Question

Question #ff017

Answer 1

Here's what I get.

Explanation:

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.

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"`