Bosch process

Ammonia factories, oil hydrogenation plants and rocket-fuel producers all need hydrogen gas by the tonne — far more than a school flask of zinc and dilute acid could ever supply. The Bosch process is the industrial answer: it manufactures hydrogen on a massive scale starting from nothing more exotic than coke (nearly pure carbon) and steam. Named after the German chemist Carl Bosch, who scaled up laboratory chemistry into working factories in the early twentieth century, the process runs in two linked stages — first making a raw fuel gas, then chemically "enriching" it — followed by a purification step that strips away every trace of carbon oxides. This article walks through both stages, shows algebraically why each carbon atom in the coke ultimately yields two molecules of hydrogen, and works a full stoichiometry problem from a mass of coke to a volume of hydrogen gas.

The core insight

The Bosch process makes hydrogen industrially by reacting red-hot coke with steam to form water gas (CO + H₂), then catalytically converting that carbon monoxide with more steam into carbon dioxide and extra hydrogen, before scrubbing out the CO₂ and CO to leave pure H₂.

The core idea

Stage 1 — water gas. Steam passed over red-hot coke at about 1000 ∘C1000\,^{\circ}\text{C} gives a mixture of carbon monoxide and hydrogen called water gas:

C+H2O→∼1000 ∘CCO+H2C + H_2O \xrightarrow{\sim 1000\,^{\circ}\text{C}} CO + H_2

where: CC is coke (solid carbon), H2OH_2O is steam, and the product mixture CO+H2CO + H_2 is water gas, not water itself. The reaction is endothermic, so the coke bed cools as it runs; industrially, air is blown through periodically to reheat the coke (C+O2→CO2C + O_2 \rightarrow CO_2), and the two steps — steam, then air, then steam again — alternate.

Stage 2 — catalytic enrichment. Water gas is mixed with more steam and passed over a finely divided iron oxide catalyst (promoted with a little chromium oxide) at about 450 ∘C450\,^{\circ}\text{C}:

CO+H2O→450 ∘CcatalystCO2+H2CO + H_2O \xrightarrow[450\,^{\circ}\text{C}]{\text{catalyst}} CO_2 + H_2

where: the catalyst does not appear in the products — it only speeds up an otherwise sluggish reaction between COCO and steam. This stage is called "enrichment" because it raises the proportion of hydrogen in the gas: every leftover COCO molecule is traded for one more H2H_2 molecule.

Purification. The gas leaving Stage 2 is mostly H2H_2, plus CO2CO_2 and a little unreacted COCO. Both must go before the hydrogen is usable.

  • The mixture is passed through water under pressure, or through sodium hydroxide or potassium carbonate solution — an alkali absorbs the acidic gas CO2CO_2: CO2+2NaOH→Na2CO3+H2OCO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O.
  • Remaining traces of COCO are removed by passing the gas through ammoniacal cuprous chloride solution, which selectively absorbs carbon monoxide.
  • What emerges is essentially pure hydrogen, ready for uses such as the Haber process (ammonia manufacture) or oil hydrogenation.
Figure
Flow diagram of the Bosch process showing coke and steam entering stage one to form water gas, water gas and more steam entering stage two over an iron oxide catalyst to form enriched gas, then a purification stage removing carbon dioxide and carbon monoxide to leave pure hydrogen
The Bosch process — two stages, then purification Stage 1 coke + steam ~1000 C, no catalyst C + H2O to CO + H2 water gas Stage 2 — enrichment water gas + more steam 450 C, iron oxide catalyst CO + H2O to CO2 + H2 enriched gas Purification NaOH or K2CO3 solution absorbs CO2 ammoniacal cuprous chloride absorbs CO pure H2 gas out used for ammonia, oil hydrogenation Net: one carbon atom from coke, combined with two steam molecules, ultimately yields two H2 molecules. C plus 2 H2O gives CO2 plus 2 H2
Derivation ANet hydrogen yield per mole of carbon

The goal is to show, by combining the two Bosch-process equations, that every mole of carbon in the coke ultimately produces exactly two moles of hydrogen gas overall — a single net equation that will drive the worked stoichiometry problem below, without needing any new chemistry beyond adding balanced equations.

Step 1. Write Stage 1 as it stands: C+H2O→CO+H2C + H_2O \rightarrow CO + H_2. One mole of carbon consumes one mole of steam and produces one mole each of carbon monoxide and hydrogen.

Step 2. Write Stage 2 as it stands: CO+H2O→CO2+H2CO + H_2O \rightarrow CO_2 + H_2. The carbon monoxide made in Stage 1 consumes a second mole of steam and produces one mole each of carbon dioxide and hydrogen.

Step 3. Add the two equations together, term by term: C+H2O+CO+H2O→CO+H2+CO2+H2C + H_2O + CO + H_2O \rightarrow CO + H_2 + CO_2 + H_2.

Step 4. Cancel the species that appear identically on both sides — one mole of COCO is made in Step 1 and consumed in Step 2, so it cancels completely (it is an intermediate, not a net product). This leaves C+2H2O→CO2+2H2C + 2H_2O \rightarrow CO_2 + 2H_2.

Step 5. Read off the result: one mole of carbon (coke) reacting with two moles of steam gives, overall, one mole of carbon dioxide and two moles of hydrogen gas. This confirms, algebraically, why each carbon atom used is worth two hydrogen molecules once both stages have run to completion — the fact stated qualitatively in the core-idea section above.

Worked example

Question: 60 g60\ \text{g} of coke is completely converted by the Bosch process (both stages, assuming complete reaction). Find the volume of hydrogen gas obtained at STP, where the molar volume of a gas at STP is 22400 mL22400\ \text{mL} and the molar mass of carbon is 12 g/mol12\ \text{g/mol}.

Step 1. Find moles of carbon: moles of C=massmolar mass=6012=5 mol\text{moles of } C = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{60}{12} = 5\ \text{mol}.

Step 2. Apply the net equation derived above, C+2H2O→CO2+2H2C + 2H_2O \rightarrow CO_2 + 2H_2: the mole ratio of CC to H2H_2 is 1:21 : 2. So moles of H2=5×2=10 molH_2 = 5 \times 2 = 10\ \text{mol}.

Step 3. Convert moles of H2H_2 to volume at STP: volume=moles×22400 mL/mol=10×22400=224000 mL=224 L\text{volume} = \text{moles} \times 22400\ \text{mL/mol} = 10 \times 22400 = 224000\ \text{mL} = 224\ \text{L}.

Answer: 60 g60\ \text{g} of coke yields 224 L224\ \text{L} of hydrogen gas at STP once both stages of the Bosch process are complete.

Competitive stretch

If instead only Stage 1 were run (no enrichment stage), the same 5 mol5\ \text{mol} of carbon would give only 5 mol5\ \text{mol} of H2H_2 — half as much, 112 L112\ \text{L} at STP — because the carbon monoxide would remain unconverted. This is exactly why the enrichment stage exists: it doubles the usable hydrogen yield from the same coke.

Common traps

  • Confusing "water gas" with actual water — water gas is the gas mixture CO+H2CO + H_2, not water vapour or liquid water.
  • Forgetting the catalyst in Stage 2 — without the iron oxide catalyst, COCO and steam do not react fast enough to be useful industrially.
  • Mixing up which stage adds extra steam — the enrichment stage (Stage 2) introduces additional steam to convert COCO into more H2H_2 and CO2CO_2; Stage 1 only passes steam over coke.
  • Assuming carbon dioxide removal and carbon monoxide removal use the same reagent — CO2CO_2 is absorbed by an alkali (NaOH or K2CO3K_2CO_3), while the last traces of COCO need a different reagent, ammoniacal cuprous chloride solution.

Ask the tutor: Want a side-by-side comparison of the Bosch process with the lab preparation of hydrogen from zinc and dilute acid?

Why does the Bosch process need two separate stages instead of one?

Stage 1 alone only gets carbon halfway to hydrogen — it produces a mixture of COCO and H2H_2 in roughly equal amounts, and COCO is a fuel gas, not hydrogen. Stage 2 exists specifically to convert that leftover COCO into extra H2H_2 using more steam and a catalyst, which is why it is called the enrichment stage. Skipping it would waste half the hydrogen locked up in the carbon monoxide, as the competitive-stretch calculation above shows directly.

Why is the iron oxide catalyst necessary in Stage 2 but not in Stage 1?

Stage 1's reaction between red-hot coke and steam at 1000 ∘C1000\,^{\circ}\text{C} is already fast enough at that temperature without a catalyst. Stage 2 runs at a much lower 450 ∘C450\,^{\circ}\text{C} (chosen partly so the catalyst stays effective and partly for practical plant design), and at that lower temperature the reaction between COCO and steam is too slow to be industrially useful on its own. The iron oxide catalyst, promoted with a little chromium oxide, speeds up just this second step without being consumed itself.

Why does the lab method with zinc and acid not work for industrial hydrogen production?

The zinc-and-acid reaction is easy to set up and control, which is exactly why it suits a classroom demonstration, but it is far too expensive and slow to scale up: zinc is a costly metal, and the reaction produces only small volumes of gas at a time. Industry instead needs a continuous, cheap-feedstock process — coke and steam are both abundant and inexpensive, and the Bosch process can run continuously in a large reactor, which is why it, not the lab method, supplies hydrogen for uses like ammonia manufacture.

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