ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 11, Problem 11B.11P
Interpretation Introduction

Interpretation:

An expression for the value of J for the most populated rotational energy level of a diatomic rotor with degeneracy of 2J+1 has to be developed.  The expression has to be evaluated for ICl.  An expression for the value of J for the most populated rotational energy level of a diatomic rotor with degeneracy of (2J+1)2 has to be developed.  The expression has to be evaluated for CH4.

Concept introduction:

A three-dimensional rotating molecule is termed as a rotor.  If the bond length of the molecule does not change during the rotation, the molecule is termed as a rigid rotor.  The population of the states depends upon the temperature.  As the temperature increases, the population of the higher states increases.

Expert Solution & Answer
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Answer to Problem 11B.11P

An expression for the value of J for the most populated rotational energy level of a diatomic rotor with degeneracy of 2J+1 is shown below.

    J=kT2hcB˜12

The value of the expression for ICl is 29.656_.

An expression for the value of J for the most populated rotational energy level of a diatomic rotor with degeneracy of (2J+1)2 is shown below.

    J=kThcB˜12

The value of the expression for CH4 is 5.796_.

Explanation of Solution

The energy of a rigid rotor (EJ) is calculated by the expression given below.

    EJ=hcB˜J(J+1)        (1)

Where,

  • h is the Planck’s constant. (6.6×1034Js).
  • c is the speed of light. (3×1010cms1)
  • B˜ is the rotational constant.
  • J is the rotational quantum number.

The degeneracy (gJ) is calculated by the formula given below.

    gJ=2J+1        (2)

The expression for population, NJ of a level J is given below.

    NJ=NgJeEJkT        (3)

Where,

  • N is the total number of molecules in the sample.
  • k is the Boltzmann constant.
  • T is the temperature.

Substitute equation (1) and (2) in (3).

    NJ=N(2J+1)ehcB˜J(J+1)kT        (4)

For maximum value of NJ (maximum population), the relation given below will hold true.

    dNJdJ=0        (5)

Substitute the value of NJ from equation (4) in (5).

    ddJ(N(2J+1)ehcB˜J(J+1)kT)=0N((2J+1)ddJehcB˜J(J+1)kT+ehcB˜J(J+1)kTddJ(2J+1))=0N((2J+1)ehcB˜J(J+1)kT(hcB˜kT)(2J+1)+2ehcB˜J(J+1)kT)=0NehcB˜J(J+1)kT((2J+1)2(hcB˜kT)+2)=0

The equation given above was solved further as shown below.

    (2J+1)2(hcB˜kT)+2=0(2J+1)2=2kThcB˜2J+1=2kThcB˜J=kT2hcB˜12        (6)

The value of B˜ for ICl is 0.1142cm1.

The value of T 25°C.

The conversion of Celsius to Kelvin is done as shown below.

  οC=273 K

Therefore, the conversion of 25°C to Kelvin is done as shown below.

  25°C=25+273 K=298K

The value of k is 1.38×1023JK1.

The value of h is 6.626×1034Js.

The value of c is 3×1010cms1.

Substitute the value of B˜, T, k, h, and c in equation (6).

    J=1.38×1023JK1×298K2×6.6×1034Js×3×1010cms1K×0.1142cm112=909.3612=30.15612=29.656_

Therefore, the value of maximum populated level of ICl is 29.656_.

The degeneracy (gJ) is calculated by the formula given below.

    gJ=(2J+1)2        (7)

Substitute equation (1) and (7) in (3).

    NJ=N(2J+1)2ehcB˜J(J+1)kT        (8)

For maximum value of NJ (maximum population), the relation given below will hold true.

    dNJdJ=0        (5)

Substitute the value of NJ from equation (8) in (5).

    ddJ(N(2J+1)2ehcB˜J(J+1)kT)=0N(ehcB˜J(J+1)kTddJ(2J+1)2+(2J+1)2ddJehcB˜J(J+1)kT)=0N(2(2J+1)2ehcB˜J(J+1)kT+(2J+1)2ehcB˜J(J+1)kT(hcB˜kT)(2J+1))=0N(2J+1)ehcB˜J(J+1)kT(4+(2J+1)2(hcB˜kT))=0

The equation given above was solved further as shown below.

    4+(2J+1)2(hcB˜kT)=0(2J+1)=4kThcB˜2J=4kThcB˜1J=kThcB˜12        (9)

The value of B˜ for CH4 is 5.24cm1.

The value of T 25°C.

The conversion of Celsius to Kelvin is done as shown below.

  οC=273 K

Therefore, the conversion of 25°C to Kelvin is done as shown below.

  25°C=25+273 K=298K

The value of k is 1.38×1023JK1.

The value of h is 6.626×1034Js.

The value of c is 3×1010cms1.

Substitute the value of B˜, T, k, h, and c in equation (6).

    J=1.38×1023JK1×298K6.6×1034Js×3×1010cms1K×5.24cm112=39.6412=6.29612=5.796_

Therefore, the value of maximum populated level of CH4 is 5.796_.

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Chapter 11 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

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