What is the molar enthalpy of formation for ammonia?

Question

Hi I'm having issues relating hess's law to this question:

The given chemical equation represent the combustion of ammonia and the combustion of hydrogen 1. 4NH3+3O2-->6H2O + 2N2. ( deltaH1 = -1516 kJ) 2. 2H2+O2 --> H2O ( deltaH2 = -572 kJ) What is the molar enthalpy of formation for ammonia? A. -100kJ B.-50kJ C. 50KJ D.100KJ 

I'm not sure how I would apply the product - reactant equation here or rearrange the equation to get my answer since I don't have an initial reaction to compare it to. How would I go about answering this question?

Thank you!

Answer 1

Eq. 1: 4NH₃ + 3O₂ ==> 6H₂O + 2N₂ ∆H = =1516 kJ

Eq. 2: 2H₂ + O₂ ==> 2H₂O ∆H = -572 kJ (note, the equation as you wrote it in the question is wrong)

Reverse Eq.1 to get 6H₂O + 2N₂ ==> 4NH₃ + 3O₂ ∆H = +1516 kJ

Multiply Eq. 2 by 3 to get 6H₂ + 3O₂ ==> 6H₂O ∆H = 3 x -572 kJ = -1716 kJ

Add the 2 equations to get 6H₂O + 2N₂ + 6H₂ + 3O₂ ==> 4NH₃ + 3O₂ + 6H₂O ∆H =-200 kJ

Cancel to get 2N₂ + 6H₂ ==> 4NH₃ ∆H = -200 kJ

This is the ∆H for formation of FOUR moles NH₃. Thus, molar ∆H = -200/4 = -50 kJ/mole (Ans. B)