Maximising Yield in a Parallel Reaction
A pharmaceutical intermediate D is produced from substrate S by a first-order reaction with rate constant kD. A competing first-order side reaction forms impurity I with rate constant kI. At 25°C, kD = 1.25 × 10−3 min−1 and kI = 5.0 × 10−5 min−1. At 55°C, kD = 6.0 × 10−2 min−1 and kI = 1.2 × 10−2 min−1. The reactor can be run for 4 hours. Product D is thermally labile and decomposes rapidly above 45°C, so the operating temperature must not exceed 45°C. The plant currently runs at 25°C and achieves only 26% isolated yield of D after 4 hours. The chemist proposes raising the temperature to 40°C to speed up the reaction. Evaluate this proposal. Use the Arrhenius equation (R = 8.314 J mol−1 K−1) and assume the rate constants follow Arrhenius behaviour over this range. The initial concentration of S is 0.10 M; D and I are the only products.
1. Calculate the activation energy for the formation of D and for the formation of I.
Read the worked answer
Use the Arrhenius equation in the form: ln(k2/k1) = (Ea/R)(1/T1 − 1/T2), with T1 = 298 K and T2 = 328 K. 1/298 − 1/328 = 3.356 × 10−3 − 3.049 × 10−3 = 3.07 × 10−4 K−1.
For D: k2/k1 = 6.0 × 10−2 / 1.25 × 10−3 = 48.0; ln 48.0 = 3.871. Ea(D) = R ln(k2/k1) / (1/T1 − 1/T2) = 8.314 × 3.871 / 3.07 × 10−4 = 1.048 × 10^5 J mol−1 = 105 kJ mol−1.
For I: k2/k1 = 1.2 × 10−2 / 5.0 × 10−5 = 240; ln 240 = 5.481. Ea(I) = 8.314 × 5.481 / 3.07 × 10−4 = 1.484 × 10^5 J mol−1 = 148 kJ mol−1.
The side reaction has a significantly higher activation energy, so it accelerates more strongly with temperature and selectivity falls as temperature rises.
2. Calculate kD and kI at 40°C (313 K) using the activation energies, and determine the selectivity ratio kD/kI at 25°C, 40°C, and 55°C.
Read the worked answer
At 40°C, 1/313 − 1/298 = 3.195 × 10−3 − 3.356 × 10−3 = −1.61 × 10−4 K−1.
For D: ln(kD40/kD25) = −Ea(D)/R × (1/313 − 1/298) = −(1.048 × 10^5 / 8.314) × (−1.61 × 10−4) = 2.03. kD40 = 1.25 × 10−3 × e^2.03 = 1.25 × 10−3 × 7.61 = 9.51 × 10−3 min−1.
For I: ln(kI40/kI25) = −(1.484 × 10^5 / 8.314) × (−1.61 × 10−4) = 2.87. kI40 = 5.0 × 10−5 × e^2.87 = 5.0 × 10−5 × 17.6 = 8.80 × 10−4 min−1.
Selectivity ratios: At 25°C: kD/kI = 1.25 × 10−3 / 5.0 × 10−5 = 25. At 40°C: 9.51 × 10−3 / 8.80 × 10−4 = 10.8. At 55°C: 6.0 × 10−2 / 1.2 × 10−2 = 5.0. As expected, selectivity decreases with temperature because Ea(I) > Ea(D).
3. Calculate the expected yield of D after 4 hours at 25°C and at 40°C, and recommend whether the chemist’s proposal should be adopted. Consider both yield and reaction time.
Read the worked answer
For parallel first-order reactions from S, the fraction of S converted to D is: fD = [kD/(kD + kI)] [1 − exp(−(kD + kI)t)].
At 25°C: kD + kI = 1.25 × 10−3 + 5.0 × 10−5 = 1.30 × 10−3 min−1. t = 240 min; (kD + kI)t = 0.312; 1 − e^−0.312 = 0.268. fD = (1.25 × 10−3 / 1.30 × 10−3) × 0.268 = 0.962 × 0.268 = 0.258, i.e. 25.8%, consistent with the plant’s 26% yield.
At 40°C: kD + kI = 9.51 × 10−3 + 8.80 × 10−4 = 1.039 × 10−2 min−1. (kD + kI)t = 2.49; 1 − e^−2.49 = 0.917. fD = (9.51 × 10−3 / 1.039 × 10−2) × 0.917 = 0.915 × 0.917 = 0.839, i.e. 83.9%.
The proposal should be adopted. At 40°C the selectivity is lower (10.8 instead of 25), but the much faster reaction more than compensates: after the same 4 hours, the yield rises from about 26% to about 84%. The temperature also remains below the 45°C thermal-lability limit for D. Operationally, 40°C is the best compromise between conversion, selectivity, and product stability.