Industrial production of ethanol requires fermentation using Saccharomyces cerev

Question

Industrial production of ethanol requires fermentation using Saccharomyces cerevisiae, a type of yeast. The elemental formula of S. cerevisiae is CH_1.83 O_0.56 N_0.17. Suppose you want to produce ethanol (C_2H_6O) in a batch process at 2.5 degree C. You plan to add 0.25 lb_n NK_3 and 5.0 lb_n glucose to the yeast for the following (unbalanced) reaction: C_6H_12O_6(s) + NH_3(g) rightarrow C_3H_8O_3(l) + C_2H_6O(l) + CH_1.83O_0.56 N_0.17(s) + CO_2(g) + H_2O(l) Note that 1 mol glucose produce 0.5 mol glycerol and that the ration of NH_3 to H_2O is 1:1. During preparation, you accidentally add too much glucose. However, you decide to left the bioreactor run anyway. After the reaction goes to completion (at least one of the starting reactants is used up), you discover that 1.4 lb_m ethanol had been produced. How much heat had been produced and subsequently removed from the system? How much glucose did you initially add to the reactor? Assume that the heat of combustion of S. cerevisiae is -21.2 kJ/g.

Explanation / Answer

[5.23]C6H12O6(s) + NH3(g) [2.61]C3H8O3(l) + [5.34]C2H6O(l) +[5.88]CH1.83O0.56N0.17(s) + [6.95]CO2(g) + H2O(l)

combustion reaction for S.cerevisae

CH(1.83)O(0.56)N(0.17) + O2 --------------- 1CO2 + 0.915 H2O +0.17 NO2   

(MW 25.17 ) enthalpy of formation of S.cerevesae = Enthalpy of formation of products - enthalpy of combustion of s.cerevisae = 0.17*33.2-0.915*241.818-1*393.509 + 25.17*21.2 = - 75.5 KJ/mol

1lb = 454 g ,

NH3 is 0.25 lb =0.25*454 =113.5 g is 6.676 moles   

ethanol is 1.4 lb = 1.4*454=635.6 g = 13.8 moles   

no of moles of glucose reacted or added = 13.5 moles , that is 13.5 *180.16 g = 5.37 lb

heat = enthalpy of reaction = enthalpy of products - enthalpy of reactants = 2.58*(-241.818)+6.95*2.58*(--393.509)+5.88*2.581(-75.5)+5.34*2.58(-235.5)+2.61*2.58(-577.9)-2.58*(--45.90)-2.58*5.23(-1271) = +1340.72 KJ

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