Question

Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
production of chlorine from electro-chemical processes, the direct catalytic oxidation of hydrochloric acid with oxygen
is used to produce chlorine. It is known as ‘Deacon process’
In production of chlorine gas by oxidation of hydrochloric acid gas, air is used 30% excess of the theoretically used.
4kmol/h of hydrochloric acid is fed to the reactor. The percentage conversion of reaction is 80%.
Reaction: 4 HCL + O2 −−→ 2 Cl2 + 2 H2O
a Calculate the composition of gas in the product stream on mole basis

Answer 1

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

• HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

How can percentage conversion give the contents of the product stream?

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$

Which gives;

$80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}$

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

• HCl: 0.4 kmol/h

• Cl₂: 1.6 kmol/h

• H₂O: 1.6 kmol/h

• O₂: 0.5 kmol/h

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