Question

Use delta H f and S to calculate delta G rxn (delta g not sys) at 25 C for the reaction below:

4KClO3 (s) --> 4KClO4 (s) + KCL (s)

Delta H not f KCLO3 = -397.7 kj/mol; KCLO4 = -432.8 kJ/mol; KCL = -436.7 kJ/mol
S KCLO3 = 143.1 kJ/mol ; KCLO4= 151.0 kJ/mol; KCL = 82.6 kJ/mol

Answer 1

Given reaction is 4KClO3(s) --> 4KClO4(s) + KCL(s)

Delta H°f (KClO3)=-397.7 kJ/mol, Delta H°f (KClO4)=-432.8 kJ/mol, Delta H°f (KCl)=-436.7 kJ/mol

Delta H°f rxn=sum of Delta H°f (products)-sum of Delta H°f (reactants)

Delta H°f rxn=[4xDelta H°f (KClO4)+Delta H°f (KCl)]-[4xDelta H°f (KClO3)]

=4x(-432.8 kJ/mol)+-436.7 kJ/mol-(4x-397.7 kJ/mol)=-577.1 kJ/mol

Delta H°f rxn=-577.1 kJ/mol

and also given Delta S°f (KClO3)=143.1 J/mol K, Delta S°f (KClO4)=151.0 J/mol K, Delta S°f (KCl)=82.6 J/mol K

Delta S°f rxn=sum of Delta S°f (products)-sum of Delta S°f (reactants)

Delta S°f rxn=[4xDelta S°f (KClO4)+Delta S°f (KCl)]-[4xDelta S°f (KClO3)]

=(4x151.0 J/mol K)+82.6 /mol K-(4x143.1 J/mol K)=114.2 J/mol K

Delta G°f rxn=Delta H°f rxn-T Delta S°f rxn where T=25°C=298 K (25+273 K)

Delta G°f rxn=-577.1 kJ/mol-(298 K)x(114.2 J/mol K)= -577.1 kJ/mol-34031.6 J/mol

Delta G°f rxn=-577.1 kJ/mol-34.0316 kJ/mol

Delta G°f rxn=-611.13 kJ/mol.