Final answer:
To determine ΔH°rxn for the given reaction, we can use the standard reaction enthalpies provided. The ΔH°rxn for the given reaction is +798 kJ/mol.
Step-by-step explanation:
To determine ΔH°rxn for the given reaction, we can use the standard reaction enthalpies provided. The reaction equation is: 4 NO(g) + 2 O₂(g) → 4 NO₂(g). First, we need to find the enthalpy change for the reaction: 4 NO(g) + O₂(g) → 4 NO₂(g). The given enthalpy change for the reaction N₂(g) + O₂(g) → 2 NO(g) is +183 kJ/mol.
Since we need 4 moles of NO(g), we can multiply the enthalpy change by 4: 4 * (+183 kJ/mol) = +732 kJ/mol. Next, we need to find the enthalpy change for the reaction 1/2 N₂(g) + O₂(g) → NO₂(g). The given enthalpy change for this reaction is +33 kJ/mol.
Since we need 2 moles of NO₂(g) in the balanced equation, we can multiply the enthalpy change by 2: 2 * (+33 kJ/mol) = +66 kJ/mol. Finally, we can add up the two enthalpy changes to find the overall enthalpy change for the reaction: +732 kJ/mol + +66 kJ/mol = +798 kJ/mol. Therefore, the ΔH°rxn for the given reaction is +798 kJ/mol.