The potential energy of attraction of the ions.

Question

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Answer 1

Question

Chapter 37, Problem 21P

(a)

To determine

The potential energy of attraction of the ions.

(a)

Expert Solution

Answer to Problem 21P

Thepotential energy of attraction of the ions is −5.39 eV .

Explanation of Solution

Given:

The equilibrium separation of the K+ and F− ions in KF is r=0.217 nm .

Formula used:

The expression for potential energy of attraction is given by,

Ue=−ke2r

Calculation:

The potential energy of attraction is calculated as,

Ue=−ke2r=−( 8.9× 10 9 N⋅ m 2 / C 2 ) ( 1.6× 10 −19 C )2( 0.217 nm)( 10 −9 m 1 nm )( 1.6× 10 −19 J eV )=−( 8.9× 10 9 N⋅ m 2 ) ( 1.6× 10 −19 )2 eV( 0.217× 10 −9 m)( 1.6× 10 −19 J)=−6.64 eV

Conclusion:

Therefore, the potential energy of attraction of the ions is −6.64 eV .

(b)

To determine

The value of dissociation energy.

(b)

Expert Solution

Answer to Problem 21P

The value of dissociation energy is 5.70 eV .

Explanation of Solution

Given:

The ionization energy of potassium is 4.34 eV

The electron affinity of Fluorine is −3.40 eV

Formula used:

The expression for total potential energy of the molecule neglecting energy of repulsion is given by,

Utot=Ue+ΔE

The expression for dissociation energy is given by,

Ed=−Utot=−(Ue+ΔE)

Here, ΔE is the difference between the ionization energy of potassium and the electron affinity of fluorine.

Calculation:

Difference between the ionization energy of potassium and the electron affinity of fluorine ΔE is calculated as,

ΔE=4.34 eV−3.40 eV=0.94 eV

The dissociation energy is calculated as,

Ed=−(Ue+ΔE)=−(−6.64 eV+0.94 eV)=5.70 eV

Conclusion:

Therefore, the value of dissociation energy is 5.70 eV .

(c)

To determine

The energy due to repulsion of the ions at the equilibrium separation.

(c)

Expert Solution

Answer to Problem 21P

The energy due to repulsion of the ions at the equilibrium separation is 0.63 eV .

Explanation of Solution

Given:

The measured dissociation energy Ed,measured=5.07 eV

Formula used:

The expression for energy due to repulsion of the ions at equilibrium separation is given by,

Urep=Ed,calculated−Ed,measured

Calculation:

The energy due to repulsion of the ions at equilibrium separation is calculated as,

Urep=Ed,calculated−Ed,measured=5.70 eV−5.07 eV=0.63 eV

Conclusion:

Therefore, the energy due to repulsion of the ions at the equilibrium separation is 0.63 eV

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Answer 2

Question

Chapter 37, Problem 21P

(a)

To determine

The potential energy of attraction of the ions.

(a)

Expert Solution

Answer to Problem 21P

Thepotential energy of attraction of the ions is −5.39 eV .

Explanation of Solution

Given:

The equilibrium separation of the K+ and F− ions in KF is r=0.217 nm .

Formula used:

The expression for potential energy of attraction is given by,

Ue=−ke2r

Calculation:

The potential energy of attraction is calculated as,

Ue=−ke2r=−( 8.9× 10 9 N⋅ m 2 / C 2 ) ( 1.6× 10 −19 C )2( 0.217 nm)( 10 −9 m 1 nm )( 1.6× 10 −19 J eV )=−( 8.9× 10 9 N⋅ m 2 ) ( 1.6× 10 −19 )2 eV( 0.217× 10 −9 m)( 1.6× 10 −19 J)=−6.64 eV

Conclusion:

Therefore, the potential energy of attraction of the ions is −6.64 eV .

(b)

To determine

The value of dissociation energy.

(b)

Expert Solution

Answer to Problem 21P

The value of dissociation energy is 5.70 eV .

Explanation of Solution

Given:

The ionization energy of potassium is 4.34 eV

The electron affinity of Fluorine is −3.40 eV

Formula used:

The expression for total potential energy of the molecule neglecting energy of repulsion is given by,

Utot=Ue+ΔE

The expression for dissociation energy is given by,

Ed=−Utot=−(Ue+ΔE)

Here, ΔE is the difference between the ionization energy of potassium and the electron affinity of fluorine.

Calculation:

Difference between the ionization energy of potassium and the electron affinity of fluorine ΔE is calculated as,

ΔE=4.34 eV−3.40 eV=0.94 eV

The dissociation energy is calculated as,

Ed=−(Ue+ΔE)=−(−6.64 eV+0.94 eV)=5.70 eV

Conclusion:

Therefore, the value of dissociation energy is 5.70 eV .

(c)

To determine

The energy due to repulsion of the ions at the equilibrium separation.

(c)

Expert Solution

Answer to Problem 21P

The energy due to repulsion of the ions at the equilibrium separation is 0.63 eV .

Explanation of Solution

Given:

The measured dissociation energy Ed,measured=5.07 eV

Formula used:

The expression for energy due to repulsion of the ions at equilibrium separation is given by,

Urep=Ed,calculated−Ed,measured

Calculation:

The energy due to repulsion of the ions at equilibrium separation is calculated as,

Urep=Ed,calculated−Ed,measured=5.70 eV−5.07 eV=0.63 eV

Conclusion:

Therefore, the energy due to repulsion of the ions at the equilibrium separation is 0.63 eV

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Answer 3

Question

Chapter 37, Problem 21P

(a)

To determine

The potential energy of attraction of the ions.

(a)

Expert Solution

Answer to Problem 21P

Thepotential energy of attraction of the ions is −5.39 eV .

Explanation of Solution

Given:

The equilibrium separation of the K+ and F− ions in KF is r=0.217 nm .

Formula used:

The expression for potential energy of attraction is given by,

Ue=−ke2r

Calculation:

The potential energy of attraction is calculated as,

Ue=−ke2r=−( 8.9× 10 9 N⋅ m 2 / C 2 ) ( 1.6× 10 −19 C )2( 0.217 nm)( 10 −9 m 1 nm )( 1.6× 10 −19 J eV )=−( 8.9× 10 9 N⋅ m 2 ) ( 1.6× 10 −19 )2 eV( 0.217× 10 −9 m)( 1.6× 10 −19 J)=−6.64 eV

Conclusion:

Therefore, the potential energy of attraction of the ions is −6.64 eV .

(b)

To determine

The value of dissociation energy.

(b)

Expert Solution

Answer to Problem 21P

The value of dissociation energy is 5.70 eV .

Explanation of Solution

Given:

The ionization energy of potassium is 4.34 eV

The electron affinity of Fluorine is −3.40 eV

Formula used:

The expression for total potential energy of the molecule neglecting energy of repulsion is given by,

Utot=Ue+ΔE

The expression for dissociation energy is given by,

Ed=−Utot=−(Ue+ΔE)

Here, ΔE is the difference between the ionization energy of potassium and the electron affinity of fluorine.

Calculation:

Difference between the ionization energy of potassium and the electron affinity of fluorine ΔE is calculated as,

ΔE=4.34 eV−3.40 eV=0.94 eV

The dissociation energy is calculated as,

Ed=−(Ue+ΔE)=−(−6.64 eV+0.94 eV)=5.70 eV

Conclusion:

Therefore, the value of dissociation energy is 5.70 eV .

(c)

To determine

The energy due to repulsion of the ions at the equilibrium separation.

(c)

Expert Solution

Answer to Problem 21P

The energy due to repulsion of the ions at the equilibrium separation is 0.63 eV .

Explanation of Solution

Given:

The measured dissociation energy Ed,measured=5.07 eV

Formula used:

The expression for energy due to repulsion of the ions at equilibrium separation is given by,

Urep=Ed,calculated−Ed,measured

Calculation:

The energy due to repulsion of the ions at equilibrium separation is calculated as,

Urep=Ed,calculated−Ed,measured=5.70 eV−5.07 eV=0.63 eV

Conclusion:

Therefore, the energy due to repulsion of the ions at the equilibrium separation is 0.63 eV

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Answer 4

Question

Chapter 37, Problem 21P

(a)

To determine

The potential energy of attraction of the ions.

(a)

Expert Solution

Answer to Problem 21P

Thepotential energy of attraction of the ions is −5.39 eV .

Explanation of Solution

Given:

The equilibrium separation of the K+ and F− ions in KF is r=0.217 nm .

Formula used:

The expression for potential energy of attraction is given by,

Ue=−ke2r

Calculation:

The potential energy of attraction is calculated as,

Ue=−ke2r=−( 8.9× 10 9 N⋅ m 2 / C 2 ) ( 1.6× 10 −19 C )2( 0.217 nm)( 10 −9 m 1 nm )( 1.6× 10 −19 J eV )=−( 8.9× 10 9 N⋅ m 2 ) ( 1.6× 10 −19 )2 eV( 0.217× 10 −9 m)( 1.6× 10 −19 J)=−6.64 eV

Conclusion:

Therefore, the potential energy of attraction of the ions is −6.64 eV .

(b)

To determine

The value of dissociation energy.

(b)

Expert Solution

Answer to Problem 21P

The value of dissociation energy is 5.70 eV .

Explanation of Solution

Given:

The ionization energy of potassium is 4.34 eV

The electron affinity of Fluorine is −3.40 eV

Formula used:

The expression for total potential energy of the molecule neglecting energy of repulsion is given by,

Utot=Ue+ΔE

The expression for dissociation energy is given by,

Ed=−Utot=−(Ue+ΔE)

Here, ΔE is the difference between the ionization energy of potassium and the electron affinity of fluorine.

Calculation:

Difference between the ionization energy of potassium and the electron affinity of fluorine ΔE is calculated as,

ΔE=4.34 eV−3.40 eV=0.94 eV

The dissociation energy is calculated as,

Ed=−(Ue+ΔE)=−(−6.64 eV+0.94 eV)=5.70 eV

Conclusion:

Therefore, the value of dissociation energy is 5.70 eV .

(c)

To determine

The energy due to repulsion of the ions at the equilibrium separation.

(c)

Expert Solution

Answer to Problem 21P

The energy due to repulsion of the ions at the equilibrium separation is 0.63 eV .

Explanation of Solution

Given:

The measured dissociation energy Ed,measured=5.07 eV

Formula used:

The expression for energy due to repulsion of the ions at equilibrium separation is given by,

Urep=Ed,calculated−Ed,measured

Calculation:

The energy due to repulsion of the ions at equilibrium separation is calculated as,

Urep=Ed,calculated−Ed,measured=5.70 eV−5.07 eV=0.63 eV

Conclusion:

Therefore, the energy due to repulsion of the ions at the equilibrium separation is 0.63 eV

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Answer 5

Chapter 37, Problem 21P

(a)

To determine

The potential energy of attraction of the ions.

(a)

Expert Solution

Answer to Problem 21P

Thepotential energy of attraction of the ions is −5.39 eV .

Explanation of Solution

Given:

The equilibrium separation of the K+ and F− ions in KF is r=0.217 nm .

Formula used:

The expression for potential energy of attraction is given by,

Ue=−ke2r

Calculation:

The potential energy of attraction is calculated as,

Ue=−ke2r=−( 8.9× 10 9 N⋅ m 2 / C 2 ) ( 1.6× 10 −19 C )2( 0.217 nm)( 10 −9 m 1 nm )( 1.6× 10 −19 J eV )=−( 8.9× 10 9 N⋅ m 2 ) ( 1.6× 10 −19 )2 eV( 0.217× 10 −9 m)( 1.6× 10 −19 J)=−6.64 eV

Conclusion:

Therefore, the potential energy of attraction of the ions is −6.64 eV .

(b)

To determine

The value of dissociation energy.

(b)

Expert Solution

Answer to Problem 21P

The value of dissociation energy is 5.70 eV .

Explanation of Solution

Given:

The ionization energy of potassium is 4.34 eV

The electron affinity of Fluorine is −3.40 eV

Formula used:

The expression for total potential energy of the molecule neglecting energy of repulsion is given by,

Utot=Ue+ΔE

The expression for dissociation energy is given by,

Ed=−Utot=−(Ue+ΔE)

Here, ΔE is the difference between the ionization energy of potassium and the electron affinity of fluorine.

Calculation:

Difference between the ionization energy of potassium and the electron affinity of fluorine ΔE is calculated as,

ΔE=4.34 eV−3.40 eV=0.94 eV

The dissociation energy is calculated as,

Ed=−(Ue+ΔE)=−(−6.64 eV+0.94 eV)=5.70 eV

Conclusion:

Therefore, the value of dissociation energy is 5.70 eV .

(c)

To determine

The energy due to repulsion of the ions at the equilibrium separation.

(c)

Expert Solution

Answer to Problem 21P

The energy due to repulsion of the ions at the equilibrium separation is 0.63 eV .

Explanation of Solution

Given:

The measured dissociation energy Ed,measured=5.07 eV

Formula used:

The expression for energy due to repulsion of the ions at equilibrium separation is given by,

Urep=Ed,calculated−Ed,measured

Calculation:

The energy due to repulsion of the ions at equilibrium separation is calculated as,

Urep=Ed,calculated−Ed,measured=5.70 eV−5.07 eV=0.63 eV

Conclusion:

Therefore, the energy due to repulsion of the ions at the equilibrium separation is 0.63 eV

Answer 6

Chapter 37, Problem 21P

(a)

To determine

The potential energy of attraction of the ions.

(a)

Expert Solution

Answer to Problem 21P

Thepotential energy of attraction of the ions is −5.39 eV .

Explanation of Solution

Given:

The equilibrium separation of the K+ and F− ions in KF is r=0.217 nm .

Formula used:

The expression for potential energy of attraction is given by,

Ue=−ke2r

Calculation:

The potential energy of attraction is calculated as,

Ue=−ke2r=−( 8.9× 10 9 N⋅ m 2 / C 2 ) ( 1.6× 10 −19 C )2( 0.217 nm)( 10 −9 m 1 nm )( 1.6× 10 −19 J eV )=−( 8.9× 10 9 N⋅ m 2 ) ( 1.6× 10 −19 )2 eV( 0.217× 10 −9 m)( 1.6× 10 −19 J)=−6.64 eV

Conclusion:

Therefore, the potential energy of attraction of the ions is −6.64 eV .

Answer 7

Chapter 37, Problem 21P

(a)

To determine

The potential energy of attraction of the ions.

Answer 8

(a)

Expert Solution

Answer to Problem 21P

Thepotential energy of attraction of the ions is −5.39 eV .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The equilibrium separation of the K+ and F− ions in KF is r=0.217 nm .

Formula used:

The expression for potential energy of attraction is given by,

Ue=−ke2r

Calculation:

The potential energy of attraction is calculated as,

Ue=−ke2r=−( 8.9× 10 9 N⋅ m 2 / C 2 ) ( 1.6× 10 −19 C )2( 0.217 nm)( 10 −9 m 1 nm )( 1.6× 10 −19 J eV )=−( 8.9× 10 9 N⋅ m 2 ) ( 1.6× 10 −19 )2 eV( 0.217× 10 −9 m)( 1.6× 10 −19 J)=−6.64 eV

Conclusion:

Therefore, the potential energy of attraction of the ions is −6.64 eV .

Answer 13

(b)

To determine

The value of dissociation energy.

(b)

Expert Solution

Answer to Problem 21P

The value of dissociation energy is 5.70 eV .

Explanation of Solution

Given:

The ionization energy of potassium is 4.34 eV

The electron affinity of Fluorine is −3.40 eV

Formula used:

The expression for total potential energy of the molecule neglecting energy of repulsion is given by,

Utot=Ue+ΔE

The expression for dissociation energy is given by,

Ed=−Utot=−(Ue+ΔE)

Here, ΔE is the difference between the ionization energy of potassium and the electron affinity of fluorine.

Calculation:

Difference between the ionization energy of potassium and the electron affinity of fluorine ΔE is calculated as,

ΔE=4.34 eV−3.40 eV=0.94 eV

The dissociation energy is calculated as,

Ed=−(Ue+ΔE)=−(−6.64 eV+0.94 eV)=5.70 eV

Conclusion:

Therefore, the value of dissociation energy is 5.70 eV .

Answer 14

(b)

To determine

The value of dissociation energy.

Answer 15

(b)

Expert Solution

Answer to Problem 21P

The value of dissociation energy is 5.70 eV .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The ionization energy of potassium is 4.34 eV

The electron affinity of Fluorine is −3.40 eV

Formula used:

The expression for total potential energy of the molecule neglecting energy of repulsion is given by,

Utot=Ue+ΔE

The expression for dissociation energy is given by,

Ed=−Utot=−(Ue+ΔE)

Here, ΔE is the difference between the ionization energy of potassium and the electron affinity of fluorine.

Calculation:

Difference between the ionization energy of potassium and the electron affinity of fluorine ΔE is calculated as,

ΔE=4.34 eV−3.40 eV=0.94 eV

The dissociation energy is calculated as,

Ed=−(Ue+ΔE)=−(−6.64 eV+0.94 eV)=5.70 eV

Conclusion:

Therefore, the value of dissociation energy is 5.70 eV .

Answer 20

(c)

To determine

The energy due to repulsion of the ions at the equilibrium separation.

(c)

Expert Solution

Answer to Problem 21P

The energy due to repulsion of the ions at the equilibrium separation is 0.63 eV .

Explanation of Solution

Given:

The measured dissociation energy Ed,measured=5.07 eV

Formula used:

The expression for energy due to repulsion of the ions at equilibrium separation is given by,

Urep=Ed,calculated−Ed,measured

Calculation:

The energy due to repulsion of the ions at equilibrium separation is calculated as,

Urep=Ed,calculated−Ed,measured=5.70 eV−5.07 eV=0.63 eV

Conclusion:

Therefore, the energy due to repulsion of the ions at the equilibrium separation is 0.63 eV

Answer 21

(c)

To determine

The energy due to repulsion of the ions at the equilibrium separation.

Answer 22

(c)

Expert Solution

Answer to Problem 21P

The energy due to repulsion of the ions at the equilibrium separation is 0.63 eV .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The measured dissociation energy Ed,measured=5.07 eV

Formula used:

The expression for energy due to repulsion of the ions at equilibrium separation is given by,

Urep=Ed,calculated−Ed,measured

Calculation:

The energy due to repulsion of the ions at equilibrium separation is calculated as,

Urep=Ed,calculated−Ed,measured=5.70 eV−5.07 eV=0.63 eV

Conclusion:

Therefore, the energy due to repulsion of the ions at the equilibrium separation is 0.63 eV

Answer 27

Ch. 37 - Prob. 1P Ch. 37 - Prob. 2P Ch. 37 - Prob. 3P Ch. 37 - Prob. 4P Ch. 37 - Prob. 5P Ch. 37 - Prob. 6P Ch. 37 - Prob. 7P Ch. 37 - Prob. 8P Ch. 37 - Prob. 9P Ch. 37 - Prob. 10P

The potential energy of attraction of the ions.

The potential energy of attraction of the ions.

Answer to Problem 21P

Explanation of Solution

The value of dissociation energy.

Answer to Problem 21P

Explanation of Solution

The energy due to repulsion of the ions at the equilibrium separation.

Answer to Problem 21P

Explanation of Solution

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Chapter 37 Solutions