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You have the following half-equation :

N2(g) + 6 H(+) + 6e- ---> 2 NH3(g) E=+0.057 (1)

O2(g) + 4H+ + 4e- ---> 2 H2O E=+1.229 (2)

If you do 3 x (2) – 2 x (1)

3 O2(g) + 4 NH3(g) ---> 6 H2O (l) + 2 N2(g)

Nh3 O2 N2 H2o

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RE:

NH3(g) + O2(g) --> N2(g) + H2O(l)?

Balance the following chemical reaction. NH3(g) + O2(g) --> N2(g) + H2O(l)

What is the enthalpy change for the reaction? Include units.

im confused.

can someone help me pleasee?

2NH3 + 3/2 O2---> N2 + 3H2O

2NH3(g) + 3O2 --> N2 + 6H2O

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