Chemistry 2006 Jamb Examination

When inert platinum electrodes are used, Cu²⁺ ions are discharged at the cathode, while OH⁻ ions (from water) are discharged at the anode. The correct anode reaction is:

\[4OH⁻ → 2H₂O + O₂ + 4e⁻\]

This corresponds to option B.

Calculate moles of HCl used.

\[ = C \times V = 0.05 \times \dfrac{10}{1000} = 5.0 \times 10^{-4} \text{ mol}\]

From equation, 2 mol HCl neutralizes 1 mol Na₂CO₃.

So moles of Na₂CO₃ = \(\dfrac{5.0 \times 10^{-4}}{2} = 2.5 \times 10^{-4}\).

That is the number of moles present in 25 cm³ solution. To find per dm³ (not asked, but check):

\[\dfrac{2.5 \times 10^{-4}}{25 \times 10^{-3}} = 0.010 \text{ mol dm}^{-3}.\]

Hence, the moles in 25 cm³ = 0.010 mol. Option C.

Heat of combustion of hydrogen refers to the reaction:

\[H₂(g) + \tfrac{1}{2}O₂(g) → H₂O(l) \]\[\quad \Delta H = -285.8 \text{ kJ}\]

This is the same as the enthalpy of formation of water from its elements. Therefore:

\[\Delta H_f (H₂O) = -285.8 \text{ kJ mol}^{-1}\]

So the correct answer is option C.

KMnO₄ is purple due to MnO₄⁻. In acidic medium, MnO₄⁻ is reduced to Mn²⁺, which is almost colourless in solution:

\[MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O\]

Thus, the purple colour disappears, leaving a colourless solution. Option D is correct.

Gas X is collected over water. For this method, the gas must be insoluble (or only slightly soluble) in water.

Hydrogen (H₂) is light, insoluble in water, and collected by upward delivery. Chlorine (Cl₂) and hydrogen chloride (HCl) are soluble in water.

Therefore, the gas is hydrogen. Correct answer: A.

2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O 

The reaction between ammonium chloride and calcium hydroxide is a double displacement reaction. However, in the reaction between ammonium chloride (NH₄Cl) and calcium hydroxide (Ca(OH)₂), quicklime (CaO) is used to dry the ammonia gas (NH₃) produced by absorbing water, preventing the formation of ammonium sulphate when sulphuric acid is used to dry the gas. 

The compound is a secondary alcohol (2-butanol).

Oxidation of secondary alcohols yields ketones:

\[ CH₃CH₂CH(OH)CH₃ → CH₃CH₂COCH₃ + H₂ \]

That product is 2-butanone. Correct option A.

2-methylbut-2-ene = a 5-carbon chain with a double bond at C-2 and a methyl substituent at C-2: $CH_3C(CH_3) = CHCH_3 $

This matches option D.

When a carboxylic acid (RCOOH) reacts with sodium bicarbonate (NaHCO3), it forms a salt (RCOONa), water (H2O), and carbon dioxide (CO2). The carbon dioxide gas produced in this reaction is what causes the bubbles or effervescence observed. 

• RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂ 

This reaction is a simple and reliable test to distinguish carboxylic acids from other organic compounds, as most other functional groups (like alcohols or phenols) are not acidic enough to react with bicarbonate and produce a gas. 

The activation energy, Ea is the energy of the transition state, (T.E) minus the reactant energy (R.E).

For Catalyzed reaction, Ea = T.E - R.E = 45 - 15 = 30

For Uncatalyzed reaction, Ea = T.E - R.E = 50 - 15 = 35