Question

What is the change in enthalpy and entropy for the reaction of 2 moles of hydrogen gas (H2) and 1 mole of oxygen gas (O2) to form 2 moles of water (H2O) at standard conditions?

Answer 1

To calculate the change in enthalpy  H  and entropy  S  for the reaction, we need to use the standard enthalpies of formation and standard entropies of the substances involved.The balanced chemical equation for the reaction is:2 H2 g  + O2 g   2 H2O l Standard enthalpies of formation  Hf  at 25C  298 K :H2 g  = 0 kJ/mol  since it's an element in its standard state O2 g  = 0 kJ/mol  since it's an element in its standard state H2O l  = -285.8 kJ/molStandard entropies  S  at 25C  298 K :H2 g  = 130.7 J/molKO2 g  = 205.2 J/molKH2O l  = 69.9 J/molKNow, we can calculate the change in enthalpy  H  and entropy  S  using the following formulas:H =  [Hf products ] -  [Hf reactants ]S =  [S products ] -  [S reactants ]For H:H = [2   -285.8 kJ/mol ] - [2  0 kJ/mol + 1  0 kJ/mol]H = -571.6 kJ/molFor S:S = [2  69.9 J/molK] - [2  130.7 J/molK + 1  205.2 J/molK]S = 139.8 - 466.6 J/molKS = -326.8 J/molKSo, the change in enthalpy  H  for the reaction is -571.6 kJ/mol, and the change in entropy  S  is -326.8 J/molK.