# Chemical Reaction Engineering Questions and Answers – Stoichiometry of Batch Systems

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

1. If species A, initial moles NA0, is being consumed in a reaction to form a product P, the number of moles of A remaining in the batch reactor after a conversion X has been achieved is given by?
a) NA0 – X
b) NA0 – NA0 X
c) NA0 – X2
d) NA0 (X – 1)

Explanation: In a reaction A → Product,
Initial number of moles of A = NA0, the number of moles consumed after time t with conversion X is given by NA0X. Thus,
Number of moles of A remaining = Initial moles – Amount of moles consumed
NA = NA0 – NA0 X.

2. For a batch reactor, with reaction system given below, with initial moles of each species represented by NA0, NB0, NC0, and ND0, which option represents the change in moles of species B?

           aA + bB → cC + dD

a) – NA0 X
b) – $$\big ( \frac {c}{a} \big )$$ NA0 X
c) – $$\big ( \frac {b}{a} \big )$$ NA0 X
d) – $$\big ( \frac {c}{d} \big )$$ NA0 X

Explanation: Considering species A as basis for calculation.
To find the change in moles of species formed, we should first find out 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 $$\big ( \frac {b}{a} \big )$$ NA0 X.
Where, NA0X represents the moles of A reacted. The negative sign appears in the option due to the fact that species B is consumed. If the species is being formed, positive sign would accompany the term.

3. For a batch reactor, with reaction system given below, with initial moles of each species represented by NA0, NB0, NC0, and ND0, which option represents the remaining number of moles of species C, at time t with conversion X?

           2A + 3B → 5C + 7D

a) NC = NA0 + NC0 X
b) NC = NC0 + $$\big ( \frac {5}{2} \big )$$ NA0 X
c) NC = NC0 + 5NA0 X
d) NC = NC0 + $$\big ( \frac {5}{7} \big )$$ NA0 X

Explanation: 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
= $$\frac {5}{2}$$ × NA0 X
Thus, adding this term to the initial moles, we get,
NC = NC0 + $$\big ( \frac {5}{2} \big )$$ NA0 X.
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4. For a batch reactor, with reaction system given below, with initial moles of each species represented by NA0, NB0, NC0, and ND0, which option represents the change in moles of species D?

           2A + 3B → 5C + 7D

a) NA0 X
b) $$\big ( \frac {5}{3} \big )$$ NA0 X
c) – $$\big ( \frac {7}{2} \big )$$ NA0 X
d) $$\big ( \frac {7}{2} \big )$$ NA0 X

Explanation: For a reaction,
aA + bB → cC + dD
The change in species D is given by $$\big ( \frac {d}{a} \big )$$ NA0 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 $$\big ( \frac {7}{2} \big )$$ NA0 X.

5. The change in total number of moles per mole of A reacted is given by $$\frac {d}{a} + \frac {c}{a} – \frac {b}{a}$$ – 1.
a) True
b) False

Explanation: For a reaction,
aA + bB → cC + dD
We can conclude that, for 1 mole of A, b / a moles of B are consumed, c / a moles ofc is formed and d / a moles of species D is formed. The above reaction can be written as,
A + $$\frac {b}{a}$$B → $$\frac {c}{a}$$C + $$\frac {d}{a}$$D
Thus, the expression, $$\frac {d}{a} + \frac {c}{a} – \frac {b}{a}$$ – 1, represents the change in total number of moles per mole of A.

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