The standard enthalpy changes at $25^\circ ext{C}$ for several fundamental gas-phase reactions are

Question

The standard enthalpy changes at $25^\circ	ext{C}$ for several fundamental gas-phase reactions are given below: (i) Formation of hydroxyl radical: $\frac{1}{2}	ext{H}_2(g) + \frac{1}{2}	ext{O}_2(g) ightarrow 	ext{OH}(g); \Delta H^\circ_1 = +10.06	ext{ kcal}$ (ii) Atomization of hydrogen molecule: $	ext{H}_2(g) ightarrow 2	ext{H}(g); \Delta H^\circ_2 = +104.18	ext{ kcal}$ (iii) Atomization of oxygen molecule: $	ext{O}_2(g) ightarrow 2	ext{O}(g); \Delta H^\circ_3 = +118.32	ext{ kcal}$. Calculate the oxygen-hydrogen ($	ext{O-H}$) bond dissociation energy (in $	ext{kcal mol}^{-1}$) in the gaseous hydroxyl radical ($	ext{OH}(g)$). (Report your answer to two decimal places)

Professor PC Thomas Classes · Answer

The $	ext{O-H}$ bond dissociation energy in the hydroxyl radical corresponds precisely to the standard enthalpy change of the bond-breaking atomization reaction: $	ext{OH}(g) ightarrow 	ext{O}(g) + 	ext{H}(g); \Delta H = E_{	ext{O-H}}$. We can synthesize this reaction using Hess's law from the three given equations: First, reverse equation (i) to dissociate $	ext{OH}(g)$: $	ext{OH}(g) ightarrow \frac{1}{2}	ext{H}_2(g) + \frac{1}{2}	ext{O}_2(g); \Delta H = -\Delta H^\circ_1 = -10.06	ext{ kcal}$. Next, add half of equation (ii) to form atomic hydrogen: $\frac{1}{2}	ext{H}_2(g) ightarrow 	ext{H}(g); \Delta H = \frac{1}{2}\Delta H^\circ_2 = \frac{1}{2}(104.18) = +52.09	ext{ kcal}$. Finally, add half of equation (iii) to form atomic oxygen: $\frac{1}{2}	ext{O}_2(g) ightarrow 	ext{O}(g); \Delta H = \frac{1}{2}\Delta H^\circ_3 = \frac{1}{2}(118.32) = +59.16	ext{ kcal}$. Summing these equations and their corresponding enthalpy changes yields: $E_{	ext{O-H}} = -10.06 + 52.09 + 59.16 = +101.19	ext{ kcal mol}^{-1}$.