Question

Calculate the standard enthalpy change (ΔH°) for the reaction:2H2(g) + O2(g) → 2H2O(g)given the following bond energies:H-H = 436 kJ/mol, O=O = 498 kJ/mol, H-O = 463 kJ/mol

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we need to find the difference between the bond energies of the reactants and the products.Reactants:2 moles of H-H bonds: 2  436 kJ/mol = 872 kJ/mol1 mole of O=O bond: 1  498 kJ/mol = 498 kJ/molTotal bond energy of reactants: 872 + 498 = 1370 kJ/molProducts:4 moles of H-O bonds  since there are 2 H-O bonds in each H2O molecule, and there are 2 H2O molecules : 4  463 kJ/mol = 1852 kJ/molTotal bond energy of products: 1852 kJ/molNow, we can calculate the standard enthalpy change  H  for the reaction:H = Bond energy of products - Bond energy of reactantsH = 1852 kJ/mol - 1370 kJ/molH = 482 kJ/molThe standard enthalpy change  H  for the reaction is 482 kJ/mol.