Question

Average bond enthalpy of water is \(464.5\ \text{kJ mol}^{-1}\). If the energy required to break the first O–H bond is \(502\ \text{kJ mol}^{-1}\), how much energy per mole is required to break the second O–H bond?

• A. 929 kJ
• B. 251 kJ
• C. 427 kJ
• D. 678 kJ

Hint:

Average bond enthalpy is the arithmetic mean of all similar bond dissociation energies in a molecule.

Answer 1

The Correct Option is C

Step 1: Understand average bond enthalpy.
Average bond enthalpy of water is the mean of energies required to break the two O–H bonds: \[ \text{Average bond enthalpy} = \frac{E_1 + E_2}{2} \]

Step 2: Substitute given values.
\[ 464.5 = \frac{502 + E_2}{2} \]

Step 3: Solve for the second bond energy.
\[ 502 + E_2 = 929 \] \[ E_2 = 929 - 502 = 427\ \text{kJ mol}^{-1} \]

Step 4: Conclusion.
The energy required to break the second O–H bond is \(427\ \text{kJ mol}^{-1}\).