Raising Reactor Temperature to Chase Conversion
The plant runs a tubular reactor for the liquid-phase consecutive reaction A → B (desired) → C (waste). Both steps are first order. At the present 120 °C, k₁ = 0.40 h⁻¹ and k₂ = 0.10 h⁻¹, and the reactor gives 2.0 h of residence time. The A conversion of 55% is below the 90% target. Operations proposes raising the reactor temperature to 150 °C, which the catalyst will tolerate, to push conversion above 90%. Activation energies are 60 kJ/mol for A → B and 90 kJ/mol for B → C. The reactor is not isothermal; the inlet half runs about 15 °F hotter than the outlet. B sells at a good margin and C must be disposed of as waste.
1. Calculate the current conversion, B yield, and selectivity to B.
Read the worked answer
For a PFR with consecutive first-order steps:
C_A/C_A0 = exp(−k₁τ) = exp(−0.40 × 2.0) = exp(−0.80) = 0.449 → conversion of A = 55.1%
C_B/C_A0 = [k₁/(k₂ − k₁)] × [exp(−k₁τ) − exp(−k₂τ)]
= (0.40 / (0.10 − 0.40)) × (0.449 − exp(−0.20))
= (−1.333) × (0.449 − 0.819) = (−1.333)(−0.370) = 0.493
C_C/C_A0 = 1 − 0.449 − 0.493 = 0.058
- Yield of B = 49.3% of the A fed
- Selectivity = 0.493 / 0.551 = 89.5%
So roughly 1 part of the A that reacts goes to waste for every 9 parts that make B.
2. Estimate the performance at 150 °C with the same 2.0 h residence time. Does the proposal work?
Read the worked answer
Use the Arrhenius ratio: k₂/k₁ rate factor = exp[−(Ea/R)(1/T₂ − 1/T₁)] with T₁ = 393 K, T₂ = 423 K.
(1/423 − 1/393) = −1.805 × 10⁻⁴ K⁻¹
A → B (60 kJ/mol): exp[+(60,000/8.314)(1.805 × 10⁻⁴)] = exp(1.303) = 3.68 → k₁ = 0.40 × 3.68 = 1.47 h⁻¹
B → C (90 kJ/mol): exp[+(90,000/8.314)(1.805 × 10⁻⁴)] = exp(1.954) = 7.06 → k₂ = 0.10 × 7.06 = 0.706 h⁻¹
The undesired step speeds up almost twice as much as the desired one.
At τ = 2.0 h: exp(−2.94) = 0.053; exp(−1.41) = 0.244
C_B/C_A0 = (1.47/(0.706 − 1.47)) × (0.053 − 0.244) = (−1.92)(−0.191) = 0.368
- Conversion of A = 1 − 0.053 = 94.7% - target met
- Yield of B = 36.8% - down from 49.3%
- Selectivity to B = 38.9% - less than half the current value
The proposal raises conversion while destroying a quarter of the B yield. It is the wrong lever at this residence time.
3. Recommend how to operate. Support the recommendation with the optimum residence time at each temperature.
Read the worked answer
Find the optimum for each temperature. For consecutive first-order kinetics the maximum B occurs at
τ_opt = ln(k₂/k₁) / (k₂ − k₁)
- At 120 °C: τ_opt = ln(0.25)/(−0.30) = 4.62 h, C_B/C_A0 = 0.63
- At 150 °C: τ_opt = ln(0.479)/(−0.766) = 0.96 h, C_B/C_A0 = 0.51
Interpretation. Higher temperature is not inherently bad for selectivity in this chemistry - it is bad only when you also hold the residence time fixed. At 150 °C the reactor should be about 1 h, not 2 h; running 2 h at 150 °C means you have already passed the B peak and are converting B into C.
Recommendation
- If the existing 2 h reactor stays as-is, hold 120 °C and accept the conversion, or push the residence time toward 4.6 h (lower feed rate, or add volume) to reach 63% B yield.
- If additional throughput is the real goal, go to 150 °C and cut the residence time to about 1 h - same reactor, twice the feed, ~51% yield. This is usually the better economic answer because it uses the capital already installed.
- Do not accept the operations proposal of 150 °C at 2 h - it is the worst of both worlds.
- Address the non-isothermal profile. The first half of the reactor at +15 °F is already behaving like the 150 °C case, so the front end makes B and the back end destroys it. Pre-cooling the feed or splitting the feed into stages flattens the profile and pushes the plant closer to the isothermal predictions above.
- Confirm the kinetics on plant data before committing - a 15 °F profile and any axial dispersion mean the real reactor is worse than the plug-flow model here.