1. Calculate AH for the reaction 2NO2(g) → N2O.(g)
delta H = ?
Given:
N2(g) + 202(g) → 2NO2(g) delta H = 67.7 kJ
N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ

- 58 kJ

Explanation:

The equation of the reaction is given as;

2NO2(g) → N2O4(g)

N2(g) + 202(g) → 2NO2(g) delta H = 67.7 kJ

Since NO2 is the reactant in the equation, we have to reverse this reaction. We now have;

2NO2(g) → N2(g) + 202(g)     delta H = - 67.7 kJ

N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ

Since NO4 is the product in the equation, we don't have to do anything to this reaction.

Adding the two reactions;

2NO2(g) → N2(g) + 202(g)     delta H = - 67.7 kJ

N2(g) + 202(g) → N2O4(g)     delta H = 9.7 kJ

N2(g) and 2O2 (g) cancels out

2NO2(g) → N2O4(g)       delta H = (-67.7 + 9.7) = - 58 kJ

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