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Multiple Reactions and Selectivity Questions and Answers

Answer: A

The competition is for the same reactant, so the relative rates decide which product predominates. Series reactions instead pass through an intermediate on the way to the final product. The competition is for the same reactant.

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2.

Match each reaction scheme with its characteristic problem.
List I
A. Parallel
B. Series
C. Series-parallel
D. Independent
List II
1. both problems at once
2. keeping the reactant on the wanted path
3. no interaction, so no selectivity question
4. stopping at the intermediate before it reacts on

Answer: B

Recognising the scheme decides what the reactor is being asked to achieve. A series scheme calls for control of residence time while a parallel scheme calls for control of concentration. The scheme decides what to control.

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Answer: C

Correct answer: the rate of the wanted reaction to the unwanted

Instantaneous selectivity is a ratio of *rates* at a point, \( r_D/r_U \), so it describes what the chemistry is doing under the conditions prevailing there and then. Because those conditions change through a reactor and through a batch, the value changes too - which is exactly why an overall figure has to be defined separately as its average.

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4.

Assertion (A): Instantaneous selectivity and overall selectivity are different quantities.
Reason (R): The instantaneous value varies through the reactor while the overall value is an average of it.

Answer: D

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

Both are true and the reason gives the relation between them. The instantaneous value is local and varies with the composition and temperature at each point; the overall value is what actually leaves the reactor, and it is the flow-weighted average of the instantaneous values along the path. Designing for selectivity means arranging the reactor so that the instantaneous value stays high everywhere.

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5.

Consider the following statements about selectivity and yield.
1. Selectivity compares the wanted product with the unwanted one.
2. Yield compares the wanted product with the theoretical maximum.
3. High conversion always implies high selectivity.

Answer: A

Correct answer: 1 and 2 only

Statements 1 and 2 define the two measures correctly. Statement 3 is the misconception the whole topic exists to correct: pushing conversion higher generally makes selectivity worse, because the wanted product is exposed to further reaction and the reactant concentration falls to where the unwanted route may compete better. The two must be traded against one another rather than pursued together.

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6.

For two parallel reactions where the wanted one has the higher order in the reactant, selectivity is improved by

Answer: B

Correct answer: keeping the reactant concentration high

With \( r_D \propto C^{a} \) and \( r_U \propto C^{b} \), the ratio goes as \( C^{a-b} \). If the wanted reaction has the higher order the exponent is positive, so selectivity rises with concentration - which argues for a batch or plug flow reactor, high pressure in the gas phase, and no dilution. Long residence time works the other way, since it reduces the concentration.

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7.

Assertion (A): A plug flow reactor is preferred when the wanted parallel reaction has the higher order.
Reason (R): It maintains a high reactant concentration through most of its length.

Answer: C

Both are true and the reason gives the mechanism: a stirred tank holds everything at the exit concentration, which is exactly where the higher order reaction is at its greatest disadvantage. Concentration is the lever when orders differ.

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Answer: D

Correct answer: a stirred tank at low concentration

If the unwanted reaction has the higher order the exponent \( a-b \) is negative, so selectivity improves as concentration falls. A stirred tank holds its entire contents at the low exit concentration, which is precisely what is wanted here - the same uniformity that costs it volume in a single-reaction duty becomes its advantage. Diluting the feed and semi-batch addition serve the same end.

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9.

Consider the following statements about controlling selectivity in parallel reactions.
1. Concentration is the lever when the two orders differ.
2. Temperature is the lever when the two activation energies differ.
3. Neither lever works if the two reactions have the same order.

Answer: A

Correct answer: 1 and 2 only

Statements 1 and 2 name the two independent levers: concentration works through the difference in reaction orders, and temperature through the difference in activation energies. Statement 3 goes too far - if the orders are equal, concentration is indeed useless, but temperature still works provided the activation energies differ, and a catalyst may discriminate where neither does.

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10.

Assertion (A): Temperature can be used to shift selectivity between two competing reactions.
Reason (R): The reaction with the higher activation energy is favoured more strongly as the temperature rises.

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. The rate ratio varies as \[ \frac{k_D}{k_U} \propto \exp\left(-\frac{E_D-E_U}{RT}\right), \] so if the wanted reaction has the higher activation energy, raising the temperature favours it. If it has the lower one, the reactor should be run as cold as the throughput allows - which is why selectivity, not rate, often sets the operating temperature.

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