ΔfH of H2Ol is –286 kJ / mole and ΔfH of H2Og is –246 kJ / mole, then the enthalpy change when 9 grams of water vapour condenses liquid water is
H2(g) + 12O2(g) → H2Ol ;ΔH=−286 kJ ……. (1)
H2(g) + 12O2(g)→ H2Og ;ΔH=−246 kJ ……..(2)
Eq(1) – Eq(2)
H2Og → H2Ol ; ΔH=−40 kJ/mole
For 18g , enthalpy change is -40kJ
For 9 grams, enthalpy change will be – 20 kJ.