Determine the molar enthalpy

Question

Using the equations N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.7 kJ/mol

Determine the molar enthalpy (in kJ/mol) for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g).

Answer 1

See Hess's Law for more information

Given:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) ... TARGET EQUATION

(1). N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol

(2). N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol

(3). 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.7 kJ/mol

Procedure:

reverse eq.(2) and multiply by 2: 4NH₃ => 2N₂ + 6H₂ ... ∆H =+91.8 x 2 = +183.6 kJ/mol

copy eq.(1) and multiply by 2: 2N₂ + 2O₂ => 4NO ... ∆H = 2x180.6 = +361.2 kJ/mol

copy eq.(3) and multiply by 3: 6H₂ + 3O₂ => 6H₂O ... ∆H = 3x-483.7 = -1451.1 kJ/mol

add all three together:

4NH₃ + 2N₂ + 2O₂ + 6H₂ + 3O₂ ==> 2N₂ + 6H₂ + 4NO + 6H₂O

combine and/or cancel like terms to get the target equation:

4NH₃ + 5O₂ ==> 4NO + 6H₂O ... TARGET EQUATION

∆Hrxn = 183.6 + 361.2 + (-1451.1) = -906.3 kJ/mol