Calculate the change in enthalpy and entropy for the reaction Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) at 25°C given the following data:- ΔH°f for MgCl2(aq) = -641.3 kJ/mol- ΔS°f for MgCl2(aq) = -200.9 J/K/mol- ΔH°f for H2(g) = 0 kJ/mol- ΔS°f for H2(g) = 130.7 J/K/mol- ΔH°f for HCl(aq) = -167 kJ/mol- ΔS°f for HCl(aq) = -18.8 J/K/mol- ΔH°f for Mg(s) = 0 kJ/mol- ΔS°f for Mg(s) = 32.7 J/K/molAssume that ΔH°f and ΔS°f are independent of temperature.

Question

Calculate the change in enthalpy and entropy for the reaction Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) at 25°C given the following data:- ΔH°f for MgCl2(aq) = -641.3 kJ/mol- ΔS°f for MgCl2(aq) = -200.9 J/K/mol- ΔH°f for H2(g) = 0 kJ/mol- ΔS°f for H2(g) = 130.7 J/K/mol- ΔH°f for HCl(aq) = -167 kJ/mol- ΔS°f fo

Answer 1

To calculate the change in enthalpy  H  and entropy  S  for the reaction, we will use the following equations:H =  Hf products  -  Hf reactants S =  Sf products  -  Sf reactants For the reaction Mg s  + 2HCl aq   MgCl2 aq  + H2 g , we have:H = [ -641.3 kJ/mol  +  0 kJ/mol ] - [ 0 kJ/mol  + 2 -167 kJ/mol ]H =  -641.3 kJ/mol  -  -334 kJ/mol H = -307.3 kJ/molS = [ -200.9 J/K/mol  +  130.7 J/K/mol ] - [ 32.7 J/K/mol  + 2 -18.8 J/K/mol ]S =  -70.2 J/K/mol  -  32.7 J/K/mol - 37.6 J/K/mol S = -70.2 J/K/mol + 4.9 J/K/molS = -65.3 J/K/molSo, the change in enthalpy  H  for the reaction is -307.3 kJ/mol, and the change in entropy  S  is -65.3 J/K/mol at 25C.