Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2(g) + O2(g)2H2O(g) ANSWER: __ kJ?

Refer to: https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation
ΔHf°[H₂O(g)] = -241.8 kJ/mol (to 1 decimal place)

The standard enthalpy change for any element in standard state = 0. Hence,
ΔHf°[H₂(g)] = ΔHf°[O₂(g)] = 0 kJ/mol

2H₂(g) + O₂(g) → 2H₂O(g) …… ΔH°

ΔH° = ΣΔH°(products) - ΣΔH°(reactants)
= 2 ΔHf°[H₂O(g)] - {2 ΔHf°[H₂(g)] + ΔHf°[O₂(g)]}
= 2 (-241.8 kJ) - (0 kJ)
= -483.6 kJ