(fHo)rxn = {(fHo)C4H6 +2*(fHo)H2O + (fHo)H2??} - 2*(fHo)C2H5OH
= {110.2 + 2*(-241.8) + 0} KJ/mol - 2*(-234.8) KJ/mol
= 96.2 KJ/mol
(fGo)rxn = {(fGo)C4H6 +2*(fGo)H2O + (fGo)H2??} - 2*(fGo)C2H5OH
= {150.7 + 2*(-228.6) + 0} KJ/mol - 2*(-167.9) KJ/mol
= 29.3 KJ/mol
(fGo)rxn = (fHo)rxn - T(fSo)rxn
i.e. 29.3 KJ/mol = 96.2 KJ/mol - 450 K * (fSo)rxn
i.e. (fSo)rxn = 148.67 J/mol-K
Here, T < (fHo)rxn / (fSo)rxn
i.e. 400 < (96.2 KJ/mol)/(0.14867 KJ/mol-K) = 647.085 K
i.e. 400 K < 647.085 K
Hence, the given reaction is non-spontaneous at the given temperature.
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