Chem E Question

Question

In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and

H2O. Sufficient air is fed to provide 35% excess oxygen, and the fractional conversion of HCl is 85%.

a) Calculate the mole fractions of the product stream components, using the extent of reaction in the

calculation. b) 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?

Answer 1

Hi Quinn, glad I can help!

Part (a): Calculating the Mole Fractions of the Product Stream

1. Balanced Chemical Reaction: 4 HCl + O₂ → 2 Cl₂ + 2 H₂O
2. Given Information:
3. Excess oxygen: 35%
4. Fractional conversion of HCl: 85%

Assume 4 moles of HCl fed (for stoichiometry):

1. Moles of HCl reacted: 4 moles × 0.85 = 3.4 moles
2. Moles of HCl unreacted: 4 - 3.4 = 0.6 moles
3. Oxygen fed and reacted:
4. Oxygen required stoichiometrically: 1 mole for 4 moles of HCl
5. Excess oxygen fed: 1 × 1.35 = 1.35 moles O₂

Oxygen reacted:

1. Moles of O₂ reacted: (3.4 moles HCl / 4) = 0.85 moles O₂
2. Moles of O₂ unreacted: 1.35 - 0.85 = 0.5 moles O₂
3. Products formed:
4. Moles of Cl₂ formed: (3.4 / 4) × 2 = 1.7 moles Cl₂
5. Moles of H₂O formed: (3.4 / 4) × 2 = 1.7 moles H₂O
6. Total moles in the product stream: Unreacted HCl + Unreacted O₂ + Cl₂ + H₂O: 0.6 moles HCl + 0.5 moles O₂ + 1.7 moles Cl₂ + 1.7 moles H₂O = 4.5 moles
7. Mole Fractions:
8. Mole fraction of HCl: 0.6 / 4.5 = 0.133
9. Mole fraction of O₂: 0.5 / 4.5 = 0.111
10. Mole fraction of Cl₂: 1.7 / 4.5 = 0.378
11. Mole fraction of H₂O: 1.7 / 4.5 = 0.378

Part (b): Speculation on Costs and Savings of Using Pure Oxygen vs Air

1. Costs:
2. Pure oxygen is more expensive due to separation costs (e.g., cryogenic distillation).
3. Additional safety costs due to handling pure oxygen.
4. Energy costs for producing and storing pure oxygen are higher than air.
5. Savings:
6. Increased reaction rates with pure oxygen could lead to faster production and reduced operating time.
7. Smaller reactor and equipment size as there's no need to handle inert nitrogen.
8. Higher yields and conversion rates, reducing the need for recycling unreacted components.
9. Simplified separation processes as there's no nitrogen to remove from the product stream.
10. Factors to Consider:
11. Scale of operation: Larger-scale processes might justify pure oxygen use if the savings outweigh the costs.
12. Process efficiency: Higher yields and conversions might offset the cost of pure oxygen.
13. Capital vs operating costs: Higher upfront costs might be offset by long-term savings in efficiency and reduced separation costs.
14. Market conditions: Availability and cost of oxygen, as well as chlorine demand, will influence the decision.