Question

The enthalpy of combustion of $H_{2\left(g\right)}$ to give $H_2{O}_{\left(g\right)}$ is $-249$ kJ mol${}^{-1}$ and bond enthalpies of $H-H$ and $O=O$ are $433$ kJ mol${}^{-1}$ and $492$ kJ mol${}^{-1}$ respectively. The bond enthalpy of $O-H$ is:

• A. $+464$ kJ mol${}^{-1}$
• B. $-464$ kJ mol${}^{-1}$
• C. $+232$ kJ mol${}^{-1}$
• D. $-232$ kJ mol${}^{-1}$

Answer 1

The correct option is A $+464$ kJ mol${}^{-1}$

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right)$

$\Delta H=-249KJ$

$\Delta H=$ B.E. of $H_2$ $+$ $\frac{1}{2}$ B.E. of $O_2$ $-$ B.E. of $H_{2}O$

where B.E. is bond enthalpy

$-249$ $=433+\frac{1}{2}\times 492-2\times B.E.(O-H)$

$-249=433+246-2\times B.E.(O-H)$

$249+433+246=2\times B.E.(O-H)$

$B.E.(O-H)=\frac{249+433+246}{2}$ kJ mol${}^{-1}$

$B.E(O-H)=464$ kJ mol ${}^{-1}$

Hence, the correct option is $A$