The value of B for an ion pair.

Question

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Answer 1

Question

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

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Answer 2

Question

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

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Answer 3

Question

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

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Answer 4

Question

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

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Answer 5

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Answer 6

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Answer 7

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Answer 8

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Answer 12

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

The value of B for an ion pair.

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The value of B for an ion pair.

Answer to Problem 2.25P

Explanation of Solution

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Chapter 2 Solutions