Pilot-Plant Methanol Reactor: Rate Versus Equilibrium
On a summer placement at a methanol pilot plant, an engineering student is asked to review a proposal to raise the reactor temperature from 520 K to 560 K. The reaction is CO(g) + 2 H₂(g) ⇌ CH₃OH(g), ΔH° = −91 kJ mol⁻¹. The feed is stoichiometric CO:H₂ = 1:2 and the reactor operates at 80 bar. Plant data give Kp = 1.6 × 10⁻² bar⁻² at 520 K and 2.1 × 10⁻³ bar⁻² at 560 K. With a fresh Cu/ZnO/Al₂O₃ catalyst and the current contact time, conversion is 12% at 520 K and 18% at 560 K. The student's supervisor argues that the higher temperature is clearly better because conversion almost doubled, and proposes removing the recycle loop to save compression costs. The plant manager wants high per-pass conversion, low energy use, and a catalyst lifetime of at least two years. The student is asked to separate equilibrium limits from kinetic effects before any change is made.
1. Using the Kp values, calculate the equilibrium conversion at 520 K and 560 K for a stoichiometric feed at 80 bar. Compare these with the measured conversions and explain what the comparison reveals about the reactor.
Read the worked answer
Let the initial amounts be 1 mol CO and 2 mol H₂, and let α be the equilibrium conversion of CO. At equilibrium:
n(CO) = 1 − α; n(H₂) = 2 − 2α; n(CH₃OH) = α; total = 3 − 2α.
The partial pressures are obtained from mole fractions multiplied by P = 80 bar. The equilibrium expression is:
Kp = p(CH₃OH) / [p(CO) p(H₂)²]
= α(3 − 2α)² / [4(1 − α)³ P²].
Rearranging:
KpP² = α(3 − 2α)² / [4(1 − α)³].
At 520 K:
KpP² = (1.6 × 10⁻²)(80²) = 102.4.
Solving gives α ≈ 0.848, so the equilibrium conversion is about 85%.
At 560 K:
KpP² = (2.1 × 10⁻³)(80²) = 13.44.
Solving gives α ≈ 0.675, so the equilibrium conversion is about 68%.
The measured conversions are only 12% at 520 K and 18% at 560 K. Both are far below the equilibrium ceiling, so the reactor is strongly kinetically limited under the current contact time and catalyst activity. The higher temperature raises the measured conversion because it increases the reaction rate, but it simultaneously lowers the equilibrium conversion from about 85% to about 68%. The supervisor has confused a rate effect with an equilibrium effect. At 520 K a longer residence time or a more active catalyst could in principle approach 85%, which is much better than the 18% observed at 560 K.
2. Evaluate the supervisor's proposal to run at 560 K and remove the recycle loop. Recommend an operating strategy consistent with high per-pass conversion, low energy use, and two-year catalyst life.
Read the worked answer
The proposal should be rejected as stated.
Temperature. Raising the temperature from 520 K to 560 K lowers Kp because the reaction is exothermic. The equilibrium conversion falls from roughly 85% to 68%. The observed increase from 12% to 18% shows that the reactor is not at equilibrium; it is limited by rate and contact time. Increasing temperature is a legitimate way to increase rate, but it must be balanced against the lower equilibrium ceiling, higher energy cost, and faster catalyst sintering. A Cu/ZnO/Al₂O₃ catalyst is sensitive to overheating, so a two-year lifetime is unlikely if the reactor is pushed to a higher temperature without other changes.
Recycle. Removing the recycle loop would be a serious mistake. Per-pass conversion is only 12–18%, so most of the feed would be wasted. Recycle returns unreacted CO and H₂ and gives a high overall conversion, even if the per-pass conversion is modest. The compression cost of recycle must be compared with the cost of lost feedstock; removing recycle simply to save compression is false economy.
Pressure. Increasing pressure above 80 bar favours methanol because the reaction reduces the number of gas moles from 3 to 1. It also increases the rate. The practical limit is compressor cost, vessel design, and safety.
A better strategy is to keep the recycle loop, operate at a moderate temperature where the catalyst lifetime is acceptable, and improve the approach to equilibrium by increasing contact time, improving heat removal, or using a more active catalyst. If throughput is the priority, a higher temperature could be justified only with a larger recycle ratio and a clear catalyst-replacement cost model. The decision should be based on per-pass conversion, overall conversion, energy per tonne of methanol, and catalyst life, not on the 12% versus 18% figures alone.
3. A colleague says the fresh catalyst has 'shifted the equilibrium to the right'. Correct this statement and explain why catalyst activity matters to the measured 12% and 18% values.
Read the worked answer
The catalyst does not shift the equilibrium position. A catalyst lowers the activation energy of the forward and reverse reactions by the same amount, so the reaction reaches equilibrium faster but the equilibrium constant and equilibrium conversion are unchanged at a given temperature. The only variables that change the equilibrium position here are temperature, pressure, and the feed composition.
The catalyst does affect the measured conversion after a fixed contact time. With a fresh, highly active catalyst, the reaction is closer to its kinetic potential, so 12% at 520 K and 18% at 560 K are measurements of rate, not of equilibrium. If the catalyst were less active or had sintered, the same reactor and contact time would give lower conversions. Conversely, a longer contact time with the same catalyst could raise the 520 K conversion towards the 85% equilibrium value.
The phrase 'shifted the equilibrium to the right' should be replaced by 'increased the rate of approach to equilibrium'. This distinction matters for the decision: if the reactor is far from equilibrium, improving catalyst activity or contact time can raise conversion without sacrificing the favourable equilibrium at 520 K.