[Dolphinsiu]

I get stucked in the following question during exam, especially in kg unit and kg/h unit!

Immobilised cells of a genetically-improved strain of Brevibaterium lactofermentum are used to convert glucose to glutamic acid for production of MSG. The immobilised cells are unable to grow, but metabolise glucose, according to th equation:

C6H12O6 + NH3 + 3/2 O2 ---> C5H9NO4 + CO2 + 3H2O

A feed stream of 4% glucose in water enters a reactor at 25? at a flow rate of 2000kg/h. The input stream from the reactor is agitated with air which contains 1500.8 kg of O2 and 518.5 kg of NH3. The product stream from the reactor contains 12.6 kg of unreacted glucose. Complete the table and estimate the cooling requirements. (hint: the heat of reaction can be estimated by determining the heat of combustion)


                                                      Mol (g mol/h)

                              Input                                                   Output

Glucose      2000000 (4%)/180 = 444.4
NH3
O2
N2
H2O          2000000 (96%)/18= 106667
CO2                           0
C5H9NO4                    0

Then I failed to complete the table as kg is not equal to kg/h!
Please *delete me* Thank you!

[eugenedakin]

Hello Dolphinsiu,

It looks like you started answering the question.. great!

Now.. continue to calculate the remaining Moles of reagents.  This way you will be able to determine the limiting reagent.

Once you determine the limiting reagent, then you can calculate the moles of products and then the heat of reaction (cooling) requirements can be determined.

After you calculate the moles, repost your version of the answer.  We can go from there.

Best of luck,

Eugene

[Donaldson Tan]

A feed stream of 4% glucose in water enters a reactor at 25? at a flow rate of 2000kg/h. The input stream from the reactor is agitated with air which contains 1500.8 kg of O2 and 518.5 kg of NH3. The product stream from the reactor contains 12.6 kg of unreacted glucose.

Just assume the underlined terms are given in kg/h. The magnitude of the number matches.

You can work the heat combustion from the standard heat of formation of each component.

You can directly deduce the extent of reaction by deducting the outlet molar flowrate of glucose leaving the reactor from the inlet molar flowrate of glucose.

Once you have the extent of reaction, you can work out the composition of your product stream.

[Dolphinsiu]

                                                  Mol (g mol/h)
                              Input                                            Output
Glucose (C6H12O6)   2000000 (4%)/180 = 444.4   12600/180 = 70
NH3                        518500/17 = 30500             30500 – 374.4 = 30125.6
O2                          1500800/32 = 46900             46900 – 3/2 (374.4) = 46338.4
N2                                0                                                    0
H2O                       2000000 (96%)/18= 106667    106667 + 3(374.4) = 107790.2
CO2                               0                                             374.4
C5H9NO4                        0                                            374.4

I have finished the table, but it seems very strange. Is it correct?

Why the remaining 70 kg glucose is unreacted although excess NH3 and O2 are used?

[Donaldson Tan]

Why the remaining 70 kg glucose is unreacted although excess NH3 and O2 are used?

You mean 70mol/h and it is 84.2% conversion - fairly high. I think a good reason for this is because this is a 2 phase reaction. The chemical reactor in question is known as the slurry reactor. It is a type of continuous stir tank reactor. The oxidation reaction only takes place in the liquid phase, which is limited by the mass transfer rate of oxygen from the air phase (bubbles) to the liquid phase. Oxygen isn't very soluble in water FYI.

[eugenedakin]

Hello Dolphinsiu,

You are getting there... Chuckle .. dont subtract your numbers yet.. it gets confusing when that happens

To help you visualize what is happening, use one of your 'Input' moles, and balance the equation.. For example, I'll use Glucose input first:

444.4 C6H12O6 + 444.4 NH3 + 666.6 O2 --->444.4 C5H9NO4 + 444.4 CO2 + 1333.2 H2O

This means that 444.4 moles of glucose and ammonia along with 666.6 moles of O2 will produce 444.4 moles of C5H9NO4 and CO2, while producing 1333.2 moles of H2O.

For arguments sake, lets say that you only put in 200 moles of Ammonia (NH4), then this reaction would not work.  We would need to place 200 moles of ammonia into the equation and then perform the calcuation.  In this case, ammonia would be our limiting-reagent, or the lowest number of moles which balances our equation. 

Ok .. now that I provided an example, its your turn to solve for the limiting reagent.

Cheers,

Eugene

[Dolphinsiu]

I truly understand what you have said!

The question is still here. Can I assume all kg into kg/h?

If not, I can find what is the limiting reagent (but I don't know how to find) . Otherwise the input NH3 and O2 are in excess!

[eugenedakin]

Hello Dolphinsiu,

Great!

As geodome has said, yes, assume all kg into kg/hr.

The limiting reagent will be glucose (444.4-70 = 374.4 mol/hr).

Now, solve the problem with heats of formation in the balanced equation with 374.4 mol/hr of glucose as your limiting reagent. 

Cheers!

Eugene

[Dolphinsiu]

Thank you! ^^

[eugenedakin]

Your Welcome