Question

Calculate the enthalpy for: 4NH3 + 5O2 → 4NO + 6H2O

Answer 1

4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(g) ∆H = ?

Sum the ∆H values for the products:

4NO = 4 mol x 90.3 kJ/mol = 361.2 kJ

6H2O = 6 mol x -241.8 = -1450.8 kJ

∑ = -1089.6

Sum the ∆H values for the reactants:

4NH3: 4mol x -46.1 kJ/mol = -184.4 kJ

6O2: 6 mol x 0 = 0 kJ (∆H for O2 is zero)

∑ = -184.4 kJ

∆Hreaction = ∑∆Hproducts - ∑∆Hreactants = -1089.6 kJ - (-184.4 kJ)

∆Hreaction = -905.2 kJ