Question

What is the enthalpy of formation of ammonia gas (NH3) at 298 K and 1 atm pressure using the given bond energies: N≡N = 418 kJ/mol, H-H = 436 kJ/mol, and N-H = 391 kJ/mol?

Answer 1

To calculate the enthalpy of formation of ammonia gas  NH3  using the given bond energies, we need to consider the balanced chemical equation for the formation of ammonia:N2 g  + 3H2 g   2NH3 g Now, we will calculate the total bond energy of reactants and products:Reactants:1 NN bond: 1  418 kJ/mol = 418 kJ/mol3 H-H bonds: 3  436 kJ/mol = 1308 kJ/molTotal bond energy of reactants: 418 kJ/mol + 1308 kJ/mol = 1726 kJ/molProducts:6 N-H bonds: 6  391 kJ/mol = 2346 kJ/molTotal bond energy of products: 2346 kJ/molNow, we can calculate the enthalpy of formation  Hf  using the following formula:Hf = Total bond energy of products - Total bond energy of reactantsHf = 2346 kJ/mol - 1726 kJ/mol = -92 kJ/mol  for 2 moles of NH3 Since we need the enthalpy of formation for 1 mole of NH3, we will divide the result by 2:Hf = -92 kJ/mol  2 = -46 kJ/molSo, the enthalpy of formation of ammonia gas  NH3  at 298 K and 1 atm pressure is -46 kJ/mol.