Chemical Reactions and Equations

Formation of a new substance is a key indicator of a chemical reaction. The equality of total mass of reactants and products in a closed container demonstrates the law of conservation of mass. Colour change alone can sometimes occur in a physical change, but the formation of a new substance here confirms a chemical change.

In a balanced equation, the number of atoms of every element is the same on both the reactant and product sides. During a chemical reaction, atoms are neither created nor destroyed; they are only rearranged. Hence, the total mass remains unchanged, demonstrating the law of conservation of mass. The law of constant proportions concerns the fixed mass ratio of elements in a compound, not the balancing of an equation.

Option A is correct. In balancing a chemical equation, coefficients change the number of molecules or formula units without changing the chemical formula or identity of a substance. For example, 2H₂O means two water molecules. In contrast, changing a subscript changes the substance itself; changing H₂O to H₂O₂ changes water into hydrogen peroxide.

In a balanced chemical equation, the number of atoms of each element is the same on the reactant and product sides, satisfying the law of conservation of mass. If the atoms of even one element are unequal, the equation is unbalanced. Reversibility and exothermic nature are different properties of a reaction; they do not describe whether atom counts are balanced.

A thin layer of magnesium oxide forms on the surface of a magnesium ribbon. This layer hinders direct contact between magnesium and oxygen in air. When the layer is removed, fresh magnesium is exposed and reacts readily with oxygen, burning with a bright flame. The amount of oxygen in the air does not increase; only the exposed magnesium surface becomes available for reaction.

Calcium oxide reacts with water to form calcium hydroxide, also called slaked lime: \(\mathrm{CaO + H_2O \rightarrow Ca(OH)_2}\). This is an exothermic reaction. Calcium carbonate is not formed by adding water; its formation requires a calcium compound and carbon dioxide.

The reaction is: \(\mathrm{CaO + H_2O \rightarrow Ca(OH)_2 + heat}\). Two reactants, calcium oxide and water, combine to form the single product calcium hydroxide; therefore, it is a combination reaction. Since heat is released, it is also exothermic. Option B describes a decomposition reaction, in which one compound breaks down.

Copper has oxidation number \(+2\) in \(\mathrm{CuO}\) and \(0\) in \(\mathrm{Cu}\), so copper(II) oxide is reduced. Hydrogen changes from \(0\) in \(\mathrm{H_2}\) to \(+1\) in \(\mathrm{H_2O}\); therefore, hydrogen is oxidised and acts as the reducing agent. Thus, option B is correct, not option A, which reverses the roles.

The thermal decomposition reaction is \(\mathrm{CaCO_3 \xrightarrow{\Delta} CaO + CO_2}\). The evolved \(\mathrm{CO_2}\) reacts with lime water, \(\mathrm{Ca(OH)_2}\), to form an insoluble white precipitate of \(\mathrm{CaCO_3}\), making the lime water milky. Oxygen supports combustion, but it does not turn lime water milky.

Lead nitrate undergoes thermal decomposition:
\[
2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}
\]
\(\mathrm{NO_2}\) is a brown gas, whereas \(\mathrm{O_2}\) is colourless. \(\mathrm{PbO}\) is a solid rather than a gas, so the correct pair is nitrogen dioxide and oxygen.

On heating, lead nitrate decomposes as \(2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). The \(\mathrm{PbO}\) formed is the yellow solid residue. Lead chloride, sulphate, or carbonate would require chloride, sulphate, or carbonate ions respectively, none of which are present in this reaction.

Thermal decomposition of lead nitrate produces nitrogen dioxide, lead oxide, and oxygen: \(2\mathrm{Pb(NO_3)_2} \rightarrow 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). \(\mathrm{NO_2}\) is a brown gas, so the brown fumes indicate its presence. Oxygen is also formed, but it is colourless.

Zinc is above copper in the reactivity series. Therefore, zinc displaces copper from copper sulphate solution: \(\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}\). The blue \(\mathrm{Cu^{2+}}\) ions are removed from the solution, producing colourless zinc sulphate, while copper is deposited on the zinc strip. Option B reverses the reactivity order.

Silver chloride undergoes photodecomposition in sunlight: \(2\mathrm{AgCl}\xrightarrow{\text{sunlight}}2\mathrm{Ag}+\mathrm{Cl}_2\). The silver metal formed is grey, so it causes the colour change. Chlorine gas is also a product, but it does not cause the grey appearance.

Some silver salts, such as silver chloride, are light-sensitive. On exposure to light, they may undergo photochemical decomposition and form metallic silver; therefore, they are stored in dark-coloured bottles to reduce exposure to light. Dark bottles do not increase their solubility in water.

The electrolysis of water is represented by \(2H_2O(l) \rightarrow 2H_2(g) + O_2(g)\). At the same temperature and pressure, gas volumes are proportional to their mole ratio, so \(H_2:O_2 = 2:1\). Therefore, for 12 mL of oxygen, the volume of hydrogen is \(2 \times 12 = 24\) mL. The 12 mL option is incorrect because hydrogen and oxygen are not produced in equal volumes.

During electrolysis of water, reduction occurs at the cathode, producing hydrogen gas, whereas oxidation at the anode produces oxygen gas. The overall reaction is \(2H_2O(l) \rightarrow 2H_2(g) + O_2(g)\). Therefore, oxygen is produced at the anode and hydrogen at the cathode. Option B reverses the gases at the two electrodes.

Iron is more reactive than copper, so it displaces copper from copper sulphate: \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). The ferrous sulphate formed makes the solution light green, while displaced copper deposits as a reddish-brown coating on the nail. A temperature rise alone, as in option D, is not sufficient evidence of a displacement reaction.

Pure water contains very few ions, so it is a poor conductor of electricity. Dilute sulfuric acid supplies ions, increasing the electrical conductivity of the solution and allowing current to pass more readily. The acid does not replace the external electrical source. During electrolysis, hydrogen is formed at the cathode and oxygen at the anode, with a volume ratio of 2:1, not 1:1.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution: \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). The iron(II) sulphate formed gives the solution a pale green colour. Copper metal is also produced, but it deposits as a reddish-brown coating on the nail; it does not make the solution green.

Iron is more reactive than copper in the reactivity series. Therefore, iron displaces copper from copper sulphate solution. The liberated copper forms a reddish-brown deposit on the nail surface. The reaction is \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). Option B is incorrect because the less reactive metal, copper, cannot displace iron from its compound.

The blue colour of copper sulphate solution is due to \(\mathrm{Cu^{2+}}\) ions. Since iron is more reactive than copper, it displaces copper in the reaction \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\), depositing copper metal. Thus, the concentration of \(\mathrm{Cu^{2+}}\) ions decreases and the blue colour fades. In contrast, \(\mathrm{Fe^{2+}}\) ions are produced rather than decreased.

In a displacement reaction, a more reactive metal displaces a less reactive metal from its salt solution. Since metal A displaces metal B and B cannot displace A, A is more reactive than B. If option B were correct, B would be able to displace A from its salt solution.

This is a double-displacement precipitation reaction: \(\mathrm{BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaCl(aq)}\). The \(\mathrm{Ba^{2+}}\) and \(\mathrm{SO_4^{2-}}\) ions form insoluble white barium sulphate, while sodium chloride remains in solution. Option B is wrong because it incorrectly replaces the sulphate ion with a sulphite ion. Exam tip: After exchanging ions in a double-displacement reaction, check the solubility of the products.

The oxidation number of \(Al\) increases from 0 to \(+3\), so aluminium is oxidised. At the same time, it reduces \(Fe^{3+}\) in \(Fe_2O_3\) to \(Fe\). A substance that causes reduction of another substance while itself being oxidised is the reducing agent; therefore, \(Al\) is the reducing agent. In contrast, \(Fe_2O_3\) is the oxidising agent because its \(Fe^{3+}\) ions are reduced.