# Chemistry

posted by
**K**
.

The following standard free energies are given for 25 degrees celcius.

1) N2(g) + 3H2(g)-> 2NH3(g) delta g = -33.0 kJ

2) 4NH3(g) + 5O2(g)-> 4NO(g) + 6H20() delta g = -1010.5 kJ

3) N2(g) + o2(g)-> 2NO(g) delta g = 173.1kJ

4) N2(g) + 2O2(g) -> 2NO2(g) delta g = 102.6 kJ

5) 2N2 + O2(g)-> 2N2O(g) delta g = 208.4kJ

Combine the reactions to obtain delta g for the following:

N2O(g) + 3/2O2(g) -> 2NO2(g)

The answer is -1.6kJ but I don't know how to solve it