Category: Objective Questions

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Liquid and Gas Phase Reactions”.

1. Which of the following is true for gas phase reactions?
A. Increase in moles of product increases the volume of the reaction mixture
B. Increase in moles of product decreases the volume of the reaction mixture
C. Decrease in moles of product increases the volume of the reaction mixture
D. Increase in moles of product does not affect the volume of the reaction mixture
Answer: A
Explanation: Increase in moles of product increases reaction mixture volume. For a reaction proceeding such that the moles of products are higher than moles of reactants, volume of reaction mixture increases.

2. For an ideal gas at any given temperature and pressure, the ideal gas law is used to express ____
A. Concentration of gas molecules
B. Mobility of gas molecules
C. Kinetic energy of gas molecules
D. Potential energy of gas molecules
Answer: A
Explanation: Ideal gas law is given as, PV = nRT,
Where, P – Pressure
T – Temperature
R – Universal gas constant
n – Number of gas moles
V – Volume
Concentration is the number of moles per unit volume = (frac{n}{V} = frac{P}{RT}. )

3. The progress of which among the following reactions results in an increase in volume?
A. A → B
B. 3A → B
C. A → 4B
D. 2A → B
Answer: C
Explanation: Increase in moles of product is depicted by the reaction, A → 4B. The number of moles of product is higher than that of reactant.

4. The rate of a gas phase reaction involving the conversion of reactant ‘i’ to product is expressed as ____
A. (-r_i)=-(frac{1}{V^2}frac{dN_i}{dt} )
B. (-r_i)=-(frac{1}{V}frac{dN_i}{dt} )
C. (-r_i)=-(frac{dN_i}{dt} )
D. (-r_i)=-V(frac{dN_i}{dt} )
Answer: B
Explanation: The rate of change of the gaseous component ‘i’ in due course of the reaction is the change in number of moles of ‘i’ per unit time per unit volume of the reaction mixture. Hence, (-r_i)=-(frac{1}{V}frac{dN_i}{dt}. )

5. State true or false.
For a gas phase reaction, the rate of the reaction, (-r_A)=-(frac{1}{V}frac{dN_A}{dt}=-frac{dC_A}{dt}) is valid only if A → B.
A. True
B. False
Answer: A
Explanation: The volume of reaction mixture remains constant as the number of moles of reactant and product is the same. Hence, the number of moles per unit volume does not vary during the progress of the reaction. Hence, the number of moles per unit volume is expressed as concentration of the reactant, A.

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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 ε_A is ____
A. -3
B. 3
C. 4
D. 2
View Answer

Answer: B
Explanation: ε_A = (frac{V_{X_{A=1}} – V_{X_{A=0}}}{V_{X_{A=0}}} )
ε_A = (frac{4 – 1}{1}) = 3.

4. Final concentration of the reactant A in terms of conversion, for a variable volume batch reactor is ____
A. C_A = (frac{C_{A0}(1+ X_A.}{(1+ ε_A X_A.} )
B. C_A = (frac{C_{A0}(1 – X_A.}{(1- ε_A X_A.} )
C. C_A = (frac{C_{A0}(1 – X_A.}{(1+ ε_A X_A.} )
D. C_A = (frac{C_{A0}(1 – X_A.}{(ε_A X_A.} )
View Answer

Answer: C
Explanation: C_A = C_A0 (1 – X_A), for a constant volume reactor. The initial number of moles for a variable volume reactor, N_A = N_A0(1 – X_A). Volume, V = V₀(1 + ε_A X_A). C_A = (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 ε_A 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, ε_A = (frac{150-100}{100}) = 0.5

6. The relationship between conversion and concentration for isothermal varying volume systems is ____
A. X_A = (frac{{1-frac{C_A}{C_{A0}}}}{1+(ε_A frac{C_A}{C_{A0}})} )
B. X_A = (frac{{1-frac{C_A}{C_{A0}}}}{1-(ε_A frac{C_A}{C_{A0}})} )
C. X_A = (frac{{1+frac{C_A}{C_{A0}}}}{1+(ε_A frac{C_A}{C_{A0}})} )
D. X_A = (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.} )
X_A = 1 – (frac{C_A}{C_{A0}}). X_A = (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, -r_A = (frac{C_{A0}}{ε_A}ln(frac{d( lnV)}{dt})) = k. Hence, (frac{C_{A0}}{ε_A} ln(frac{V}{V_0})) = kt.

8. State true or false.
For negative values of ε_A, 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 ε_A 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.

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Tanks in Series Model”.

1. If τ is the average residence time and σ² is the standard deviation, then the number of tanks necessary to model a real reactor as N ideal tanks in series is ____
A. N = (frac{tau^2}{σ^2} )
B. N = (frac{σ^2}{τ^2} )
C. N = σ²
D. N = (frac{1}{τ^2} )
Answer: A
Explanation: τ² = 1 and σ² = (frac{1}{N}.) The standard deviation is obtained as, σ² = (int_0^∞)(t-τ)²E(t)dt.

2. State true or false.
The tank in series model is a single parameter model.
A. False
B. True
Answer: B
Explanation: The tank in series model is used to represent non – ideal flow in PFR. It is a one parameter model and the parameter is the number of tanks.

3. State true or false.
The tank in series model depicts a non – ideal tubular reactor as a series of equal sized CSTRs.
A. True
B. False
Answer: A
Explanation: A number of tanks in series represent a PFR. Any CSTR behaves like a PFR if its volume is reduced. Infinite CSTRs are connected in series to approach PFR behaviour.

4. For a first order reaction, where k is the first order rate constant, the conversion for N tanks in series is obtained as ____
A. X_A = 1-(frac{1}{(1+frac{τk}{N})^N} )
B. X_A = 1+(frac{1}{(1+frac{τk}{N})^N} )
C. X_A = (frac{1}{(1+frac{τk}{N})^N} )
D. X_A = (frac{1}{(1+frac{τk}{N})^N} )– 1
Answer: A
Explanation: For N tanks in series, the combination approaches non – ideal PFR behaviour. The concentration in the N^th CSTR is given as C_N = (frac{C_0}{(1+ τk)^N}.frac{C_N}{C_0} = frac{1}{(1+ τk)^N}.) X_A = 1 – (frac{C_N}{C_0}.) For N – tanks, X_A = 1-(frac{1}{(1+frac{τk}{N})^N}. )

5. Which of the following correctly represents the Damkohler number for a first order reaction? (Where, τ is the space time)
A. k
B. τ
C. (frac{1}{kτ})
D. k τ
Answer: D
Explanation: Damkohler number is the product of first order rate constant and space time. It is the measure of the degree of completion of reaction.

6. According to tanks in series model, the spread of the tracer curve is proportional to ____
A. Square of distance from the tracer origin
B. Square root of distance from the tracer origin
C. Cube of distance from the tracer origin
D. Inverse square of distance from the tracer origin
Answer: B
Explanation: (σ_{tracer , curve}^2) α Distance from point of origin
Spread of the curve α (sqrt{Distance , from , origin}. )

7. If τ² = 100 and σ² = 10, the number of tanks necessary to model a real reactor as N ideal tanks in series is ____
A. 1
B. 10
C. 5
D. 100
Answer: B
Explanation: N = (frac{τ^2}{σ^2} )
N = (frac{100}{10}) = 10.

8. If τ = 5 s, first order rate constant, k = 0.25 sec^-1 and the number of tanks, N is 5, then the conversion is ____
A. 67.2%
B. 75%
C. 33%
D. 87.45%
Answer: A
Explanation: X_A = 1-(frac{1}{(1+frac{τk}{N})^N} )
1-(frac{1}{(1+frac{5×0.25}{5})^5}) = 67.2%.

9. The exit age distribution as a function of time is ____
A. E = (frac{t^{N-1}}{τ^N}frac{N^N}{(N-1)!}e^frac{-tN}{τ})
B. E = (frac{t^{N-1}}{τ^N}frac{N}{(N-1)!}e^frac{-tN}{τ})
C. E = (frac{t^N}{τ^N}frac{N^N}{(N-1)!}e^frac{-tN}{τ})
D. E = (frac{t^{N-1}}{τ^2}frac{N^N}{(N-1)!}e^frac{-tN}{τ})
Answer: A
Explanation: For N tanks in series, the exit age distribution is, E = (frac{t^{N-1}}{τ^N}frac{N^N}{(N-1)!}e^frac{-tN}{τ}.) The number of tanks in series, N = (frac{τ^2}{σ^2}. )

10. There are 5 tanks connected in series. If the average residence time is 5 sec, first order rate constant is 0.5 sec^-1, the initial concentration is 5(frac{mol}{m^3},) then the conversion at the exit of 5^th reactor in ((frac{mol}{m^3})) is ____
A. 0.34
B. 0.51
C. 0.65
D. 0.81
Answer: C
Explanation: C_N = C₀ (frac{1}{(1+frac{τk}{N})^N} )
C_N = 5×(frac{1}{(1+frac{5×0.5}{5})^5} )
C_N = 0.65(frac{mol}{m^3}. )

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Molecular Biology Multiple Choice Questions on “Separation of DNA Fragments by Electrophoresis”.

1. Which of the following cannot be used for the separation of nucleic acids?
A. SDS – PAGE
B. PAGE
C. Northern blotting
D. PAGE

Answer: A
Explanation: Sodium dodecyl sulphate is a detergent, often used in biochemical preparations, binds to proteins and causes them to form a rod like structure. Most proteins bind SDS in the same ratio (1.4g per g of protein). Thus, the electrophoresis of proteins in an SDS – containing polyacrylamide gel separates them in order of their molecular masses. It is not known to have a similar effect on nucleic acids.

2. The polymerization of the gel used in PAGE occurs between polyacrylamide and ____________
A. N, N – acrylamide
B. Bisacrylamide
C. N – methyleneacrylamide
D. N, N – methylene bisacrylamide

Answer: D
Explanation: In PAGE, polyacrylamide gel electrophoresis, gel is made by free radical by polymerizing the monomers together. The two major monomers between which the polymerization occurs are polyacrylamide and N, N – methylene bisacrylamide in the buffer of choice.

3. If DNA is digested by endonucleases in four sites giving rise to fragments of which two are equal in length how many bands would be seen after electrophoresis?
A. 3
B. 4
C. 5
D. 6

Answer: B
Explanation: Digestion at four sites gives rise to five fragments. Two fragments are of same size thus they will form a single band. Therefore only four bands is observed.

4. The fluorescent dye such Ethidium is used for visualizing DNA. How do ethidium binds to DNA?
A. Stacked between histone molecules
B. Binds to the nucleotide base
C. Intercalated between the stacked bases
D. Binds to the phosphodiester backbone

Answer: C
Explanation: Ethidium is intercalated between the stacked bases. This increases the spacing of successive base pairs, distorts the regular sugar phosphate backbone and decreases the twist of the helix.

5. Pulse field gel electrophoresis separates DNA molecules of size ___________
A. 10 – 20 bp
B. 20 – 30 Kb
C. 30 – 50 Kb
D. 40 – 50 bp

Answer: C
Explanation: DNA molecules ranging from 30 – 50 kb migrate in a snake like manner thus, cannot be readily be resolved. These long DNAs can be separated by changing the orientation of the electric field. This process of separation is known as PFGE.

6. Which of the following will migrate faster? The condition is the molecular weight of the following is equal.
A. Supercoiled circular DNA
B. Nicked circular DNA
C. Single stranded DNA
D. Double stranded DNA

Answer: A
Explanation: Supercoiled circular DNA has a less effective volume than the others. Thus, it migrates more rapidly when subjected to electrophoresis due to its compact structure.

7. Agarose can be extracted from which of the following?
A. Gracilaria esculentus
B. Lycazusican esculentum
C. Ficum benghalensis
D. Agrostis stolonifera

Answer: A
Explanation: As we know agarose is extracted from seaweeds. Precisely two types of species are used for the extraction; they are the Gracilaria esculentus and Gelidium nudifrons.

8. Electrophoresis cannot be used to separate _______________
A. DNA
B. RNA
C. Amino acid
D. Protein

Answer: C
Explanation: DNA, RNA and protein can be separated by the method of electrophoresis as it can be separate charged molecules. Amino acids are generally separated by the process of chromatography.

9. Which of the following is not a character of polyacrylamide gel?
A. Inert
B. Ionic strength
C. Stable over a wide range of pH
D. Separate upto a few 100 bp of DNA

Answer: D
Explanation: The pore size of polyacrylamide gel is very small due to high resolution power. Thus, it can separate DNA fragments upto a few 100 base pairs only.

10. Pulse field gel electrophoresis was developed by ____________
A. Collins and John
B. Kary Mullis
C. Patrick O’ Farrell
D. Schwartz and Cantor

Answer: D

Molecular Biology Multiple Choice Questions on “Nucleotides are the Precursors for DNA Synthesis”.

1. Which of the following is not true about nucleotides?
A. Monomeric units
B. Ubiquitous substances
C. Energy rich molecules
D. Non enzymatic molecules

Answer: D
Explanation: The nucleotides are known to be enzymatic molecules. For example, ribozymes are catalytic molecules. Also certain derivatives of nucleotides such as ATP, coenzyme A, nicotinamide adenine dinucleotide, etc., are important catalytic molecules that participate in various enzymatic processes.

2. The nitrogenous base is covalently linked to the _________ carbon of the pentose sugar.
A. C1
B. C2
C. C3
D. C4

Answer: A
Explanation: Nucleotides are phosphate esters of a five carbon sugar, either ribose of 2’-deoxyribose. The nitrogenous base is covalently linked to the C1 carbon of this pentose sugar to form the nucleotide.

3. In which carbon do the deoxyribonucleotides lack an –OH molecule?
A. C1
B. C2
C. C3
D. C4

Answer: B
Explanation: A deoxyribonucleotides lack an –OH molecule at the C2 position of the ribose sugar ring. This is the substrate for DNA synthesis and is known as the 2’-deoxyribonucleotide.

4. Which of the following is not a part of a nucleotide?
A. Ester linkage
B. Phosphate group
C. Base
D. Hydrogen bond

Answer: D
Explanation: Nucleotides are phosphate esters of a five carbon sugar, either ribose of 2’-deoxyribose. The nitrogenous base is covalently linked to the C1 carbon of this pentose sugar to form the nucleotide. Hydrogen bond is made by the bases to hold the two strands of DNA together and is not a part of the nucleotide.

5. Which of the following is not a part of a nucleoside?
A. Deoxyribose sugar
B. Glycosidic linkage
C. Phosphate
D. Base

Answer: C
Explanation: A nucleoside is the deoxyribose sugar linked to the base with a glycosidic linkage. Addition of phosphate at the 5’-carbone leads to the formation of the nucleotide.

6. The linkage between the sugar and the base is a covalent linkage.
A. True
B. False

Answer: A
Explanation: A glycosidic bond is one which links a type of covalent linkage which forms between a sugar molecule and another molecule which may or may not be a sugar. In case of nucleotide/nucleoside the sugar and the base is a glycosidic linkage.

7. Which of the following is not a nucleotide?
A. AMP
B. TMP
C. GMP
D. CMP

Answer: B
Explanation: TMP is not a nucleotide. Thymine is not present in form of Thymidine monophosphate as it is not used in RNAs. Thymine is present in the form of dTMP deoxythymidine monophosphate as it is used in the synthesis of DNA only.

8. Purines and pyrimidines form glycosidic linkage with the purines and pyrimidines at the N6 position.
A. True
B. False

Answer: B

Molecular Biology Multiple Choice Questions on “Processing of Ribosomal and Transfer RNA”.

1. How many pre – rRNA transcript in eukaryotes are used to produce all the different mature rRNAs?
A. 1
B. 2
C. 3
D. 4

Answer: B
Explanation: Eukaryotes have 4 species of ribosomal RNA molecules. Three of these rRNAs are derived by the cleavage of a single precursor whereas the last rRNA is synthesized as a single molecule.

2. In bacterial cell the 3 rRNAs 23S, 16S and 5S are derived from different transcripts.
A. True
B. False

Answer: B
Explanation: No in bacterial cell the 3 rRNAs 23S, 16S and 5S are not derived from different transcripts. As in eukaryotes, the bacterial rRNAs 23S, 16S and 5S are derived by the cleavage of one single pre – rRNA transcript.

3. Which one of the following rRNA undergoes least post-transcriptional processing?
A. 28S
B. 18S
C. 5.8S
D. 5S

Answer: D
Explanation: The rRNA undergoes least post-transcriptional processing is the 5S rRNA. This is because it is synthesized separately and thus do not undergo any kind of cleavage. The rest three are derived by the cleavage of one single pre – rRNA transcript.

4. The bacterial pre – rRNA undergo __________ cleavage.
A. 1
B. 2
C. 3
D. 4

Answer: B
Explanation: The bacterial pre – rRNA undergo 2 cleavages. As three rRNA molecules are flanked together the first cleavage separates the precursors of the three rRNAs. A further cleavage known as the secondary cleavage ensures the production of the mature rRNA molecules.

5. Which of the following types of processing is not used to prepare mature rRNA?
A. Splicing
B. Methylation
C. Glycosylation
D. Conversion of uridine

Answer: A
Explanation: The preparation of mature rRNA does not include splicing in most of the cases. They only include methylation of specific bases and some specific sugar moieties and conversion of uridines to pseudouridines.

6. Each tRNA is produced from different transcripts.
A. True
B. False

Answer: B
Explanation: Like rRNAs, tRNAs in both bacteria and eukaryotes are synthesized as a longer precursor molecule. Some of these pre – tRNAs contain several individual tRNA sequences. Also in bacteria, some of the tRNAs are included in pre – rRNA transcripts.

7. The pre – tRNA involves cleavage by __________
A. tRNA itself
B. RNase H
C. RNase P
D. RNA polymerase

Answer: C
Explanation: The processing of the 5’ end of the pre – tRNAs involves cleavage by an enzyme called RNase P. RNase P consists of RNA and in molecules, both of which are required for maximal activity.

8. RNase P is a ribozyme.
A. True
B. False

Answer: A
Explanation: In 1983 Sidney Altman and his colleagues demonstrated that the isolated RNA component of the RNase P is itself capable of catalyzing pre – tRNA cleavage. These experiments established that RNase P is a ribozyme – an enzyme in which RNA rather than protein is responsible for catalytic activity.

9. The common sequence for all tRNAs at their 3’ end is __________
A. CCC
B. CCT
C. GGA
D. CCA

Answer: D
Explanation: All tRNAs have the common sequence CCA at their 3’ end. This sequence is the site of amino acid attachment, so it is required for tRNA function during protein synthesis.

10. The 3’ CCA terminus is exceptionally added by the RNA polymerase during transcription.
A. True
B. False

Answer: B
Explanation: The 3’ CCA terminus is encoded in the DNA of some tRNA genes, but in others it is not. Thus the 3’ CCA terminus is added during RNA processing by an enzyme which recognizes and adds CCA to the 3’ end of all tRNAs that lack this sequence.

11. Though unusual, pre – tRNA also undergo splicing. It is achieved by the help of __________
A. tRNA itself
B. Splicosome
C. RNA polymerase exonuclease activity
D. Endonuclease

Answer: D