Find the enthapy change for the reaction in kJ . 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g) ΔH= ? Given : (NH3(g)) = -46.1, (NO(g)) = + 90.3 kJ/mol and (H2O(g)) = -241.8 kJ/mol.

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

∆H_reaction = ∑∆H_products - ∑∆H_reactants = -1089.6 kJ - (-184.4 kJ)

∆H_reaction = -905.2 kJ