Question

Lime is a term that includes calcium oxide (cao, also called quicklime) and calcium hydroxide (ca(oh)2 also called slaked lime). it is used in the steel industry to remove acidic impurities, in air-pollution control to remove acidic oxides such as so2, and in water treatment. slaked lime is produced by treating quicklime with water: cao(s) + h2o(l) → ca(oh)2(s)δh o = −65.2 kj/mol given the standard enthalpies of formation of cao and h2o are −635.6 kj/mol and −285.8 kj/mol, respectively, calculate the standard enthalpy of formation of ca(oh)2.

Answer 1

ΔH = ΔHfo Ca(OH)2 - (Δ Hfo CaO + ΔHfo H2O)
-65.2kJmol = ΔHfoCa(OH)2 - (-635.6kJ/mol + -285.8kJ/mol)
ΔHfo Ca(OH)2 = -986.6kJ/mol

Answer 2

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.