The answer is in the photo
Answer: - 986.6 kj/mol
Explanation:
1) Equation given:
CaO(s) + H₂O (l) → Ca (OH)₂ (s) δh⁰ = −65.2 kj/mol
2) Standard enthalpies of formation given:
CaO, δhf⁰ = −635.6 kj/mol
H₂O, δhf⁰ = −285.8 kj/mol
3) Calculate the standard enthalpy of formation of Ca(OH)₂.
δh⁰ = ∑δh⁰f of products - ∑ δh⁰f of reactants
Using the mole coefficients of the balanced chemical equation:
δh⁰ = δh⁰f Ca(OH)₂ - δh⁰f CaO - δh⁰f H₂O
⇒ δh⁰f Ca(OH)₂ = δh⁰ + δh⁰f CaO + δh⁰f H₂O
⇒ δh⁰f Ca(OH)₂ = - 65.2 kj/mol − 635.6 kj/mol) − 285.8 kj/mol) = - 986.6 kj/mol.
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