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1.	The time required to convert to a fraction of 0.7 for a first order reaction is ____(a) t0.7 = \(\frac{(0.7)}{C_{A0} k}\)(b) t0.7 = \(\frac{(-0.7)}{C_{A0} k}\)(c) t0.7 = \(\frac{(0.3)}{C_{A0} k}\)(d) t0.7 = \(\frac{(-0.3)}{C_{A0} k}\)The question was asked in an online interview.Enquiry is from Rate Data Collection & Analysis topic in division Collection and Analysis of Rate Data of Chemical Reaction Engineering
Answer» The correct choice is (d) t0.7 = \(\frac{(-0.3)}{C_{A0} k}\) The explanation is: The TIME for converting to any FRACTION, F is, TF = \(\frac{C_{A0}^{1-n}(F^{n-1}-1)}{(n-1)k}.\) At F = 0.7, t0.7 = \(\frac{C_{A0}^{1-2}(0.7^1- 1)}{k}.\)

2.	For a zero order reaction, the slope of t0.5 and CAo is ____(a) 2(b) 3(c) 1(d) 0I got this question during an interview.I want to ask this question from Rate Data Collection & Analysis topic in chapter Collection and Analysis of Rate Data of Chemical Reaction Engineering
Answer» CORRECT answer is (C) 1 The explanation is: Slope is (1-n). At n=0, slope = 1.

3.	If t0.5 = 100 minutes and initial concentration is 10, then the order of the reaction is ____(a) 4(b) -1(c) -2(d) 1This question was addressed to me in final exam.This interesting question is from Rate Data Collection & Analysis in chapter Collection and Analysis of Rate Data of Chemical Reaction Engineering
Answer» CORRECT answer is (b) -1 For explanation: t0.5 = \(C_{A0}^{1-n}.\)100 = 10^(1 – n). HENCE, n = -1.

4.	The fractional conversion is expressed as____(a) \(\frac{C_A}{C_{A0}} \)(b) \(\frac{C_{A0}}{C_A} \)(c) \(\frac{1}{C_{A0}} \)(d) \(\frac{C_A}{kC_{A0}} \)I have been asked this question in exam.I need to ask this question from Rate Data Collection & Analysis topic in chapter Collection and Analysis of Rate Data of Chemical Reaction Engineering
Answer» Right choice is (a) \(\FRAC{C_A}{C_{A0}} \) For explanation: Fractional conversion is the TIME REQUIRED to reduce the CONCENTRATION to any fraction. It is the ratio of FINAL to initial concentration.

5.	The relationship between t.0.5 and CAo for a zero order reaction is ____(a) t0.5 = \(\frac{1}{kC_{A0}} \)(b) t0.5 = \(\frac{1}{C_{A0}} \)(c) t0.5 = \(\frac{C_{A0}}{k}\)(d) t0.5 = \(\frac{C_{A0}}{2k}\)The question was asked at a job interview.The doubt is from Rate Data Collection & Analysis topic in chapter Collection and Analysis of Rate Data of Chemical Reaction Engineering
Answer» Right CHOICE is (d) t0.5 = \(\frac{C_{A0}}{2k}\) To explain I would SAY: For zero order REACTION, \(\frac{-dC_A}{dt}\) = k. t0.5 = \(\frac{C_{A0}- \frac{C_{A0}}{2}}{k}.\) HENCE, t0.5 = \(\frac{C_{A0}}{2k}.\)

6.	What is the slope of the plot of ln (t0.5) and ln (CAo)? (Where n is the reaction order)(a) 1-n(b) 2-n(c) n(d) \(\frac{n}{2} \)I have been asked this question during an interview.My question is from Rate Data Collection & Analysis in division Collection and Analysis of Rate Data of Chemical Reaction Engineering
Answer» The CORRECT OPTION is (a) 1-N The explanation is: The plot is a straight line of slope (1-n). The equation representing the relationship is ln (t0.5) = ln\((\FRAC{(2^{n-1}-1)}{(n-1)k})\) + (1 – n) ln (CAo).