ChemicalReaction EngineeringmediumFE ExamCourseworkPE / Advanced

CSTR Volume for a First-Order Reaction

A liquid-phase first-order reaction $A \rightarrow B$ has $k = 0.10\ \text{min}^{-1}$ .

A CSTR operating at steady state must achieve $90\%$ conversion at a volumetric feed rate of $v_0 = 10\ \text{L/min}$ . Assume constant density.

Find the required reactor volume.

Given

k	0.1 1/min	Rate constant
X	0.9 -	Fractional conversion
v0	10 L/min	Volumetric feed rate

Hint 1

A CSTR is perfectly mixed — at what concentration does the reaction actually proceed?

Hint 2

The whole vessel sits at the exit concentration, C_A0(1−X).

Hint 3

V = v₀X/(k(1−X)).

Worked solution — try the problem first

CSTR design equation:

$V = \frac{F_{A0}X}{-r_A}$

For first order, $-r_A = kC_A$ , and in a CSTR the reaction occurs at the exit concentration $C_A = C_{A0}(1-X)$ . With $F_{A0} = v_0C_{A0}$ :

$V = \frac{v_0C_{A0}X}{kC_{A0}(1-X)} = \frac{v_0 X}{k(1-X)}$

$C_{A0}$ cancels — the volume is independent of feed concentration for a first-order reaction.

$V = \frac{10 \times 0.90}{0.10 \times 0.10} = \frac{9.0}{0.010} = 900\ \text{L}$

$\boxed{V = 900\ \text{L}}$

Why so large. A CSTR is perfectly mixed, so the entire vessel sits at the low exit concentration where the rate is slowest. A PFR achieving the same conversion needs only 230 L — under a third of the volume. The penalty grows sharply with conversion: at 99% the CSTR would need 9900 L.

Concepts:CSTR designReactor sizingFirst-order kinetics

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