The calorific value of H₂(g) at STP is 12.78 kJ/L hence approximate standard enthalpy of formation of H₂O(l) is

$1 {\mathrm{LH}}_2\left(\mathrm{g}\right)\text{ at }\mathrm{STP}=\frac{1}{22.4}\mathrm{mol}$

.'. Heat released due to combustion of $\frac{1}{22.4}$mol of H₂(g) = 12.78 KJ

Heat released due to combustion of I mol of H₂(g) = 12.78 × 22.4 = 286.27 kJ

$\therefore$approximate standard enthalpy of formation of H₂O(l) = 286 KJ