Question

N2 (g) + 2O2(g) = 2NO2 (g) ΔH = 66.4 kJ 2NO (g) + O2 (g) = 2NO2 (g) ΔH = -114.2 kJ the enthalpy of the reaction of the nitrogen to produce nitric oxide N2 (g) + O2 (g) = 2NO (g) is __________ kJ.

Answer 1

Answer:

ΔH  = 180.6 kJ

Explanation:

Given that:

N2 (g) + 2O2(g) = 2NO2 (g)           ΔH = 66.4 kJ

2NO (g) + O2 (g) = 2NO2 (g)         ΔH = -114.2 kJ                    

N2 (g) + O2 (g) = 2NO (g)              ΔH  = ????

The subtraction of both equations would yield the unknown ΔH , therefore:

ΔH = 66.4 - ( - 114.2 kJ)

ΔH  = 180.6 kJ

Answer 2

The enthalpy of the reaction of the nitrogen to produce nitric oxide is;

ΔH_rxn = 180.2 KJ

We are given the reactions;

N₂ (g) + 2O₂ (g) = 2NO₂ (g)     ΔH = 66.4 kJ

2NO (g) + O₂ (g) = 2NO₂ (g)    ΔH = -114.2 kJ

According to hess's law, we need to subtract equation 2 from eq 1 to get;

N₂ (g) + 2O₂ (g) - 2NO (g) - O₂ (g) = 2NO₂ (g) - 2NO₂ (g)

N₂ (g) + O₂ (g) - 2NO (g) = 0

N₂ (g) + O₂ (g) = 2NO (g)

Since we subtracted equation 2 from equation 1, then the enthalpy of the reaction of the nitrogen to produce nitric oxide is;

ΔH_rxn = 66.4 - (-114.2)

ΔH_rxn = 66.4 + 114.2

ΔH_rxn = 180.2 KJ

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