Question

What is the change in enthalpy and entropy for the reaction of 2 moles of sodium hydroxide with 1 mole of sulfuric acid to form 2 moles of water and 1 mole of sodium sulfate, assuming standard conditions?

Answer 1

To calculate the change in enthalpy  H  and entropy  S  for the given reaction, we need to know the standard enthalpies of formation  Hf  and standard entropies  S  for each substance involved in the reaction. The reaction is as follows:2 NaOH  aq  + H2SO4  aq   2 H2O  l  + Na2SO4  aq The standard enthalpies of formation  Hf  and standard entropies  S  for each substance are as follows  in kJ/mol for Hf and J/molK for S :NaOH  aq : Hf = -469.15 kJ/mol, S = 57.2 J/molKH2SO4  aq : Hf = -909.3 kJ/mol, S = 156.9 J/molKH2O  l : Hf = -285.83 kJ/mol, S = 69.9 J/molKNa2SO4  aq : Hf = -1387.1 kJ/mol, S = 149.1 J/molKFirst, let's calculate the change in enthalpy  H :H = [2  Hf H2O  + Hf Na2SO4 ] - [2  Hf NaOH  + Hf H2SO4 ]H = [2   -285.83  +  -1387.1 ] - [2   -469.15  +  -909.3 ]H =  -571.66 - 1387.1  -  -938.3 - 909.3 H = -1958.76 + 1847.6H = -111.16 kJ/molNow, let's calculate the change in entropy  S :S = [2  S H2O  + S Na2SO4 ] - [2  S NaOH  + S H2SO4 ]S = [2  69.9 + 149.1] - [2  57.2 + 156.9]S =  139.8 + 149.1  -  114.4 + 156.9 S = 288.9 - 271.3S = 17.6 J/molKSo, the change in enthalpy  H  for the reaction is -111.16 kJ/mol, and the change in entropy  S  is 17.6 J/molK.