Q.Bond enthalpies of $H-H$ and $O=O$ are $436$ and $498 \text{ kJ/mol}$ respectively. If enthalpy of formation of $H_2O(g)$ is $-242 \text{ kJ/mol}$ , the bond enthalpy of $O-H$ is:

464 \text{ kJ/mol}

928 \text{ kJ/mol}

232 \text{ kJ/mol}

436 \text{ kJ/mol}

The correct solution involves applying the fundamental concept to derive the final value step by step...

Already a member? Login here

Q.Bond enthalpies of H−HH-HH−H and O=OO=OO=O are 436436436 and 498 kJ/mol498 \text{ kJ/mol}498 kJ/mol respectively. If enthalpy of formation of H2O(g)H_2O(g)H2​O(g) is −242 kJ/mol-242 \text{ kJ/mol}−242 kJ/mol, the bond enthalpy of O−HO-HO−H is: