Respuesta :
Answer:
Enthalpy of formation of ClO (ΔHf⁰) = 96.5 Kj/mole
Explanation:
Given that
chlorine-oxygen bond has an enthalpy of 243 kJ/mole , an oxygen-oxygen bond has an enthalpy of 498 kJ/mole.
[tex]\frac{1}{2}[/tex][tex]Cl_{2}[/tex](g) + [tex]\frac{1}{2} O_{2}[/tex](g) → ClO(g)
ΔHr = ∑ Bond energies
= [tex]\frac{1}{2}[/tex] x (Bond enthalpy of oxygen) - (Bond enthalpy of Cl - O bond)
= [tex]\frac{1}{2}[/tex] x 498 kj/mole - 243 kj/mole
= + 6 kj/mole
ClO(g) + [tex]\frac{1}{2} O_{2}[/tex](g) → ClO₂ (g)
ΔHr = Enthalpy of formation of ClO₂ - Enthalpy of formation of ClO
⇒Enthalpy of formation of ClO = Enthalpy of formation of ClO₂ - ΔHr
⇒Enthalpy of formation of ClO = 102.5 kj/mole - 6 kj/mole
⇒Enthalpy of formation of ClO = 96.5 kj/mole
