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.

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Answer 1

Question

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

Answer to Problem 14B.8P

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Explanation of Solution

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

PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH), 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{δOδHrO−H+δOδHrO⋅⋅⋅H+δO2rO−O} (1)

Where,

NA is the Avogadro's number. (6.022×1023 mol−1)
ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
δ is the charge.
r is the distance.
e is the charge on an electron. (1.6×10−19 C)

Equation (1) is solved further as shown below.

Vm=NAe24πε0{δOδH(1rO−H+1rO⋅⋅⋅H)+δO2rO−O} (2)

The value of rO⋅⋅⋅H is calculated by the formula given below.

rO⋅⋅⋅H=(rO−H2+rO−O2−2rO−HrO−Ocosθ)1/2 (3)

The value of rO−H is 95.7 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

95.7 pm=95.7×10−12 m

The value of rO−O is 200 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

200 pm=200×10−12 m

Substitute the value of rO−O and rO−H in equation (3).

rO⋅⋅⋅H=((95.7×10−12 m)2+(200×10−12 m)2−(2×95.7×10−12 m×200×10−12 m×cosθ))1/2=(9.16×10−21 m2+4×10−20 m2−(3.828×10−20cosθ m2))1/2=(4.916×10−20−(3.828×10−20cosθ))1/2 m

The value of δO is −0.83e.

The value of δH is +0.45e.

Substitute the value of NA, ε0, δO, δH, rO−O, rO−H, and rO⋅⋅⋅H in equation (2).

Vm=(6.022×1023 mol−1 ×e24×31.4×8.854×10−12 J−1 C2 m−1×{(−0.83e)(+0.45e)(195.7×10−12 m+1(4.916×10−20−3.828×10−20cosθ)1/2 m)+(−0.83e)2200×10−12 m})

Substitute the value of e in the above equation.

Vm=(6.022×1023 mol−1×(1.6×10−19 C)24×3.14×8.854×10−12 J−1 C2 m−1×{(−0.83×1.6×10−19 C)(+0.45×1.6×10−19 C)(195.7×10−12 m+1(4.916×10−20−3.828cosθ)1/2 m)+(−0.83×1.6×10−19 C)2200×10−12 m})=1.386×10−4 J mol−1 m(−9.562×10−39 m−1(1.045×1010+1(4.916×10−20−3.828cosθ)1/2 )+8.81792×10−29m−1)=1.386×10−4 J mol−1(−9.99×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 +8.81792×10−29)=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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Answer 2

Question

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

Answer to Problem 14B.8P

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Explanation of Solution

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

PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH), 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{δOδHrO−H+δOδHrO⋅⋅⋅H+δO2rO−O} (1)

Where,

NA is the Avogadro's number. (6.022×1023 mol−1)
ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
δ is the charge.
r is the distance.
e is the charge on an electron. (1.6×10−19 C)

Equation (1) is solved further as shown below.

Vm=NAe24πε0{δOδH(1rO−H+1rO⋅⋅⋅H)+δO2rO−O} (2)

The value of rO⋅⋅⋅H is calculated by the formula given below.

rO⋅⋅⋅H=(rO−H2+rO−O2−2rO−HrO−Ocosθ)1/2 (3)

The value of rO−H is 95.7 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

95.7 pm=95.7×10−12 m

The value of rO−O is 200 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

200 pm=200×10−12 m

Substitute the value of rO−O and rO−H in equation (3).

rO⋅⋅⋅H=((95.7×10−12 m)2+(200×10−12 m)2−(2×95.7×10−12 m×200×10−12 m×cosθ))1/2=(9.16×10−21 m2+4×10−20 m2−(3.828×10−20cosθ m2))1/2=(4.916×10−20−(3.828×10−20cosθ))1/2 m

The value of δO is −0.83e.

The value of δH is +0.45e.

Substitute the value of NA, ε0, δO, δH, rO−O, rO−H, and rO⋅⋅⋅H in equation (2).

Vm=(6.022×1023 mol−1 ×e24×31.4×8.854×10−12 J−1 C2 m−1×{(−0.83e)(+0.45e)(195.7×10−12 m+1(4.916×10−20−3.828×10−20cosθ)1/2 m)+(−0.83e)2200×10−12 m})

Substitute the value of e in the above equation.

Vm=(6.022×1023 mol−1×(1.6×10−19 C)24×3.14×8.854×10−12 J−1 C2 m−1×{(−0.83×1.6×10−19 C)(+0.45×1.6×10−19 C)(195.7×10−12 m+1(4.916×10−20−3.828cosθ)1/2 m)+(−0.83×1.6×10−19 C)2200×10−12 m})=1.386×10−4 J mol−1 m(−9.562×10−39 m−1(1.045×1010+1(4.916×10−20−3.828cosθ)1/2 )+8.81792×10−29m−1)=1.386×10−4 J mol−1(−9.99×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 +8.81792×10−29)=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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Answer 3

Question

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

Answer to Problem 14B.8P

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Explanation of Solution

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

PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH), 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{δOδHrO−H+δOδHrO⋅⋅⋅H+δO2rO−O} (1)

Where,

NA is the Avogadro's number. (6.022×1023 mol−1)
ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
δ is the charge.
r is the distance.
e is the charge on an electron. (1.6×10−19 C)

Equation (1) is solved further as shown below.

Vm=NAe24πε0{δOδH(1rO−H+1rO⋅⋅⋅H)+δO2rO−O} (2)

The value of rO⋅⋅⋅H is calculated by the formula given below.

rO⋅⋅⋅H=(rO−H2+rO−O2−2rO−HrO−Ocosθ)1/2 (3)

The value of rO−H is 95.7 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

95.7 pm=95.7×10−12 m

The value of rO−O is 200 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

200 pm=200×10−12 m

Substitute the value of rO−O and rO−H in equation (3).

rO⋅⋅⋅H=((95.7×10−12 m)2+(200×10−12 m)2−(2×95.7×10−12 m×200×10−12 m×cosθ))1/2=(9.16×10−21 m2+4×10−20 m2−(3.828×10−20cosθ m2))1/2=(4.916×10−20−(3.828×10−20cosθ))1/2 m

The value of δO is −0.83e.

The value of δH is +0.45e.

Substitute the value of NA, ε0, δO, δH, rO−O, rO−H, and rO⋅⋅⋅H in equation (2).

Vm=(6.022×1023 mol−1 ×e24×31.4×8.854×10−12 J−1 C2 m−1×{(−0.83e)(+0.45e)(195.7×10−12 m+1(4.916×10−20−3.828×10−20cosθ)1/2 m)+(−0.83e)2200×10−12 m})

Substitute the value of e in the above equation.

Vm=(6.022×1023 mol−1×(1.6×10−19 C)24×3.14×8.854×10−12 J−1 C2 m−1×{(−0.83×1.6×10−19 C)(+0.45×1.6×10−19 C)(195.7×10−12 m+1(4.916×10−20−3.828cosθ)1/2 m)+(−0.83×1.6×10−19 C)2200×10−12 m})=1.386×10−4 J mol−1 m(−9.562×10−39 m−1(1.045×1010+1(4.916×10−20−3.828cosθ)1/2 )+8.81792×10−29m−1)=1.386×10−4 J mol−1(−9.99×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 +8.81792×10−29)=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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Answer 4

Question

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

Answer to Problem 14B.8P

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Explanation of Solution

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

PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH), 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{δOδHrO−H+δOδHrO⋅⋅⋅H+δO2rO−O} (1)

Where,

NA is the Avogadro's number. (6.022×1023 mol−1)
ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
δ is the charge.
r is the distance.
e is the charge on an electron. (1.6×10−19 C)

Equation (1) is solved further as shown below.

Vm=NAe24πε0{δOδH(1rO−H+1rO⋅⋅⋅H)+δO2rO−O} (2)

The value of rO⋅⋅⋅H is calculated by the formula given below.

rO⋅⋅⋅H=(rO−H2+rO−O2−2rO−HrO−Ocosθ)1/2 (3)

The value of rO−H is 95.7 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

95.7 pm=95.7×10−12 m

The value of rO−O is 200 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

200 pm=200×10−12 m

Substitute the value of rO−O and rO−H in equation (3).

rO⋅⋅⋅H=((95.7×10−12 m)2+(200×10−12 m)2−(2×95.7×10−12 m×200×10−12 m×cosθ))1/2=(9.16×10−21 m2+4×10−20 m2−(3.828×10−20cosθ m2))1/2=(4.916×10−20−(3.828×10−20cosθ))1/2 m

The value of δO is −0.83e.

The value of δH is +0.45e.

Substitute the value of NA, ε0, δO, δH, rO−O, rO−H, and rO⋅⋅⋅H in equation (2).

Vm=(6.022×1023 mol−1 ×e24×31.4×8.854×10−12 J−1 C2 m−1×{(−0.83e)(+0.45e)(195.7×10−12 m+1(4.916×10−20−3.828×10−20cosθ)1/2 m)+(−0.83e)2200×10−12 m})

Substitute the value of e in the above equation.

Vm=(6.022×1023 mol−1×(1.6×10−19 C)24×3.14×8.854×10−12 J−1 C2 m−1×{(−0.83×1.6×10−19 C)(+0.45×1.6×10−19 C)(195.7×10−12 m+1(4.916×10−20−3.828cosθ)1/2 m)+(−0.83×1.6×10−19 C)2200×10−12 m})=1.386×10−4 J mol−1 m(−9.562×10−39 m−1(1.045×1010+1(4.916×10−20−3.828cosθ)1/2 )+8.81792×10−29m−1)=1.386×10−4 J mol−1(−9.99×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 +8.81792×10−29)=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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Answer 5

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

Answer to Problem 14B.8P

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Explanation of Solution

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

PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH), 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{δOδHrO−H+δOδHrO⋅⋅⋅H+δO2rO−O} (1)

Where,

NA is the Avogadro's number. (6.022×1023 mol−1)
ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
δ is the charge.
r is the distance.
e is the charge on an electron. (1.6×10−19 C)

Equation (1) is solved further as shown below.

Vm=NAe24πε0{δOδH(1rO−H+1rO⋅⋅⋅H)+δO2rO−O} (2)

The value of rO⋅⋅⋅H is calculated by the formula given below.

rO⋅⋅⋅H=(rO−H2+rO−O2−2rO−HrO−Ocosθ)1/2 (3)

The value of rO−H is 95.7 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

95.7 pm=95.7×10−12 m

The value of rO−O is 200 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

200 pm=200×10−12 m

Substitute the value of rO−O and rO−H in equation (3).

rO⋅⋅⋅H=((95.7×10−12 m)2+(200×10−12 m)2−(2×95.7×10−12 m×200×10−12 m×cosθ))1/2=(9.16×10−21 m2+4×10−20 m2−(3.828×10−20cosθ m2))1/2=(4.916×10−20−(3.828×10−20cosθ))1/2 m

The value of δO is −0.83e.

The value of δH is +0.45e.

Substitute the value of NA, ε0, δO, δH, rO−O, rO−H, and rO⋅⋅⋅H in equation (2).

Vm=(6.022×1023 mol−1 ×e24×31.4×8.854×10−12 J−1 C2 m−1×{(−0.83e)(+0.45e)(195.7×10−12 m+1(4.916×10−20−3.828×10−20cosθ)1/2 m)+(−0.83e)2200×10−12 m})

Substitute the value of e in the above equation.

Vm=(6.022×1023 mol−1×(1.6×10−19 C)24×3.14×8.854×10−12 J−1 C2 m−1×{(−0.83×1.6×10−19 C)(+0.45×1.6×10−19 C)(195.7×10−12 m+1(4.916×10−20−3.828cosθ)1/2 m)+(−0.83×1.6×10−19 C)2200×10−12 m})=1.386×10−4 J mol−1 m(−9.562×10−39 m−1(1.045×1010+1(4.916×10−20−3.828cosθ)1/2 )+8.81792×10−29m−1)=1.386×10−4 J mol−1(−9.99×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 +8.81792×10−29)=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Answer 6

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

Answer to Problem 14B.8P

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Explanation of Solution

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

PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH), 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{δOδHrO−H+δOδHrO⋅⋅⋅H+δO2rO−O} (1)

Where,

NA is the Avogadro's number. (6.022×1023 mol−1)
ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
δ is the charge.
r is the distance.
e is the charge on an electron. (1.6×10−19 C)

Equation (1) is solved further as shown below.

Vm=NAe24πε0{δOδH(1rO−H+1rO⋅⋅⋅H)+δO2rO−O} (2)

The value of rO⋅⋅⋅H is calculated by the formula given below.

rO⋅⋅⋅H=(rO−H2+rO−O2−2rO−HrO−Ocosθ)1/2 (3)

The value of rO−H is 95.7 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

95.7 pm=95.7×10−12 m

The value of rO−O is 200 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

200 pm=200×10−12 m

Substitute the value of rO−O and rO−H in equation (3).

rO⋅⋅⋅H=((95.7×10−12 m)2+(200×10−12 m)2−(2×95.7×10−12 m×200×10−12 m×cosθ))1/2=(9.16×10−21 m2+4×10−20 m2−(3.828×10−20cosθ m2))1/2=(4.916×10−20−(3.828×10−20cosθ))1/2 m

The value of δO is −0.83e.

The value of δH is +0.45e.

Substitute the value of NA, ε0, δO, δH, rO−O, rO−H, and rO⋅⋅⋅H in equation (2).

Vm=(6.022×1023 mol−1 ×e24×31.4×8.854×10−12 J−1 C2 m−1×{(−0.83e)(+0.45e)(195.7×10−12 m+1(4.916×10−20−3.828×10−20cosθ)1/2 m)+(−0.83e)2200×10−12 m})

Substitute the value of e in the above equation.

Vm=(6.022×1023 mol−1×(1.6×10−19 C)24×3.14×8.854×10−12 J−1 C2 m−1×{(−0.83×1.6×10−19 C)(+0.45×1.6×10−19 C)(195.7×10−12 m+1(4.916×10−20−3.828cosθ)1/2 m)+(−0.83×1.6×10−19 C)2200×10−12 m})=1.386×10−4 J mol−1 m(−9.562×10−39 m−1(1.045×1010+1(4.916×10−20−3.828cosθ)1/2 )+8.81792×10−29m−1)=1.386×10−4 J mol−1(−9.99×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 +8.81792×10−29)=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Answer 7

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.

Answer 8

Expert Solution & Answer

Answer to Problem 14B.8P

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

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

PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH), 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{δOδHrO−H+δOδHrO⋅⋅⋅H+δO2rO−O} (1)

Where,

NA is the Avogadro's number. (6.022×1023 mol−1)
ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
δ is the charge.
r is the distance.
e is the charge on an electron. (1.6×10−19 C)

Equation (1) is solved further as shown below.

Vm=NAe24πε0{δOδH(1rO−H+1rO⋅⋅⋅H)+δO2rO−O} (2)

The value of rO⋅⋅⋅H is calculated by the formula given below.

rO⋅⋅⋅H=(rO−H2+rO−O2−2rO−HrO−Ocosθ)1/2 (3)

The value of rO−H is 95.7 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

95.7 pm=95.7×10−12 m

The value of rO−O is 200 pm.

The conversion of pm to m is shown below.

1 pm=10−12 m

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

200 pm=200×10−12 m

Substitute the value of rO−O and rO−H in equation (3).

rO⋅⋅⋅H=((95.7×10−12 m)2+(200×10−12 m)2−(2×95.7×10−12 m×200×10−12 m×cosθ))1/2=(9.16×10−21 m2+4×10−20 m2−(3.828×10−20cosθ m2))1/2=(4.916×10−20−(3.828×10−20cosθ))1/2 m

The value of δO is −0.83e.

The value of δH is +0.45e.

Substitute the value of NA, ε0, δO, δH, rO−O, rO−H, and rO⋅⋅⋅H in equation (2).

Vm=(6.022×1023 mol−1 ×e24×31.4×8.854×10−12 J−1 C2 m−1×{(−0.83e)(+0.45e)(195.7×10−12 m+1(4.916×10−20−3.828×10−20cosθ)1/2 m)+(−0.83e)2200×10−12 m})

Substitute the value of e in the above equation.

Vm=(6.022×1023 mol−1×(1.6×10−19 C)24×3.14×8.854×10−12 J−1 C2 m−1×{(−0.83×1.6×10−19 C)(+0.45×1.6×10−19 C)(195.7×10−12 m+1(4.916×10−20−3.828cosθ)1/2 m)+(−0.83×1.6×10−19 C)2200×10−12 m})=1.386×10−4 J mol−1 m(−9.562×10−39 m−1(1.045×1010+1(4.916×10−20−3.828cosθ)1/2 )+8.81792×10−29m−1)=1.386×10−4 J mol−1(−9.99×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 +8.81792×10−29)=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

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

Vm=1.386×10−4 J mol−1(−1.17×10−29+−9.562×10−39(4.916×10−20−3.828cosθ)1/2 )

Answer 12

Ch. 14 - Prob. 14A.1ST Ch. 14 - Prob. 14A.2ST Ch. 14 - Prob. 14B.1ST Ch. 14 - Prob. 14C.1ST Ch. 14 - Prob. 14C.2ST Ch. 14 - Prob. 14D.1ST Ch. 14 - Prob. 14D.2ST Ch. 14 - Prob. 14E.1ST Ch. 14 - Prob. 14A.1DQ Ch. 14 - Prob. 14A.2DQ

Answer to Problem 14B.8P

Explanation of Solution

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