Answer:
Δ[tex]rH=-869kJ/mol[/tex]
Explanation:
Hello, for this chemical reaction, the overall enthalpy of reaction is defined by:
Δ[tex]rH=3H_{f_{H_2O}}-H_{f_{O_3}}[/tex]
Since the enthalpy of formation of water is 0 as long as it is a pure element. Now, applying the formula with the proper values (extracted from the NIST database), we get:
Δ[tex]rH=3*(-242kJ/mol)-(143kJ/mol)[/tex]
Δ[tex]rH=-869kJ/mol[/tex]
Best regards.
