Enthalpy Change Given 3 Reactions & Their Enthalpy Changes

Question

Use the enthalpy changes given in the data below to calculate the enthalpy change for this reaction:

Ca(OH)₂(s) → CaO(s) + H₂O(g)

Data:

Reaction no.	Reaction	ΔrH° / kJ mol-1 at 298K, standard state = 1 bar.
1	Ca(s)+ 1/2 O2(g) → CaO(s)	-635
2	Ca(OH)2(s) → Ca(s) + O2(g) + H2(g)	+986
3	H2O(g) → H2(g) + 1/2 O2(g)	+242

Answer 1

Hess' Law:

Target equation is Ca(OH)₂(s) ==> CaO(s) + H₂O(g)

copy rx.1: Ca(s) + 1/2O₂(g) ===> CaO ∆H = -635

copy rx.2: Ca(OH)₂(s) ===> Ca(s) + O₂(g) + H₂(g) ∆H = +986

reverse rx3. H₂(g) + 1/2O₂(g) ===> H₂O(g) ∆H = -242 (NOTE change of sign since reversed the rx)

Add them together to get...

Ca(s) + 1/2O₂(g) + Ca(OH)₂(s) + H₂(g) + 1/2O₂(g) ===> CaO(s) + Ca(s) + O₂(g) + H₂(g) + H₂O(g)

Cancel like terms on both sides of equation to end up with the target equation...

Ca(OH)₂(s) ===> CaO(s) + H₂O(g) ∆H = -635 + 986 + (-242) = +109 kJ/mole