Ideal Flow Reactors CSTR and PFR Questions and Answers

Answer: A

Correct answer: 2 - 4 - 1 - 3

A single CSTR is worst, because all of it sits at the exit composition. Splitting the same volume into two tanks in series lets the first operate at a higher concentration and so a higher rate, and more tanks improve it further. In the limit of infinitely many tanks the series approaches plug flow exactly, which is why the PFR heads the order and the sequence runs downward as the number of stages falls.

Answer: B

A large recycle mixes the contents thoroughly and brings the whole reactor towards the exit composition. With no recycle at all it behaves as a plug flow reactor, so the ratio spans both ideals. It is a useful laboratory reactor for this reason.

Answer: C

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the two limits. With no recycle the fluid makes a single pass and the vessel is a PFR. With very large recycle the feed is diluted into a stream circulating many times, so the composition becomes almost uniform and the vessel behaves as a stirred tank. Varying the ratio therefore sweeps the whole range between the two ideals, which makes the recycle reactor useful for measuring kinetics.

Answer: D

The volumetric flow rises as the reaction proceeds, so the fluid moves through faster than the entering flow alone would suggest. Space time and residence time coincide only at constant density. Expansion accelerates the flow through the tube.

Answer: A

Correct answer: 1 and 2 only

Statements 1 and 2 are the standard trade: the PFR is usually smaller and the CSTR easier to control thermally. Statement 3 is wrong - the order matters greatly. At zero order the two volumes are identical, at first order the CSTR is somewhat larger, and at higher orders or high conversion the penalty grows sharply, while autocatalysis reverses the comparison altogether.

Answer: B

A higher order makes the rate fall away more steeply with conversion, which penalises operating at the exit concentration. For a zero order reaction the two reactors need the same volume. The steeper the fall, the greater the penalty.

Answer: C

Correct answer: equal, the rate being independent

A zero order rate does not depend on concentration, so it is the same everywhere in either reactor and the integral collapses: \[ \tau = \frac{C_{A0}X}{k} \] for both. The CSTR’s usual disadvantage comes entirely from operating at the exit concentration, and if the rate does not depend on concentration there is no disadvantage to have.