Calculate the standard enthalpy change and the standard entropy change for the reaction: 2HCl(g) + Mg(s) → MgCl2(s) + H2(g) Given the following standard enthalpy of formation values: - HCl(g) = -92.31 kJ/mol - MgCl2(s) = -641.8 kJ/mol - H2(g) = 0 kJ/mol - Mg(s) = 0 kJ/mol Also given the following standard molar entropy values:- HCl(g) = 186.9 J/Kmol - MgCl2(s) = 138.53 J/Kmol - H2(g) = 130.7 J/Kmol - Mg(s) = 32.85 J/Kmol

Question

Calculate the standard enthalpy change and the standard entropy change for the reaction: 2HCl(g) + Mg(s) → MgCl2(s) + H2(g) Given the following standard enthalpy of formation values: - HCl(g) = -92.31 kJ/mol - MgCl2(s) = -641.8 kJ/mol - H2(g) = 0 kJ/mol - Mg(s) = 0 kJ/mol Also given the following standard molar entropy values:- HCl(g) = 186.9 J/Kmol - MgCl2(s) = 138.53 J/Km

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the formula:H =  [Hf products ] -  [Hf reactants ]H = [ -641.8 kJ/mol  +  0 kJ/mol ] - [2 -92.31 kJ/mol  +  0 kJ/mol ]H =  -641.8 kJ/mol  -  -184.62 kJ/mol H = -457.18 kJ/molTo calculate the standard entropy change  S  for the reaction, we can use the formula:S =  [S products ] -  [S reactants ]S = [ 138.53 J/Kmol  +  130.7 J/Kmol ] - [2 186.9 J/Kmol  +  32.85 J/Kmol ]S =  269.23 J/Kmol  -  373.8 J/Kmol + 32.85 J/Kmol S = 269.23 J/Kmol - 406.65 J/KmolS = -137.42 J/KmolSo, the standard enthalpy change  H  for the reaction is -457.18 kJ/mol, and the standard entropy change  S  is -137.42 J/Kmol.