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Bond dissociation enthalpy of H₂, Cl₂, and HCl are 434, 242, and 431 kJ mol⁻¹ respectively. Enthalpy of formation of HCl is:

1. 93 kJ mol⁻¹
2. −245 kJ mol⁻¹
3. −93 kJ mol⁻¹
4. 245 kJ mol⁻¹

Correct answer: −93 kJ mol⁻¹

Solution

The enthalpy of formation of HCl can be calculated using the bond dissociation enthalpies: ΔH = [Bond enthalpy of H₂ + Bond enthalpy of Cl₂] - [2 × Bond enthalpy of HCl]. Substituting values: ΔH = [434 + 242] - [2 × 431] = 676 - 862 = -186 kJ for 2 moles of HCl. For 1 mole, ΔH = -93 kJ mol⁻¹.

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