calculate molar enthalpy

Question

Using the equations

2 H₂ (g) + O₂ (g) → 2 H₂O (l) ∆H° = -572 kJ/mol

Ca (s) + ½ O₂ (g) → CaO (s) ∆H° = -635 kJ/mol

CaO (s) + H₂O (l) → Ca(OH)₂ (s) ∆H° = -64 kJ/mol

Determine the molar enthalpy (in kJ/mol) for the reaction Ca (s) + 2 H₂O (l) → Ca(OH)₂ (s) + H₂ (g).

Answer 1

This is an example of Hess' Law:

Ca(s) + 2H₂O(l) ==> Ca(OH)₂(s) + H₂(g) ... TARGET EQUATION

Given:

eq.1: 2H₂ (g) + O₂ (g) → 2H₂O (l) ... ∆H° = -572 kJ

eq. 2: Ca (s) + ½O₂ (g) → CaO (s ...) ∆H° = -635 kJ

eq. 3: CaO (s) + H₂O (l) → Ca(OH)₂ (s) ... ∆H° = -64 kJ

Copy eq.2: Ca (s) + ½O₂ (g) → CaO (s ) ... ∆H° = -635 kJ

Copy eq.3: CaO (s) + H₂O (l) → Ca(OH)₂ (s) ... ∆H° = -64 kJ

Reverse eq.1 and divide by 2: H₂O(l) ==> H₂(g) + 1/2 O₂(g) ... ∆Hº = +286 kJ

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Ca (s) + ½O₂ (g) + CaO (s) + H₂O (l) + H₂O(l) ==> CaO (s) + Ca(OH)₂ (s) + H₂(g) + 1/2 O₂(g)

Ca(s) + 2H₂O(l) ==> Ca(OH)₂(s) + H₂(g) ... TARGET EQUATION

∆H = -635 + -64 + 286 = -413 kJ