Answer to Question #100953 in General Chemistry for Asif Sultan

Question

For the reaction:
3N2O(g) + 2NH3(g) ---------- 4N2(g) + 3H2O(g) deltaH=-879.6kj/mol
If enthapy of formation of NH3(g) is -45.9kj/mol & for H2O(g) is -241.8kj/mol, then enthaply of formation of N2O(g) will be:

Answer 1

deltaH(of reaction) = 4*deltaH(formation)N2 + 3*deltaH(formation)H2O - 2*deltaH(formation)NH3 - 3*deltaH(formation) N2O;

deltaH(formation)N2 = 0;

deltaH(formation) N2O = (3*deltaH(formation)H2O - 2*deltaH(formation)NH3 - deltaH(of reaction))/3 = -274KJ/mol