Calculate ∆G° using these ∆G°f values (3 of 3)

Question

Also calculate ∆G° again but using ∆H° and ∆S°

NH4Cl (s) ------> NH3 (g) HCl (g)

∆G°f

NH4Cl= -203.0 kJ/mol

NH3= -16 kJ/mol

HCl= -95.30 kJ/mol

∆H°

NH4Cl= -314.4 kJ/mol

NH3= -45.9 kJ/mol

HCl= -92.31 kJ/mol

∆S°

NH4Cl= 94.6 J/mol*K

NH3= 193 J/mol*K

HCl= 186.79 J/mol*K

Answer 1

NH4Cl (s) ------> NH3 (g) HCl (g)

prod - reactants:

d G = (-95.30 & -16) - (-203.6)

your answer is: dG = +92.3 kJ

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NH4Cl (s) ------> NH3 (g) HCl (g)

Also calculate  Ã¢ÂˆÂ†H°

d H = prod - reactants:

d H = (-45.9 & -92.31) - (-314.4)

your answer is: dH  = +176.2 kJ

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NH4Cl (s) ------> NH3 (g) HCl (g)

Also calculate  Ã¢ÂˆÂ†S°

d S = prod - reactants:

d S = (193 & + 186.79) - (94.6)

your answer is: dS  = +285.2 J

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Also calculate ∆G° again but using ∆H° and ∆S°

dG = dH - TdS

dG = 176,5 KJ-  298.15 (0.2852kJ)

dG = 176,5 KJ-  (85.0)

dG = 91.47

your answer is 91.5 kJ