Question

What is the enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(g), given that the bond enthalpy of H-H, O=O, and H-O are 436 kJ/mol, 498 kJ/mol, and 464 kJ/mol, respectively?

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.For the reactants, we have:2 moles of H-H bonds and 1 mole of O=O bond.For the products, we have:4 moles of H-O bonds  since there are 2 moles of H2O, each with 2 H-O bonds .Now, we can calculate the total bond enthalpy for the reactants and products:Reactants bond enthalpy =  2 moles H-H  436 kJ/mol  +  1 mole O=O  498 kJ/mol Reactants bond enthalpy =  872 kJ  +  498 kJ  = 1370 kJProducts bond enthalpy =  4 moles H-O  464 kJ/mol Products bond enthalpy = 1856 kJNow, we can find the enthalpy change for the reaction:H = Products bond enthalpy - Reactants bond enthalpyH = 1856 kJ - 1370 kJH = 486 kJThe enthalpy change for the reaction is 486 kJ.