Question

Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ 12N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ

Answer 1

i guess its 1/2N2 not 12N2 because the reaction isn't possible that way.so, i would answer taking 1/2N2.

The Hess law is used to discover the reaction's entalphy and doing it, you must cancel the same substances, if they are in differents sides, to make the given reaction. Remember that if you multiply or divide the reaction, the same happens with the entalphy and when you invert the reaction, the entalphy's sign must be also invent the reaction.

The reaction you have is: 2 NO + O2 -> 2 NO2
It means that 2 NO and O2 must be on the left side and 2 NO2, on the right one, so you will look for the entalphies that were given and invert and/or multiply to copy this order, look:

1/2 N2(g) + O2(g) --> NO2(g), deltaH = 33.2 kJ (should be multiplied by 2)

N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ

1/2 N2(g) + 1/2O2(g) --> NO(g), delta H = 90.2 kJ (should be multiplied by 2 and inverted)

2 NO -> N2 + O2..............deltaH = -180,4 kJ

Now you have:
N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ
2 NO -> N2 + O2..............deltaH = -180,4 kJ
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canceling both N2 and 2 O2 with O2, you will find the given reaction:
2 NO + O2 -> 2 NO2........deltaH = 66,4 - 180,4 = -114,0 kJ

i hope it helps!!