Question

Calculate the standard enthalpy change (∆H°) for the reaction: 2NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)Given the following bond dissociation energies (in kJ/mol):N≡N = 941, N-H = 391, O=O = 495, O-H = 464, N-O = 201, H-O-H = 464, N-H + O=O = 632, N=O = 607, and H-N≡N = 945.

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we need to find the difference between the bond dissociation energies of the reactants and products.Reactants:2NH3 g : There are 2 N-H bonds in each NH3 molecule, so there are 4 N-H bonds in 2NH3. The bond dissociation energy for N-H is 391 kJ/mol, so the total energy for 2NH3 is 4 * 391 = 1564 kJ/mol.5O2 g : The bond dissociation energy for O=O is 495 kJ/mol, so the total energy for 5O2 is 5 * 495 = 2475 kJ/mol.Total energy for reactants = 1564 + 2475 = 4039 kJ/molProducts:4NO g : There are 1 N=O bond in each NO molecule, so there are 4 N=O bonds in 4NO. The bond dissociation energy for N=O is 607 kJ/mol, so the total energy for 4NO is 4 * 607 = 2428 kJ/mol.6H2O l : There are 2 O-H bonds in each H2O molecule, so there are 12 O-H bonds in 6H2O. The bond dissociation energy for O-H is 464 kJ/mol, so the total energy for 6H2O is 12 * 464 = 5568 kJ/mol.Total energy for products = 2428 + 5568 = 7996 kJ/molNow we can find the standard enthalpy change  H  for the reaction:H = Total energy for products - Total energy for reactantsH = 7996 - 4039H = 3957 kJ/mol