Category: Objective Questions

Bioprocess Engineering Multiple Choice Questions on “Estimating Oxygen Solubility”.

1. The partial pressure of oxygen at 1atm is ________________
A. 0.2000 atm
B. 0.2098 atm
C. 0.2099 atm
D. 0.2096 atm

Answer: C
Explanation: The mole fraction of oxygen in air is 0.2099, so the partial pressure of oxygen at 1 atm air pressure is 0.2099 atm.

2. What is the unit of oxygen solubility “C*_AL”?
A. mgl^-1
B. mg^-1l^-1
C. m^-1g^-1l^-1
D. mgl

Answer: A
Explanation: Respiration is expressed in molar units related to biochemical stoichiometries. C*_AL is oxygen solubility in units of mg l^-1.

3. The addition of ions and sugars added to the fermentation increases the oxygen solubility?
A. True
B. False

Answer: B
Explanation: Oxygen solubility is decreased by the ions and sugars normally added to fermentation media. In a typical fermentation medium, oxygen solubility is between 5% and 25% lower than in water as a result of solute effects. In simple systems, with all the components being dissolved, C₀ represents the oxygen solubility and linearly decreases with increasing solute concentrations. In complex solutions with multi-phase structure an increase in C₀ can be detected. It suggests that C₀ consists of two components — one being oxygen solubility, the other being determined by the amount of oxygen adsorbed on the interphase and bound by macromolecules. The presence of biomass leads to a decrease in C₀.

4. The solubility of oxygen in water is temperature and pressure dependent?
A. True
B. False

Answer: A
Explanation: The solubility of oxygen in water is temperature and pressure dependent. About twice as much (14.6 mg•L^-1) dissolves at 0 °C than at 20 °C (7.6 mg•L^-1), this estimates that at high temperature oxygen solubility is low. Less oxygen dissolves at high elevations compared to low elevations because the atmospheric pressure is less and thus the partial pressure is lower, this estimates that at low pressure the oxygen solubility is also low.

5. “The amount of air dissolved in a fluid is proportional to the pressure in the system”, which law is applicable to this statement?
A. Raoult’s law
B. Fick’s law
C. Henry’s law
D. Newton’s law

Answer: C
Explanation: Solubility of air in water follows Henry’s Law – “the amount of air dissolved in a fluid is proportional to the pressure in the system” – and can be expressed as:
c = p_g / k_H
where, c = solubility of dissolved gas
k_H = proportionality constant depending on the nature of the gas and the solvent
p_g = partial pressure of gas (Pa, psi)
The solubility of oxygen in water is higher than the solubility of nitrogen. Air dissolved in water contains approximately 35.6% oxygen compared to 21% in air.

6. Henry Law’s Constants at a system temperature of 25°C (77°F) of oxygen is 756.7 atm/(mol/litre). Molar Weight of O₂ is 31.9988 g/mol and partial fraction in air is ~ 0.21. Calculate the Oxygen dissolved in the Water at atmospheric pressure.
A. 0.0090 g/liter
B. 0.0089 g/liter
C. 0.0080 g/liter
D. 0.0099 g/liter

Answer: B
Explanation: Oxygen dissolved in the Water at atmospheric pressure can be calculated as:
c = p_g / k_H
c_o = (1 atm) 0.21 / (756.7 atm/(mol/litre)) (31.9988 g/mol)
= 0.0089 g/liter.

7. Henry Law’s Constants at a system temperature of 25°C (77°F) of nitrogen is 1600 atm/(mol/litre). Molar weight of N₂ is 28.0134 g/mol and partial fraction in air is ~ 0.79. Calculate the Nitrogen dissolved in the Water at atmospheric pressure.
A. 0.0138 g/liter
B. 0.0130 g/liter
C. 0.0132 g/liter
D. 0.0134 g/liter

Answer: A
Explanation: Nitrogen dissolved in the Water at atmospheric pressure can be calculated as:
c = p_g / k_H
c_n = (1 atm) 0.79 / (1600 atm/(mol/litre)) (28.0134 g/mol)
= 0.0138 g/liter.

8. Refer to Q6 and Q7, and calculate the air dissolved in water?
A. 0.0228 g/liter
B. 0.0223 g/liter
C. 0.0227 g/liter
D. 0.0222 g/liter

Answer: C
Explanation: Since air mainly consists of Nitrogen and Oxygen – the air dissolved in the water can be calculated as:
c_a = (0.0089 g/litre) + (0.0138 g/litre)
= 0.0227 g/liter.

9. The dissolved oxygen decreases when?
A. The temperature is increased
B. The pressure is increased
C. The salinity is decreased
D. The salinity is increased

Answer: D
Explanation: Dissolved oxygen decreases exponentially as salt levels increase. That is why, at the same pressure and temperature, saltwater holds about 20% less dissolved oxygen than freshwater.

10. The Henry’s law constant for O₂ in water at 25°C is 1.27×10⁻³M/atm and the mole fraction of O₂ in the atmosphere is 0.21. Calculate the solubility of O₂ in water at 25°C at an atmospheric pressure of 1.00 atm.
Strategy: ▪ Use Dalton’s law of partial pressures to calculate the partial pressure of oxygen.
▪ Use Henry’s law to calculate the solubility, expressed as the concentration of dissolved gas.
A. 2.5×10^-4 M
B. 2.1×10^-4 M
C. 2.3×10^-4 M
D. 2.7×10^-4M

Answer: D
Explanation: According to Dalton’s law, the partial pressure of O₂ is proportional to the mole fraction of O₂:
P_A=X_AP_t=(0.21)(1.00atm)=0.21atm
From Henry’s law, the concentration of dissolved oxygen under these conditions is:
CO₂ = Kp_O2 = (1.27×10⁻³M/atm) (0.21atm) = 2.7×10^-4 M.

11. The value of Henry’s law constant increases with increasing temperature?
A. True
B. False

Answer: A
Explanation: The value of the Henry’s law constant is found to be temperature dependent. The value generally increases with increasing temperature. As a consequence, the solubility of gases generally decreases with increasing temperature. The decrease in solubility of gases with increasing temperature is an example of the operation of Le Chatelier’s Principle. The heat or enthalpy change of the dissolution reaction of most gases is negative, which is to say the reaction is exothermic. As a consequence, increasing the temperature leads to gas evolution.

12. Does yeast need oxygen in fermentation process?
A. True
B. False

Answer: A
Explanation: Oxygen is a critical additive in brewing. Oxygen is the only necessary nutrient not naturally found in wort. Adding adequate oxygen to wort requires a fundamental understanding of why yeast need oxygen, how much oxygen they need, and how to get oxygen into solution and the factors affecting solubility of oxygen. Yeast use oxygen for cell membrane synthesis. Without oxygen, cell growth will be extremely limited. Yeast can only produce sterols and certain unsaturated fatty acids necessary for cell growth in the presence of oxygen.

13. Which type of homebrewer is best for 8 ppm of dissolved oxygen in solution?
A. Siphon sprays
B. Whipping
C. Splashing and shaking
D. Pure air through a stone with an aquarium pump

Answer: C
Explanation: Homebrewers have several aeration/oxygenation methods available to them: siphon sprays, whipping, splashing, shaking, pumping air through a stone with an aquarium pump, and injecting pure oxygen through a sintered stone. Pumping compressed air through a stone is not an efficient way to provide adequate levels of DO. Traditional splashing and shaking, although laborious, is fairly efficient at dissolving up to 8 ppm oxygen. To increase levels of oxygen, the carboy headspace can be purged with pure oxygen prior to shaking.

14. The atmospheric pressure is 1.0 atm and Henry’s law constant for O₂ is 1.66 x 10⁻⁶ M/mm Hg at 25 °C. Assume air contains 21% oxygen. Calculate the partial pressure of oxygen. (21% of air is oxygen and the mole fraction of O₂ is 0.21).
A. 180 mm Hg
B. 130 mm Hg
C. 120 mm Hg
D. 160 mm Hg

Answer: D
Explanation: The partial pressure of oxygen is:
P (O₂) = (1.0 atm) ((760 mm Hg)/(1 atm)) (0.21) = 160 mm Hg.

15. Refer to Q14 and, Calculate the solubility of oxygen in units of grams of oxygen per liter of water.
A. 0.0080 g/L
B. 0.0082 g/L
C. 0.0083 g/L
D. 0.0085 g/L

Bioprocess Engineering Multiple Choice Questions on “Internal Mass Transfer and Reaction”.

1. The heat diffusion is which type of process?
A. Conduction
B. Convection
C. Radiation
D. Osmosis

Answer: B
Explanation: Convection is usually the dominant form of heat transfer in liquids and gases. Although often discussed as a distinct method of heat transfer, convective heat transfer involves the combined processes of unknown conduction (heat diffusion) and advection (heat transfer by bulk fluid flow).

2. The immobilized cells have different temperature gradient.
A. True
B. False

Answer: B
Explanation: Kinetic parameters for enzyme and cell reactions are strong functions of temperature. If temperature in the particle varies, different values of the kinetic parameters must be applied. However, as temperature gradients generated by immobilised cells and enzymes are generally negligible, assuming constant temperature throughout the particle is reasonable.

3. The effective diffusivity of substrate in the solid, (mathcal{D}_{Ac}) is dependent of substrate.
A. True
B. False

Answer: B
Explanation: The value of (mathcal{D}_{Ac}) is a complex function of the molecular – diffusion characteristics of the substrate, the tortuousness of the diffusion path within the solid, and the fraction of the particle volume available for diffusion. (mathcal{D}_{Ac}) is constant and independent of substrate concentration in the particle; this means that (mathcal{D}_{Ac}) does not change with position.

4. Enzyme is immobilised in 8 mm diameter agarose beads at a concentration of 0.018 kg protein m^-3 gel. Ten beads are immersed in a well-mixed solution containing 3.2 ×10^-3 kg m^-3 substrate. The effective diffusivity of substrate in agarose gel is 2.1 × 10^-9 m² s^-1. Kinetics of the enzyme can be approximated as first order with specific rate constant 3.11 × 10⁵ s^{– 1} per kg protein. Mass transfer effects outside the particles are negligible. Estimate the volume per bead.
A. 2.65×10^-7 m³
B. 2.66×10^-7 m³
C. 2.67×10^-7 m³
D. 2.68×10^-7 m³

Answer: D
Explanation: R = 4×10^-3 m; (mathcal{D}_{Ac}) = 2.1×10^-9 m² s^-1. In the absence of external mass-transfer effects, C_As = 3.2×10^-3 kg m^-3.
Volume per bead = 4/3 π R³ = 4/3 π (4×10^-3 m)³= 2.68×10^-7 m³.

5. Refer to Q4 and Estimate the amount of enzyme present.
A. 4.81×10^-8 kg
B. 4.83×10^-8 kg
C. 4.85×10^-8 kg
D. 4.87×10^-8 kg

Answer: B
Explanation: 10 beads have volume 2.68×10^-6 m³. The amount of enzyme present is:
2.68 ×10^-6 m³ (0.018 kgm^-3) = 4.83 × 10^-8 kg.

6. Refer to Q4 and Q5, and Estimate the intrinsic first-order rate constant “k₁”.
A. 0.005 s^-1
B. 0.015 s^-1
C. 0.025 s^-1
D. 0.035 s^-1

Answer: B
Explanation: k I = 3.11×10⁵ s^-1kg^-1(4.83 x 10^-8 kg) = 0.015 s^-1.

7. The unit of intrinsic factor “k” remains constant in every order.
A. True
B. False

Answer: B
Explanation: k₁ is the intrinsic first-order rate constant with dimensions T^-1, whereas, k₀ is the intrinsic zero-order rate constant with units of, for example, g mol s^-1 m^-3particle.

8. Non-viable yeast cells are immobilised in alginate beads. The beads are stirred in glucose medium under anaerobic conditions. The effective diffusivity of glucose in the beads depends on cell density according to the relationship:
(mathcal{D}_{Ac}) = 6.33-7.17yC
where (mathcal{D}_{Ac}) is effective diffusivity ×10¹⁰ m² s^-1 and y_C is the weight fraction of yeast in the gel. Rate of glucose uptake can be assumed to be zero order; the rate constant at a yeast density in alginate of 15 wt% is 0.5 g l^-1 min^-1. For maximum reaction rate, the concentration of glucose inside the particles should remain above zero. Estimate the value of k₀, Converting k₀ to units of kg, m and s?
A. 8.33 x 10^-3 kg m^-3 s^-1
B. 6.33 x 10^-3 kg m^-3 s^-1
C. 4.33 x 10^-3 kg m^-3 s^-1
D. 2.33 x 10^-3 kg m^-3 s^-1

9. What is the concept of overall coefficient of heat transfer is used in heat transfer problems?
A. Conduction
B. Convection
C. Radiation
D. Conduction and convection

Answer: D
Explanation: The overall heat transfer by combined Modes is usually expressed on terms of an overall conductance or overall heat transfer coefficient ‘U’. There are numerous methods for calculating the heat transfer coefficient in different heat transfer modes, different fluids, flow regimes, and under different thermohydraulic conditions. Often it can be estimated by dividing the thermal conductivity of the convection fluid by a length scale.

10. The natural convection air cooled condensers are used in ________
A. Water coolers
B. Domestic refrigerator
C. Room air conditioners
D. Water boilers

Answer: B
Explanation: Air cooled condensers are of two types: natural convection and forced convection. In the natural convection type, the air flows over it in natural a way depending upon the temperature of the condenser coil. In the forced air type, a fan operated by a motor blows air over the condenser coil.

11. Concentration at the surface of the spherical catalyst will be higher than the inside of the catalyst.
A. True
B. False

Answer: A
Explanation: Concentration at the pore mouth will be higher than that inside the pore, .i.e., internal diffusion which is diffusion of reactants or products from particle surface (pore mouth) to pellet interior. And Mole balance over the shell thickness Δr is: IN – OUT + GEN = ACCUM.

12. To increase the overall rate of a rxn limited by internal diffusion the reaction should not ______
A. decrease the radius R
B. increase the concentration of A
C. increase the radius R
D. increase the temperature

Answer: C
Explanation: Internal diffusion limits the observed rate when decreasing d_p increases –r’_A . If the rate increases when d_p decreases but does not change with F_T0, then larger d_p is limited by internal diffusion. (where, F_T0 is the particle size, d_p is the diameter and –r’_A is the reaction rate).

Bioprocess Engineering Multiple Choice Questions & Answers focuses on “Methods Used for the Cultivation of Animal Cells”.

1. Primary plant cells are better formed aggregates than animal cells.
A. True
B. False
Answer: A
Explanation: Unlike plant cells, primary mammalian cells do not normally form aggregates, but grow in the form of monolayers on support surfaces such as glass surfaces of flasks. Using the proteolytic enzyme trypsin, individual cells in a tissue can be separated to form single-cell cultures.

2. What does “T” refers to in Tissue culture flask (T-Flask)?
A. Total volume of the flask
B. Total weight of the flask
C. Total surface area of the flask
D. Total mass of the flask
Answer: C
Explanation: The “T” referred to the total surface area of the flask that was available for cell growth: thus a T-25 flask had a 25cm² growth area. By the 1960s, straight neck T-flasks was available molded from polystyrene that was treated to enhance cell attachment.

3. Which of the following is not a type of basic T-Flask?
A. T-25
B. T-55
C. T-75
D. T-175
Answer: B
Explanation: Flasks are divided into 6 types, depending on the culture scale and the cap type. Cell growth area for T25, T75 and T175 are 25 cm², 75 cm², and 175 cm², respectively. Each flask can be provided with plug or filter caps.

4. EDTA is carcinogenic.
A. True
B. False
Answer: B
Explanation: The typical concentration of use of EDTA is less than 2%, with the other salts in current use at even lower concentrations. The lowest dose reported to cause a toxic effect in animals was 750 mg/kg/day. These chelating agents are cytotoxic and weakly genotoxic, but not carcinogenic.

5. Transformed cell lines are immortal.
A. True
B. False
Answer: A
Explanation: Cells that can be propagated indefinitely are called continuous, immortal, or transformed cell lines. Cancer cells are naturally immortal. All cancerous cell lines are transformed, although it is not clear whether all transformed cell lines are cancerous.

6. Nontransformed cell lines form monolayer whereas transformed cell lines form multilayer.
A. True
B. False
Answer: A
Explanation: One particular characteristic is contact inhibition. In two-dimensional culture on a surface nontransformed cells form only a monolayer, as cell division is inhibited when a cell’s surface is in contact with other cells. Transformed cells do not “sense” the presence of other cells and keep on dividing to form multilayer structures.

7. Baculovirus infects insect cell lines and are also pathogenic to humans.
A. True
B. False
Answer: B
Explanation: The baculovirus that infects insect cells is an ideal vector for genetic engineering, because it is nonpathogenic to humans and has a very strong promoter that encodes for a protein that is not essential for virus production in cell culture. The insertion of a gene under the control of this promoter can lead to high expression levels (40% of the total protein as the target protein).

8. Hybridoma cells have an application to produce:
A. Antigens
B. Antibodies
C. Cancer cells
D. Cell lines
Answer: B
Explanation: Hybridoma cells are obtained by fusing lymphocytes (normal blood cells that make antibodies) with myeloma (cancer) cells. Lymphocytes producing antibodies grow slowly and are mortal. After fusion with myeloma cells, hybridomas become immortal, can reproduce indefinitely, and produce antibodies. Using hybridoma cells, highly specific, monoclonal (originating from one cell) antibodies can be produced against specific antigens.

9. Vaccine is not a serum.
A. True
B. False
Answer: A
Explanation: Both antisera and antitoxins are means of proactively combating infections. The introduction of compounds to which the immune system responds is an attempt to build up protection against microorganisms or their toxins before the microbes actually invade the body.

10. Serum does not require sterilization.
A. True
B. False
Answer: B
Explanation: Serum must be filtered sterilized, and contamination with viruses and possibly mycoplasma (a wall less bacterium) are potential problems. Contamination by prions (agents that cause diseases such as “mad cow” disease) is a real concern, and source animals cannot come from regions known to have contaminated animals.

11. What is the full form of MEM in Eagle’s cell culture medium?
A. Maximum evaporating medium
B. Maximum essential medium
C. Minimum essential medium
D. Minimum evaporating medium
Answer: C
Explanation: Eagle’s minimal essential medium (EMEM) is a cell culture medium developed by Harry Eagle that can be used to maintain cells in tissue culture.

12. Transferrin is present in serum media.
A. True
B. False
Answer: B
Explanation: A simpler serum-free medium may contain insulin, transferrin, and selenium as serum replacement components, in addition to glucose, glutamine, other amino acids, and salts.

13. Small bubbles are more shear sensitive than large bubbles.
A. True
B. False
Answer: B
Explanation: Animal cells are very shear sensitive, and rising air bubbles may cause shear damage to cells, particularly at the point of bubble rupture. Very small bubbles are less damaging.

14. Pluronic F-68 is ____________
A. Ionic
B. Non-ionic
C. Covalent
D. Metallic
Answer: B
Explanation: Poloxamers are nonionic triblock copolymers composed of a central hydrophobic chain of polyoxypropylene (poly(propylene oxide)) flanked by two hydrophilic chains of polyoxyethylene (poly(ethylene oxide)). Poloxamers are also known by the trade names Synperonics, Pluronics, and Kolliphor.

15. Spinner flask contains:
A. Spindle shaped agitators
B. Cylindrical agitators
C. Simple agitators
D. Spoon-like agitators
Answer: D
Explanation: Spinner flasks contain a magnetically driven impeller or spoonlike agitators that operate at 10 to 60 rpm. Aeration is usually by surface aeration using 5% CO₂-enriched and filtered air for mammalian cell lines. Spinner flasks are set on a magnetic stirrer plate in a CO₂ incubator.

To practice all areas of Bioprocess Engineering, About Us.

Privacy Policy

Chemical Reaction Engineering Questions and Answers for Freshers focuses on “Reaction Kinetics Elements – Modelling the Rate Coefficient – 2”.

1. The reaction rate constant is related to reaction rate as ____
A. k α (-r_A)
B. k α (frac{1}{(-r_A.} )
C. k α e^(-r_A)
D. k is independent of (-r_A)

Answer: A
Explanation: The rate of reaction, (-r_A) = kC_Aⁿ
Where, n is the reaction order. The rate increases as the value of k increases.

2. State true or false.
The Arrhenius law plot of lnk vs 1/ T gives a straight line with large slope for large activation energy.
A. True
B. False

Answer: A
Explanation: The plot of lnk vs 1/ T gives a straight line of slope –(frac{E_a}{(R×T)}). As E_a increases, slope increases.

3. The unit of activation energy is ____
A. mol
B. mol.K
C. K
D. J/ mol

Answer: D
Explanation: Activation energy is expressed in J/ mol or kJ/ mol. Activation energy is the energy required for a chemical reaction to occur.

4. State true or false.
The value of frequency factor affects the temperature sensitivity of reaction.
A. True
B. False

Answer: B
Explanation: Frequency factor in Arrhenius equation does not affect the temperature sensitivity of the reaction. Frequency factor is independent of temperature. It takes the units of rate constant.

5. The rate constant and half life for reactions are related as ____
A. k α t_0.5
B. k α (frac{1}{(t_{0.5})} )
C. k α e^(-t_0.5)
D. k is independent of t_0.5

Answer: B

Chemical Reaction Engineering Multiple Choice Questions & Answers on “ Reversible Reactions Rate Laws”.

1. Choose the correct rate law for the elementary reversible reaction 3A ⇌ B + 2C.
A. –r_A = kC_A³ – C_BC_C²/K_C
B. –r_A = k[C_A³ – C_BC_C²K_C]
C. –r_A = k[C_A³ – C_BC_C²]/K_C
D. –r_A = k[C_A³ – C_BC_C²/K_C]
Answer: D
Explanation: Let’s say the rate constant for the forward reaction is k1 and for the backward reaction it is k₂.
Rate of disappearance of A = k₁C_A³
Rate of formation of A = k₂C_BC_C²
-r_A = k₁C_A³ – k₂C_BC_C². We know that K_C = k₁/k₂. So, -r_A = k[C_A³ – C_BC_C²/K_C].

2. Choose the correct rate law for the set of elementary reactions.
2A ⇌ B + C [k₁,-k₂]
A + 5B → 6D [k₃]
3C + 2D + B ⇌ 6E [k₄, -k₅]
5A + 3E ⇌ 2R [k₆]
A. –r_A = k₁C_A² – k₂C_B²C_C – k₃C_AC_B⁵ – k₆C_A5C_E³
B. –r_A = k₁C_A² – k₂C_B²C_C + k₃C_AC_B⁵ – k₆C_A5C_E³
C. –r_A = k₁k₄C_AC_CC_D – k₂k₅C_E³ + k₃k₆C_A⁶C_B⁵C_E³
D. –r_A = k₁k₄C_AC_CC_D – k₂k₅C_E³ – k₃k₆C_A⁶C_B⁵C_E³
Answer: B
Explanation: A appears in reactions 1, 2 and 4.
Rate of disappearance of A = k₁C_A² + k₃C_AC_B⁵ + k₆C_A5C_E³
Rate of formation of A = k₂C_BC_C

3. The equilibrium constant increases when _____________
A. Initial concentration of reactant increases in endothermic reactions
B. Initial concentration of reactant decreases in endothermic reactions
C. Temperature increases in exothermic reactions
D. Temperature decreases in exothermic reactions
Answer: D
Explanation: Equilibrium constant is independent of the initial reactant concentration. It only depends on temperature. As temperature rises, equilibrium constant increases for an endothermic reaction and decreases for an exothermic reaction. In both these cases the reverse happens when the temperature falls.

4. For the reaction ½ A + 3B → 2C
the specific reaction rate will be maximum for which species?
A. A
B. B
C. C
D. Equal for all species
Answer: B
Explanation: -r_A/( ½ ) = -r_B/3 = r_C/2
k_B = 3k_C/2 = 6k_A.
Thus, k_B has the biggest of the three rate constant values.

5. Which of the following is true for the rate law of a reversible reaction?
A. It reduces to irreversible form when product concentration is zero
B. It satisfies thermodynamic relationships at all times
C. Rate law can be written only in terms of partial pressures
D. Rate law can be written only in terms of concentrations
Answer: A
Explanation: Rate law can be written in terms of both concentrations or partial pressures. The thermodynamic relationships are satisfied only at equilibrium. If the product concentration is zero, the rate of the backward reaction becomes zero and the rate law reduces to irreversible form.

6. Which of the following constants is a type of equilibrium constant?
A. Dissociation constant
B. Association constant
C. Solubility
D. Decay constant
Answer: D
Explanation: Decay constant is actually a type of rate constant, for a given decay reaction. Solubility, association and dissociation constants are equilibrium constants. These are reaction quotients calculated at the stag of ionic equilibrium w.r.t various ionic and non-ionic species present in a system.

7. Choose the correct statement.
A. A + C → B
B + C → D
D → A + 2C
The above set represents a reversible reaction system
B. An irreversible reaction is just an extreme case of reversible reaction
C. A reversible reaction always reaches completion
D. Reversible reactions are spontaneous but very slow
Answer: B
Explanation: As strange as it may sound, it is actually true. All chemical systems have a natural tendency to attain an equilibrium. All reactions are technically reversible to some degree. The ones which we generally name irreversible reactions are nothing but extremely one-sided cases of reversible reactions, thus, making it almost impossible to observe the equilibrium stage due to only trace presence of certain species remaining.

About Us.

Privacy policy

Chemical Reaction Engineering Questions and Answers for Experienced people focuses on “Non Ideal Flow Basics – Properties of C, F and E Curves – 2”.

1. Which of the following is not a characteristic of tracer?
A. Inertness
B. Non – reactive
C. Easily detectable
D. Adsorbance onto the reactor surface
Answer: D
Explanation: The tracer should not stick to the walls or other surfaces of reactor. The physical properties of the tracer should be similar to that of the system.

2. If the exit age distribution is given as δ(t – 4), then (int_0^∞) t² E(t)dt is ____
A. 16
B. 4
C. 8
D. 20
Answer: A
Explanation: Exit age distribution is represented by dirac delta function. (int_0^∞) δ(t-τ)f(t)dt = f(τ). Hence, (int_0^∞) t² E(t)dt = 42 = 16.

3. If C=2 and area under the C – curve is 4 for a pulse input, then the value of E is ____
A. 2
B. 0.5
C. 1
D. 8
Answer: B
Explanation: E = (frac{C}{int_0^∞ C dt} )
(int_0^∞) C dt is the area under the C – curve. Hence, E = (frac{2}{4}) = 0.5.

4. The F(t) function for a PFR is _____
A. F(t) = (begin{cases}
0 & for & 0 < t < overline{t}, \
∞ & for & t > overline{t}\
end{cases} )
B. F(t) = (begin{cases}
1 & for & 0 < t < overline{t}, \
0 & for & t > overline{t}\
end{cases} )
C. F(t) = (begin{cases}
0 & for & 0 < t < overline{t}, \
1 & for & t > overline{t}\
end{cases} )
D. F(t) = (begin{cases}
∞ & for & 0 < t < overline{t}, \
1 & for & t > overline{t}\
end{cases} )
Answer: C
Explanation: The plug flow reactor has no longitudinal mixing. Hence, all fluid elements leaving the reactor have the same age (t).

5. Which of the following represents area under the C – curve? (Where, M is the total amount of tracer injected and
v is the volumetric flow rate of the effluent)
A. (frac{M}{v} )
B. (frac{v}{M} )
C. M × v
D. M × V
Answer: A
Explanation: C_area = (frac{M}{v} ). The unit is (kg.s)/m³.

To practice all areas of Chemical Reaction Engineering for Experienced people About Us.

Privacy policy