Final answer:
To determine ∆H°rxn for 2 NO(g) + O2(g) → 2 NO2(g), we use Hess's law and the enthalpy changes of provided formation reactions. By subtracting the enthalpy change for the formation of 2 NO(g) from the doubled enthalpy change for the formation of 2 NO2(g), we find that the enthalpy change for the target reaction is -117 kJ.
Step-by-step explanation:
To calculate the ∆H°rxn for the reaction 2 NO(g) + O2(g) → 2 NO2(g), we can apply Hess's law. This law states that the total enthalpy change for a reaction is the same, no matter how the reaction occurs, as long as the initial and final conditions are the same. In this case, we are given the enthalpy changes for the formation of NO and NO2 from elemental nitrogen and oxygen.
The molecules or compounds with a standard enthalpy of formation (∆H°f) of zero at 298 K are N2(g) and O2(g), as they are elemental forms in their most stable states under standard conditions.
Step 1: The given reactions are as follows:
-
- N2(g) + O2(g) → 2 NO(g) ∆H°rxn = +183 kJ
-
- 1/2 N2(g) + O2(g) → NO2(g) ∆H°rxn = +33 kJ
To find the enthalpy change for the formation of 2 NO2(g), we need to double the second reaction:
-
- N2(g) + 2 O2(g) → 2 NO2(g) ∆H° = +66 kJ
The enthalpy change for the target reaction can now be calculated by subtracting the enthalpy change of the first reaction (
formation of 2 NO(g)
) from the enthalpy change of the doubled second reaction (
formation of 2 NO2(g)
).
Step 2: Perform the subtraction:
∆H°rxn (2 NO2(g)) - ∆H°rxn (2 NO(g)) = +66 kJ - (+183 kJ) = -117 kJ
Therefore, the enthalpy change for the reaction 2 NO(g) + O2(g) → 2 NO2(g) is -117 kJ.