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

Given reactions and enthalpies:

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

The reaction for \(H_2(g)\) and \(O_2(g)\) to form \(H_2O(g)\) is given by:

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

From the given data:

• 2 \(\times 220 = 440 \, \text{kJ/mol}\) (for \(H-H\) bonds)
• 250  kJ/mol (for \(O-O\) bonds)
• \(\Delta H(H_2O(g)) = -242 \, \text{kJ/mol}\) (for the formation of \(H_2O\))

The bond energy formula is:

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

Substituting the values:

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

Solving for \(\text{B.E.} (O-H)\):

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