The formation enthalpies, \( \Delta H_f^\circ \) for \( \text{H}_2 \) and \( \text{O}_2 \) are 220.0 and 250.0 kJ mol\(^{-1}\), respectively, at 298.15 K, and \( \Delta H_f^\circ \) for \( \text{H}_2\text{O} \) (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:
Show Hint
Correct Answer: 114
Solution and Explanation
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.
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