Question 4.1: The average bond enthalpy of the O—H bond in water is define...
The average bond enthalpy of the O—H bond in water is defined as one-half of the enthalpy change for the reaction H_{2}O\left( g \right)\longrightarrow 2 H\left( g \right)+O\left( g \right). The formation enthalpies, \Delta H°_{f},for H(g) and O\left( g \right) are 218.0 and 249.2 KJ mol^{-1}, respectively, at 298.15 K, and \Delta H°_{f} for H_{2}O\left( g \right) is -241.8 KJ mol^{-1} at the same temperature.a. Use this information to determine the average bond enthalpy of the O—H bond in water at 298.15 K. b. Determine the average bond energy \Delta U of the O—H bond in water at 298.15 K. Assume ideal gas behavior.
a. We consider the sequence
\begin{array}{r c}\begin{matrix}H_{2}O\left( g \right)\longrightarrow H_{2}\left( g \right)+1/2 O_{2}\left( g \right)& \Delta H°=241.8 kJ mol^{-1}\\ H_{2}O\left( g \right)\longrightarrow 2H\left( g \right) & \Delta H°= 2× 218.0 kJ mol^{-1} \\ 1/2 O_{2}\left( g \right)\longrightarrow O\left( g \right) & \Delta H°= 249.2 kJ mol^{-1} \end{matrix}& \\ \hline \begin{matrix} H_{2}O\left( g \right)\longrightarrow H_{2}O\left( g \right)+O\left( g \right)& \Delta H°=927.0 KJ mol^{-1}\end{matrix} \end{array}This is the enthalpy change associated with breaking both O—H bonds under standard conditions. We conclude that the average bond enthalpy of the O—H bond in water is \frac{1}{2} × 927.0 KJ mol^{-1}=463. KJmol^{-1}. We emphasize that this is the average value because the values of \Delta H for the transformations H_{2}O\left( g \right)\longrightarrow H\left( g \right)+OH\left( g \right) and OH\left( g \right) \longrightarrow O\left( g \right)+H\left( g \right) differ.
b. \Delta U° =\Delta H° – \Delta\left( PV \right)=\Delta H° -\Delta nRT
=927.0 KJ mol^{-1}-2× 8.314 J mol^{-1} K^{-1} × 298.15 K
=922.0 KJ mol^{-1}
The average value for \Delta U° for the O—H bond in water is \frac{1}{2} × 922.0 KJ mol^{-1}=461. KJ mol^{-1}. The bond energy and the bond enthalpy are nearly identical.