Question

The formation enthalpies, \( \Delta H_f^\circ \) for \( \text{H}_2 \) and \( \text{O}_2 \) are 220.0 and 250.0 kJ mol\(^{-1}\), respectively, at 298.15 K, and \( \Delta H_f^\circ \) for \( \text{H}_2\text{O} \) (g) is -242.0 kJ mol\(^{-1}\) at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is: 

Answer 1

Correct Answer: 466

The enthalpies for the formation of individual atoms from diatomic molecules are provided:

• \(\frac{1}{2} H_2(g) \rightarrow H(g); \Delta H(H(g)) = 220 \, \text{kJ/mol}\)
• \(\frac{1}{2} O_2(g) \rightarrow O(g); \Delta H(O(g)) = 250 \, \text{kJ/mol}\)

The enthalpy of formation for gaseous water from its elements is:

\(H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g); \Delta H(H_2O(g)) = -242 \, \text{kJ/mol}\)

Based on the provided data:

• The energy required to break 2 moles of \(H-H\) bonds in \(H_2(g)\) is 2 \(\times 220 = 440 \, \text{kJ/mol}\).
• The energy required to break 1 mole of \(O=O\) bonds in \(\frac{1}{2} O_2(g)\) is 250 kJ/mol.
• The enthalpy change for the formation of \(H_2O(g)\) is \(\Delta H(H_2O(g)) = -242 \, \text{kJ/mol}\).

The relationship between enthalpy change and bond energies is:

\(\Delta H(H_2O(g)) = (\text{Bond Energy of Reactants}) - (\text{Bond Energy of Products})\)

Applying this to the formation of \(H_2O(g)\):

\(\Delta H(H_2O(g)) = [2 \times (\text{B.E.} (H-H)) + \frac{1}{2} \times (\text{B.E.} (O=O))] - [2 \times (\text{B.E.} (O-H))]\)

The provided data simplifies this to:

\(\Delta H(H_2O(g)) = 440 + 250 - 2 (\text{B.E.} (O-H))\)

Substituting the known enthalpy change:

\(-242 = 440 + 250 - 2 (\text{B.E.} (O-H))\)

Solving for the bond energy of the \(O-H\) bond:

\(\text{B.E.} (O-H) = 466 \, \text{kJ/mol}\)