Question

Calculate the standard enthalpy change of the following chemical reaction at 298 K given the following bond energies: 2H2(g) + O2(g) --> 2H2O(l) Bond energies: H-H: 432 kJ/mol O=O: 498 kJ/mol O-H: 464 kJ/mol

Answer 1

To calculate the standard enthalpy change of the reaction, we need to find the difference between the bond energies of the reactants and 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 O-H bonds  since there are 2 moles of H2O, each with 2 O-H bonds .First, let's calculate the total bond energy of the reactants:Reactants bond energy =  2 moles H-H  432 kJ/mol H-H  +  1 mole O=O  498 kJ/mol O=O Reactants bond energy =  864 kJ  +  498 kJ  = 1362 kJNow, let's calculate the total bond energy of the products:Products bond energy =  4 moles O-H  464 kJ/mol O-H Products bond energy = 1856 kJNow, we can find the standard enthalpy change of the reaction:H = Products bond energy - Reactants bond energyH = 1856 kJ - 1362 kJ = 494 kJThe standard enthalpy change of the reaction at 298 K is 494 kJ.