CH 1206 (b) Calculate the molar and mass flow rates of all constituents leaving
ID: 701233 • Letter: C
Question
CH 1206 (b) Calculate the molar and mass flow rates of all constituents leaving the bioreactor including the products and the excess reactants. (c) Confirm that total mass, but not total moles, is conserved. 3.23 During cellular metabolism, glucose is combusted to carbon dioxide and water. One of the many steps in glycolysis is the Krebs cycle. A simplified bio chemical summary of several steps in the Krebs cycle is as follows: 1 C6H507 (citrate) + aH2O + bPO4- pC4H20s (oxaloacetate) +qH rCO2 +sPO Note that this equation captures only the exchange of chemical species, no the exchange of their respective charges. It is known through biochemical ex periments that for every molecule of citrate consumed, one molecule of ox- aloacetate is generated. A tissue mass is comprised of many cells, each which conducts the process of glycolysis including the Krebs cycle. Assume a molar flow rate of 0.10 mol/day of C&H;,07 into this tissue. (a) Balance the above equation. Determine the stoichiometric coefficients a (b) What is the minimum flow rate of water needed for the fractional conver (e) Assume that the fractional conversion of water is 0.80 and the fractiona 4, r,and s sion of C&HsO;, to be 1.0 conversion of C&H;,O, is 1.0. What is the limiting reactant? Calculate te reaction rate (R) and the inlet molar flow rate of H20. Deine ermine theExplanation / Answer
a) C6H5O7 + H2O + PO4 ----> C4H2O5 + 7H + 2CO2 + PO3
Given that for one molecule of C6H5O7, one molecule of C4H2O5is formed. Thus p = 1. Now balancing for C gives, r = 2.
Balancing equation for H ---> 5 + 2a = 2 + q
Balancing equation for O ---> 7 + a + 4b = 5 + 4 + 3s
Balancing equation for P ---> b = s = 1. ( as only two species have P in them)
Solving this gives,
a = 1 ; b = 1; q = 7 and r = 2 ; s = 1.
b) Given molar flow rate of C6H5O7 = 0.1 mol/day
For a fractional conversion of 1.0, it means that all of C6H5O7 is converted. From equation it is evident that for one mol of C6H5O7 , one mole of water is required. Thus for 0.1 mol/day of C6H5O7require 0.1 mol/day of water.
c) Given that the fractional conversion of water = 0.8 and that of C6H5O7 is 1.0. The limiting reactant is the one that gets consumed first. As in our case all of C6H5O7 gets consumed as conversion is 1.0 but water doesn't. Hence the limiting reactant is C6H5O7 .
Fractional conversion of water = 0.8 = Amount of water converted/ Total amount water at inlet
Amount of water converted = 0.1 mol/day ( from reaction stoichiometry).
Total amount at inlet = 0.1/0.8 = 0.125 mol/day
Rate of reaction = - dC/dt = - {Concentration difference between exit and inlet / Time taken to reach exit concentration}
Rate of reaction = -[0 (mol) - 0.1 (mol) ]/ 1 (day) = 0.1 mol/day
The negative sign implies that concentration decreases as the time increases.
Exit flow rate of water = inlet flow rate - consumption rate = 0.125 - 0.1 = 0.025 mol/day
Exit flow rate of C4H2O5 = 0.1 mol/day ( from stoichiometry, one mole of C6H5O7 gives one mole of C4H2O5)
Exit flow rate of H = 7*0.1 = 0.7 mol/day ( from stoichiometry, one mole of C6H5O7 gives 7 moles of H)