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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

User Kevin Qi
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4.9k points

1 Answer

2 votes

Answer:

- 58 kJ

Step-by-step 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

User Donald Miner
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6.2k points