Chemistry 2013 Jamb Examination

During electrolysis of acidified water, oxygen is liberated at the anode according to the equation:\[ 2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- \] Four moles of electrons produce one mole of O₂.

Calculate total charge passed:\[ Q = I \times t = 5 \times 193 = 965 \; \text{C} \]Moles of electrons:\[ n(e^-) = \dfrac{Q}{F} \]\[= \dfrac{965}{96500} = 0.01 \; \text{mol of electrons} \]Moles of oxygen produced:\[ n(O_2) = \dfrac{0.01}{4} = 0.0025 \; \text{mol} \]Volume of oxygen at s.t.p:\[ V = n \times 22.4 \]\[= 0.0025 \times 22.4 = 0.056 \; \text{dm}^3 \]\[ \therefore \; 0.056 \; \text{dm}^3 = 56 \; \text{cm}^3 \] Thus, 0.056 dm³ of oxygen gas is evolved.

Cobalt-60 emits high-energy gamma rays which can penetrate metals and thick materials. These rays expose photographic films placed on the opposite side of a metal object, revealing flaws or internal cracks without damaging the material. This non-destructive testing technique is essential in quality control during welding and casting. Other isotopes such as Iodine-131 and Phosphorus-32 are used in medicine and biological studies, while Carbon-14 is used in dating ancient remains, not in industrial radiography.

The reaction is exothermic (ΔH = −395.7 kJ mol⁻¹). According to Le Chatelier’s principle, a decrease in temperature favours the formation of products in an exothermic reaction because heat is considered a product. Lowering the temperature shifts equilibrium towards SO₃ formation. Pressure also affects equilibrium, but since temperature directly influences exothermic balance, decreasing temperature yields more SO₃. Catalysts only speed up the rate but do not affect the equilibrium position.

The atomic number of fluorine is 9. Its electronic configuration is 1s² 2s² 2p⁵.

In the 2p subshell, there are 3 orbitals which can hold up to 6 electrons. For fluorine, the 5 electrons fill as follows: one electron in each of the three orbitals, then pairing begins. This leaves one unpaired electron in the p-orbitals.

Hence, fluorine has one unpaired electron in its p-orbitals.

Gamma (γ) rays are high-energy electromagnetic radiations with no mass or charge, allowing them to penetrate deeply through materials but making them the least ionising. Alpha particles are the most ionising due to their double positive charge and large mass, followed by beta particles and X-rays. Gamma rays, however, can travel long distances and require dense materials such as lead for shielding, reflecting their weak ionising but strong penetrating ability.

Oxidation involves loss of electrons while reduction involves gain. Zinc displaces copper from CuSO₄ because zinc is more electropositive.

Zn → Zn²⁺ + 2e⁻ (oxidation)
Cu²⁺ + 2e⁻ → Cu (reduction)

The substance reduced (Cu²⁺ in CuSO₄) acts as the oxidising agent since it gains electrons. Therefore, CuSO₄(aq) is the oxidising agent.

   

The structure contains a carbonyl group (C=O) bonded to two alkyl groups, characteristic of ketones (alkanones). Alkanals have the C=O group at the end of the carbon chain, while alkanones have it within the chain. The compound thus represents propanone (CH₃COCH₃), making it an alkanone.

The activation energy represents the minimum energy required for reactant molecules to undergo a successful reaction. In the graph, the catalysed reaction has a lower peak compared to the uncatalysed one, indicating a reduction in activation energy. The difference corresponds to 100 kJ, showing that the catalyst provides an alternative pathway with a lower energy barrier, thereby increasing the rate of reaction without being consumed.

In equilibrium expressions, only gaseous and aqueous species are included; solids are excluded because their concentrations are constant. For the reaction:
3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)
Only H₂O and H₂ appear in the equilibrium expression:\[ K = \dfrac{[H_2]^4}{[H_2O]^4} \]This shows that option C correctly represents the equilibrium constant.