This set of Chemical Reaction Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Stoichiometry – Equations for Batch Concentrations”.

1. Species A with number of initial moles denoted by N_{A0} is consumed in a batch reactor to form a product. Which of these options correctly represents the concentration of species A in the reactor after a conversion X has been achieved?

a) \(\frac {N_{A0} – X}{V} \)

b) \(\frac {N_{A0} – N_{A0} X}{V} \)

c) \(\frac {N_{A0} – X^2}{V} \)

d) \(\frac {N_{A0} (X – 1)}{V} \)

View Answer

Explanation: In a reaction A → Product,

Initial number of moles of A = N

_{A0}, the number of moles consumed after time t with conversion X is given by N

_{A0}X. Thus,

Number of moles of A remaining = Initial moles – Amount of moles consumed

N

_{A}= N

_{A0}– N

_{A0}X

The concentration of A is given by number of moles of A per unit volume, which is represented by

C

_{A}= \(\frac {N_A}{V} \)

Therefore, substituting the term N

_{A}, we get,

C

_{A}= \(\frac {N_{A0} – N_{A0} X}{V} \).

2. For a batch reactor, with reaction system given below, which option represents the concentration of species B?

aA + bB → cC + dD

a) C_{B} = \(\frac {N_{A0} (1 – X)}{V} \)

b) C_{B} = \(\frac {N_{B0} – ( \frac {c}{a} ) N_{A0} X}{V} \)

c) C_{B} = \(\frac {N_{B0} – ( \frac {b}{a} ) N_{A0} X}{V} \)

d) C_{B} = \(\frac {N_{B0} – ( \frac {d}{a} ) N_{A0} X}{Vd} \)

View Answer

Explanation: Considering species A as basis for calculation.

To find the change in moles of species formed, we should first find our how much species B is reacted, which will correspond to the change in species B. Therefore, for every mole of A, b / a moles of B must be consumed. Thus, moles of B reacted is given by \( ( \frac {b}{a} )\) N

_{A0}X.

The moles of species B is therefore is given by,

N

_{B}= N

_{B0}– \( ( \frac {b}{a} )\) N

_{A0}X

And the concentration is given by number of moles per unit volume. Thus, the concentration term comes out to be,

C

_{B}= \(\frac {N_{B}}{V} \)

C

_{B}= \(\frac {N_{B0} – ( \frac {b}{a} ) N_{A0} X}{V} \)

Where, N

_{A0}X represents the moles of A reacted. The negative sign appears in the option due to the fact that species B is consumed.

3. For a batch reactor, with reaction system given below, which of these options below represents concentration of species C, at time t with conversion X?

2A + 3B → 5C + 7D

a) C_{C} = \(\frac {N_{A0} + N_{C0} X}{V} \)

b) C_{C} = \(\frac {N_{C0} + ( \frac {5}{2} ) N_{A0} X}{V} \)

c) C_{C} = \(\frac {N_{C0} + 5N_{A0} X}{V} \)

d) C_{C} = \(\frac {N_{C0} + ( \frac {5}{7} ) N_{A0} X}{V} \)

View Answer

Explanation: The concentration of any species is given by number of moles of that species per unit volume.

To calculate the remaining number of moles of species C after a certain time and conversion X, we need to find out the number of moles C reacted or is being created by the reaction. Since, the species C is being created by the reaction, the amount created would be added to the initial moles. From the reaction it is evident that, for every one mole of A, 5 / 2 moles of C is formed. Thus,

Moles of C formed = \(\frac {5}{2}\) × moles of A consumed

= \(\frac {5}{2}\) × N

_{A0}X

Thus, adding this term to the initial moles, we get,

N

_{C}= N

_{C0}+ \( ( \frac {5}{2} )\) N

_{A0}X

Concentration of species B is then given by,

C

_{C}= \(\frac {N_{C}}{V} \)

C

_{C}= \(\frac {N_{C0} + ( \frac {5}{2} ) N_{A0} X}{V} \).

4. For a batch reactor, with reaction system given below, which of these options below represents concentration of species D, at time t with conversion X?

2A + 3B → 5C + 7D

a) C_{D} = \(\frac {N_{D0} + N_{A0} X}{V} \)

b) C_{D} = \(\frac {N_{D0} + ( \frac {5}{2} ) N_{A0} X}{V} \)

c) C_{D} = \(\frac {N_{D0} + 5N_{A0} X}{V} \)

d) C_{D} = \(\frac {N_{D0} + ( \frac {7}{2} ) N_{A0} X}{V} \)

View Answer

Explanation:

The concentration of any species is given by number of moles of that species per unit volume.

C

_{D}= \(\frac {N_D}{V} \)

For a reaction,

aA + bB → cC + dD

The change in species D is given by \( ( \frac {d}{a} )\) N

_{A0}X. Here species A is selected as basis. The term has positive sign, due to the fact that D is being formed in the reaction and thus, its number of moles is going to increase as the reaction proceeds. For the reaction in question, we find that for every one mole of A, 7 / 2 moles of D are formed. Thus, the amount of change in number of moles of species D is \( ( \frac {7}{2} )\) N

_{A0}X.

Thus, adding this term to the initial moles, we get,

N

_{D}= N

_{D0}+ \( ( \frac {5}{2} )\) N

_{A0}X

Thereafter, the concentration of species D is calculated by by substituting ND0 in the concentration formula.

C

_{D}= \(\frac {N_{D0} + ( \frac {7}{2} ) N_{A0} X}{V}\).

5. For a reaction given below, calculate the concentration of species Z where the initial moles of Y is 10mol / dm^{3} with conversion of Y is 0.5 at a certain time. There is 30 mol initial amount of species Z present in the batch reactor. The volume of the reactor is 10 dm^{3}.

Y + 2Z → W + P

a) 2 mol / dm^{3}

b) 7 mol / dm^{3}

c) 3 mol / dm^{3}

d) 5 mol / dm^{3}

View Answer

Explanation: The concentration is given by number of moles per unit volume. Thus, the concentration term comes out to be,

C

_{Z}= \(\frac {N_Z}{Volume}\)

The number of moles of species Z is given by,

N

_{Z}= N

_{Z0}– 2N

_{Y0}X

Where X is the conversion. The concentration is then given by,

C

_{Z}= \(\frac {N_{Z0} – 2N_{Y0}X}{V}\)

Substituting the values given in the question, given that,

N

_{Z0}= 30 mol

N

_{Y0}= 10 mol

X = 0.5

V = 10 dm

^{3}

We get the concentration of species Z to be,

C

_{Z}= \(\frac {30 – 2(10)(0.5)}{10}\)

C

_{Z}= \( \frac {20}{10}\) = 2 \( \frac {mol}{dm^3}\).

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