Question

Calculate the standard enthalpy and entropy change for the reaction, given the following information:2 H2(g) + O2(g) → 2 H2O(g)∆H°f [H2O(g)] = -241.8 kJ/mol∆H°f [H2(g)] = 0 kJ/mol∆H°f [O2(g)] = 0 kJ/mol∆S° [H2O(g)] = 188.8 J/Kmol∆S° [H2(g)] = 130.7 J/Kmol∆S° [O2(g)] = 205.2 J/KmolAssuming that all heat capacities are independent of temperature, calculate ∆S°rxn and ∆H°rxn for this reaction.

Answer 1

To calculate the standard enthalpy change  Hrxn  for the reaction, we can use the following equation:Hrxn =  Hf  products  -  Hf  reactants For this reaction, the equation becomes:Hrxn = [2  Hf  H2O g  ] - [2  Hf  H2 g   + Hf  O2 g  ]Plugging in the given values:Hrxn = [2   -241.8 kJ/mol ] - [2   0 kJ/mol  +  0 kJ/mol ]Hrxn = -483.6 kJ/molNow, to calculate the standard entropy change  Srxn  for the reaction, we can use the following equation:Srxn =  S  products  -  S  reactants For this reaction, the equation becomes:Srxn = [2  S  H2O g  ] - [2  S  H2 g   + S  O2 g  ]Plugging in the given values:Srxn = [2   188.8 J/Kmol ] - [2   130.7 J/Kmol  +  205.2 J/Kmol ]Srxn =  377.6 J/Kmol  -  467.6 J/Kmol Srxn = -90 J/KmolSo, the standard enthalpy change for the reaction is -483.6 kJ/mol, and the standard entropy change for the reaction is -90 J/Kmol.