The formation enthalpies, ΔH°_f for H₍g₎ and O₍g₎ are 220.0 and 250.0 kJ mol⁻¹, respectively, at 298.15 K, and ΔH°_f for H₂O₍g₎ is -242.0 kJ mol⁻¹ at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is ______ kJ mol⁻¹ (nearest integer).
Answer: 466
Solution
The atomisation steps are given as 21H2(g)→H(g), ΔH=220 kJ/mol, and 21O2(g)→O(g), ΔH=250 kJ/mol. Combining these, H2(g)+21O2(g)→2H(g)+O(g) has ΔH=2(220)+250=690 kJ/mol, and forming the two O–H bonds of water from these free atoms releases 2×BE(O−H). Since ΔHf(H2O,g)=−242 kJ/mol, −242=690−2BE(O−H), so BE(O−H)=2690+242=466 kJ/mol.
JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Integer answer
