OpenSpecEngineering

Finds how long a batch reactor must run to reach a target conversion, for a zero-, first- or second-order reaction. It also gives the half-life and the concentration left at the end. Process engineers and chemists use it to set batch cycle times and to compare kinetics measured in the lab.

Equations

For a single reactant A in an isothermal, constant-volume batch reactor, the design equation integrates to [1], [2]:

Zero order: t = C_A0·X/k
t_½ = C_A0/(2k)

First order: t = ln(1/(1 − X))/k
t_½ = ln 2/k

Second order: t = X/(k·C_A0·(1 − X))
t_½ = 1/(k·C_A0)

t is the reaction time, X the fraction of A converted, from 0 to 1, C_A0 the starting concentration of A in mol/L, and k the rate constant. The final concentration is C_A = C_A0·(1 − X). The half-life t_½ is the time to convert half of A. Only for first order is it independent of C_A0.

Inputs and limits

• k must use mol, L and s. Its units follow the order: mol/(L·s) for zero order, 1/s for first order, L/(mol·s) for second order. A k quoted per minute must be divided by 60.
• The second-order form is for −r_A = k·C_A², or for A + B with equal starting concentrations. Unequal feeds need a different form [1].
• The time is reaction time only. Filling, heating, cooling and emptying add to the batch cycle.
• The reactor is isothermal. If the temperature drifts, k changes and the result does not hold.
• Liquids keep a near-constant volume. A gas-phase reaction that changes the number of moles at constant pressure does not.

Worked example

Ethyl acetate is saponified with sodium hydroxide in a stirred batch vessel at 25 °C, with both at 0.1 mol/L. The reaction is second order with k = 0.11 L/(mol·s). How long until 95 % of the ester has reacted?

1. Choose Reaction Order Second order (−rA = k·CA²).
2. Enter Rate Constant k (consistent units: mol,L,s) 0.11, Initial Concentration CA0 (mol/L) 0.1 and Target Conversion (X, 0–1) 0.95.
3. Press Solve. The Solution shows: Reaction time: 1,727.27 s
   Reaction time: 28.7879 min
   Reaction time: 0.479798 hr
   Half-life (t½): 90.9091 s
   Final concentration CA: 0.005 mol/L

Check by hand: k·C_A0 = 0.011 s⁻¹, so t_½ = 1/0.011 = 90.9 s, and t = 0.95/(0.011 × 0.05) = 1727 s. For second order, reaching 95 % takes 19 half-lives, against about 4.3 for a first-order reaction, so the last few percent cost most of the time.