Question for Chemical Engineering Analysis. I\'m struggling a lot with part a (n

Question

Question for Chemical Engineering Analysis. I'm struggling a lot with part a (not sure if I'm setting up my material balance correctly, whether or not I should be including Nitrogen, whether or not I'm supposed to account for consumption, etc.) but help with all parts would be much appreciated. Thanks!

In the Deacon process for the manufacture of chlorine, HC1 and O2 react to form Cl2 and H20. Sufficient air (21mole% O2, 79% N2 ) is fed to provide 35% excess oxygen, and the fractional conversion of HCl is 85%.Calculate the mole fractions of the product stream components, using atomic species balances in your calculation.Again calculate the mole fractions of the product stream components, only this time use the extent of reaction in the calculation.An alternative to using air as the oxygen source would be to feed pure oxygen to the reactor. Running with oxygen imposes a significant extra process cost relative to running with air, but also offers the potential for considerable savings. Speculate on what the cost and savings might be. What would determine which way the process should be run?

Explanation / Answer

4HCl + O2 + 3.76N2 ---> 2Cl2 + 2H2O + 3.76N2

(a) total air consumption = 1 + 3.76 = 4.76 vol of air

with 35% excess = 0.35 x 4.76 + 4.76 = 7.93 vol of air is the fuel added

excess air will appear as O2 = 0.2 x 1.67 = 0.35 O2

and N2 = 1.67 - 0.35 = 1.32

total moles on product side = 2 + 2 + 1.32 + 3.76 + 0.35 = 9.43

so, proportion of components in product mixture

Cl2 = 2/9.43 = 0.212

H2O = 2/9.43 = 0.212

N2 = 5.08/9.43 = 0.539

O2 = 0.35/9.43 = 0.037

moles of Cl2 = 0.212/71 = 0.002986 mols

moles of H2O = 0.212/18.015 = 0.0118 mols

moles of N2 = 0.539/28 = 0.01925 mols

moles of O2 = 0.037/32 = 0.00116 mols

total moles = 0.0352 mols

moles fraction of,

Cl2 = 0.002986/0.0352 = 0.085

H2O = 0.0118/0.0352 = 0.335

N2 = 0.01925/0.0352 = 0.547

O2 = 0.00116/0.0352 = 0.0330

(b) The extent of reaction = 85% HCl conversion

moles of Cl2 = 0.85 x 0.002986 = 0.00254 mols

moles of H2O = 0.85 x 0.0118 = 0.01003 mols

moles of N2 = 0.85 x 0.01925 = 0.0164 mols

moles of O2 = 0.85 x 0.00116 = 0.00099 mols

total moles = 0.02996 mols

moles fraction of,

Cl2 = 0.00254/0.02996 = 0.085

H2O = 0.01003/0.02996 = 0.335

N2 = 0.0164/0.02996 = 0.547

O2 = 0.00099/0.02996 = 0.0330

(c) If we use pure O2 instead of air, the process cost goes down. This is due to reduction is the poduct separation step of Cl2 from N2. the toal energy of the process would be thus lesser then when air is used as the oxidant for the reaction.

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.