30 + Interview Questions in Kinetics Of Homogeneous Reactions in Chemical Reaction Engineering Page 1 InterviewSolution

1.	The performance equation of a CSTR is ____(a) \(\frac{F_{A_0}}{V} = \frac{(-r_A)}{X_A} \)(b) \(\frac{F_{A_0}}{V} = \frac{X_A}{(r_A)} \)(c) \(\frac{V}{X_A} = \frac{(-r_A)}{F_{A_0}} \)(d) \(\frac{F_{A_0}}{V} = \frac{X_A}{-r_A} \)I had been asked this question in an interview.This intriguing question comes from Kinetics of Homogeneous Reactions in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The CORRECT choice is (a) \(\FRAC{F_{A_0}}{V} = \frac{(-r_A)}{X_A} \) Explanation: For a CSTR, Molar FLOW rate in = Molar flow rate out + accumulation FA0 = FA + (-rA)V FA0 = FA0(1- XA) + (-rA)V Hence, FA0XA = (-rA)V.

Discussion

2.	For a PFR, the area under the curve \(\frac{1}{-r_A}\) vs XA gives ____(a) \(\frac{F_{A_0}}{dV} \)(b) \(\frac{dV}{F_{A_0}} \)(c) dVCA0(d) dVFA0The question was posed to me in an interview for internship.My question comes from Kinetics of Homogeneous Reactions topic in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct option is (b) \(\FRAC{dV}{F_{A_0}} \) For EXPLANATION I would say: For a PFR, FA0dXA = (-RA)dV. \(\frac{dXA}{(-rA)}\) gives . Area under the CURVE gives \(\frac{dV}{FA0}. \)

Discussion

3.	For an endothermic reaction, the average energy of products is ____(a) Equal to the average energy of reactants(b) Lower than or equal to the average energy of reactants(c) Lower than the average energy of reactants(d) Greater than the average energy of reactantsI had been asked this question in exam.This intriguing question comes from Kinetics of Homogeneous Reactions topic in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» CORRECT choice is (d) GREATER than the AVERAGE energy of reactants Explanation: In an endothermic REACTION, reactants absorb energy to form products. HENCE, the average energy of products is greater than the average energy of reactants.

Discussion

4.	The equation representing the variation of rate constant with respect to temperature by Arrhenius equation is ____(a) ln(\(\frac{k_2}{k_1}\)) = –\(\frac{E}{R}(\frac{1}{T_2} – \frac{1}{T_1}) \)(b) ln(\(\frac{k_2}{k_1}\)) = \(\frac{E}{R}(\frac{1}{T_2} – \frac{1}{T_1}) \)(c) ln(\(\frac{k_2}{k_1}\)) = –\(\frac{E}{R}(\frac{1}{T_1}- \frac{1}{T_2}) \)(d) ln(\(\frac{k_2}{k_1}\)) = –\(\frac{E}{R}(\frac{1}{T_2} + \frac{1}{T_1}) \)I had been asked this question in semester exam.Asked question is from Kinetics of Homogeneous Reactions topic in portion Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct answer is (a) ln(\(\frac{k_2}{k_1}\)) = –\(\frac{E}{R}(\frac{1}{T_2} – \frac{1}{T_1}) \) For EXPLANATION I would say: By Arrhenius equation, k = \(Ae^{\frac{-EA}{RT}},\) where A is the frequency factor. For TWO different temperatures, T1 and T2, the Arrhenius equation is reduced as ln(\(\frac{k_2}{k_1}\)) = –\(\frac{E}{R}(\frac{1}{T_2} – \frac{1}{T_1}). \)

Discussion

5.	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 moleculesI had been asked this question by my school teacher while I was bunking the class.The above asked question is from Liquid and Gas Phase Reactions in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» CORRECT choice is (a) Concentration of gas molecules 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 = \(\FRAC{n}{V} = \frac{P}{RT}. \)

Discussion

6.	The minimum energy required by the reactants to effect the product formation is termed as _____(a) Threshold energy(b) Potential energy(c) Activation energy(d) Kinetic energyThe question was asked in examination.I would like to ask this question from Kinetics of Homogeneous Reactions in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct OPTION is (a) Threshold ENERGY The best I can explain: Threshold energy is the energy required so that the REACTION occurs. The energy SUPPLIED to the reactants to attain the threshold energy is TERMED as Activation energy.

Discussion

7.	State true or false.Transition state theory postulates that the reacting molecules form unstable intermediates, which are subsequently decomposed to products.(a) Transition state theory postulates that the reacting molecules form unstable intermediates, which are subsequently decomposed to products.(b) True(c) FalseI got this question in examination.The above asked question is from Kinetics of Homogeneous Reactions topic in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct option is (a) TRANSITION STATE theory POSTULATES that the reacting molecules FORM unstable intermediates, which are subsequently decomposed to products. Explanation: Transition state theory proposes that reactants form intermediates CALLED activated complex, which then stabilizes by product formation. A + B → AB* → Products

Discussion

8.	The temperature dependency of rate constant as given by Collision theory is ____(a) k α T(b) k α T^2(c) k α T^3(d) k α T^0.5This question was posed to me in a job interview.The origin of the question is Kinetics of Homogeneous Reactions topic in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct CHOICE is (d) k α T^0.5 The explanation: According to COLLISION THEORY, k α T^0.5\(e^{\frac{Ea}{RT}}\) Where, Ea is the activation energy. Hence the RATE CONSTANT, k α T^0.5.

Discussion

9.	Collision theory states that ____(a) Molecules do not collide during the reaction(b) Reactant molecules collide to effect product formation(c) Reactants form complexes(d) Products formed are unstableI got this question in a job interview.Asked question is from Kinetics of Homogeneous Reactions topic in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct option is (b) Reactant MOLECULES collide to EFFECT PRODUCT formation The best I can explain: COLLISION Theory states that there occur collisions between the reactant molecules. The collisions in which the molecules possess a minimum ENERGY are effective.

Discussion

10.	State true or false.=-\(\frac{1}{V}\frac{dN_A}{dt}=-\frac{dC_A}{dt}\) is valid only if A → B.(a) =-\(\frac{1}{V}\frac{dN_A}{dt}=-\frac{dC_A}{dt}\) is valid only if A → B.(b) True(c) FalseThe question was posed to me by my school principal while I was bunking the class.This interesting question is from Liquid and Gas Phase Reactions in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct CHOICE is (a) =-\(\frac{1}{V}\frac{dN_A}{dt}=-\frac{dC_A}{dt}\) is valid only if A → B. For EXPLANATION I would say: 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.

Discussion

11.	The activation energy for a reaction is 122137.5 J/ mol. If the rate constant at 330K is 0.5 min^-1, then the value of rate constant (in min^-1) at 360K by Arrhenius law is ____(a) 20.42(b) 18.65(c) 30.12(d) 12.32I have been asked this question during an online interview.My question is based upon Kinetics of Homogeneous Reactions topic in chapter Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct answer is (a) 20.42 For EXPLANATION I would say: ln(\(\FRAC{k_2}{k_1}) = -\frac{E}{R} (\frac{1}{T_2} – \frac{1}{T_1}) \) ln(\(\frac{k_2}{0.5}) = -\frac{E}{8.314} (\frac{1}{360}- \frac{1}{330}) \) K2 = 20.42 min^-1.

Discussion

12.	The rate of a gas phase reaction involving the conversion of reactant ‘i’ to product is expressed as ____(a) (-ri)=-\(\frac{1}{V^2}\frac{dN_i}{dt} \)(b) (-ri)=-\(\frac{1}{V}\frac{dN_i}{dt} \)(c) (-ri)=-\(\frac{dN_i}{dt} \)(d) (-ri)=-V\(\frac{dN_i}{dt} \)I had been asked this question by my college professor while I was bunking the class.I want to ask this question from Liquid and Gas Phase Reactions in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» CORRECT answer is (b) (-ri)=-\(\frac{1}{V}\frac{dN_i}{dt} \) The best 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, (-ri)=-\(\frac{1}{V}\frac{dN_i}{dt}. \)

Discussion

13.	Which one of the following combinations gives the highest conversion for second order reaction?(a) PFR followed by smaller CSTR followed by a bigger CSTR(b) PFR followed by bigger CSTR followed by a bigger CSTR(c) Smaller CSTR followed by PFR followed by a bigger CSTR(d) Bigger CSTR followed by a PFR followed by smaller CSTRThis question was addressed to me by my college professor while I was bunking the class.Question is taken from Kinetics of Homogeneous Reactions in chapter Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct answer is (a) PFR followed by smaller CSTR followed by a bigger CSTR To explain I would say: For reactions of order greater than ONE, HIGHER conversion is achieved in a PFR. Smaller the volume of a CSTR, it tends to BEHAVE as a PFR. Hence, for higher CONVERSIONS, the reactors are arranged in the order PFR followed by smaller CSTR followed by a bigger CSTR.

Discussion

14.	If the conversion of a first order liquid phase reaction occurring in a CSTR is 75%, molar feed rate is 5 mol/min, the rate of the reaction is 5 \(\frac{mol}{litre.min} \) then the volume of the reactor (in litre) is ____(a) 0.5(b) 0.75(c) 0.33(d) 0.4The question was asked during an interview for a job.The above asked question is from Kinetics of Homogeneous Reactions topic in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Right answer is (b) 0.75 To explain: For a CSTR, the performance EQUATION is FA0XA = (-rA)V FA0 is MOLAR FEED rate and XA is the fractional conversion 5 × 0.75 = 5 × V V = 0.75 L

Discussion

15.	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 mixtureI got this question in an internship interview.Query is from Liquid and Gas Phase Reactions topic in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct choice is (a) Increase in moles of product INCREASES the volume of the reaction MIXTURE The BEST I can explain: 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.

Discussion

16.	The combination of ideal reactors among the following is ____(a) Plug flow reactor and batch reactor(b) Batch reactor and mixed flow reactor(c) Plug flow reactor and mixed flow reactor(d) Batch reactor onlyThis question was addressed to me during an online interview.Enquiry is from Kinetics of Homogeneous Reactions in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct OPTION is (C) Plug flow reactor and mixed flow reactor For explanation: Batch reactor is an unsteady STATE reactor. Plug flow reactor and mixed flow REACTORS operate under steady state CONDITIONS.

Discussion

17.	If kB is Boltzmann constant and h is Plank’s constant, the rate of decomposition of complex formed during product formation, as given by Transition state theory is ____(a) \(\frac{k_B T}{h} \)(b) \(\frac{k_B h}{T} \)(c) \(\frac{k_B T}{2h} \)(d) \(\frac{k_B T^2}{h} \)The question was asked in class test.My question comes from Kinetics of Homogeneous Reactions in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct choice is (a) \(\frac{k_B T}{h} \) To ELABORATE: The rate of DECOMPOSITION of the complex is the same for all reactions. Decomposition rate is GIVEN as, R = \(\frac{k_B T}{h} \).

Discussion

18.	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 → BI got this question at a job interview.The question is from Liquid and Gas Phase Reactions in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct CHOICE is (c) A → 4B The 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.

Discussion

19.	State true or false.The average concentration of product is low inside a CSTR.(a) The average concentration of product is low inside a CSTR.(b) True(c) FalseI got this question by my school principal while I was bunking the class.My query is from Kinetics of Homogeneous Reactions in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct ANSWER is (a) The average concentration of product is low INSIDE a CSTR. The best explanation: There is continuous MIXING of all fluid MOLECULES inside a CSTR. As there is continuous recycle, the average concentration is low in the reactor.

Discussion

20.	The activation energy of a reaction is 155326 J/ mol. The rate constant of the reaction at and 300K as a function of rate constant at 400K, obtained by the Collision theory is ____(a) k1 = 1.5 × 10^-7k2(b) k1 = 1.2 × 10^-6 k2(c) k1 = 1.5 × 10^-6 k2(d) k1 = 1.2 × 10^-7 k2This question was addressed to me at a job interview.I need to ask this question from Kinetics of Homogeneous Reactions in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Right choice is (a) K1 = 1.5 × 10^-7k2 The explanation: By COLLISION THEORY, ln(\(\frac{k_2}{k_1}) = -\frac{E}{R} (\frac{1}{T_2} – \frac{1}{T_1}) + 0.5ln(\frac{T_2}{T_1}) \) ln(\(\frac{k_2}{k_1}) = -\frac{155326}{8.314}(\frac{1}{400} – \frac{1}{300}) + 0.5ln(\frac{400}{300}) \) Hence, k1 = 1.5 × 10^-7k2.

Discussion

21.	State true or false.There is no axial mixing in a plug flow reactor.(a) There is no axial mixing in a plug flow reactor.(b) True(c) FalseThe question was posed to me during an interview for a job.The origin of the question is Kinetics of Homogeneous Reactions topic in chapter Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct answer is (a) There is no axial mixing in a PLUG flow REACTOR. The best I can explain: There is no mixing of early and LATER entering FLUID in a plug flow reactor. There is no overtaking between the fluid molecules.

Discussion

22.	The temperature dependency of rate constant as given by Transition state theory is ____(a) k α T(b) k α T^2(c) k α T^3(d) k α T^0.5I got this question in a national level competition.My doubt stems from Kinetics of Homogeneous Reactions in chapter Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Right answer is (a) k α T Explanation: According to Transition STATE THEORY, k α T\(e^{\FRAC{Ea}{RT}}\). k is a linear FUNCTION of temperature.

Discussion

23.	The performance equation of a PFR is ____(a) \(\frac{F_{A_0}}{dV} = \frac{dX_A}{(-r_A)} \)(b) \(\frac{dV}{F_{A_0}} = \frac{(-r_A)}{dX_A} \)(c) \(\frac{F_{A_0}}{dV} = \frac{(-r_A)}{dX_A} \)(d) \(\frac{F_{A_0}}{dX_A} = \frac{(-r_A)}{dV} \)This question was addressed to me in exam.Query is from Kinetics of Homogeneous Reactions topic in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct answer is (c) \(\frac{F_{A_0}}{dV} = \frac{(-r_A)}{dX_A} \) Easiest explanation: FA0dXA = (-rA)dV The REACTION is INTEGRATED in the entire volume of REACTOR. The material BALANCE is carried out in a small differential volume dV.

Discussion

24.	If the rate constant of a reaction at 600K is 100 times the rate constant at 500K, then the value of the activation energy obtained by Transition state theory is ____(a) 120987.12(b) 167435.15(c) 110319.28(d) 156435.54The question was posed to me during an interview for a job.The origin of the question is Kinetics of Homogeneous Reactions in chapter Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct choice is (c) 110319.28 For explanation: By Transition state theory, LN(\(\frac{k_2}{k_1}) = -\frac{E}{R} (\frac{1}{T_2} – \frac{1}{T_1}) + ln(\frac{T_2}{T_1}) \) ln(\(\frac{100k_1}{k_1}) = -\frac{E}{8.314} (\frac{1}{600} – \frac{1}{500}) + ln(\frac{600}{500}) \) E = 110319.28.

Discussion

25.	Which of the following is true for a reaction occurring in a batch reactor?(a) The accumulation term is zero(b) Only the component input term is zero(c) Only the component output term is zero(d) Both the component input and output terms are zeroThe question was posed to me by my school teacher while I was bunking the class.My question is from Kinetics of Homogeneous Reactions in portion Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct option is (d) Both the component INPUT and output terms are zero Easiest explanation: REACTION is carried out in batches. As and when a new batch of reactant is fed to the reactor, the PRECEDING batch product is WITHDRAWN.

Discussion

26.	If CA is the final concentration and CA0 is the initial concentration, the conversion of a reaction is expressed as ____(a) \(\frac{C_{A0}-C_A}{(C_{A0})} \)(b) \(\frac{C_{A0}-C_A}{(-r_A)}\)V(c) \(\frac{C_{A0}-C_A}{(-r_A)V} \)(d) \(\frac{C_A}{(C_{A0})} \)I had been asked this question in an interview for job.The above asked question is from Kinetics of Homogeneous Reactions in chapter Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» RIGHT choice is (a) \(\FRAC{C_{A0}-C_A}{(C_{A0})} \) Best explanation: Conversion in a chemical reaction is the ratio of the amount of REACTANT consumed to form products to the amount of reactant fed. If the entire reactant is CONVERTED to product, then the conversion is 100%.

Discussion

27.	If the rate constant of a reaction at 275K is 1 min^-1 and the rate constant at 300K is 2 min^-1, what is the activation energy (in J/ mol) as obtained by Arrhenius law?(a) 24655(b) 19019.14(c) 366543.2(d) 18989.32The question was asked during an online interview.Query is from Kinetics of Homogeneous Reactions topic in chapter Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct answer is (b) 19019.14 For explanation I would say: ln(\(\frac{k_2}{k_1}) = -\frac{E}{R} (\frac{1}{T_2} – \frac{1}{T_1}) \) ln(2) = –\(\frac{E}{8.314} (\frac{1}{360}- \frac{1}{275}) \) E = 19019.14 J/ mol.

Discussion

28.	Which of the following is incorrect about Transition state theory?(a) It is based on statistical mechanics(b) The formation of activated complex to be rapid(c) The decomposition of activated complex is slow(d) The formation of activated complex is the rate determining stepI had been asked this question in an online interview.Question is taken from Kinetics of Homogeneous Reactions topic in chapter Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» The correct answer is (d) The FORMATION of activated COMPLEX is the rate determining step To explain I would say: The activated complex formation step is the fastest. The rate determining step is the slowest of all the STEPS involved in a reaction. Hence, the DECOMPOSITION of activated complex is the rate determining step.

Discussion

29.	Which of the following pairing is incorrect? (Where A is the frequency factor)(a) Arrhenius equation; k = A\(e^{\frac{-Ea}{RT}} \)(b) Transition state theory; k = AT\(e^{\frac{-Ea}{RT}} \)(c) Collision theory; k = AT^0.5\(e^{\frac{-Ea}{RT}} \)(d) Arrhenius equation; k = AT^2\(e^{\frac{-Ea}{RT}} \)The question was posed to me during an interview for a job.My doubt is from Kinetics of Homogeneous Reactions topic in section Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» CORRECT choice is (d) Arrhenius EQUATION; k = AT^2\(e^{\frac{-Ea}{RT}} \) To explain I WOULD say: Arrhenius equation: k = A\(e^{\frac{-Ea}{RT}} \). By collision theory, k α T^0.5 By transition STATE theory, k α T

Discussion

30.	State true or false.Transition state theory shows that there exists no equilibrium between activated complex and reacting molecules at all times.(a) True(b) FalseI got this question in an interview for job.This is a very interesting question from Kinetics of Homogeneous Reactions topic in division Kinetics of Homogeneous Reactions of Chemical Reaction Engineering
Answer» Correct option is (b) False Explanation: There exists equilibrium between activated COMPLEX and REACTING molecules at all times. The RATE of DECOMPOSITION of the complex is same for all reactions.

Discussion

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