Important Questions for CSIR NET Chemical Science - Physical Chemistry-Part IV

By Astha Singh|Updated : April 7th, 2022

Are you preparing for CSIR-NET 2022 and looking for some short and authentic study notes for Chemical Sciences to smoothen your preparation journey? We have got you covered!

Candidates preparing for their upcoming CSIR NET 2022 exam can really make their preparation journey easier with the help of some reliable study notes that cover the topics in the most simple way. We at BYJU'S Exam Prep have come up with the idea of providing Topic Wise Important Questions on Physical Chemistry-Part IV in the Chemical Science syllabus. 

The important questions on Physical Chemistry-Part IV are developed by our experienced subject-matter experts to provide you with the most standard and authentic set of study materials to be focused upon. The students need the best resources for their preparation to clear the CSIR NET 2022 examination, Here are the most reliable important questions to make the topics easier for you and also help you to save your time to do preparations for the upcoming CSIR-NET 2022 exam.

Most Important Questions Physical Chemistry-Part IV

1. According to the transition state theory, a reaction between an atom and a linear molecule forms a non-linear transition state. The temperature dependence of pre-exponential factor(A) is

  1. T0
  2. T0.5
  3. T2
  4. T1.5

2. For a 2-component system, calculate the degree of freedom from the reduced phase rule equation.

  1. F’ = 2 – P
  2. F’ = 1 – P
  3. F’ = 3 + P
  4. F’ = 3 – P

3. For the hypothetical reaction: 2B(g) → B2(g), ΔCP (in joules) = 6.0 + 2.0 × 10–3 T and ΔH°298 = – 20.0 kJ mol–1. Calculate the temperature at which ΔH° = 0 for this reaction.

  1. 2557 K
  2. 3225 K
  3. 4250 K
  4. 2678 K

4. 1× 10-6 moles of the enzymes carbonic anhydrase dehydrate H2CO3 and produces 0.6 mol of CO2 per second. Calculate the turnover number of the enzyme.

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5. The molarity of MgSO4 solution is 3M. The value of ionic strength will be:

  1. 12
  2. 24
  3. 36
  4. 48

6. Determine the temperature at which the average velocity of oxygen equals that of hydrogen at 20 K.

  1. 400 K
  2. 300 K
  3. 320 K
  4. 530 K

7. What will be the point group of XeO2(CF2)N(SO2F)OSO2F?

  1. C2v
  2. Cs
  3. C3h
  4. D2

8. How many normal modes of vibration are there for C60?

  1. 174
  2. 180
  3. 60
  4. 75

9. Calculate the length of a cubic unit cell having a 4.8 Å wavelength. The (1,1,1) plane showing 1st order deflection at 60ᵒ is:

  1. 8.9 Å
  2. 2.8 Å
  3. 0.8 Å
  4. 4.8 Å

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Answer Key:

  1. A
  2. D
  3. A
  4. A
  5. A
  6. C
  7. B
  8. A
  9. D
  10. A

Solutions:

Solution 1. According to the transition state theory,

A-B + C → non-linear T.S.

A = NAkT/h [ Q*/QAB QC]

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Solution 2. For a two-component system, the phase rule becomes: 

F = 4 – P

Since the minimum no. of phases, P, in any system is 1, the maximum no. of degrees of freedom, F, is 3 Thus, three variables would be necessary to describe a system. Since three variables are difficult to plot in a graph, it is customary to hold one of them, say, the pressure, constant on a diagram of temperature plotted versus concentration. This reduces the degree of freedom of the system by one and the phase rule equation is then written as:

F’ = 3 – P

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This is a quadratic equation in T which on solving gives two roots – one positive and the other negative. The negative root has no physical significance. The positive root gives T = 2557 K.

Solution 4. Turnover number: Number of molecules converted in unit time by one molecule of enzyme. 

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Solution 7. Bicoordinated derived Species = CF2

Monovalent derived Species

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Solution 8. C60 is a non-linear molecule and is composed of 60 atoms. In total it has 3N degrees of freedom and with 60 atoms, this means there are 180 degrees of freedom in total. 3 of these correspond to translation and three correspond to rotation.

The degrees of freedom left over are the vibrational degrees of freedom and so we have

= 3N – 6

= 180 – 6

= 174 vibrational modes.

Solution 9. Bragg equation can be represented as:

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