Answer:
ΔH = -162.5 kJ.
Explanation:
Hello.
In this case, we first rearrange the reactions:
ClO(g) + O₃(g) ⇒ Cl(g) + 2O₂(g); ΔH =-122.8 kJ
2O₃(g) ⇒ 3O₂(g); ΔH=-285.3 kJ
O₃(g) + Cl(g) ⇒ ClO(g) + O₂(g); ΔH= ?
Thus, we are going to use the Hess law, as an strategy to rearrange the known chemical reactions and thereby compute the enthalpy of reaction of the unknown one.
1. The first reaction must be inverted in order to obtain chlorine as a reactant in the third one, therefore, the enthalpy of reaction becomes positive:
Cl(g) + 2O₂(g) ⇒ ClO(g) + O₃(g); ΔH = 122.8 kJ
2. Second reaction remains the same:
2O₃(g) ⇒ 3O₂(g); ΔH=-285.3 kJ
Then, we add them to obtain:
Cl(g) + 2O₂(g) + 2O₃(g) ⇒ ClO(g) + O₃(g) + 3O₂(g)
Whereas we can subtract both oxygen and ozone to obtain the third one:
O₃(g) + Cl(g) ⇒ ClO(g) + O₂(g)
Therefore, the enthalpy of reaction turns out:
ΔH = 122.8 kJ + (-285.3 kJ
)
ΔH = -162.5 kJ.
Best regards.
