The formation enthalpies, $$\Delta H_{f}\ominus$$ for $$H_{g}$$ and $$O_{g}$$ are 220.0 and $$250.0 kJ mol^{-1}$$, respectively, at 298.15 K , and $$\Delta H_{f}\ominus$$ for $$H_{2}O_{g}$$ is $$-242.5 kJ mol^{-1}$$ at the same temperature. The average bond enthalpy of the $$O-H$$ bond in water at 298.15 K is __________$$kJ mol^{-1}$$(nearest integer).

The average bond enthalpy of the O-H bond in water is calculated using the given formation enthalpies and the concept of bond dissociation.

The formation reaction for water vapor is:

$$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g), \quad \Delta H_f^\ominus = -242.5 \text{kJ mol}^{-1}$$

This reaction can be broken down into steps involving the dissociation of the reactants into atoms and the formation of water from these atoms.

First, the dissociation of H₂(g) into hydrogen atoms:

$$\text{H}_2(g) \rightarrow 2\text{H}(g), \quad \Delta H = 2 \times \Delta H_f^\ominus(\text{H})$$

Given $$\Delta H_f^\ominus(\text{H}) = 220.0 \text{kJ mol}^{-1}$$,

$$\Delta H = 2 \times 220.0 = 440.0 \text{kJ mol}^{-1}$$

Next, the dissociation of $$\frac{1}{2}\text{O}_2(g)$$ into an oxygen atom:

$$\frac{1}{2}\text{O}_2(g) \rightarrow \text{O}(g), \quad \Delta H = \Delta H_f^\ominus(\text{O})$$

Given $$\Delta H_f^\ominus(\text{O}) = 250.0 \text{kJ mol}^{-1}$$,

$$\Delta H = 250.0 \text{kJ mol}^{-1}$$

The formation of H₂O(g) from the atoms is:

$$2\text{H}(g) + \text{O}(g) \rightarrow \text{H}_2\text{O}(g), \quad \Delta H_{\text{atom}}$$

The overall enthalpy change for the formation reaction is the sum of these steps:

$$\Delta H_f^\ominus(\text{H}_2\text{O}) = [\Delta H \text{ for } \text{H}_2(g) \rightarrow 2\text{H}(g)] + [\Delta H \text{ for } \frac{1}{2}\text{O}_2(g) \rightarrow \text{O}(g)] + \Delta H_{\text{atom}}$$

Substituting the values:

$$-242.5 = 440.0 + 250.0 + \Delta H_{\text{atom}}$$

$$-242.5 = 690.0 + \Delta H_{\text{atom}}$$

Solving for $$\Delta H_{\text{atom}}$$:

$$\Delta H_{\text{atom}} = -242.5 - 690.0 = -932.5 \text{kJ mol}^{-1}$$

The enthalpy change $$\Delta H_{\text{atom}}$$ for the reaction $$2\text{H}(g) + \text{O}(g) \rightarrow \text{H}_2\text{O}(g)$$ represents the energy released when two O-H bonds are formed. Therefore, it is equal to the negative of twice the average bond enthalpy of the O-H bond (since two identical O-H bonds are formed):

$$\Delta H_{\text{atom}} = -2 \times \text{(O-H bond enthalpy)}$$

Substituting the value:

$$-932.5 = -2 \times \text{(O-H bond enthalpy)}$$

Solving for the O-H bond enthalpy:

$$2 \times \text{(O-H bond enthalpy)} = 932.5$$

$$\text{O-H bond enthalpy} = \frac{932.5}{2} = 466.25 \text{kJ mol}^{-1}$$

Rounding to the nearest integer:

$$\text{Average O-H bond enthalpy} = 466 \text{kJ mol}^{-1}$$