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Answer 1 · 2016-11-11T13:24:31

Here's what I get.

Your text may have slightly different values for the enthalpies of formation.

For $"H"_2"O(g)"$ : $Δ_text(f)H^° = "-241.83 kJ/mol"$

For $"MgO(s)"$ : $Δ_"f"H^° = "-601.24 kJ/mol"$

Here's the enthalpy diagram I created in Excel.

Enthalpy Diagram

Step 1 is the decomposition of water into its elements (the green arrow).

$"H"_2"O(g)" → "H"_2"(g)" + "½O"_2("g)"; ΔH = "+241.83 kJ"$

Step 2 is the formation of MgO(s) from its elements (the red arrow).

$"Mg(s)" color(white)(l)+ "½O"_2("g)" → "MgO(s)"; ΔH = "-601.24 kJ"$

$Δ_"rxn"H = "+241.83 kJ - 601.24 kJ"= "-359.41 kJ"$