Question

1) What is the enthalpy of reaction, ΔH_rxn for the reaction of nitrogen gas with oxygen gas to produce NO₂(g), based on the following information? These reactions are not at standard state or at 298 K.

N₂(g) + O₂(g) → 2 NO(g); ΔH = 83.1 kJ

2 NO₂(g) → 2 NO(g) + O₂(g); ΔH = -141.7 kJ

Report your answer in kJ to 1 decimal place.

2)What is the value of ΔH_rxn if the reaction below is reversed? In other words, PCl₅ becomes the reactant. Report your answer in kJ to 1 decimal place.
PCl₃(g) + Cl₂(g) → PCl₅(g); ΔH_rxn = 414.7

Answer 1

1.

N₂(g) + O₂(g) → 2 NO(g); ΔH = 83.1 kJ           ------------ (1)

2 NO₂(g) → 2 NO(g) + O₂(g); ΔH = -141.7 kJ   ------------ (2)

If we reverse the reaction (2)

2 NO(g) + O₂(g) → 2 NO₂(g)  ΔH = 141.7 kJ        ------------ (3)

(1) + (3)

N₂(g) + O₂(g) → 2 NO(g); ΔH = 83.1 kJ           ------------ (1)

2 NO(g) + O₂(g) → 2 NO₂(g)  ΔH = 141.7 kJ        ------------ (3)

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N₂(g) + 2 O₂(g) → 2 NO₂(g)          ΔH = 83.1 kJ + 141.7 kJ = 224.8 J

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2.

PCl₃(g) + Cl₂(g) → PCl₅(g); ΔH_rxn = 414.7kJ

If we reverse the reaction, only the sign of ΔH_rxn will be changed

PCl₅(g) → PCl₃(g) + Cl₂(g) ΔH_rxn = - 414.7 kJ