kept in open it absorbs water vapours from atmosphere and its level goes up.
(iv) When hot concentrated sulphuric acid is added to sodium chloride crystals then white dense fumes are seen if a rod dipped in ammonia solution is brought near it.
(v) No visible change is observed.

📝 Teacher's Note: Demonstrate the hygroscopic nature using a small amount of concentrated acid in open air - students can see the level rising over time.

🎯 Exam Tip: For part (iv), remember the white fumes are HCl gas reacting with NH₃ to form NH₄Cl - this identifies the presence of chloride ions.

Solution 6:
(i) In contact process sulphur trioxide formed is dissolved in sulphuric acid and not water because sulphur trioxide when directly dissolved in water gives highly exothermic reaction.
(ii) Sulphur dioxide gets oxidized when exposed to air to form sulphur trioxide which reacts with water vapours to form sulphuric acid.
\( SO_2 + H_2O + O_2 \rightarrow 2H_2SO_4 \)
(iii) When water is added to concentrated sulphuric acid the heat evolved may be sufficient to raise the temperature of water to its boiling point. This may throw acid violently out of the container. Thus concentrated sulphuric acid is diluted by adding a small amount of sulphuric acid to large amount of water.
(iv) When concentrated sulphuric acid is exposed to air, it absorbs water vapours from the atmosphere thus increasing its volume and becoming dilute.
(v) Sulphuric acid when reacts with sodium chloride it forms sodium bisulphate and hydrochloric acid because sulphuric acid when treated with salts of more volatile acids displaces the more volatile acid.
\( NaCl + H_2SO_4 \xrightarrow{heat} NaHSO_4 + HCl \)
(vi) When barium chloride is added to dilute sulphuric acid a white precipitate of barium sulphate is
\( BaCl_2 + H_2SO_4 \rightarrow BaSO_4 \downarrow + 2HCl \)
(vii) Hot concentrated sulphuric acid act as a powerful oxidizing agent. Due to heat it decomposes to form nascent oxygen which helps in oxidation of carbon to carbon dioxide.
\( H_2SO_4 \rightarrow H_2O + SO_2 + [O] \)
\( C + 2[O] \rightarrow CO_2 \)
(viii) Ammonia gas being basic in nature cannot be dried by using concentrated sulphuric acid.

📝 Teacher's Note: For dilution safety, use the mnemonic "Add acid to water, never water to acid" - teach this rule first before any practical work.

🎯 Exam Tip: The white precipitate test with BaCl₂ is the standard test for sulphate ions - always mention that BaSO₄ is insoluble in water.

Solution 6 (continued):
(viii) Ammonia gas being basic in nature cannot be dried by using concentrated sulphuric acid.
(ix) The carbohydrates contain carbon, hydrogen and oxygen. The hydrogen and oxygen are always in ratio of 2:1 which is absorbed by acid in the form of water, thus leaving carbon behind. This is also called as charring.
(x) Concentrated sulphuric acid should not be added to oxalic acid or formic acid as it removes water and forms Carbon monoxide. Carbon monoxide is harmful for health so this addition must be done in an open laboratory.
(xi) When concentrated sulphuric acid is added to blue crystalline copper sulphate, it removes water from salt and turns it into powdery white.
(xii) Concentrated sulphuric acid must be stored in air tight bottles as it gains water from air and gets slightly diluted thus resulting in increase in volume.
(xiii) Cotton contains cellulose. When cotton is treated with concentrated sulphuric acid it removes water from cotton and carbon is left behind. This way cotton clothes get burnt.
\( (C_6H_{10}O_5)_n \xrightarrow{conc.H_2SO_4} 5(C)_n + 6(H_2O)_n \)
(xiv) As sulphuric acid is non volatile, it is used to prepare volatile acids like HCl and \( HNO_3 \). Thus these acids cannot be used to prepare sulphuric acid.

📝 Teacher's Note: Demonstrate the charring effect using a small piece of paper or sugar cube - it's a dramatic way to show the dehydrating power.

🎯 Exam Tip: Remember that H₂SO₄ acts as dehydrating agent with organic compounds and as oxidizing agent with metals - know when to apply which property.

Solution 7:
(i) Sulphur trioxide to sulphur acid: There is absorption of sulphur trioxide in 98% sulphuric acid to form oleum.
\( SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)
Dilution of oleum to get sulphuric acid of desired concentration.
\( H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4 \)
(ii) Sulphur trioxide to oleum:
Absorption of sulphur trioxide in 98% sulphuric acid to form oleum.
\( SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)
(iii) Dilute sulphuric acid to hydrogen: Dilute sulphuric acid react with metals above hydrogen in the activity series to form sulphate salts and Hydrogen.
\( Mg + H_2SO_4 \rightarrow MgSO_4 + H_2 \)
(iv) Aqueous barium chloride to barium sulphate: Dilute sulphuric acid when added to the aqueous solution of barium form their insoluble sulphates.
\( BaCl_2 + H_2SO_4 \rightarrow BaSO_4 \downarrow + 2HCl \)
(v) Aqueous lead nitrate to lead sulphate: Dilute sulphuric acid when added to the aqueous solution of lead form their insoluble sulphates.
\( Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4 \downarrow + 2HNO_3 \)

📝 Teacher's Note: For conversions, always teach students to identify what property of H₂SO₄ is being used - whether it's acting as an acid, dehydrating agent, or oxidizing agent.

🎯 Exam Tip: In conversion questions, write all intermediate steps clearly and balance all equations - partial credit is often given for correct intermediates even if the final step is wrong.

Solution 8:

Answer:
(vi) Sodium chloride to hydrogen chloride: Sulphuric acid when reacts with sodium chloride it forms sodium bisulphate and hydrochloric acid because sulphuric acid when treated with salts of more volatile acids displaces more volatile acids.

\( \text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{200°C} \text{NaHSO}_4 + \text{HCl} \)

(vii) Sucrose to sugar charcoal: Carbohydrates contain carbon, hydrogen and oxygen. Thus sulphuric acid removes hydrogen and oxygen in form of water leaving behind carbon.

\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \xrightarrow{\text{Conc.}\text{H}_2\text{SO}_4} 12\text{C} + 11\text{H}_2\text{O} \)

(viii) Oxalic acid to carbon monoxide: Sulphuric acid is used to remove moisture from oxalic acid.

\( 2\text{COOH} + \text{H}_2\text{SO}_4 \rightarrow 2\text{CO} + \text{H}_2\text{SO}_4\text{H}_2\text{O} \)

In simple words: Concentrated sulphuric acid acts as a dehydrating agent, removing water from compounds and displacing weaker acids from their salts.

📝 Teacher's Note: Use the word equation method first, then show the balanced chemical equations. Students should understand that H₂SO₄ acts differently in each case - as a dehydrating agent, displacing agent, or drying agent.

🎯 Exam Tip: Always mention the specific property of sulphuric acid being used in each reaction - displacement, dehydration, or oxidation. This earns extra marks.

Solution 8:

Answer:
(i) Concentrated sulphuric acid is non-volatile hence it displaces more volatile acids.

\( \text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{200°C} \text{NaHSO}_4 + \text{HCl} \)

(ii) Concentrated sulphuric acid act as dehydrating agent as it removes water of crystallization of hydrated salts and renders them anhydrous.

\( \text{CuSO}_4\text{5H}_2\text{O} \xrightarrow{\text{H}_2\text{SO}_4} \text{CuSO}_4 + 5\text{H}_2\text{O} \)

(iii) Concentrated sulphuric acid is an oxidizing agent as sulphuric acid decomposes to form nascent oxygen which helps in the oxidation of sulphur and carbon.

\( \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{O} + \text{SO}_2 + [\text{O}] \)
\( \text{C} + 2[\text{O}] \rightarrow \text{CO}_2 \)

(iv) Concentrated sulphuric acid ionize in two steps. Hence it is dibasic.

\( \text{H}_2\text{SO}_4 + \text{H}_2\text{O} \rightleftharpoons \text{HSO}_3^- + \text{H}_3\text{O}^+ \)
\( \text{H}_2\text{SO}_4 + \text{H}_2\text{O} \rightleftharpoons \text{SO}_4^{2-} + \text{H}_3\text{O}^+ \)

In simple words: Concentrated sulphuric acid has four main properties - it pushes out weaker acids, removes water from substances, adds oxygen to other compounds, and can donate two hydrogen ions.

📝 Teacher's Note: Demonstrate the dehydrating property using copper sulphate crystals - students can see the blue crystals turn white. This visual demonstration helps reinforce the concept.

🎯 Exam Tip: For each property, write one balanced equation as evidence. The examiner looks for specific chemical evidence to support each property mentioned.

Solution 9:

Answer: Sulphuric acid is known as king of chemicals because in almost all industries it is used directly or indirectly. In 8th century it was obtained by distillation of green vitriol (\( \text{FeSO}_4\text{.7H}_2\text{O} \)). It is called oil of green vitriol because of its oily appearance and because of the fact that it was present in vitreous or glassy substances like ferrous sulphate etc.

In simple words: Sulphuric acid is called the "king of chemicals" because it's used in almost every industry, and it got its old name "oil of green vitriol" because it looked oily and came from green-colored iron compounds.

📝 Teacher's Note: Ask students to research and list 5-6 industries where sulphuric acid is used. This helps them understand why it's called the "king of chemicals."

🎯 Exam Tip: Always mention both the historical name "oil of green vitriol" and the modern importance in industries to get full marks for this question.

Solution 10:

Answer:
(i) In manufacturing of fertilizers sulphuric acid act as an electrolyte.
(ii) In chemical industry for the manufacturing of hydrochloric acid & nitric acid it acts as a non-volatile acid.
(iii) In petroleum industry sulphuric acid act as oxidizing agent.

In simple words: Sulphuric acid is useful in many industries - it helps make fertilizers, produces other acids like HCl and HNO₃, and helps clean petroleum products.

📝 Teacher's Note: Connect this to everyday products - fertilizers for farming, car batteries (electrolyte), and gasoline refining. This makes the industrial uses more relatable.

🎯 Exam Tip: Mention the specific role of H₂SO₄ in each industry (electrolyte, non-volatile acid, oxidizing agent) rather than just listing the industries.

Solution 11:

Answer:
(i) Difference between dilute \( \text{H}_2\text{SO}_4 \) and Conc. \( \text{H}_2\text{SO}_4 \):

In simple words: Dilute sulphuric acid is tested by adding barium or lead salts which form white precipitates, while concentrated sulphuric acid is tested by its reactions with copper (gives SO₂ gas) or salt (gives HCl gas).

📝 Teacher's Note: Perform these tests as demonstrations. The white precipitates and gas evolution are excellent visual confirmations that students remember easily.

🎯 Exam Tip: Always write the balanced chemical equations for each test. Mention the color and solubility of precipitates and the smell/color of gases evolved.

Solution 1993-1:

Answer: Concentrated sulphuric acid absorbs water vapours from the atmosphere. Hence it should be kept in air tight bottles.

In simple words: Concentrated sulphuric acid pulls water from the air around it, so we must store it in tightly sealed bottles to prevent it from getting diluted.

📝 Teacher's Note: Show students how concentrated H₂SO₄ becomes warm when exposed to air moisture. This hygroscopic property is why laboratory bottles are always kept tightly closed.

🎯 Exam Tip: Use the term "hygroscopic" - it shows you understand the scientific property behind the storage requirement.

Solution 1994-1:

Answer: Sulphuric acid when treated with sulphites gives sulphate salts, water and sulphur dioxide gas. Sulphur dioxide turns potassium dichromate solution green. Thus the negative ion is Sulphite ion.

In simple words: When sulphuric acid reacts with sulphites, it produces SO₂ gas which turns orange dichromate solution green - this is how we identify sulphite ions.

📝 Teacher's Note: Emphasize the color change test with K₂Cr₂O₇ paper. This is a standard qualitative test that students should remember for identifying sulphites.

🎯 Exam Tip: Mention both the gas evolved (SO₂) and the confirmatory test (dichromate paper turning green) to get full marks.

Solution 1994-2:

Answer: Oxalic acid reacts with sulphuric acid to produce carbon monoxide

In simple words: When oxalic acid meets sulphuric acid, it breaks down and produces carbon monoxide gas.

📝 Teacher's Note: This reaction shows sulphuric acid acting as a dehydrating agent, removing water and breaking down the oxalic acid molecule.

🎯 Exam Tip: Write the balanced equation: (COOH)₂ + H₂SO₄ → 2CO + H₂SO₄.H₂O to support your answer.

Solution 1994-3:

Answer:
(i) The purpose of contact process is to manufacture sulphuric acid.
(ii) Two gases that are combined during contact process are sulphur dioxide and sulphur trioxide.
(iii) Vanadium pentoxide is the catalyst used in the process.
(iv) \( \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \)

In simple words: The contact process is like a factory method to make sulphuric acid by combining sulphur gases with the help of a vanadium catalyst.

📝 Teacher's Note: Draw a simple flow chart of the contact process. Visual representation helps students understand the sequence of reactions better.

🎯 Exam Tip: Remember the three main steps: burning sulphur, catalytic oxidation with V₂O₅, and absorption in water. Mention optimal temperature (450°C) if asked for details.

Solution 1995-4:

Answer: When crystals of \( \text{CuSO}_4\text{.5H}_2\text{O} \) is placed in concentrated sulphuric acid, it removes the water of crystallization of hydrated salt and renders them anhydrous. Its colour change to white.

In simple words: Concentrated sulphuric acid sucks out all the water from blue copper sulphate crystals, turning them white and completely dry.

📝 Teacher's Note: This is an excellent demonstration of the dehydrating property. Students can visually see the blue crystals turn white as water is removed.

🎯 Exam Tip: Mention both the physical change (blue to white) and the chemical reason (removal of water of crystallization) for complete marks.

Solution 1995-5:

Answer: Balanced equations for the chemical reactions that take place during the conversion of sulphur dioxide to sulphuric acid are:

(i) \( 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 + \text{Heat} \)
(ii) \( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \)
(iii) \( \text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4 \)

Vanadium pentoxide is used as a catalyst.

In simple words: Making sulphuric acid from SO₂ involves three steps - first adding oxygen with a catalyst, then mixing with existing acid, and finally adding water to get the final product.

📝 Teacher's Note: Emphasize that step (ii) uses oleum formation to prevent the violent reaction that would occur if SO₃ directly contacted water.

🎯 Exam Tip: Always mention the catalyst (V₂O₅) and write all three equations in the correct sequence to show the complete process.

Solution 1995-1:

Answer: Two other acids other than sulphuric acid which can be prepared by using sulphuric acid are hydrochloric acid and nitric acid.

In simple words: Sulphuric acid can be used to make two other important acids - hydrochloric acid (from salt) and nitric acid (from nitrates).

📝 Teacher's Note: Show the displacement reactions: H₂SO₄ + NaCl → HCl and H₂SO₄ + NaNO₃ → HNO₃. This demonstrates the non-volatile nature of sulphuric acid.

🎯 Exam Tip: Write the reactions showing how H₂SO₄ displaces these acids from their salts due to its non-volatile nature.

Solution 1995-2:

Answer: Sulphuric acid is non-volatile. So when it is treated with salts of more volatile acids, and heated, concentrated sulphuric acid displaces the more volatile acids.

In simple words: Since sulphuric acid doesn't evaporate easily, it can push out other acids that do evaporate easily when heated with their salts.

📝 Teacher's Note: Compare boiling points: H₂SO₄ (338°C) vs HCl (-85°C) vs HNO₃ (83°C). This physical property difference explains the displacement reactions.

🎯 Exam Tip: Use the term "displacement reaction" and mention that the property responsible is the "non-volatile nature" of sulphuric acid.

Solution 1995-3:

Answer:
(i) Hydrogen:

\( \text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2 \)

(ii) Carbon dioxide

\( \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \)

(iii) Sulphur dioxide

\( \text{Na}_2\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{SO}_2 \)

In simple words: Sulphuric acid can produce three different gases - hydrogen from metals, carbon dioxide from carbonates, and sulphur dioxide from sulphites.

📝 Teacher's Note: These are standard acid reactions. Emphasize the patterns - metals give H₂, carbonates give CO₂, and sulphites give SO₂.

🎯 Exam Tip: Write balanced equations and identify the gas evolved in each case. This shows understanding of different types of acid reactions.

Solution 1996-1:

Answer: When barium chloride solution is added to dilute sulphuric acid a white precipitate of barium sulphate is formed. The precipitate is insoluble in dil. Hydrochloric acid.

\( \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2\text{HCl} \)

In simple words: When you mix barium chloride with dilute sulphuric acid, you get a white powder (barium sulphate) that doesn't dissolve even in other acids.

📝 Teacher's Note: This is the standard test for sulphate ions. The insolubility of BaSO₄ in acids confirms the presence of sulphate ions specifically.

🎯 Exam Tip: Always mention that the white precipitate is insoluble in dilute HCl - this distinguishes sulphate test from other white precipitates.

Solution 1998-1:

Answer: Balanced equation for the reaction between iron and dilute sulphuric acid is:

\( \text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2 \)

In simple words: When iron meets dilute sulphuric acid, it forms iron sulphate salt and hydrogen gas bubbles.

📝 Teacher's Note: This is a typical metal-acid reaction. Students should recognize the pattern: Metal + Acid → Salt + Hydrogen gas.

🎯 Exam Tip: Ensure the equation is balanced and identify the products as salt (FeSO₄) and gas (H₂).

Solution 1998-2:

Answer: Oxide of sulphur which reacts with water to give sulphuric acid is sulphur dioxide.

\( \text{SO}_2 + 2\text{H}_2\text{O} + \text{O}_2 \rightarrow 2\text{H}_2\text{SO}_4 \)

In the contact process Oxide of sulphur react with sulphuric acid to form Oleum.

\( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \)

In simple words: Sulphur dioxide can form sulphuric acid with water and oxygen, while in industry, sulphur trioxide is mixed with existing sulphuric acid to make oleum (extra strong acid).

📝 Teacher's Note: Clarify the difference between SO₂ (needs oxidation) and SO₃ (directly forms acid). The contact process uses SO₃ for efficiency.

🎯 Exam Tip: Write both reactions to show complete understanding - the natural process (SO₂) and the industrial process (SO₃).

Solution 1998-3:

Answer: Balanced equations are:

(i) Copper carbonate:

\( \text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2 \)

(ii) Lead nitrate solution:

\( \text{Pb(NO}_3\text{)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{PbSO}_4 \downarrow + 2\text{HNO}_3 \)

(iii) Zinc hydroxide:

\( \text{Zn(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + 2\text{H}_2\text{O} \)

In simple words: Sulphuric acid reacts with carbonates to give CO₂ gas, with nitrates to give white precipitates, and with hydroxides to give water.

📝 Teacher's Note: These represent three different types of reactions - acid-carbonate, acid-salt (precipitation), and acid-base neutralization.

🎯 Exam Tip: Identify the type of reaction and the products formed in each case to show complete understanding.

Solution 1999-1:

Answer: Sulphuric acid removes water of crystallization from Hydrated copper sulphate.

In simple words: Sulphuric acid acts like a water absorber, sucking out water molecules that are stuck to copper sulphate crystals.

📝 Teacher's Note: This demonstrates the dehydrating property of concentrated sulphuric acid. The blue CuSO₄·5H₂O crystals become white anhydrous CuSO₄.

🎯 Exam Tip: Mention the color change from blue to white as evidence of water removal from the hydrated salt.

Solution 1999-2:

Answer: Balanced equation for the reaction between Zinc and dilute sulphuric acid is:

\( \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \)

(i) The purpose of contact process is to manufacture sulphuric acid.
(ii) Vanadium pentoxide is the catalyst used in the contact process.
(iii) Balanced equation for the reaction in the contact process which takes place in the presence of catalyst is:

\( 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 + \text{Heat} \)

In simple words: Zinc reacts with sulphuric acid to give zinc sulphate and hydrogen gas, while the contact process uses a vanadium catalyst to combine sulphur dioxide with oxygen to make sulphur trioxide for acid production.

📝 Teacher's Note: Connect the two parts - showing both a simple metal-acid reaction and the industrial process for making the acid itself.

🎯 Exam Tip: Write the balanced equations and mention the catalyst (V₂O₅) and the reversible nature of the contact process reaction.

Solution 2000-1:

Answer: When concentrated sulphuric acid is added to copper sulphate -5water, its colour change to white.

In simple words: Concentrated sulphuric acid removes water from blue copper sulphate crystals, making them turn white.

📝 Teacher's Note: This is the same dehydration reaction as shown earlier. The consistent color change from blue to white demonstrates the dehydrating property clearly.

🎯 Exam Tip: Always mention the specific color change (blue to white) when discussing dehydration of copper sulphate crystals.

Solution 2002-1

Equations are:

(i) \( 2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \)

(ii) \( Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4 \downarrow + 2HNO_3 \)

Solution 2003-1:

Hot concentrated nitric acid oxidizes sulphur directly into sulphuric acid.

\( S + 6HNO_3 \rightarrow H_2SO_4 + 6NO_2 + 2H_2O \)

The name of the process by which sulphuric acid is manufactured is contact process.

Vanadium pentoxide is the catalyst used during the process.

Solution 2003-2:

Less volatile.

Solution 2003-3:

Equations:

(i) \( Fe + H_2SO_4 \rightarrow FeSO_4 + H_2 \)

(ii) \( Cu + H_2SO_4 \rightarrow CuSO_4 + H_2 \)

(iii) \( Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4 \downarrow + 2HNO_3 \)

(iv) \( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)

Solution 2004-1:

(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) \( H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4 \)

(iii) In stepE, 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
\( 2 NaOH + SO_2 \rightarrow Na_2SO_3 + H_2O \)

Solution 2005-3:

Balanced equation:

\( KHCO_3 + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O + 2CO_2 \)

Solution 2005-2:

(i) B

(ii) C

(iii) A

Solution 2006-1:

(a) Contact process is used for the large scale manufacture of sulphuric acid.

(b) Sulphuric acid has great affinity for water hence it is used as dehydrating agent.

(c) As an oxidizing agent:
\( H_2SO_4 \rightarrow H_2O + SO_3 + [O] \)
\( C + 2[O] \rightarrow CO_2 \)

As a non-volatile acid:
\( NaCl + H_2SO_4 \xrightarrow{200°C} NaHSO_4 + HCl \)

Solution 2007-1:

\( Pb(NO_3)_2 + H_2SO_4 \rightarrow PbSO_4 \downarrow + 2HNO_3 \quad Cu + 2H_2SO_4 \rightarrow CuSO_4 + 2H_2O + SO_2 \quad 2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4 \)

Solution 2007-2:

(a) B

(b) D

(c) C

(d) A

(e) A

Solution 2007-3:

(a) HCl has higher boiling point where as sulphuric acid has lower boiling point.

(b) When barium chloride is added to dilute sulphuric acid, white precipitate of barium sulphate is formed but with dilute hydrochloric acid no change is observed.

Solution 2008-1:

Lead nitrate

Solution 2008-2:

(i) Zinc, Dilute sulphuric
\( Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2 \)

(ii) Sodium sulphite, Dilute sulphuric acid
\( NaSO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + SO_2 \)

(iii) Sodium carbonate, Dilute sulphuric acid
\( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)

(iv) Zinc, calcium carbonate, dilute sulphuric acid
\( Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2 \)
\( ZnSO_4 + CuCO_3 \rightarrow ZnCO_3 + CuSO_4 \)

Solution 2009-1:

Hydrogen chloride is a colourless pungent acidic gas produced by the action of concentrated sulphuric acid on sodium chloride.