Ammonia-Synthesis Loop with Argon Purge: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2015, Q3

Question 3: Ammonia-Synthesis Loop with Argon Purge (Part A — 25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A closed synthesis loop (mixer → reactor → condenser → split into recycle + purge) with a stoichiometric 3:1 H₂:N₂ feed, 20% single-pass conversion, and an inert (argon) that must be bled through a purge. Basis: 100 mol fresh H₂+N₂ (which carries 0.31 mol Ar). Because H₂–N₂ is consumed as the gas passes through the reactor while argon is not, the Ar : H₂–N₂ ratio is highest at the reactor exit; that exit gas becomes the recycle and purge. Holding the whole reactor at or below the limit therefore sets the recycle (and hence purge) at the ceiling 4 mol Ar / 100 mol H₂–N₂ (the reactor inlet is then only 3.2 mol Ar / 100).

Find. (a) the purge (vent) amount and the recycle rate; (b) the ammonia produced — all per 100 mol fresh H₂+N₂.

Figure 3 — Synthesis loop: fresh feed + recycle → reactor (20%/pass) → condenser removes NH₃; the gas splits into recycle and an argon-controlling purge.

Approach. Argon enters only in the fresh feed and leaves only in the purge, so an inert (argon) balance fixes the purge; the single-pass conversion then closes the recycle rate, and an overall H₂–N₂ balance gives the ammonia made.

Argon balance sets the purge. At steady state, Ar in (fresh) = Ar out (purge), and the purge carries 4 mol Ar per 100 mol H₂–N₂: $$0.31=0.04\,(n_{HN_2}^{purge})\;\Rightarrow\;n_{HN_2}^{purge}=\boxed{7.75\ \text{mol H}_2\text{–N}_2}\;(+\,0.31\text{ mol Ar}=8.06\text{ mol total vented}).$$
Overall H₂–N₂ balance gives ammonia. Of the 100 mol fresh H₂–N₂, whatever is not vented is converted; since $N_2+3H_2\to2NH_3$ turns 4 mol of reactant mixture into 2 mol NH₃: $$n_{HN_2}^{react}=100-7.75=92.25\ \text{mol}\;\Rightarrow\;n_{NH_3}=\tfrac12(92.25)=\boxed{46.1\ \text{mol NH}_3.}$$
Single-pass conversion sizes the recycle. The condenser exit gas ($=$ recycle $R$ + purge) is the 80% unconverted part of the reactor feed $(100+R)$: $$R+7.75=0.80\,(100+R)\;\Rightarrow\;0.20R=72.25\;\Rightarrow\;R=\boxed{361\ \text{mol H}_2\text{–N}_2}\;(≈376\text{ mol with its Ar}).$$ The vent is therefore only $7.75/369=2.1\%$ of the gas leaving the condenser — a small bleed suffices because argon is dilute.

Assumption

The exam says the limit applies “in the reactor” without naming inlet or outlet. The answer above caps the reactor exit (the highest-argon point, so the limit holds everywhere in the bed). If the limit is instead read as applying to the reactor feed, the exit and recycle gas reach $0.04/0.80=5$ mol Ar / 100 mol H₂–N₂. The argon balance then gives a purge of $0.31/0.05=6.20$ mol H₂–N₂ + 0.31 mol Ar $=6.51$ mol, ammonia $\tfrac12(100-6.20)=46.9$ mol and recycle $R=(80-6.20)/0.20=369$ mol H₂–N₂. The method is identical; only the stream where the 4/100 ceiling applies changes.

Quantity (per 100 mol fresh H₂+N₂)	Result
(a) Purge / vent	7.75 mol H₂–N₂ + 0.31 mol Ar = 8.06 mol (2.1% of loop gas)
(a) Recycle rate	361 mol H₂–N₂ (≈ 376 mol incl. Ar)
(b) Ammonia produced	46.1 mol NH₃

Concept & Theory

Topic. Recycle-and-purge material balance with inert accumulation.

Key relations. Inert balance (in = purge) at the specified ceiling concentration; single-pass conversion $X$ linking reactor feed and unconverted gas; overall reactant balance; stoichiometry $N_2+3H_2\to2NH_3$ (4 mol reactant → 2 mol NH₃).

Why this works. A purge is the only exit for a chemically inert species, so its rate is pinned entirely by the argon balance — independent of the per-pass conversion. The single-pass conversion is what determines how large the recycle must be to achieve the high overall conversion, and the overall H₂–N₂ balance (fresh minus purge) gives the ammonia without tracking the loop internals.

Common pitfall. Confusing single-pass conversion (20%) with overall conversion (here 92%): the recycle multiplies the loop throughput so a modest per-pass figure still yields near-complete overall conversion. Also, the purge composition equals the recycle composition (a common split), not the fresh-feed composition.

Question 3 of 6: Ammonia-Synthesis Loop with Argon Purge

Question 3: Ammonia-Synthesis Loop with Argon Purge (Part A — 25 marks)