NVAEasyJEE 2025Enthalpies (Bond, Combustion, Formation…)

JEE Chemistry 2025 Question with Solution

The formation enthalpies, ΔHf\Delta H_f^\circ for H2\text{H}_2 and O2\text{O}_2 are 220.0kJ mol1220.0 \, \text{kJ mol}^{-1} and 250.0kJ mol1250.0 \, \text{kJ mol}^{-1}, respectively, at 298.15K298.15 \, \text{K}, and ΔHf\Delta H_f^\circ for H2O\text{H}_2\text{O} (g)\text{(g)} is 242.0kJ mol1-242.0 \, \text{kJ mol}^{-1} at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15K298.15 \, \text{K} is:

Answer

Correct answer:114

Step-by-step solution

Standard Method

Given: ΔHf(H2)=220.0kJ mol1\Delta H_f^\circ(\text{H}_2) = 220.0 \, \text{kJ mol}^{-1}, ΔHf(O2)=250.0kJ mol1\Delta H_f^\circ(\text{O}_2) = 250.0 \, \text{kJ mol}^{-1}, and ΔHf(H2O)=242.0kJ mol1\Delta H_f^\circ(\text{H}_2\text{O}) = -242.0 \, \text{kJ mol}^{-1} at 298.15K298.15 \, \text{K}.

Find: The average bond enthalpy of the O-H bond in water.

Using Hess's law:

ΔHf(H2O)=Bond enthalpy of O-H×2(ΔHf(H2)+ΔHf(O2))\Delta H_f^\circ(\text{H}_2\text{O}) = \text{Bond enthalpy of O-H} \times 2 - \left( \Delta H_f^\circ(\text{H}_2) + \Delta H_f^\circ(\text{O}_2) \right)

Substituting the given values:

242=2×Bond enthalpy of O-H(220+250)-242 = 2 \times \text{Bond enthalpy of O-H} - (220 + 250) 242=2×Bond enthalpy of O-H470-242 = 2 \times \text{Bond enthalpy of O-H} - 470 2×Bond enthalpy of O-H=2282 \times \text{Bond enthalpy of O-H} = 228 Bond enthalpy of O-H=114kJ/mol\text{Bond enthalpy of O-H} = 114 \, \text{kJ/mol}

Therefore, the average bond enthalpy of the O-H bond in water is 114kJ/mol114 \, \text{kJ/mol}.

Common mistakes

  • Using the formation enthalpy values without applying Hess's law correctly. This is wrong because the solution explicitly relates product and reactant enthalpies through a balance equation. Instead, write the Hess's law relation first and then substitute the values.

  • Forgetting that water contains two O-H bonds. This is wrong because the bond enthalpy term must be multiplied by 22. Instead, use 2×2 \times bond enthalpy of O-H before solving.

  • Making a sign error with ΔHf(H2O)=242.0kJ mol1\Delta H_f^\circ(\text{H}_2\text{O}) = -242.0 \, \text{kJ mol}^{-1}. This is wrong because changing the negative sign changes the final numerical value completely. Instead, substitute the value with its correct negative sign.

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