Question

Calculate the enthalpy for the following reaction: 4 NH3 (g) + 5 O2 (g) into 4 NO (g) + 6 H2O (g).
You may only use the following information:
N2 (g) + O2 (g) into 2 NO (g); Delta Hf = 180.6 kJ
N2 (g) + 3 H2 (g) into 2 NH3 (g); Delta Hf = -91.8 kJ
2 H2 (g) + O2 (g) into 2 H2O (g); Delta Hf = -483.7 kJ

My answer was -910.6 KJ but the real answer was -918.3 KJ. I need to know what I am doing wrong.

Answer 1

I didn't get either answer.
Look at your equation to make sure it adds to the desired equation in the problem. Here is what I did.
equation 1 x 2
equation 2 reversed and x 2.
equation 3 x 3.
Add it. First look at the equation.
(2N2) + 2O2 + 4NH3 + (6H2) +3O2 ==>
(2N2) + + (6H2) + 4NO + 6H2O

The 2N2 and 6H2 cancel since they are on opposite sides of the equation and we are left with
4NH3 + 5O2 ==>4NO + 6H2O which is exactly what we want.
Then (180.6*2)+(91.8*2)+(-483.7*3) = -906.3 kJ.
Check my work.

Answer 2

Determine the enthalpy of formation of the reaction: [4 NH₃ + 5 O₂ -> 4 NO + 6 H₂O]