Reactors in Series and Parallel Questions and Answers

Answer: A

Each tank steps the concentration down a little, and many small steps approximate the continuous gradient of plug flow. An infinite number of infinitesimal tanks is exactly a plug flow reactor. Each tank steps the concentration down.

Answer: B

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

Both are true and the reason gives the mechanism. A single stirred tank holds all its contents at the exit concentration, where the rate is lowest. Splitting the same volume into a cascade means the first tanks sit at higher concentrations and therefore higher rates, and only the last works at the exit condition - so the same volume does more work.

Answer: C

Correct answer: 75 per cent

For a first order reaction each equal tank multiplies the concentration by \( 1/(1+k\tau) \), and here \( k\tau = 0.5\times2 = 1 \): \[ \frac{C_2}{C_0} = \left(\frac{1}{1+1}\right)^{2} = \frac{1}{4}, \qquad X = 1-\tfrac14 = 75\%. \] A single tank of the same total space time, \( \tau=4 \), would give \( 1-1/3 = 67 \) per cent - which is the distractor offered, and the measure of what the split is worth.

Answer: D

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: conversion rises with the number of tanks at fixed total volume, and each addition helps less than the last because the cascade is approaching the plug flow limit it cannot pass. Statement 3 mistakes a diminishing return for a proportional one - conversion is bounded by 100 per cent, so it cannot double however many tanks are added.

Answer: A

Correct answer: True

The cascade converges on plug flow behaviour, so the gap left to close shrinks with every tank added and so does the benefit of the next one. In practice the first two or three splits capture most of what is available, and beyond about five or six tanks the extra vessels, pumps and control cost more than the conversion they buy.

Answer: B

For first order kinetics the product of the two factors is unchanged by their order. For other orders the sequence does matter, and the optimum depends on which way the rate curve bends. First order is the one order that is indifferent.

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 algebra. For first order the outlet ratio is \[ \frac{1}{(1+k\tau_1)(1+k\tau_2)}, \] which is symmetric in the two space times, so exchanging the reactors changes nothing. For any higher order the expression is no longer symmetric, and the smaller vessel should come first - where the concentration and therefore the rate are highest, so the small volume is used where it does most good.

Answer: D

Correct answer: All of them

All three are correct. For first order the ordering is immaterial, as the previous item shows. For an order above one the smaller reactor should be placed first, where the higher concentration makes its limited volume most productive. And ordering cannot arise in parallel at all, since every branch receives the same fresh feed simultaneously rather than one after another.

Answer: A

Equal space times mean each reactor reaches the same conversion and the combined product is uniform. Unequal space times give streams of different conversion that dilute one another on mixing. Volume proportion is the rule to follow.

Answer: B

Both are true and the reason gives the objective: a badly split feed produces one over-converted and one under-converted stream, which mix to give a poorer result than either would give alone. Uniformity of the product is the objective.