Question

Calculate the standard enthalpy change for the given reaction using the given data:2H2(g) + O2(g) --> 2H2O(l)Given:ΔHf°(H2O(l)) = -285.83 kJ/molΔHf°(O2(g)) = 0 kJ/molΔHf°(H2(g)) = 0 kJ/mol

Answer 1

To calculate the standard enthalpy change  H  for the given reaction, we can use the formula:H =  Hf products  -  Hf reactants For the products, we have 2 moles of H2O l , and for the reactants, we have 2 moles of H2 g  and 1 mole of O2 g . Using the given standard enthalpies of formation  Hf , we can plug in the values:H = [2  Hf H2O l  ] - [2  Hf H2 g   + Hf O2 g  ]H = [2   -285.83 kJ/mol ] - [2   0 kJ/mol  +  0 kJ/mol ]H =  -571.66 kJ  -  0 kJ H = -571.66 kJThe standard enthalpy change for the given reaction is -571.66 kJ.