▶️ Answer/Explanation
(a) \(H_2 (g) + \frac{1}{2}O_2 (g) → H_2O(l)\)
Explanation: The standard enthalpy change of formation (\(\Delta H_f^\theta\)) of \(H_2O\) is defined as the enthalpy change when 1 mole of \(H_2O(l)\) is formed from its elements in their standard states (\(H_2(g)\) and \(O_2(g)\)).
(b) \(\Delta H = +230 \, \text{kJ mol}^{-1}\)
Explanation: Using Hess’s Law, \(\Delta H = \sum \Delta H_f^\theta (\text{products}) – \sum \Delta H_f^\theta (\text{reactants})\):
\(\Delta H = [2 \times (-134) + 3 \times (-286)] – [(-1264) + 2 \times (-46)] = -1074 + 1304 = +230 \, \text{kJ mol}^{-1}\).
(c)(i) Giant covalent structure.
Explanation: Boron carbide’s high melting point (>2000°C) and hardness suggest a giant covalent lattice with strong covalent bonds between boron and carbon atoms.
(c)(ii) Empirical formula: \(B_4C\).
Explanation: – Moles of boron: \(\frac{78.26 \, \text{g}}{10.8 \, \text{g/mol}} = 7.246 \, \text{mol}\).
– Moles of carbon: \(\frac{21.74 \, \text{g}}{12 \, \text{g/mol}} = 1.812 \, \text{mol}\).
– Simplest ratio: \(\frac{7.246}{1.812} \approx 4\) (Boron) : \(1\) (Carbon).
Thus, the empirical formula is \(B_4C\).