Question

Calculate the enthalpy change (ΔH) for the following reaction: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)using the following bond enthalpies (in kJ/mol):NH3(g): N-H = 391, N ≡ N = 941O2(g): O=O = 498NO(g): N=O = 628H2O(g): O-H = 463

Answer 1

To calculate the enthalpy change  H  for the given reaction, we need to find the difference between the bond enthalpies of the reactants and products.Reactants:4NH3 g : 4 x  3 N-H bonds  = 4 x 3 x 391 kJ/mol = 4692 kJ/mol5O2 g : 5 x  1 O=O bond  = 5 x 498 kJ/mol = 2490 kJ/molTotal bond enthalpy of reactants = 4692 + 2490 = 7182 kJ/molProducts:4NO g : 4 x  1 N=O bond  = 4 x 628 kJ/mol = 2512 kJ/mol6H2O g : 6 x  2 O-H bonds  = 6 x 2 x 463 kJ/mol = 5556 kJ/molTotal bond enthalpy of products = 2512 + 5556 = 8068 kJ/molNow, we can calculate the enthalpy change  H  for the reaction:H = Bond enthalpy of products - Bond enthalpy of reactantsH = 8068 kJ/mol - 7182 kJ/mol = 886 kJ/molThe enthalpy change  H  for the given reaction is 886 kJ/mol.