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Chemistry · CHEMICAL BONDING AND MOLECULAR STRUCTURE

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For water vapor, the stepwise bond dissociation enthalpies are: \( \text{H}_2\text{O}(\text{g}) \rightarrow \text{H}(\text{g}) + \text{OH}(\text{g});\ \Delta_{\text{a}}H_1^\ominus = 502\text{ kJ mol}^{-1} \) \( \text{OH}(\text{g}) \rightarrow \text{H}(\text{g}) + \text{O}(\text{g});\ \Delta_{\text{a}}H_2^\ominus = 427\text{ kJ mol}^{-1} \) What is the mean (average) \( \text{O}-\text{H} \) bond enthalpy in \( \text{H}_2\text{O} \)?

  1. A
    \( 929.0\text{ kJ mol}^{-1} \)
  2. B
    \( 427.0\text{ kJ mol}^{-1} \)
  3. C
    \( 502.0\text{ kJ mol}^{-1} \)
  4. D
    \( 464.5\text{ kJ mol}^{-1} \)

Correct answer · Option D

\( 464.5\text{ kJ mol}^{-1} \)

Explanation

The mean bond enthalpy is obtained by dividing the total bond dissociation enthalpy by the number of bonds broken: \( \frac{502 + 427}{2} = 464.5\text{ kJ mol}^{-1} \).
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