Question

Calculate the average bond enthalpy of the $O-H$ bond in water at 298 K using the data / information given below :-
1. ${\Delta }_f{H}^0\left[H\left(g\right)\right]=218$ kJ / mol
2. ${\Delta }_f{H}^0\left[O\left(g\right)\right]=249.2$ kJ / mol
3. ${\Delta }_f{H}^0\left[H_{2}O\left(g\right)\right]=-241.8$ kJ / mol
The average bond enthalpy of the $-$ bond in water is defined as one - half of the enthalpy change for
the reaction $H_{2}O\left(g\right)\to 2H\left(g\right)+O\left(g\right).$
Also, determine the $\Delta U$ of the $O-H$ bond in water at 298 K. Assume ideal gas behaviour.

• A. $E(O-H)=462$ kJ/mol; $\Delta U=461$ kJ/mol
• B. $E(O-H)=463.5$ kJ/mol; $\Delta U=463.5$ kJ/mol
• C. $E(O-H)=461$ kJ/mol; $\Delta U=461$ kJ/mol
• D. $E(O-H)=463.5$ kJ/mol; $\Delta U=461$ kJ/mol

Answer 1

Answer: (D)

$E(O-H)=463.5$ kJ/mol; $\Delta U=461$ kJ/mol

$\Delta H_r$ $=2$$\times$ ${\Delta }_f{H}^0\left[H\left(g\right)\right]$ + ${\Delta }_f{H}^0\left[O\left(g\right)\right]$ - ${\Delta }_f{H}^0\left[H_{2}O\left(g\right)\right]$
$\Delta H_r$ = 2 $\times$ $218+249.2+241.8=927$ kJ/mol
Therefore, average bond enthalpy of the OH is $\frac{{\Delta H}_r}{2}$ = $\frac{927}{2}$ $=463.5$ kJ/mol
For ideal gas, $H=U+nRT$
i.e, $U=H-nRT$ ($n=1$, as there are $2O-H$ in $H_{2}O$)
Therefore, $U=463.5-$($\frac{8.314\times 298}{1000}$) $=461$ kJ/mol. (Correct Option:D)