Chemical Reaction Engineering Questions and Answers – Stoichiometry – Batch Reactors with Variable Volume

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This set of Chemical Reaction Engineering Question Paper focuses on “Stoichiometry – Batch Reactors with Variable Volume”.

1. State true or false.
A variable volume reactor is the one in which the reaction proceeds by a change in number of moles at a given pressure and temperature.
a) True
b) False
View Answer

Answer: a
Explanation: The pressure is held constant as the volume varies. The variable volume systems are also termed as constant pressure systems.

2. The fractional change in volume of a system for variable volume systems, expressed in terms of the number of moles is ____
a) ε = \(\frac{Change \, in \, number \, of \, moles \, of \, the \, reaction \, system \, when \, the \, reaction \, is \, complete}{Total \, number \, of \, moles \, fed} \)
b) ε = \(\frac{Total \, number \, of \, moles \, fed}{Change \, in \, number \, of \, moles \, of \, the \, reaction \, system \, when \, the \, reaction \, is \, complete} \)
c) ε = \(\frac{Number \, of \, moles \, left \, when \, the \, reaction \, is \, complete}{Total \, number \, of \, moles \, fed} \)
d) ε = \(\frac{Total \, number \, of \, moles \, fed}{Number \, of \, moles \, left \, when \, the \, reaction \, is \, complete} \)
View Answer

Answer: a
Explanation: ε is the fractional change in volume of the reaction system between no conversion and complete conversion of the reactant. It is the ratio of the change in moles of the reaction mixture to achieve complete conversion to the number of moles fed initially.

3. For the reaction A → 4R, the value of εA is ____
a) -3
b) 3
c) 4
d) 2
View Answer

Answer: b
Explanation: εA = \(\frac{V_{X_{A=1}} – V_{X_{A=0}}}{V_{X_{A=0}}} \)
εA = \(\frac{4 – 1}{1}\) = 3.
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4. Final concentration of the reactant A in terms of conversion, for a variable volume batch reactor is ____
a) CA = \(\frac{C_{A0}(1+ X_A)}{(1+ ε_A X_A)} \)
b) CA = \(\frac{C_{A0}(1 – X_A)}{(1- ε_A X_A)} \)
c) CA = \(\frac{C_{A0}(1 – X_A)}{(1+ ε_A X_A)} \)
d) CA = \(\frac{C_{A0}(1 – X_A)}{(ε_A X_A)} \)
View Answer

Answer: c
Explanation: CA = CA0 (1 – XA), for a constant volume reactor. The initial number of moles for a variable volume reactor, NA = NA0(1 – XA). Volume, V = V0(1 + εA XA). CA = \(\frac{N_A}{V}\) = \(\frac{C_{A0}(1 – X_A)}{(1+ ε_A X_A)}.\)

5. For the reaction A → 2R containing 50% moles initially, the value of εA is ____
a) 1
b) 0.75
c) 0.5
d) 0.9
View Answer

Answer: c
Explanation: Initially, there are 50 moles of inerts and 50 moles of A, 100 moles in total. On complete conversion, A forms (50×2) 100 moles of product, R. 1 mole reactant forms 2 moles of product. As the inerts do not get converted during the reaction, there are 150 moles in total at the end of the reaction. Hence, εA = \(\frac{150-100}{100}\) = 0.5

6. The relationship between conversion and concentration for isothermal varying volume systems is ____
a) XA = \(\frac{{1-\frac{C_A}{C_{A0}}}}{1+(ε_A \frac{C_A}{C_{A0}})} \)
b) XA = \(\frac{{1-\frac{C_A}{C_{A0}}}}{1-(ε_A \frac{C_A}{C_{A0}})} \)
c) XA = \(\frac{{1+\frac{C_A}{C_{A0}}}}{1+(ε_A \frac{C_A}{C_{A0}})} \)
d) XA = \(\frac{{1+\frac{C_A}{C_{A0}}}}{1-(ε_A \frac{C_A}{C_{A0}})} \)
View Answer

Answer: a
Explanation: \(\frac{C_A}{C_{A0}} = \frac{(1 – X_A)}{(1+ ε_A X_A)} \)
XA = 1 – \(\frac{C_A}{C_{A0}}\). XA = \(\frac{{1-\frac{C_A}{C_{A0}}}}{1+(ε_A \frac{C_A}{C_{A0}})}. \)

7. The relation between rate and time for a zero order reaction in a variable volume reactor is expressed as ____
a) \(\frac{C_{A0}}{ε_A}ln(\frac{V}{V_0})\) = t
b) \(\frac{1}{ε_A} ln(\frac{V}{V_0})\) = kt
c) \(\frac{C_{A0}}{ε_A} ln(\frac{V}{V_0})\) = kt
d) \(\frac{C_{A0}}{ε_A} ln(\frac{V}{CC_A})\) = kt
View Answer

Answer: c
Explanation: For a zero order reaction in a variable volume reactor, -rA = \(\frac{C_{A0}}{ε_A}ln(\frac{d( lnV)}{dt})\) = k. Hence, \(\frac{C_{A0}}{ε_A} ln(\frac{V}{V_0})\) = kt.
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8. State true or false.
For negative values of εA, there is a reduction in volume of the reaction mixture as the reaction proceeds.
a) True
b) False
View Answer

Answer: a
Explanation: Negative value of εA implies that the initial volume is greater than the volume at complete conversion. The number of moles of the reactant is greater than the number of moles of product.

9. The plot of ln\((\frac{V}{V_0})\) and time for a zero order reaction in a varying volume reactor for εA > 0 is ____
a)

Answer: a
Explanation: \(\frac{C_{A0}}{ε_A}ln(\frac{V}{V_0})\) = kt. The plot of \(ln(\frac{V}{V_0})\) and time gives a straight line of slope \(\frac{kε_A}{C_{A0}}\). For εA > 0, the plot is increasing.
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10. The plot of -ln(1-\(\frac{∆V}{ε_A V_0}) \) and time for first order reaction in a varying volume reactor is ____
a) Plot of -ln(1-ΔVεAV0) & time for first order reaction in volume reactor - option a
b) Plot of -ln(1-ΔVεAV0) & time for first order reaction in volume reactor - option b
c) Plot of -ln(1-ΔVεAV0) & time for first order reaction in volume reactor - option c
d) Plot of -ln(1-ΔVεAV0) & time for first order reaction in volume reactor - option d
View Answer

Answer: b
Explanation: For the first order reaction at constant pressure and variable volume, -ln(1-\(\frac{∆V}{ε_A V_0}) \) = kt. The plot is linear with slope k.

Sanfoundry Global Education & Learning Series – Chemical Reaction Engineering.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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