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Le Chatelier in an Industrial Ammonia Reactor

Praxis Chemistry 5245 · Put your knowledge to work in context.

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Le Chatelier in an Industrial Ammonia Reactor

A simplified Haber-Bosch reactor runs N2(g) + 3 H2(g) ⇌ 2 NH3(g), ΔH = −92 kJ/mol. The plant currently operates at 450 °C and 200 atm with a catalyst, achieving about 15% conversion per pass. An engineer proposes three changes to raise per-pass yield: (1) lower the temperature to 350 °C, (2) raise the pressure to 300 atm, and (3) remove NH3 continuously as it forms. Management wants higher yield but also cares about reaction rate and energy cost. The catalyst is active only above roughly 400 °C; below that, the rate becomes impractically slow with the current catalyst.

1. For each of the three proposed changes, predict the effect on equilibrium position using Le Chatelier's principle, and state whether the change also affects the reaction rate.

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(1) Lower T to 350 °C: The forward reaction is exothermic, so removing heat shifts equilibrium toward products (higher equilibrium yield). However, lowering T also decreases the rate constant (Arrhenius), and the catalyst is reported inactive below ~400 °C, so the practical rate collapses. Net: higher equilibrium yield but slower approach - a trade-off.

(2) Raise P to 300 atm: There are 4 mol gas on the left and 2 mol on the right. Increasing pressure shifts equilibrium toward the side with fewer gas moles (products). Rate also increases because higher partial pressures raise collision frequency. Net: favorable for both yield and rate, but higher pressure raises equipment/energy costs.

(3) Remove NH3 continuously: Lowering product concentration shifts equilibrium to the right (products replenished), increasing conversion. Removing product does not change the forward rate constant, though it can affect the net rate. Net: favorable for yield; this is why real plants condense ammonia out and recycle unreacted gas.

2. The engineer argues that because the reaction is exothermic, running as cold as possible is always best. Critique this reasoning in terms of the rate–yield trade-off and identify the operating-point compromise.

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The argument conflates thermodynamics with kinetics. Exothermicity means K increases as T decreases (van't Hoff: ln(K2/K1) = −ΔH/R (1/T2 − 1/T1); with ΔH negative, lowering T raises K). So colder gives a higher equilibrium conversion.

But equilibrium yield is only useful if the reaction reaches it in reasonable time. The rate constant falls exponentially with 1/T, and the catalyst has a minimum working temperature (~400 °C). At 350 °C the mixture may be thermodynamically favorable but kinetically frozen.

The compromise operating point balances these: a moderate temperature (here ~450 °C) hot enough for acceptable rate and catalyst activity, but not so hot that K collapses; combined with high pressure and continuous NH3 removal to push conversion. The judgment the exam tests is recognizing that 'more favorable equilibrium' ≠ 'better process,' and that the optimum is a kinetic-thermodynamic compromise, not an extreme.

3. If the plant raises pressure to 300 atm and removes NH3, the engineer claims conversion will exceed the equilibrium conversion at 200 atm with no removal. Is this claim coherent, and how would you interpret 'conversion' in a flow reactor?

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The claim is loosely worded but defensible if 'equilibrium conversion' is defined for a fixed set of conditions. Raising pressure shifts the equilibrium constant expressed in partial pressures toward products, so the equilibrium conversion at 300 atm is genuinely higher than at 200 atm. Continuous removal of NH3 keeps the reaction quotient Q below K, so the system never reaches the static equilibrium conversion - it keeps converting as product is withdrawn. Thus per-pass conversion can exceed the single-pass equilibrium conversion computed for the no-removal, 200 atm case.

In a flow reactor, 'conversion' means fraction of limiting reactant consumed per pass through the reactor. With recycle of unreacted gas, overall conversion can approach completion even though per-pass conversion is modest. The exam-relevant judgment: distinguish per-pass conversion, equilibrium conversion, and overall conversion with recycle - they are not the same number, and comparisons are only meaningful when the basis is stated.