Chemistry · Lesson

Major industrial chemicals and reaction systems

Read major-chemical reaction systems through inventory bounds, conversion, global nets and thermodynamic quotients.

By the end, you can

Calculate a limiting formal feed inventory.
Separate reaction conversion from product recovery.
Interpret a global industrial net and its selected atom economy.
Use dimensionless activities to assess an instantaneous reaction tendency.

Industrial chemistry transforms feedstocks into useful materials

Large chemical systems connect feedstocks to intermediates and products such as fertilizers, polymers, construction materials and other substances. A product’s chemical identity and required specification determine which transformations and separations matter.

A net equation describes a stoichiometric relation. It does not describe every catalyst, intermediate, separation, energy input or byproduct in the production system. Different routes can deliver the same specified product with different wider consequences.

A feed bound is different from an attained output

The formal ammonia calculation below starts with complete assigned reactant amounts and a balanced net. It gives a maximum allowed by that inventory, before equilibrium, kinetics or separations are considered. An actual reaction system needs evidence for those additional constraints. A product name alone does not specify a production route.

In the separate ethene hydration case, reaction extent and recovery are assigned. Conversion counts the share of feed reactant consumed, while isolated yield counts collected desired product relative to the declared feed basis. Unisolated ethanol remains chemically present in the account.

Desired-product selectivity needs its own consumed-reactant denominator and treatment of competing products. Retain unreacted feed, auxiliary substances and relevant output streams rather than using a product-only ledger.

Reaction progress and isolation retain different quantities

Quantity	Recovery 0.850	Recovery 0.650
Ethene conversion	0.400	0.400
Ethanol formed / kmol	2.72	2.72
Ethanol isolated / kmol	2.312	1.768
Ethanol unisolated / kmol	0.408	0.952
Isolated ethene-based yield	0.340	0.260

Original assigned ethene hydration case with extent 2.72 kmol. Changing only recovery leaves the full reaction inventory unchanged; the ethanol split changes. No actual process yield is reported.

An equilibrium limit and a production rate are different constraints

A reversible transformation has an equilibrium relation under specified conditions. Reaction kinetics and transport affect how quickly a system approaches an admissible composition. A catalyst changes reaction routes and rates without changing the same reaction’s thermodynamic equilibrium under unchanged conditions.

Removing products or changing a maintained boundary can affect overall conversion. This does not mean a catalyst alone makes an unfavorable equilibrium disappear. A process account must include separation and its material and energy requirements.

A net equation and an equilibrium relation have bounded roles

A global industrial net can summarize a multi-step process after intermediate and regenerated participants cancel. It is not necessarily a direct reaction between its left-hand substances. The cited soda-ash account makes that distinction explicit; the original mass example below evaluates only its formal inventory.

Atom economy compares a selected desired-product mass with the stoichiometric reactant mass in that written net. It excludes many process inputs and does not establish actual yield, resource use, toxicity or coproduct benefit. Its chosen boundary should accompany the number.

A reaction quotient compares relative activities with a dimensionless equilibrium constant under defined standard states. At fixed temperature Q/K determines the thermodynamic tendency. It does not calculate a kinetic rate. Compressing an assigned ideal mixture changes its instantaneous quotient; subsequent reaction and real-gas behaviour require a further model.

Producing a compound and obtaining a suitable product are different steps

A reaction stream can contain unconverted feed, desired product and other components. Product specification can require separation, purification or formulation. Trace constituents can matter even when a bulk composition looks similar.

A separation yield and reaction yield should be applied to their respective inventories. The product of compatible sequential yields gives an overall recovered fraction, while adding those percentages generally does not.

Compare routes within a common boundary

Feedstock origin, energy supply, reaction selectivity, separations and waste streams can affect a route’s wider performance. A lower input for one unit can shift a requirement elsewhere in the process.

The original reaction and recovery ledgers below use assigned inputs and balanced transformations. They are neither current industry production statistics nor recipes for synthesizing industrial chemicals.

Worked examples

A formal feed inventory sets an upper bound

For the formal net N₂+3H₂→2NH₃, assign 8.30 kmol N₂ and 21.9 kmol H₂. Use rounded teaching molar masses 28,2,17 kg kmol⁻¹. Find the maximum inventory conversion under this net.

Extent ξ cannot exceed either 8.30 or 21.9/3=7.30 kmol. Thus the formal maximum is 7.30 kmol, producing 14.6 kmol NH₃ with 1.00 kmol N₂ and no H₂ remaining.
Nitrogen atoms close: 16.6=14.6+2.00 kmol atom amounts. Hydrogen atoms close: 43.8=3×14.6. The written neutral reaction has 20 electrons on each side per extent: N₂ contributes 14 and three H₂ six; two NH₃ also total 20.
On the supplied rounded mass basis, inputs 8.30×28+21.9×2=276.2 kg equal 14.6×17+1.00×28=276.2 kg outputs.
This is a theoretical inventory bound. No equilibrium conversion, rate, catalyst, temperature, pressure or actual production yield is assigned.

At most 14.6 kmol NH₃ under the formal net; 1.00 kmol N₂ remains.

Conversion and isolation answer different questions

For formal neutral ethene C₂H₄ + H₂O → ethanol CH₃−CH₂−OH, assign initial amounts 6.80 and 15.0 kmol, net extent 2.72 kmol and ethanol recovery 0.850. Compare conversion with isolated yield.

Ethene conversion is 2.72/6.80=0.400. Formed ethanol is 2.72 kmol; unreacted ethene 4.08 kmol and water 12.28 kmol.
Each written extent contains C₂H₆O on both sides, charge zero and 26 electrons: ethene 16 plus water ten equals ethanol 26. The initial C/H/O atom amounts are 13.6/57.2/15.0 kmol and the full post-reaction inventory retains them.
Isolated ethanol is 2.72×0.850=2.312 kmol and unisolated ethanol 0.408 kmol. Isolated yield relative to initial ethene is 2.312/6.80=0.340.
The reaction extent and recovery fraction are stipulated independently. They do not predict phase behaviour or actual suitability of a particular process. Unisolated product remains in the inventory rather than disappearing.

Ethene conversion 0.400; isolated ethene-based ethanol yield 0.340.

A global net does not specify a direct reaction

Consider the formal global soda-ash net 2NaCl+CaCO₃→Na₂CO₃+CaCl₂ for extent 100 mol. Use rounded atomic masses Na23, Cl35.5, Ca40, C12, O16 g mol⁻¹ and select Na₂CO₃ as the desired product.

Formula-unit molar masses are 58.5,100,106,111 g mol⁻¹. Inputs are 11.7 kg NaCl and 10.0 kg CaCO₃; outputs are 10.6 kg Na₂CO₃ and 11.1 kg CaCl₂, closing total mass 21.7 kg.
Na200, Cl200, Ca100, C100 and O300 mol atom amounts close across the formal net. Neutral formula-unit electron counts close as 2×28+50=106 and 52+54=106 per extent; this is not a claim of molecular salt entities or dissolved speciation.
The selected formal atom economy is 106/(2×58.5+100)=106/217=0.488479263, or about 48.85%. It does not count auxiliaries, actual yield, separation, energy or coproduct usefulness.
The cited industrial source explicitly treats this as a global net that cannot be obtained by simply reacting those two left-hand compounds directly. Actual intermediate steps and regeneration belong to the process account.

Mass closes at 21.7 kg; selected formal atom economy 106/217, without a direct-route claim.

The quotient uses relative activities

For ideal N₂(g)+3H₂(g)⇌2NH₃(g) at one fixed unspecified temperature, assign dimensionless activities a_N₂=0.200,a_H₂=0.600,a_NH₃=0.100 relative to p°=1 bar, and dimensionless K=0.180. Find Q and the instantaneous tendency.

Q=a_NH₃²/(a_N₂a_H₂³)=0.100²/(0.200×0.600³)=25/108=0.231481481.
Q>K, so Δ_rG=RTln(Q/K)>0 for the written forward reaction, with molar gas constant R and absolute temperature T; the thermodynamic tendency is reverse. This does not give a rate or final composition.
Before reaction, uniform scaling of all partial pressures by f multiplies Q by f^(2−1−3)=f⁻². K stays the same at the stipulated temperature and standard states.
Real gases may need fugacities instead of ideal relative pressures. A catalyst changes kinetic paths without changing the equilibrium constant for those fixed standard states. No actual industrial pressure or equilibrium datum is given.

Dimensionless Q=0.231481481>K; the instantaneous thermodynamic tendency is reverse.

Try it yourself

Change only hydrogen to 27.3 kmol. Recompute the limiting inventory, product bound and remaining reactants on the same rounded mass basis.
Show the answer and reasoning
ξ_max=8.30 kmol; NH₃ bound 16.6 kmol, H₂ remainder 2.40 kmol, N₂ remainder zero; rounded mass 287.0 kg on both sides.
The larger H₂ inventory makes N₂ limiting; reactant and product atom inventories still close.
Change only ethanol recovery to 0.65, retaining the same stipulated extent. Which amount and yield change, and which reaction inventory stays fixed?
Show the answer and reasoning
Isolated ethanol 1.768 kmol, unisolated 0.952 kmol and yield 0.260; formed 2.72 kmol and conversion 0.400 are unchanged.
Recovery changes the isolated share, not the stipulated reaction progress.
Scale the formal extent to 80 mol with the same net, desired product and rounded atomic-mass basis. Recompute masses and atom economy.
Show the answer and reasoning
Inputs 9.36+8.00=17.36 kg; outputs 8.48+8.88=17.36 kg; atom economy remains 106/217.
All extent-based masses scale by 0.800; the formal mass ratio is unchanged.
Instantaneously multiply all three partial pressures by 2 at the same temperature before reaction occurs. Recompute Q and its tendency relative to the same K.
Show the answer and reasoning
Q=0.0578703704=original Q/4<K; forward thermodynamic tendency at the same assigned temperature.
The written reaction has gas mole-count change −2; use the same dimensionless standard-state quotient.

What to carry forward

A stoichiometric feed bound does not establish an attained yield.
Reaction conversion and isolation recovery use different denominators.
A global net can omit essential intermediate process steps.
Q/K gives a thermodynamic tendency under the declared standard states.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.