Final answer:
The standard reaction enthalpy (ΔH°rxn) for the reaction 4 NO(g) + 2 O2(g) → 4 NO2(g) is calculated to be -234 kJ using Hess's Law and the given standard reaction enthalpies. The correct answer is A.-234 kj.
Step-by-step explanation:
To calculate the standard reaction enthalpy (ΔH°rxn) for the reaction 4 NO(g) + 2 O2(g) → 4 NO2(g), we need to use Hess's Law and manipulate the given reactions to match the target equation while taking into account their respective reaction enthalpies.
The given reactions are:
- N2(g) + O2(g) → 2 NO(g) with ΔH°rxn = +183 kJ
- 1/2 N2(g) + O2(g) → NO2(g) with ΔH°rxn = +33 kJ
To form 4 NO2 according to the target equation, we need to multiply the second given reaction by 4:
- 2 N2(g) + 4 O2(g) → 4 NO2(g) with ΔH°rxn = 4 × 33 kJ = +132 kJ
Next, we will consider the first given reaction. To form 4 NO(g), we need to multiply the reaction by 2:
- 2 N2(g) + 2 O2(g) → 4 NO(g) with ΔH°rxn = 2 × 183 kJ = +366 kJ
To find the overall ΔH°rxn, we need to consider the enthalpy change for the formation of 4 NO(g) from the first reaction and reverse this reaction's enthalpy change because we are consuming this reactant in the target reaction:
The enthalpy change for the target reaction is therefore:
ΔH°rxn = ΔH°formation of 4 NO2 - ΔH°formation of 4 NO
ΔH°rxn = +132 kJ - (+366 kJ)
ΔH°rxn = -234 kJ
Therefore, the standard reaction enthalpy (ΔH°rxn) for the reaction is -234 kJ, which corresponds to option A.