Question

What is the enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(l) given the following bond enthalpy values: H-H: 436 kJ/mol, O=O: 495 kJ/mol, H-O: 464 kJ/mol (assume that all gases are at standard conditions)?

Answer 1

To calculate the enthalpy change for the reaction, we need to find the difference between the bond enthalpies 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  495 kJ/mol = 495 kJ/molTotal bond enthalpy of reactants: 872 + 495 = 1367 kJ/molProducts:4 moles of H-O bonds  since there are 2 H-O bonds in each H2O molecule and we have 2 H2O molecules : 4  464 kJ/mol = 1856 kJ/molTotal bond enthalpy of products: 1856 kJ/molEnthalpy change  H  = Bond enthalpy of products - Bond enthalpy of reactantsH = 1856 kJ/mol - 1367 kJ/mol = 489 kJ/molThe enthalpy change for the reaction is 489 kJ/mol.