MCQ
The enthalpy change at $298\,K$ for decomposition is given in the following two steps

Step $I\,\,:$ ${H_2}O(g)\, \to \,H(g)\, + \,OH(g)\,;$              $\Delta H = 498\,kJ\,mol^{-1}$

Step $II\,\,:$ $OH(g)\, \to \,H(g)\, + \,O(g)\,;$              $\Delta H = 428\,kJ\,mol^{-1}$

The bond enthalpy of the $O-H$ bond is....$kJ\,mol^{-1}$

  • A
    $498$
  • $463$
  • C
    $428$
  • D
    $70$

Answer

Correct option: B.
$463$
b
Given dissociation enthalpy of $H - OH =498 KJ / mol$

Given dissociation enthalpy of $O - H =428 KJ / mol$ $\therefore$ Bond enthalpy of $O - H =\frac{498+428}{2}=\frac{926}{2}=463 KJ / mol$

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