Question

What is the change in enthalpy and entropy when 2 moles of Hydrogen gas reacts with 1 mole of Oxygen gas to form 2 moles of water vapor at a constant pressure of 1 atm and a temperature of 298 K?

Answer 1

To calculate the change in enthalpy  H  and entropy  S  for the reaction, we first need to write the balanced chemical equation:2 H g  + O g   2 HO g Next, we need to find the standard enthalpies of formation  Hf  and standard molar entropies  S  for each substance involved in the reaction. These values can be found in tables or online databases. Here are the values we need:Hf  H  = 0 kJ/mol  since it is an element in its standard state Hf  O  = 0 kJ/mol  since it is an element in its standard state Hf  HO, g  = -241.8 kJ/molS  H  = 130.7 J/molKS  O  = 205.2 J/molKS  HO, g  = 188.8 J/molKNow we can calculate the change in enthalpy  H  for the reaction using the following equation:H =  [Hf  products ] -  [Hf  reactants ]H = [2   -241.8 kJ/mol ] - [2  0 kJ/mol + 1  0 kJ/mol]H = -483.6 kJThe change in enthalpy  H  for the reaction is -483.6 kJ.Next, we can calculate the change in entropy  S  for the reaction using the following equation:S =  [S  products ] -  [S  reactants ]S = [2  188.8 J/molK] - [2  130.7 J/molK + 1  205.2 J/molK]S = 377.6 J/molK - 466.6 J/molKS = -89.0 J/molKThe change in entropy  S  for the reaction is -89.0 J/molK.