Question #ff017

1 Answer
Nov 11, 2016

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.

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"#