Batch Reactor Design

From a mole balance to a usable design equation for time-to-conversion.

reaction engineeringmole balancesdesign equations

⏱️ About 18 min

If you can predict how long a batch needs to run, you can size schedules, campaigns, and production capacity.

💡

The big idea: For a batch reactor, conversion grows with time because moles are consumed; the design equation links time to the rate law.

🎯 By the end, you'll be able to

Write a mole balance for a batch reactor
Express the design equation in terms of conversion
Use a first-order rate law to compute batch time
Recognize how slowing rates at high conversion stretch batch time

📎 Helpful to know first

Kinetics Applications

What makes batch design different?

In a batch reactor there is no flow in or out during the reaction step. That means the number of moles of species A inside the vessel changes only because of reaction. Instead of solving for a reactor volume, we usually solve for the required reaction time to reach a desired conversion.

🔑 Conversion is the natural progress variable

Batch design is easiest when you express the changing amount of A using conversion X, where X = (N_A0 - N_A)/N_A0. Then N_A = N_A0(1 - X).

Start from the batch mole balance

A mole balance on species A in a batch reactor is:

accumulation = generation by reaction.

Because A is being consumed, the reaction term is negative for A. We write the rate of disappearance as (-r_A) so it is positive.

\[ \frac{dN_A}{dt} = -r_A V \quad \Rightarrow \quad t = \int_{0}^{X}\frac{N_{A0}\,dX}{V(-r_A)} \]

Batch reactor design equation written in terms of conversion X.

Constant-volume, first-order kinetics

For many liquid-phase batch problems, volume is approximately constant. If the reaction is first-order in A, then (-r_A) = k C_A. With constant volume, C_A = C_A0(1 - X).

\[ t = \frac{1}{k}\ln\!\left(\frac{1}{1-X}\right) \]

Closed-form time to reach conversion X for an irreversible first-order reaction at constant volume.

✨ Why time blows up near complete conversion

As X approaches 1, the remaining concentration of A becomes small and the reaction slows. The logarithm grows without bound, which captures the practical reality: reaching 99.9% conversion can take much longer than reaching 90%.

📝 Worked example: A liquid-phase batch reactor operates at constant volume. The reaction is first-order in A with k = 0.25 min<sup>-1</sup>. How long is required to reach X = 0.80?

Use the first-order batch expression: t = (1/k) ln(1/(1 - X)).
Compute 1 - X = 1 - 0.80 = 0.20, so 1/(1 - X) = 1/0.20 = 5.
Compute ln(5) = 1.6094 (check: exp(1.6094) = 5.00).
Compute t = (1/0.25) × 1.6094 = 4 × 1.6094 = 6.4376 min.
Round sensibly to t = 6.44 min.

✓ 6.44 min

✏️ Practice: A constant-volume batch reactor runs a first-order reaction with k = 0.12 min<sup>-1</sup>. Find the time needed to reach X = 0.70.

min

Solution

Use t = (1/k) ln(1/(1 - X)).
Compute 1 - X = 0.30, so 1/(1 - X) = 3.3333.
Compute ln(3.3333) = 1.2040 (check: exp(1.2040) = 3.333).
Compute t = (1/0.12) × 1.2040 = 8.3333 × 1.2040 = 10.033 min.
Report t = 10.03 min.

Check your understanding

1. In a batch reactor during the reaction step, which term is zero?

There is no flow into or out of a closed batch reactor during the reaction step, so inlet/outlet terms are zero.

2. For a first-order batch reaction at constant volume, what happens to the time required as X approaches 1?

The expression t = (1/k) ln(1/(1 - X)) diverges as (1 - X) goes to zero.

3. If k doubles (all else unchanged), the time to reach a given conversion will:

For first-order batch kinetics, t is proportional to 1/k.

✅ Key takeaways

Batch design usually solves for time, not volume.
A batch mole balance leads to an integral relationship between time and conversion.
For first-order, constant-volume systems: t = (1/k) ln(1/(1 - X)).
High conversion requires disproportionately more time because the rate slows as A is depleted.

➡️ Next we shift from time-to-conversion to steady-flow sizing: the CSTR design equation.

Want to test yourself on this? Try the Chemical Aptitude test →