ATKINS' PHYSICAL CHEMISTRY
ATKINS' PHYSICAL CHEMISTRY
11th Edition
ISBN: 9780190053956
Author: ATKINS
Publisher: Oxford University Press
Question
Book Icon
Chapter 11, Problem 11C.15P
Interpretation Introduction

Interpretation:

The value of B˜0 and B˜1 have to be calculated.  The wave numbers of the transitions originating from J=3 of the P branches and R branches have to be predicted.  The dissociation energy of the IF molecule has to be predicted.

Concept introduction:

The internal mode of vibration of molecule leads to a periodic change in the internal internuclear separation of a molecule. Due to this change, the molecules are considered as harmonic oscillators. The strength of a bond in a bond is measured by force constant. The force constant of a molecule is related to the effective mass and vibrational frequency.  The vibrational energy of a molecule is given by the equation as shown below.

  Eν=(ν+12)ω.

Expert Solution & Answer
Check Mark

Answer to Problem 11C.15P

The value of B˜0 is 0.278775cm-1_.

The value of B˜1 is 0.276905cm-1_.

The wave numbers of the transitions originating from J=3 of the P branches is 601.724cm-1_.

The wave numbers of the transitions originating from J=3 of the R branches is 605.626cm-1_.

The dissociation energy of the IF molecule is 29641.42cm-1_.

Explanation of Solution

The rotational constant for a diatomic molecule in vibrational state with quantum number v is given by the expression shown below.

    B˜v=B˜ea(v+12)        (1)

Where,

  • B˜e is the rotational constant corresponding to the equilibrium bond length.
  • v is the quantum number.
  • a is morse potential energy.

The values of B˜e and a for IF molecule are 0.27971cm1 and 0.187m1 respectively.

Substitute the value of of B˜e and a in equation (1) to calculate the value of B˜0 at v=0, as shown below.

    B˜v=B˜ea(v+12)B˜0=0.27971cm10.187m1×(1m100cm)(0+12)=0.27971cm10.000935cm1=0.278775cm-1_

Substitute the value of of B˜e and a in equation (1) to calculate the value of B˜1 at v=1, as shown below.

    B˜v=B˜ea(v+12)B˜1=0.27971cm10.187m1×(1m100cm)(1+12)=0.27971cm10.002805cm1=0.276905cm-1_

The wave numbers of the transitions originating from J=3 of the P branches is calculated by the formula given below.

    v˜p(J)=v˜(B˜1+B˜0)J+(B˜1B˜0)J2        (2)

Where,

  • v˜ is wave number.
  • B˜0 is rotational constant at v=0.
  • B˜1 is rotational constant at v=1.
  • J is vibrational state.

When anharmonicities are present v˜ is replaced by the term ΔG˜(v).

    ΔG˜(v)=v˜2(v+1)xev˜

Where,

  • xe is anharmonicity constant.

For v=0,

    ΔG˜(v)=v˜2(v+1)xev˜=v˜2(0+1)xev˜=v˜2xev˜

In equation (2) v˜ is replaced by the term ΔG˜(v), which is equal to v˜2xev˜.  Therefore, the equation (2) is written as shown below.

    v˜p(J)=v˜2xev˜(B˜1+B˜0)J+(B˜1B˜0)J2        (3)

For P branches, the value of v˜, xev˜, J, B˜1 and B˜0 are 610.258cm1, 3.141cm1, 4, 0.276905cm1, and 0.278775cm1.

Substitute the value of v˜, xev˜, J, B˜1 and B˜0 in equation (3).

    v˜p(J)=(610.258cm12×3.141cm1(0.276905cm1+0.278775cm1)4+(0.276905cm10.278775cm1)42)=610.258cm16.282cm12.222cm10.02999cm1=610.258cm17.4004cm1=601.724cm-1_

The wave numbers of the transitions originating from J=3 of the R branches is calculated by the formula given below.

    v˜R(J)=v˜2xev˜+(B˜1+B˜0)(J+1)+(B˜1B˜0)(J+1)2        (4)

For R branches, the value of v˜, xev˜, J, B˜1 and B˜0 are 610.258cm1, 3.141cm1, 2, 0.276905cm1, and 0.278775cm1.

Substitute the value of v˜, xev˜, J, B˜1 and B˜0 in equation (4).

  v˜R(J)=(610.258cm12×3.141cm1+(0.276905cm1+0.278775cm1)(2+1)+(0.276905cm10.278775cm1)(2+1)2)=610.258cm16.282cm1+1.667cm10.0168cm1=610.258cm17.9658cm1=605.626cm-1_

The dissociation energy of the IF molecule is calculated by the formula given below.

    De=v˜24xev˜        (5)

Where,

  • xe is anharmonicity constant.
  • v˜ is wave number.

Substitute the value of v˜ and xev˜, in equation (5).

    De=v˜24xev˜=(610.258cm1)24×3.141cm1=372414.82612.564cm1=29641.42cm-1_

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Please help me calculate the undiluted samples ppm concentration. My calculations were 280.11 ppm. Please see if I did my math correctly using the following standard curve. Link: https://mnscu-my.sharepoint.com/:x:/g/personal/vi2163ss_go_minnstate_edu/EVSJL_W0qrxMkUjK2J3xMUEBHDu0UM1vPKQ-bc9HTcYXDQ?e=hVuPC4
Provide an IUPAC name for each of the compounds shown. (Specify (E)/(Z) stereochemistry, if relevant, for straight chain alkenes only. Pay attention to commas, dashes, etc.) H₁₂C C(CH3)3 C=C H3C CH3 CH3CH2CH CI CH3 Submit Answer Retry Entire Group 2 more group attempts remaining Previous Next
Arrange the following compounds / ions in increasing nucleophilicity (least to most nucleophilic) CH3NH2 CH3C=C: CH3COO 1 2 3 5 Multiple Choice 1 point 1, 2, 3 2, 1, 3 3, 1, 2 2, 3, 1 The other answers are not correct 0000

Chapter 11 Solutions

ATKINS' PHYSICAL CHEMISTRY

Ch. 11 - Prob. 11E.2STCh. 11 - Prob. 11F.2STCh. 11 - Prob. 11A.1DQCh. 11 - Prob. 11A.2DQCh. 11 - Prob. 11A.3DQCh. 11 - Prob. 11A.1AECh. 11 - Prob. 11A.1BECh. 11 - Prob. 11A.2AECh. 11 - Prob. 11A.2BECh. 11 - Prob. 11A.3AECh. 11 - Prob. 11A.3BECh. 11 - Prob. 11A.4AECh. 11 - Prob. 11A.4BECh. 11 - Prob. 11A.5AECh. 11 - Prob. 11A.5BECh. 11 - Prob. 11A.6AECh. 11 - Prob. 11A.6BECh. 11 - Prob. 11A.7AECh. 11 - Prob. 11A.7BECh. 11 - Prob. 11A.8AECh. 11 - Prob. 11A.8BECh. 11 - Prob. 11A.9AECh. 11 - Prob. 11A.9BECh. 11 - Prob. 11A.10AECh. 11 - Prob. 11A.10BECh. 11 - Prob. 11A.11AECh. 11 - Prob. 11A.11BECh. 11 - Prob. 11A.12AECh. 11 - Prob. 11A.12BECh. 11 - Prob. 11A.1PCh. 11 - Prob. 11A.2PCh. 11 - Prob. 11A.3PCh. 11 - Prob. 11A.4PCh. 11 - Prob. 11A.5PCh. 11 - Prob. 11A.7PCh. 11 - Prob. 11A.8PCh. 11 - Prob. 11A.9PCh. 11 - Prob. 11A.11PCh. 11 - Prob. 11B.1DQCh. 11 - Prob. 11B.2DQCh. 11 - Prob. 11B.3DQCh. 11 - Prob. 11B.4DQCh. 11 - Prob. 11B.5DQCh. 11 - Prob. 11B.6DQCh. 11 - Prob. 11B.7DQCh. 11 - Prob. 11B.8DQCh. 11 - Prob. 11B.1AECh. 11 - Prob. 11B.1BECh. 11 - Prob. 11B.3AECh. 11 - Prob. 11B.3BECh. 11 - Prob. 11B.4BECh. 11 - Prob. 11B.5AECh. 11 - Prob. 11B.5BECh. 11 - Prob. 11B.6AECh. 11 - Prob. 11B.6BECh. 11 - Prob. 11B.7AECh. 11 - Prob. 11B.7BECh. 11 - Prob. 11B.8AECh. 11 - Prob. 11B.8BECh. 11 - Prob. 11B.9AECh. 11 - Prob. 11B.9BECh. 11 - Prob. 11B.10AECh. 11 - Prob. 11B.10BECh. 11 - Prob. 11B.11AECh. 11 - Prob. 11B.11BECh. 11 - Prob. 11B.12AECh. 11 - Prob. 11B.12BECh. 11 - Prob. 11B.13AECh. 11 - Prob. 11B.13BECh. 11 - Prob. 11B.14AECh. 11 - Prob. 11B.14BECh. 11 - Prob. 11B.1PCh. 11 - Prob. 11B.2PCh. 11 - Prob. 11B.3PCh. 11 - Prob. 11B.4PCh. 11 - Prob. 11B.5PCh. 11 - Prob. 11B.6PCh. 11 - Prob. 11B.7PCh. 11 - Prob. 11B.8PCh. 11 - Prob. 11B.9PCh. 11 - Prob. 11B.10PCh. 11 - Prob. 11B.11PCh. 11 - Prob. 11B.12PCh. 11 - Prob. 11B.13PCh. 11 - Prob. 11B.14PCh. 11 - Prob. 11C.1DQCh. 11 - Prob. 11C.2DQCh. 11 - Prob. 11C.3DQCh. 11 - Prob. 11C.4DQCh. 11 - Prob. 11C.1AECh. 11 - Prob. 11C.1BECh. 11 - Prob. 11C.2AECh. 11 - Prob. 11C.2BECh. 11 - Prob. 11C.3AECh. 11 - Prob. 11C.3BECh. 11 - Prob. 11C.4AECh. 11 - Prob. 11C.4BECh. 11 - Prob. 11C.5AECh. 11 - Prob. 11C.5BECh. 11 - Prob. 11C.6AECh. 11 - Prob. 11C.6BECh. 11 - Prob. 11C.7AECh. 11 - Prob. 11C.7BECh. 11 - Prob. 11C.8AECh. 11 - Prob. 11C.8BECh. 11 - Prob. 11C.2PCh. 11 - Prob. 11C.3PCh. 11 - Prob. 11C.4PCh. 11 - Prob. 11C.5PCh. 11 - Prob. 11C.6PCh. 11 - Prob. 11C.7PCh. 11 - Prob. 11C.8PCh. 11 - Prob. 11C.9PCh. 11 - Prob. 11C.10PCh. 11 - Prob. 11C.11PCh. 11 - Prob. 11C.12PCh. 11 - Prob. 11C.13PCh. 11 - Prob. 11C.15PCh. 11 - Prob. 11C.17PCh. 11 - Prob. 11C.18PCh. 11 - Prob. 11C.19PCh. 11 - Prob. 11D.1DQCh. 11 - Prob. 11D.2DQCh. 11 - Prob. 11D.3DQCh. 11 - Prob. 11D.1AECh. 11 - Prob. 11D.1BECh. 11 - Prob. 11D.2AECh. 11 - Prob. 11D.2BECh. 11 - Prob. 11D.3AECh. 11 - Prob. 11D.3BECh. 11 - Prob. 11D.4AECh. 11 - Prob. 11D.4BECh. 11 - Prob. 11D.5AECh. 11 - Prob. 11D.5BECh. 11 - Prob. 11D.6AECh. 11 - Prob. 11D.6BECh. 11 - Prob. 11D.7AECh. 11 - Prob. 11D.7BECh. 11 - Prob. 11D.2PCh. 11 - Prob. 11E.1DQCh. 11 - Prob. 11E.1AECh. 11 - Prob. 11E.1BECh. 11 - Prob. 11E.2AECh. 11 - Prob. 11E.2BECh. 11 - Prob. 11E.3AECh. 11 - Prob. 11E.3BECh. 11 - Prob. 11E.1PCh. 11 - Prob. 11E.2PCh. 11 - Prob. 11F.1DQCh. 11 - Prob. 11F.2DQCh. 11 - Prob. 11F.3DQCh. 11 - Prob. 11F.4DQCh. 11 - Prob. 11F.5DQCh. 11 - Prob. 11F.6DQCh. 11 - Prob. 11F.1AECh. 11 - Prob. 11F.1BECh. 11 - Prob. 11F.2AECh. 11 - Prob. 11F.2BECh. 11 - Prob. 11F.3AECh. 11 - Prob. 11F.3BECh. 11 - Prob. 11F.4AECh. 11 - Prob. 11F.4BECh. 11 - Prob. 11F.5AECh. 11 - Prob. 11F.5BECh. 11 - Prob. 11F.6AECh. 11 - Prob. 11F.6BECh. 11 - Prob. 11F.7AECh. 11 - Prob. 11F.7BECh. 11 - Prob. 11F.8AECh. 11 - Prob. 11F.8BECh. 11 - Prob. 11F.9AECh. 11 - Prob. 11F.9BECh. 11 - Prob. 11F.10AECh. 11 - Prob. 11F.10BECh. 11 - Prob. 11F.11AECh. 11 - Prob. 11F.11BECh. 11 - Prob. 11F.12AECh. 11 - Prob. 11F.12BECh. 11 - Prob. 11F.13AECh. 11 - Prob. 11F.13BECh. 11 - Prob. 11F.1PCh. 11 - Prob. 11F.2PCh. 11 - Prob. 11F.3PCh. 11 - Prob. 11F.4PCh. 11 - Prob. 11F.5PCh. 11 - Prob. 11F.6PCh. 11 - Prob. 11F.7PCh. 11 - Prob. 11F.8PCh. 11 - Prob. 11F.9PCh. 11 - Prob. 11F.10PCh. 11 - Prob. 11F.11PCh. 11 - Prob. 11F.12PCh. 11 - Prob. 11G.1DQCh. 11 - Prob. 11G.2DQCh. 11 - Prob. 11G.3DQCh. 11 - Prob. 11G.4DQCh. 11 - Prob. 11G.5DQCh. 11 - Prob. 11G.1AECh. 11 - Prob. 11G.1BECh. 11 - Prob. 11G.2AECh. 11 - Prob. 11G.2BECh. 11 - Prob. 11G.1PCh. 11 - Prob. 11G.2PCh. 11 - Prob. 11G.3PCh. 11 - Prob. 11G.4PCh. 11 - Prob. 11G.5PCh. 11 - Prob. 11G.6PCh. 11 - Prob. 11.1IACh. 11 - Prob. 11.5IACh. 11 - Prob. 11.8IA
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY