Answer:
[tex]\Delta H_{rxn3}=-162.5 kJ/mol[/tex]
Explanation:
The reaction we need to calculate:
[tex]O_3 (g) + Cl (g) \longrightarrow ClO (g) + O_2 (g)[/tex]
1) [tex]O_3 (g) + ClO (g) \longrightarrow Cl (g) +2 O_2 (g)[/tex]
[tex]\Delta H_{rxn}=-122.8 kJ/mol[/tex]
We need the ClO in the products side, so we use the inverse of this reaction:
[tex]Cl (g) +2 O_2 (g) \longrightarrow O_3 (g) + ClO (g) [/tex]
[tex]\Delta H_{rxn1}=122.8 kJ/mol[/tex]
2) [tex]2 O_3 (g) \longrightarrow 3 O_2 (g)[/tex]
[tex]\Delta H_{rxn2}=-285.3 kJ/mol[/tex]
Now we need to combine this two:
[tex]Cl (g) +2 O_2 (g) + 2 O_3 (g)\longrightarrow O_3 (g) + ClO (g) + 3 O_2 (g) [/tex]
[tex]Cl (g) + O_3 (g)\longrightarrow ClO (g) + O_2 (g) [/tex]
The enthalpy of reaction:
[tex]\Delta H_{rxn3}=\Delta H_{rxn1}+ \Delta H_{rxn2}=122.8 kJ/mol-285.3 kJ/mol[/tex]
[tex]\Delta H_{rxn3}=-162.5 kJ/mol[/tex]