Student Solutions Manual to accompany Atkins' Physical Chemistry 11th  edition
Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition
11th Edition
ISBN: 9780198807773
Author: ATKINS
Publisher: Oxford University Press
Question
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Chapter 14, Problem 14B.8P
Interpretation Introduction

Interpretation:

The dependence of the molar potential energy of interaction on the angle θ has to be calculated.

Concept introduction:

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to oxygen, nitrogen or fluorine and it develops some partial charge on both the atom.  Hydrogen bonds can also be formed between the lone pair of an electronegative atom and hydrogen atom attached with an electronegative atom.

Expert Solution & Answer
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Answer to Problem 14B.8P

The dependence of the molar potential energy of interaction on the angle θ is shown below.

    Vm=1.386×104Jmol1(1.17×1029+9.562×1039(4.916×10203.828cosθ)1/2)

Explanation of Solution

The arrangement of a system containing an OH group and O group is shown below.

Student Solutions Manual to accompany Atkins' Physical Chemistry 11th  edition, Chapter 14, Problem 14B.8P

Figure 1

The expression for the molar potential energy (Vm) of an electrostatic model of hydrogen bonding is shown below.

    Vm=NAe24πε0{δOδHrOH+δOδHrOH+δO2rOO}        (1)

Where,

  • NA is the Avogadro's number. (6.022×1023mol1)
  • ε0 is the vacuum permittivity. (8.854×1012J1C2m1)
  • δ is the charge.
  • r is the distance.
  • e is the charge on an electron. (1.6×1019C)

Equation (1) is solved further as shown below.

    Vm=NAe24πε0{δOδH(1rOH+1rOH)+δO2rOO}        (2)

The value of rOH is calculated by the formula given below.

    rOH=(rOH2+rOO22rOHrOOcosθ)1/2        (3)

The value of rOH is 95.7pm.

The conversion of pm to m is shown below.

    1pm=1012m

Therefore, the conversion of 95.7pm to m is shown below.

    95.7pm=95.7×1012m

The value of rOO is 200pm.

The conversion of pm to m is shown below.

    1pm=1012m

Therefore, the conversion of 200pm to m is shown below.

    200pm=200×1012m

Substitute the value of rOO and rOH in equation (3).

    rOH=((95.7×1012m)2+(200×1012m)2(2×95.7×1012m×200×1012m×cosθ))1/2=(9.16×1021m2+4×1020m2(3.828×1020cosθm2))1/2=(4.916×1020(3.828×1020cosθ))1/2m

The value of δO is 0.83e.

The value of δH is +0.45e.

Substitute the value of NA, ε0, δO, δH, rOO, rOH, and rOH in equation (2).

    Vm=(6.022×1023mol1×e24×31.4×8.854×1012J1C2m1×{(0.83e)(+0.45e)(195.7×1012m+1(4.916×10203.828×1020cosθ)1/2m)+(0.83e)2200×1012m})

Substitute the value of e in the above equation.

    Vm=(6.022×1023mol1×(1.6×1019C)24×3.14×8.854×1012J1C2m1×{(0.83×1.6×1019C)(+0.45×1.6×1019C)(195.7×1012m+1(4.916×10203.828cosθ)1/2m)+(0.83×1.6×1019C)2200×1012m})=1.386×104Jmol1m(9.562×1039m1(1.045×1010+1(4.916×10203.828cosθ)1/2)+8.81792×1029m1)=1.386×104Jmol1(9.99×1029+9.562×1039(4.916×10203.828cosθ)1/2+8.81792×1029)=1.386×104Jmol1(1.17×1029+9.562×1039(4.916×10203.828cosθ)1/2)

Therefore, the dependence of the molar potential energy of interaction on the angle is shown below.

    Vm=1.386×104Jmol1(1.17×1029+9.562×1039(4.916×10203.828cosθ)1/2)

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Chapter 14 Solutions

Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition

Ch. 14 - Prob. 14A.3DQCh. 14 - Prob. 14A.1AECh. 14 - Prob. 14A.1BECh. 14 - Prob. 14A.2AECh. 14 - Prob. 14A.2BECh. 14 - Prob. 14A.3AECh. 14 - Prob. 14A.3BECh. 14 - Prob. 14A.4AECh. 14 - Prob. 14A.4BECh. 14 - Prob. 14A.5AECh. 14 - Prob. 14A.5BECh. 14 - Prob. 14A.6AECh. 14 - Prob. 14A.6BECh. 14 - Prob. 14A.7AECh. 14 - Prob. 14A.7BECh. 14 - Prob. 14A.8AECh. 14 - Prob. 14A.8BECh. 14 - Prob. 14A.9AECh. 14 - Prob. 14A.9BECh. 14 - Prob. 14A.1PCh. 14 - Prob. 14A.2PCh. 14 - Prob. 14A.3PCh. 14 - Prob. 14A.4PCh. 14 - Prob. 14A.5PCh. 14 - Prob. 14A.6PCh. 14 - Prob. 14A.7PCh. 14 - Prob. 14A.8PCh. 14 - Prob. 14A.10PCh. 14 - Prob. 14A.12PCh. 14 - Prob. 14A.13PCh. 14 - Prob. 14B.1DQCh. 14 - Prob. 14B.2DQCh. 14 - Prob. 14B.3DQCh. 14 - Prob. 14B.4DQCh. 14 - Prob. 14B.5DQCh. 14 - Prob. 14B.1AECh. 14 - Prob. 14B.1BECh. 14 - Prob. 14B.2AECh. 14 - Prob. 14B.2BECh. 14 - Prob. 14B.3AECh. 14 - Prob. 14B.3BECh. 14 - Prob. 14B.4AECh. 14 - Prob. 14B.4BECh. 14 - Prob. 14B.5AECh. 14 - Prob. 14B.5BECh. 14 - Prob. 14B.6AECh. 14 - Prob. 14B.6BECh. 14 - Prob. 14B.1PCh. 14 - Prob. 14B.2PCh. 14 - Prob. 14B.3PCh. 14 - Prob. 14B.4PCh. 14 - Prob. 14B.5PCh. 14 - Prob. 14B.6PCh. 14 - Prob. 14B.7PCh. 14 - Prob. 14B.8PCh. 14 - Prob. 14B.10PCh. 14 - Prob. 14C.1DQCh. 14 - Prob. 14C.2DQCh. 14 - Prob. 14C.1AECh. 14 - Prob. 14C.1BECh. 14 - Prob. 14C.2AECh. 14 - Prob. 14C.2BECh. 14 - Prob. 14C.3AECh. 14 - Prob. 14C.3BECh. 14 - Prob. 14C.4AECh. 14 - Prob. 14C.4BECh. 14 - Prob. 14C.1PCh. 14 - Prob. 14C.2PCh. 14 - Prob. 14D.1DQCh. 14 - Prob. 14D.2DQCh. 14 - Prob. 14D.3DQCh. 14 - Prob. 14D.4DQCh. 14 - Prob. 14D.5DQCh. 14 - Prob. 14D.1AECh. 14 - Prob. 14D.1BECh. 14 - Prob. 14D.2AECh. 14 - Prob. 14D.2BECh. 14 - Prob. 14D.3AECh. 14 - Prob. 14D.3BECh. 14 - Prob. 14D.4AECh. 14 - Prob. 14D.4BECh. 14 - Prob. 14D.5AECh. 14 - Prob. 14D.5BECh. 14 - Prob. 14D.6AECh. 14 - Prob. 14D.6BECh. 14 - Prob. 14D.8AECh. 14 - Prob. 14D.8BECh. 14 - Prob. 14D.9AECh. 14 - Prob. 14D.9BECh. 14 - Prob. 14D.2PCh. 14 - Prob. 14D.3PCh. 14 - Prob. 14D.4PCh. 14 - Prob. 14D.6PCh. 14 - Prob. 14D.7PCh. 14 - Prob. 14D.8PCh. 14 - Prob. 14D.9PCh. 14 - Prob. 14D.10PCh. 14 - Prob. 14E.1DQCh. 14 - Prob. 14E.2DQCh. 14 - Prob. 14E.3DQCh. 14 - Prob. 14E.4DQCh. 14 - Prob. 14E.5DQCh. 14 - Prob. 14E.1AECh. 14 - Prob. 14E.1BECh. 14 - Prob. 14E.1PCh. 14 - Prob. 14E.3PCh. 14 - Prob. 14.1IACh. 14 - Prob. 14.2IACh. 14 - Prob. 14.6IACh. 14 - Prob. 14.8IA
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