Question

The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning the formation and decomposition of ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons (CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas.

Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone.

(1)ClO(g)+O3(g)⟶Cl(g)+2O2(g)Δ𝐻∘rxn=−122.8 kJ(2)2O3(g)⟶3O2(g)Δ𝐻∘rxn=−285.3 kJ(3)O3(g)+Cl(g)⟶ClO(g)+O2(g) Δ𝐻∘rxn= ?

Answer 1

1)ClO(g)+O3(g)⟶Cl(g)+2O2(g)Δ𝐻∘rxn=−122.8 kJ(2)2O3(g)⟶3O2(g)Δ𝐻∘rxn=−285.3 kJ(3)O3(g)+Cl(g)⟶ClO(g)+O2(g)
You have
1) ClO(g) + O3(g)⟶Cl(g) + 2O2(g) Δ𝐻∘rxn = −122.8 kJ
(2) 2O3(g)⟶3O2(g) Δ𝐻∘rxn = −285.3 kJ and you want this one:
(3) O3(g)+Cl(g)⟶ ClO(g) + O2(g) dH - ?
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Reverse equation 1 and reverse delta H like this
1) reversed is Cl(g) + 2O2(g) --> ClO(g) + O3(g) and add to 2 like this:
2)................... 2O3(g)⟶3O2(g)
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1 rev + 2 is.........Cl + 2O2 + 2O3 ==> ClO + O3 + 3O2
Now cancel the common items to both sides to get this:
Cl + O3 ==> ClO + O2 Now add the dH values together.
equation reversed is +122.8 kJ and equation 2 is -285.3 kJ =
+122.8 kJ + (-285.3) = ? kJ.