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:

Explanation:

Equation of reactions:

1. N2 (g) + 2O2(g) --> 2NO2 (g)

ΔH = 66.4 kJ

2. 2NO (g) + O2 (g) --> 2NO2 (g)

ΔH = -114.2 kJ

Final equation

N2 (g) + O2 (g) --> 2NO (g)

Inverting equation 2, note that the ΔH will now be positive.

3. 2NO2 (g) --> 2NO (g) + O2 (g)

ΔH = 114.2 kJ

Adding equation 1 and 3 together as well as their enthalpies,

N2 (g) + 2O2(g) --> 2NO2 (g)

2NO2 (g) --> 2NO (g) + O2 (g)

---------------------------------------

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

ΔH = 114.2 kJ + 66.4 kJ

= 180.6 kJ

N2 (g) + O2 (g) --> 2NO (g)

ΔH = 180.6 kJ

Answer 2

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