Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: 4 NO(g) + 2 O2(g) → 4 NO2(g). Given: N2(g) + O2(g) → 2 NO(g), ΔH°rxn = +183 kJ. What is the ΔH°rxn for the reaction 4 NO(g) + 2 O2(g) → 4 NO2(g)?

• A. -366 kJ
• B. +114 kJ
• C. -114 kJ
• D. +366 kJ

Answer: (C)

Verified step by step guidance

1. Understand that the problem involves using Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction.
2. Identify the given reaction and the target reaction. The given reaction is: \( \text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{NO}(g) \) with \( \Delta H^\circ_{rxn} = +183 \text{kJ} \). The target reaction is: \( 4 \text{NO}(g) + 2 \text{O}_2(g) \rightarrow 4 \text{NO}_2(g) \).
3. Recognize that the target reaction can be achieved by manipulating the given reaction. Specifically, you need to double the given reaction to match the stoichiometry of the target reaction: \( 2 \times (\text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{NO}(g)) \). This results in \( 2 \text{N}_2(g) + 2 \text{O}_2(g) \rightarrow 4 \text{NO}(g) \) with \( \Delta H^\circ = 2 \times 183 \text{kJ} \).
4. Consider the enthalpy change for the formation of \( \text{NO}_2 \) from \( \text{NO} \) and \( \text{O}_2 \). The reaction \( 4 \text{NO}(g) + 2 \text{O}_2(g) \rightarrow 4 \text{NO}_2(g) \) can be seen as the reverse of the formation of \( \text{NO} \) from \( \text{N}_2 \) and \( \text{O}_2 \).
5. Apply Hess's Law to find the enthalpy change for the target reaction by summing the enthalpy changes of the steps involved. Since the target reaction is the reverse of the doubled given reaction, the enthalpy change will be the negative of the doubled enthalpy change: \( \Delta H^\circ_{rxn} = -2 \times 183 \text{kJ} \).