Question

What is the change in enthalpy for the following reaction at room temperature: 2H2(g) + O2(g) -> 2H2O(l) given the bond dissociation energies: H-H: 436 kJ/mol O=O: 498 kJ/mol H-O: 463 kJ/mol

Answer 1

To calculate the change in enthalpy  H  for the reaction, we can use the bond dissociation energies given. The formula for this is:H =   bond energies of reactants  -   bond energies of products For the reactants, we have 2 moles of H-H bonds and 1 mole of O=O bond:2 moles H-H: 2  436 kJ/mol = 872 kJ1 mole O=O: 1  498 kJ/mol = 498 kJFor the products, we have 4 moles of H-O bonds in 2 moles of H2O:4 moles H-O: 4  463 kJ/mol = 1852 kJNow, we can calculate the change in enthalpy:H =  872 kJ + 498 kJ  - 1852 kJH = 1370 kJ - 1852 kJH = -482 kJThe change in enthalpy for the reaction is -482 kJ.