Question

What is the enthalpy change for the following chemical reaction?2H2 (g) + O2 (g) → 2H2O (l)Given the enthalpy of formation values:ΔHf°(H2O) = -285.8 kJ/molΔHf°(H2) = 0 kJ/molΔHf°(O2) = 0 kJ/mol

Answer 1

To calculate the enthalpy change  H  for the reaction, we can use the formula:H =  [Hf products ] -  [Hf reactants ]For this reaction, the products are 2 moles of H2O and the reactants are 2 moles of H2 and 1 mole of O2.H = [2  Hf H2O ] - [2  Hf H2  + Hf O2 ]Given the enthalpy of formation values:Hf H2O  = -285.8 kJ/molHf H2  = 0 kJ/molHf O2  = 0 kJ/molSubstitute the values into the formula:H = [2   -285.8 kJ/mol ] - [2   0 kJ/mol  +  0 kJ/mol ]H =  -571.6 kJ  -  0 kJ H = -571.6 kJThe enthalpy change for the reaction is -571.6 kJ.