Question

Calculate the enthalpy change for the following chemical reaction:2H2(g) + O2(g) -> 2H2O(l) given the following bond enthalpy values:H-H bond energy = 436 kJ/mol O=O bond energy = 498 kJ/mol H-O bond energy = 464 kJ/mol

Answer 1

To calculate the enthalpy change for the reaction, we need to find the difference between the bond energies of the reactants and products.Reactants:2 moles of H-H bonds  2H2  = 2 * 436 kJ/mol = 872 kJ/mol1 mole of O=O bonds  O2  = 498 kJ/molTotal bond energy of reactants = 872 kJ/mol + 498 kJ/mol = 1370 kJ/molProducts:4 moles of H-O bonds  2H2O  = 4 * 464 kJ/mol = 1856 kJ/molTotal bond energy of products = 1856 kJ/molEnthalpy change  H  = Total bond energy of products - Total bond energy of reactantsH = 1856 kJ/mol - 1370 kJ/mol = 486 kJ/molThe enthalpy change for the reaction is 486 kJ/mol.