Question

In the Lebedev's process 1,3-butadiene is produced from ethanol according to the following chemical equation 20,H,OH(g)->CH.-CH-CH-CH2(g)+2H,O(g)+H2(g) Calculate the butadiene percent yield of the reaction at T-4 lXata 50 K andp 2 atm. CH,OH2348 110.2 -241.8 C4H, H20 167.9 150.7 -228.6

Answer 1

($\Delta$_fH^o)_rxn = {($\Delta$_fH^o)_C4H6 +2*($\Delta$_fH^o)_H2O + ($\Delta$_fH^o)_H2??} - 2*($\Delta$_fH^o)_C2H5OH

= {110.2 + 2*(-241.8) + 0} KJ/mol - 2*(-234.8) KJ/mol

= 96.2 KJ/mol

($\Delta$_fG^o)_rxn = {($\Delta$_fG^o)_C4H6 +2*($\Delta$_fG^o)_H2O + ($\Delta$_fG^o)_H2??} - 2*($\Delta$_fG^o)_C2H5OH

= {150.7 + 2*(-228.6) + 0} KJ/mol - 2*(-167.9) KJ/mol

= 29.3 KJ/mol

($\Delta$_fG^o)_rxn = ($\Delta$_fH^o)_rxn - T($\Delta$_fS^o)_rxn

i.e. 29.3 KJ/mol = 96.2 KJ/mol - 450 K * ($\Delta$_fS^o)_rxn

i.e. ($\Delta$_fS^o)_rxn = 148.67 J/mol-K

Here, T < ($\Delta$_fH^o)_rxn / ($\Delta$_fS^o)_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.