Question

The formation enthalpies, \( \Delta H_f^\circ \) for \( H_2 \) and \( O_2 \) are 220.0 and 250.0 kJ mol\(^{-1}\), respectively, at 298.15 K, and \( \Delta H_f^\circ \) for \( H_2O \) (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:

Hint:

Bond enthalpies can be determined using Hess's law by relating the formation enthalpies of products and reactants.

Answer 1

The bond enthalpy can be calculated using the following equation based on Hess's law:

\[ \Delta H_f^\circ(H_2O) = \text{Bond enthalpy of O-H} \times 2 - \left( \Delta H_f^\circ(H_2) + \Delta H_f^\circ(O_2) \right) \]

\[ -242 = 2 \times \text{Bond enthalpy of O-H} - (220 + 250) \]

\[ -242 = 2 \times \text{Bond enthalpy of O-H} - 470 \]

\[ 2 \times \text{Bond enthalpy of O-H} = 228 \]

\[ \text{Bond enthalpy of O-H} = 114 \, \text{kJ/mol} \]

Final Conclusion: The average bond enthalpy of the O-H bond in water is 114 kJ/mol.