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.

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

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.

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