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

To determine the molar enthalpy for the reaction Ca (s) + 2 H₂O (l) → Ca(OH)₂ (s) + H₂ (g), we can use Hess's law and combine the given reactions. The molar enthalpy is -1271 kJ/mol.

Step-by-step explanation:

To determine the molar enthalpy for the reaction Ca (s) + 2 H₂O (l) → Ca(OH)₂ (s) + H₂ (g), we can use Hess's law. For this, we need to find a combination of reactions that will add up to the desired reaction. Here is one possible combination:

1. Ca (s) + ½ O₂ (g) → CaO (s) ∆H° = -635 kJ/mol
2. CaO (s) + H₂O (l) → Ca(OH)₂ (s) ∆H° = -64 kJ/mol
3. 2 H₂ (g) + O₂ (g) → 2 H₂O (l) ∆H° = -572 kJ/mol

Now, we can add these reactions together to obtain the desired reaction:

• Ca (s) + ½ O₂ (g) + 2 H₂ (g) + O₂ (g) → Ca(OH)₂ (s) + H₂ (g)
• ∆H° = -635 kJ/mol + -572 kJ/mol + -64 kJ/mol = -1271 kJ/mol

Therefore, the molar enthalpy for the reaction Ca (s) + 2 H₂O (l) → Ca(OH)₂ (s) + H₂ (g) is -1271 kJ/mol.