Answer:
-162,5 kJ/mol
Step-by-step explanation:
Cl(g) + 2O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ/mol (as we used the reaction in the opposite direction, it will turn the enthalpy from exothermic to endothermic)
2O3(g) --> 3O2(g) ΔH = -285.3 kJ/mol
Cl(g) + O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ
+ 2O3 (g) --> 3O2(g) ΔH = - 285.3 kJ
O3(g) + Cl(g) --> ClO(g) + 2O2(g) ΔH = 122.8 + (-285.3) = -162,5 kJ