Hess's Law Paper

Question

Calculate the enthalpy change in the reaction below with the provided data

4 NH₃ (g) + 5 O₂ (g) ---> 4 NO (g) + 6 H₂O (l)

ΔH_f° NH₃ (g) = -45.5 kJ mol^-1

ΔH_f° H₂O (l) = -286 kJ mol^-1

ΔH_f° NO (g) = 91 kJ mol^-1

Answer 1

∆Hrxn = ∑∆H_fº products - ∑∆H_fºreactants

∆H_fº products = 4 mols NO x 91 kJ / mol NO + 6 mols H2O x -286 kJ/mol = 364 kJ + (-1716 kJ) = -1352 kJ

∆H_fº reactants = 4 mols NH3 x -45.5 kJ/mol + 5 mols O2 x 0 kJ/mol = - 182 kJ (note ∆H_fº for O2 is zero

∑∆H_fº products - ∑∆H_fºreactants = -1352 - (-182) = -1170 kJ