Question

Calculate the enthalpy change for the reaction: 2H2O(l) → 2H2(g) + O2(g) given the following bond enthalpies: H-H (436 kJ/mol), O=O (498 kJ/mol), H-O (463 kJ/mol). Is the reaction exothermic or endothermic?

Answer 1

To calculate the enthalpy change for the reaction, we need to find the difference between the bond enthalpies of the reactants and products.Reactants: 2H2O l There are 2 H-O bonds in each H2O molecule, so there are a total of 4 H-O bonds in 2H2O.Total bond enthalpy of reactants = 4 * H-O bond enthalpy = 4 * 463 kJ/mol = 1852 kJ/molProducts: 2H2 g  + O2 g There are 1 H-H bond in each H2 molecule, so there are a total of 2 H-H bonds in 2H2.There is 1 O=O bond in O2 molecule.Total bond enthalpy of products =  2 * H-H bond enthalpy  + O=O bond enthalpy =  2 * 436 kJ/mol  + 498 kJ/mol = 872 kJ/mol + 498 kJ/mol = 1370 kJ/molEnthalpy change  H  = Total bond enthalpy of products - Total bond enthalpy of reactants = 1370 kJ/mol - 1852 kJ/mol = -482 kJ/molSince the enthalpy change is negative, the reaction is exothermic.