Answer – We are given standard enthalpy and entropy of each in given reaction and need to calculate the standard free energy for the reaction at temperature 25oC –
N2(g) + 3 H2(g) <-----> 2 NH3(g)
ΔHorxn= ∑ ΔHof of product – ∑ ΔHof of reactant
= [ 2* ΔHof NH3(g)] – [ΔHof N2(g) + 3*ΔHof H2(g)]
= (2*46.1) – ( 0.00 + 3*0.00)
= 92.2 kJ/mol
ΔSorxn= ∑ ΔSof of product – ∑ ΔSof of reactant
= [ 2* ΔSof NH3(g)] – [ΔSof N2(g) + 3*ΔSof H2(g)]
= (2*192.3) – ( 191.5 + 3*130.6)
= -200.7 J/mol.K
we know formula for standard free energy
ΔGorxn= ΔHorxn -T ΔSorxn
= 92.2 kJ/mol – 298K * (-0.2007 kJ/mol.K)
= 152.0 kJ/mol
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