Question 1 of 6: Monochlorodecane Process — Recycle Structure and Minimum Selectivity

Question 1: Monochlorodecane Process — Recycle Structure and Minimum Selectivity (20 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.

(a) Alternative recycle structures [12 points]

The two reactions form a series (consecutive) system, DEC → MCD → DCD, in which the desired product is the intermediate, MCD. Chlorine is consumed by both steps; the wanted byproduct HCl is co-produced by both and is saleable, while DCD is worthless. The two synthesis levers the question asks for are the single-pass conversion of the raw materials and the ratio (excess) of the two reactants. Because the boiling points are widely and monotonically separated (HCl 188 < Cl₂ 239 < DEC 447 < MCD 488 < DCD 514 K), a simple distillation train can recover HCl overhead for sale, take Cl₂ and unreacted decane for recycle, deliver MCD as the product, and reject DCD as the heavy bottoms — so any unconverted reactant can be recycled to extinction and only selectivity, not per-pass conversion, limits the raw-material economics.

Enumerating the alternatives along the two levers:

• Excess decane, low per-pass conversion (recycle DEC). Feed chlorine as the limiting reactant and hold a large excess of decane in the reactor. A high DEC : low MCD concentration ratio favours the first reaction and starves the second; unreacted decane is recovered and recycled. Chlorine is driven to near-complete conversion so little or no Cl₂ recycle is needed.
• Excess chlorine (recycle Cl₂). A chlorine excess drives decane conversion high but keeps Cl₂ abundant in the reactor, which promotes the second chlorination of MCD to DCD — poor for selectivity. Unreacted Cl₂ is recovered overhead and recycled.
• Near-stoichiometric feed. An intermediate case: moderate conversion, both reactants partly recycled, selectivity between the two extremes.
• High vs. low conversion (each of the above). Pushing single-pass conversion up (less recycle, smaller separation load) forces more of the already-formed MCD onward to DCD; keeping conversion low (large recycle, larger columns) preserves selectivity. The trade-off is selectivity/raw-material cost against recycle and separation capital.

The representative structure — excess decane, low per-pass conversion, with decane (and any chlorine) recycled — is shown below.

(b) Structure that best suppresses the side reaction [3 points]

The excess-decane, low-conversion structure with decane recycle is the most effective. The side reaction MCD + Cl₂ → DCD is fed by both a high MCD concentration and a high Cl₂ concentration; holding a large excess of decane keeps the MCD mole fraction low and, because chlorine is the limiting reactant, keeps Cl₂ low as well. Both drivers of the consecutive reaction are simultaneously suppressed, so decane selectivity to MCD is maximised. The penalty — a large decane recycle and correspondingly larger recovery columns — is acceptable because decane is recovered essentially completely by distillation.

(c) Minimum selectivity for profitable operation [5 points]

Given. Molecular weights and values from the table; two reactions sharing chlorine; DCD worthless, HCl saleable. Define the selectivity $S$ as the fraction of reacted decane that ends as MCD (the remainder, $1-S$, is carried on to DCD).

Find. The value of $S$ at which the economic potential — value of saleable products minus cost of raw materials, per unit of decane reacted — falls to zero.

Approach

On a basis of one mole of decane reacted, write chlorine consumption and HCl/MCD/DCD production in terms of $S$ from reaction stoichiometry, price each stream, and set the economic potential $EP(S)=0$. Below this selectivity the raw materials cost more than the products are worth.

1. Stoichiometric material balance per mole of decane reacted. A fraction $S$ follows only the first reaction (1 mol Cl₂, 1 mol HCl, 1 mol MCD each); the fraction $1-S$ is chlorinated twice (2 mol Cl₂, 2 mol HCl, 1 mol DCD each). Hence, per mole decane: $$n_{\text{MCD}}=S,\quad n_{\text{DCD}}=1-S,\quad n_{\text{Cl}_2}=n_{\text{HCl}}=S+2(1-S)=2-S.$$
2. Value of products (per mole decane). DCD is worthless, so only MCD and HCl earn revenue: $$V_{\text{prod}}=M_{\text{MCD}}p_{\text{MCD}}\,S+M_{\text{HCl}}p_{\text{HCl}}(2-S)=176(0.45)S+36(0.35)(2-S).$$ With $176(0.45)=79.2$ and $36(0.35)=12.6$, this is $V_{\text{prod}}=79.2\,S+12.6(2-S)=66.6\,S+25.2$ ($ per mol decane).
3. Cost of raw materials (per mole decane). One mole of decane plus $2-S$ moles of chlorine: $$C_{\text{raw}}=M_{\text{DEC}}p_{\text{DEC}}+M_{\text{Cl}_2}p_{\text{Cl}_2}(2-S)=142(0.27)+71(0.21)(2-S).$$ With $142(0.27)=38.34$ and $71(0.21)=14.91$, this is $C_{\text{raw}}=38.34+14.91(2-S)=68.16-14.91\,S$.
4. Economic potential and the break-even selectivity. Subtract: $$EP(S)=V_{\text{prod}}-C_{\text{raw}}=(66.6\,S+25.2)-(68.16-14.91\,S)=81.51\,S-42.96.$$ Setting $EP(S)=0$ gives $S_{\min}=\dfrac{42.96}{81.51}=0.527$. Minimum decane selectivity for profitable operation: $\boxed{S_{\min}=0.527\ \ (52.7\%)}$
5. Interpret. Below about 53 % selectivity, so much decane and chlorine are consumed making worthless DCD that the MCD and HCl revenue cannot pay for the feed — the process loses money on raw materials alone, before any capital or utility charge. This is precisely why part (b)'s excess-decane, low-conversion structure (which pushes selectivity well above this floor) is chosen.