4.62. Oxygen consumed by a living organism in aerobic reactions is used in addin

Question

4.62. Oxygen consumed by a living organism in aerobic reactions is used in adding mass to the organism and/or the production of chemicals and carbon dioxide. Since we may not know the molecular compositions of all species in such a reaction, it is common to define the ratio of moles of cO2 produced per mole of O2 consumed as the respiratory quotient, RQ, where nco2 RQ= Since it generally is impossible to predict values of RQ, they must be determined from operating data. Mammalian cells are used in a bioreactor to convert glucose to glutamic acid by the reaction The feed to the bioreactor comprises 1 .00 × 102 mol C6HnOdday, 1 .20 × 102 mol NHyday, and 1.10 × 102 mol 02/day. Data on the system show that RQ = 0.45 mol CO2 produced/mol O2 consumed.

Explanation / Answer

Solution

a) determining the stoichiometric coefficients and limiting reactant.

C6H12O6 + aNH3 + bO2 à pC5H9NO4 + qCO2 + rH2O

Complete reaction is

C6H12O6 + NH3 + 3/2 O2 à C5H9NO4 + CO2 + 3H2O

a = 1, b = 3/2, p = 1, q = 1, r = 3

Now we will determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation.

For C6H12O6 = 100 moles/ 1

                         = 100

For NH3 = 120 moles/1 (a=1)

                = 120

For O2 = 110 moles/(3/2)   (b=3/2)

             = 73.33

As we can see 73.33<100<120

Therefore, O2 is limiting reactant.

b) Calculating molar and mass flow rate of all species leaving the reactor.

As shown above in calculation O2 is limiting factor and therefore only same amount of other reactants will react and excess will remain.

C6H12O6 =

Out of 100 only 73.33 will react.

=100-73.33 = 26.67 moles C6H12O6/day = 4804.75 g/day

NH3 =

Out of 120 only 73.33 will react

=120-73.33 = 46.67 moles NH3/day = 794.83 g/day

ii)Now, products will be

73.33 moles of C5H9NO4/day = 10788.9 g/day

32.99 moles of CO2/day = 1451.88 g/day (as 73.33*0.45 Given in RQ)

73.33 * 3 = 219.99 moles of H2O/day = 3963.1g/day

Fractional conversions of non limiting reactant

Moles of C6H12O6/moles of C5H9NO4 = 100/73.33 = 1.36

Moles of NH3/ moles of C5H9NO4= 120/73.33 = 1.63

Moles of C6H12O6/moles of CO2= 100/32.99 = 3.03

   

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