Question

Rewrite equation with the delta H value included in either reactants or products and identify the reaction as either endo- or exo- thermic: 2 NO₂ (g) --> 2 NO (g) + O₂ (g)   H= 114.2 kJ/mol

Answer 1

The reaction you have is: 2 NO2 --------> 2 NO + O2

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

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

NO(g)------------> 1/2 N2(g) + 1/2O2(g), delta H = -90.3 kJ (should be multiplied by 2 and inverted)

N2(g) + O2(g)-------------> 2NO(g)............deltaH = +180.6 kJ

Now you have:
2NO2(g) ----------> N2(g) + 2 O2(g).........deltaH = -66.4 kJ
N2(g) + O2(g)-------------> 2NO(g)......... .deltaH = +180.6 kJ
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canceling both N2 and 2 O2 with O2, you will find the given reaction:
2 NO2----------->2 NO + O2 ........deltaH = -66.4 + 180.6 = +114.2 kJ