When NH3 is treated with oxygen gas, the products obtained are N2(g) and H2O(l). If standard enthalpies of formation at 298 K for NH3(g) and H2O(l) are –46.00 kJ/mol and –286.0 kJ/mol respectively, calculate the enthalpy change of the reaction.

Answer 1

4NH3 + 3O2 ==> 2N2 + 6H2O

dHrxn = (n*dHf products) - (n*dHf reactants)