Consider the following reactions:
(a) H+(aq)+OH−(aq)→H2O(l) ΔH=−x1 kJ mol−1
(b) H2(g)+12O2(g)→H2O(l) ΔH=−x2 kJ mol−1
(c) CO2(g)+H2(g)→CO(g)′+H2O(l) ΔH=−x3 kJ mol−1
(d) C2H2(g)+52O2(g)→2CO2(g)+H2O(l) ΔH=x4 kJ mol−1
Enthalpy of formation of H2O(l) is
−x2 kJ.mol−1
This question, or rather the answer to this question, is based on the very definition of the enthalpy (or heat) of formation:
It is the amount of heat evolved or absorbed during the formation of 1 mole of a compound from its constituent elements (standard state).
H2(g)+12O2(g)⟶H2O(l)
Hence the heat of formation of H2O(l)=−x2 kJ.mol−1.