We are given the two equations
1. #"BaCl"_2"(s)" → "BaCl"_2"(aq)";color(white)(mmmmmmmmml) Δ_text(sol)H = "-20.6 kJ/mol"#
2. #"BaCl"_2·"2H"_2"O(s)" → "BaCl"_2"(aq)" + "2H"_2"O(l)"; Δ_text(sol)H = color(white)(l)"8.8 kJ/mol"#
From these, we must derive the target equation
#"BaCl"_2"(s)" + "2H"_2"O(l)" → "BaCl"_2·"2H"_2"O"; color(white)(mmm)Δ_text(hyd)H = ?"#
The target equation has #"BaCl"_2# on the left, so we start with equation 1.
It also has #"BaCl"_2·2"H"_2"O"# on the right, so we reverse equation 2 to get equation 3.
When we reverse an equation, we change the sign of its #ΔH#.
Finally, we add equations 1 and 3, canceling species that appear on opposite sides of the reaction arrow.
When we add two equations, we add their #ΔH# values.
This gives the target equation.
1. #"BaCl"_2"(s)" → color(red)(cancel(color(black)("BaCl"_2"(aq)"))); color(white)(mmmmmmmmml)Δ_text(sol)H =color(white)(l) "-20.6 kJ/mol"#
3. #color(red)(cancel(color(black)("BaCl"_2"(aq)"))) + "2H"_2"O(l)" → "BaCl"_2·"2H"_2"O(s)"; Δ_text(sol)H =color(white)(ll) "-8.8 kJ/mol"#
#"BaCl"_2"(s)" + "2H"_2"O(l)" → "BaCl"_2·"2H"_2"O"; color(white)(mmmll)Δ_text(hyd)H = "-29.4 kJ/mol"#
Thus, for the hydration of #"BaCl"_2#, #Δ_text(hyd)H = "-29.4 kJ/mol"#.
