Chemical Reactions and Equations

On heating, lead nitrate undergoes thermal decomposition: \(2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). Therefore, lead oxide, \(\mathrm{PbO}\), is the solid product, whereas nitrogen dioxide and oxygen are gases. Lead chloride cannot form because no chloride source is present in the reaction.

In sunlight, silver chloride decomposes: \(2\mathrm{AgCl}(s)\xrightarrow{\text{sunlight}}2\mathrm{Ag}(s)+\mathrm{Cl}_2(g)\). The metallic silver formed is grey, so the white silver chloride gradually appears grey. Chlorine gas is also produced, but metallic silver is the main cause of the colour change.

Silver chloride and silver bromide are photosensitive salts. On exposure to light, they decompose to form metallic silver, which appears dark grey or black and helps form a photographic image. For example: \(2\mathrm{AgBr}\xrightarrow{\text{light}}2\mathrm{Ag}+\mathrm{Br}_2\). Option C is incorrect because these salts are not chemically stable in light; they undergo a chemical change when exposed to it.

The governing concept is electrolysis and electrode reactions. In acidified-water electrolysis, oxidation occurs at the anode, producing oxygen; reduction at the cathode produces hydrogen. The overall equation is 2H₂O → 2H₂ + O₂, so option A is correct. Hydrogen is the cathode gas, while nitrogen and chlorine are not produced under these conditions.

The balanced equation for electrolysis of water is \(2\mathrm{H_2O} \rightarrow 2\mathrm{H_2} + \mathrm{O_2}\). At the same temperature and pressure, gas volumes are proportional to their number of moles. Therefore, the volume ratio of hydrogen to oxygen is \(2:1\). The ratio \(1:2\) is the reverse ratio, so it is not correct.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution: \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). Therefore, iron(II) sulphate and copper are the products. Iron oxide and hydrogen are not formed because oxygen and an acid are not involved in this reaction.

Iron is more reactive than copper. Therefore, iron displaces copper from copper sulphate solution, and copper gets deposited on the surface of the nail. The reaction is \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). Iron is not displaced; it takes the place of copper in copper sulphate.

In a displacement reaction, a more reactive metal displaces a less reactive metal from its salt solution. Since metal A cannot displace metal B, A is less reactive than B. Option B is the opposite conclusion: if A were more reactive, it should displace B.

The reaction is \(\mathrm{BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)}\). Barium sulphate, \(\mathrm{BaSO_4}\), is insoluble in water, so it appears as a white solid precipitate. Evolution of a brown gas is not a feature of this double-displacement precipitation reaction.

In a double displacement reaction, the ions of two ionic compounds interchange. In general, \(AB + CD \rightarrow AD + CB\). Therefore, the products are identified by forming the new cation–anion pairs. Option D describes a single displacement reaction, not a double displacement reaction.

In a chemical equation, a downward arrow indicates a precipitate formed in a solution. A precipitate is an insoluble solid that separates from the liquid, so option A is correct. Gas evolution is commonly shown by an upward arrow, whereas heat and light are energy changes and are not represented by the downward precipitate symbol.

In the reaction between copper oxide \(\text{CuO}\) and hydrogen \(\text{H}_2\), the products are copper and water. Therefore, the substance formed as water is \(\text{H}_2\text{O}\, .\)

Copper oxide contains copper combined with oxygen. When it changes into copper, oxygen is removed, as in CuO + H₂ → Cu + H₂O. Removal of oxygen is the school-level definition of reduction, so option A is correct. Oxidation involves addition of oxygen; neutralisation is acid-base reaction, and evaporation is a physical change.

Thermal decomposition of lead nitrate produces lead oxide \(PbO\), brown nitrogen dioxide \(NO_2\) gas, and oxygen \(O_2\). In option A, both sides contain \(2\) Pb, \(4\) N, and \(12\) O atoms, so the equation is balanced. In option B, oxygen atoms are not balanced: there are \(6\) on the left but \(7\) on the right.

In a combination reaction, two or more reactants combine to form one product. Here, \(\mathrm{CaO}\) and \(\mathrm{H_2O}\) combine to form only \(\mathrm{Ca(OH)_2}\), so option A is correct. In contrast, in option B one compound breaks down into \(\mathrm{KCl}\) and \(\mathrm{O_2}\), so it is a decomposition reaction.

A redox reaction involves transfer of electrons. One substance is oxidised by losing electrons, while another substance is reduced by gaining those same electrons. Therefore, oxidation and reduction occur simultaneously; a reaction involving only oxidation or only reduction is not a redox reaction.

Rusting occurs when iron reacts slowly with oxygen in the presence of moisture, producing hydrated iron(III) oxide. Oxygen is added to iron, so the process is oxidation; because it develops gradually rather than suddenly, it is specifically slow oxidation and option A is correct. The other choices describe physical changes, not rust formation.

Rusting requires iron to be in contact with oxygen and water/moisture. A layer of oil forms a protective barrier on the iron surface and prevents air and moisture from reaching it, so rusting is reduced. In contrast, option B suggests increasing reaction with water, which would not reduce rusting.

Galvanisation protects iron by covering it with a layer of zinc. This barrier prevents oxygen and moisture from reaching the iron surface, thereby stopping or greatly slowing the electrochemical reactions responsible for rusting. Therefore option A is correct. Light and sound do not cause rust, and soil alone is not the relevant chemical factor; moisture and oxygen are essential.

Oils and fats react with oxygen in air and undergo oxidation. This produces substances with an unpleasant smell and taste, a process called rancidity. In reduction, a substance loses oxygen, so it does not cause this type of spoilage of oils and fats.

Rancidity is caused by the oxidation of fats and oils, producing an unpleasant smell and taste. An airtight container limits the food's contact with oxygen, so the rate of oxidation decreases and rancidity is reduced. Increased moisture does not prevent rancidity; it can instead spoil food in other ways.

The governing concept is oxidation and rancidity. Nitrogen is comparatively unreactive, so it replaces much of the oxygen inside a sealed food packet. With less oxygen available, oils and fats oxidise more slowly, delaying rancidity and helping preserve flavour. Therefore, option A is correct. Nitrogen does not add moisture, make food acidic, or form a precipitate, so options B, C, and D are unsuitable.

During curd formation, bacteria convert lactose in milk into lactic acid. The lactic acid coagulates casein protein, forming curd with properties different from those of milk. Because new substances are formed, this is a chemical change. In contrast, separation of fat is only a physical separation of components of a mixture and does not form a new substance.

The governing concept is an endothermic energy change. If the container becomes cooler during a reaction, the reacting system may have absorbed heat from the surroundings, lowering the surroundings’ temperature. Therefore option A is correct. A reaction that releases heat would generally warm its surroundings, while burning and metal deposition are specific processes that cannot be concluded merely from cooling. The word “can” appropriately indicates an inference, not proof of one particular reaction.

In school chemical equations, an upward arrow is used to indicate that a gaseous product has been evolved and escapes from the reaction mixture. Thus option A correctly explains the symbol’s purpose. The arrow is not a general rule that every solid or liquid moves upward, and a precipitate is normally represented as an insoluble solid, often by a downward arrow when appropriate. Therefore options B, C, and D are incorrect.