The net potential energy for a Ba 2+ S 2- ion pair.

Question

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Answer 1

Question

Chapter 2.9, Problem 59AAP

To determine

The net potential energy for a Ba2+S2- ion pair.

Expert Solution & Answer

Answer to Problem 59AAP

The net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Explanation of Solution

The net force acting between two atoms is,

Fnet=Fattraction−Frepulsion

Refer Equation (2.8),

Write the formula for repulsion force (Frepulsion).

Frepulsion=−nban+1 (I)

Here, the constants are n, b, and inter-ionic distance is a.

Write the formula for net potential energy (Enet).

Enet=Fattraction+Frepulsion=z1z2e24πεoa+ban (II)

Here, the electrons removed from Ba is z1, the electrons added to S is z2, the electron charge is e, the permittivity of space is εo and the inter-ionic distance is a.

When, the bond is formed, the net potential energy become zero.

0=Fattraction+FrepulsionFrepulsion=−Fattraction

Write the expression for inter-ionic distance.

a=r1+r2 (III)

Here the ionic distance of Ba2+ is r1 and the ionic radius of S2- is r2.

Conclusion:

Substitute 0.143 nm for r1 and 0.174 nm for r2 in equation (II).

a=0.143 nm+0.174 nm=(0.317 nm)(10−9 m1 nm)=3.17×10−10 m

Refer problem 2.58. The force of attraction between Ba2+ and S2- ions is,

Fattraction=9.1628×10−9 N

Here,

Frepulsion=−Fattraction=−(9.1628×10−9 N)=−9.1628×10−9 N

Substitute −9.1628×10−9 N for Frepulsion, 10.5 for n, and 3.17×10−10 m for a in

Equation (I).

−9.1628×10−9 N=−(10.5)b(3.17×10−10 m)10.5+1b=(3.17×10−10 m)11.5(9.1628×10−9 N)10.5b=5.2992×10−11810.5b=5.0468×10−119 N⋅m11.5

Substitute 2 for z1, −2 for z2, 1.6×10−19 C for e, 8.85×10−12 C2/N⋅m2 for εo, 3.17×10−10 m for a, 5.0468×10−119 N⋅m11.5 for b, and 10.5 for n in Equation (III).

Enet=(2)(−2)(1.6×10−19 C)24π(8.85×10−12 C2/N⋅m2)(3.17×10−10 m)+5.0468×10−119 N⋅m11.5(3.17×10−10 m)10.5=−(2.9046×10−18 N⋅m)+(2.7663×10−19N⋅m)=−2.6279×10−18 N⋅m=−2.6279×10−18J

Thus, the net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

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Answer 2

Question

Chapter 2.9, Problem 59AAP

To determine

The net potential energy for a Ba2+S2- ion pair.

Expert Solution & Answer

Answer to Problem 59AAP

The net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Explanation of Solution

The net force acting between two atoms is,

Fnet=Fattraction−Frepulsion

Refer Equation (2.8),

Write the formula for repulsion force (Frepulsion).

Frepulsion=−nban+1 (I)

Here, the constants are n, b, and inter-ionic distance is a.

Write the formula for net potential energy (Enet).

Enet=Fattraction+Frepulsion=z1z2e24πεoa+ban (II)

Here, the electrons removed from Ba is z1, the electrons added to S is z2, the electron charge is e, the permittivity of space is εo and the inter-ionic distance is a.

When, the bond is formed, the net potential energy become zero.

0=Fattraction+FrepulsionFrepulsion=−Fattraction

Write the expression for inter-ionic distance.

a=r1+r2 (III)

Here the ionic distance of Ba2+ is r1 and the ionic radius of S2- is r2.

Conclusion:

Substitute 0.143 nm for r1 and 0.174 nm for r2 in equation (II).

a=0.143 nm+0.174 nm=(0.317 nm)(10−9 m1 nm)=3.17×10−10 m

Refer problem 2.58. The force of attraction between Ba2+ and S2- ions is,

Fattraction=9.1628×10−9 N

Here,

Frepulsion=−Fattraction=−(9.1628×10−9 N)=−9.1628×10−9 N

Substitute −9.1628×10−9 N for Frepulsion, 10.5 for n, and 3.17×10−10 m for a in

Equation (I).

−9.1628×10−9 N=−(10.5)b(3.17×10−10 m)10.5+1b=(3.17×10−10 m)11.5(9.1628×10−9 N)10.5b=5.2992×10−11810.5b=5.0468×10−119 N⋅m11.5

Substitute 2 for z1, −2 for z2, 1.6×10−19 C for e, 8.85×10−12 C2/N⋅m2 for εo, 3.17×10−10 m for a, 5.0468×10−119 N⋅m11.5 for b, and 10.5 for n in Equation (III).

Enet=(2)(−2)(1.6×10−19 C)24π(8.85×10−12 C2/N⋅m2)(3.17×10−10 m)+5.0468×10−119 N⋅m11.5(3.17×10−10 m)10.5=−(2.9046×10−18 N⋅m)+(2.7663×10−19N⋅m)=−2.6279×10−18 N⋅m=−2.6279×10−18J

Thus, the net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

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Answer 3

Question

Chapter 2.9, Problem 59AAP

To determine

The net potential energy for a Ba2+S2- ion pair.

Expert Solution & Answer

Answer to Problem 59AAP

The net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Explanation of Solution

The net force acting between two atoms is,

Fnet=Fattraction−Frepulsion

Refer Equation (2.8),

Write the formula for repulsion force (Frepulsion).

Frepulsion=−nban+1 (I)

Here, the constants are n, b, and inter-ionic distance is a.

Write the formula for net potential energy (Enet).

Enet=Fattraction+Frepulsion=z1z2e24πεoa+ban (II)

Here, the electrons removed from Ba is z1, the electrons added to S is z2, the electron charge is e, the permittivity of space is εo and the inter-ionic distance is a.

When, the bond is formed, the net potential energy become zero.

0=Fattraction+FrepulsionFrepulsion=−Fattraction

Write the expression for inter-ionic distance.

a=r1+r2 (III)

Here the ionic distance of Ba2+ is r1 and the ionic radius of S2- is r2.

Conclusion:

Substitute 0.143 nm for r1 and 0.174 nm for r2 in equation (II).

a=0.143 nm+0.174 nm=(0.317 nm)(10−9 m1 nm)=3.17×10−10 m

Refer problem 2.58. The force of attraction between Ba2+ and S2- ions is,

Fattraction=9.1628×10−9 N

Here,

Frepulsion=−Fattraction=−(9.1628×10−9 N)=−9.1628×10−9 N

Substitute −9.1628×10−9 N for Frepulsion, 10.5 for n, and 3.17×10−10 m for a in

Equation (I).

−9.1628×10−9 N=−(10.5)b(3.17×10−10 m)10.5+1b=(3.17×10−10 m)11.5(9.1628×10−9 N)10.5b=5.2992×10−11810.5b=5.0468×10−119 N⋅m11.5

Substitute 2 for z1, −2 for z2, 1.6×10−19 C for e, 8.85×10−12 C2/N⋅m2 for εo, 3.17×10−10 m for a, 5.0468×10−119 N⋅m11.5 for b, and 10.5 for n in Equation (III).

Enet=(2)(−2)(1.6×10−19 C)24π(8.85×10−12 C2/N⋅m2)(3.17×10−10 m)+5.0468×10−119 N⋅m11.5(3.17×10−10 m)10.5=−(2.9046×10−18 N⋅m)+(2.7663×10−19N⋅m)=−2.6279×10−18 N⋅m=−2.6279×10−18J

Thus, the net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

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Answer 4

Question

Chapter 2.9, Problem 59AAP

To determine

The net potential energy for a Ba2+S2- ion pair.

Expert Solution & Answer

Answer to Problem 59AAP

The net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Explanation of Solution

The net force acting between two atoms is,

Fnet=Fattraction−Frepulsion

Refer Equation (2.8),

Write the formula for repulsion force (Frepulsion).

Frepulsion=−nban+1 (I)

Here, the constants are n, b, and inter-ionic distance is a.

Write the formula for net potential energy (Enet).

Enet=Fattraction+Frepulsion=z1z2e24πεoa+ban (II)

Here, the electrons removed from Ba is z1, the electrons added to S is z2, the electron charge is e, the permittivity of space is εo and the inter-ionic distance is a.

When, the bond is formed, the net potential energy become zero.

0=Fattraction+FrepulsionFrepulsion=−Fattraction

Write the expression for inter-ionic distance.

a=r1+r2 (III)

Here the ionic distance of Ba2+ is r1 and the ionic radius of S2- is r2.

Conclusion:

Substitute 0.143 nm for r1 and 0.174 nm for r2 in equation (II).

a=0.143 nm+0.174 nm=(0.317 nm)(10−9 m1 nm)=3.17×10−10 m

Refer problem 2.58. The force of attraction between Ba2+ and S2- ions is,

Fattraction=9.1628×10−9 N

Here,

Frepulsion=−Fattraction=−(9.1628×10−9 N)=−9.1628×10−9 N

Substitute −9.1628×10−9 N for Frepulsion, 10.5 for n, and 3.17×10−10 m for a in

Equation (I).

−9.1628×10−9 N=−(10.5)b(3.17×10−10 m)10.5+1b=(3.17×10−10 m)11.5(9.1628×10−9 N)10.5b=5.2992×10−11810.5b=5.0468×10−119 N⋅m11.5

Substitute 2 for z1, −2 for z2, 1.6×10−19 C for e, 8.85×10−12 C2/N⋅m2 for εo, 3.17×10−10 m for a, 5.0468×10−119 N⋅m11.5 for b, and 10.5 for n in Equation (III).

Enet=(2)(−2)(1.6×10−19 C)24π(8.85×10−12 C2/N⋅m2)(3.17×10−10 m)+5.0468×10−119 N⋅m11.5(3.17×10−10 m)10.5=−(2.9046×10−18 N⋅m)+(2.7663×10−19N⋅m)=−2.6279×10−18 N⋅m=−2.6279×10−18J

Thus, the net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

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Answer 5

Chapter 2.9, Problem 59AAP

To determine

The net potential energy for a Ba2+S2- ion pair.

Expert Solution & Answer

Answer to Problem 59AAP

The net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Explanation of Solution

The net force acting between two atoms is,

Fnet=Fattraction−Frepulsion

Refer Equation (2.8),

Write the formula for repulsion force (Frepulsion).

Frepulsion=−nban+1 (I)

Here, the constants are n, b, and inter-ionic distance is a.

Write the formula for net potential energy (Enet).

Enet=Fattraction+Frepulsion=z1z2e24πεoa+ban (II)

Here, the electrons removed from Ba is z1, the electrons added to S is z2, the electron charge is e, the permittivity of space is εo and the inter-ionic distance is a.

When, the bond is formed, the net potential energy become zero.

0=Fattraction+FrepulsionFrepulsion=−Fattraction

Write the expression for inter-ionic distance.

a=r1+r2 (III)

Here the ionic distance of Ba2+ is r1 and the ionic radius of S2- is r2.

Conclusion:

Substitute 0.143 nm for r1 and 0.174 nm for r2 in equation (II).

a=0.143 nm+0.174 nm=(0.317 nm)(10−9 m1 nm)=3.17×10−10 m

Refer problem 2.58. The force of attraction between Ba2+ and S2- ions is,

Fattraction=9.1628×10−9 N

Here,

Frepulsion=−Fattraction=−(9.1628×10−9 N)=−9.1628×10−9 N

Substitute −9.1628×10−9 N for Frepulsion, 10.5 for n, and 3.17×10−10 m for a in

Equation (I).

−9.1628×10−9 N=−(10.5)b(3.17×10−10 m)10.5+1b=(3.17×10−10 m)11.5(9.1628×10−9 N)10.5b=5.2992×10−11810.5b=5.0468×10−119 N⋅m11.5

Substitute 2 for z1, −2 for z2, 1.6×10−19 C for e, 8.85×10−12 C2/N⋅m2 for εo, 3.17×10−10 m for a, 5.0468×10−119 N⋅m11.5 for b, and 10.5 for n in Equation (III).

Enet=(2)(−2)(1.6×10−19 C)24π(8.85×10−12 C2/N⋅m2)(3.17×10−10 m)+5.0468×10−119 N⋅m11.5(3.17×10−10 m)10.5=−(2.9046×10−18 N⋅m)+(2.7663×10−19N⋅m)=−2.6279×10−18 N⋅m=−2.6279×10−18J

Thus, the net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Answer 6

Chapter 2.9, Problem 59AAP

To determine

The net potential energy for a Ba2+S2- ion pair.

Expert Solution & Answer

Answer to Problem 59AAP

The net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Explanation of Solution

The net force acting between two atoms is,

Fnet=Fattraction−Frepulsion

Refer Equation (2.8),

Write the formula for repulsion force (Frepulsion).

Frepulsion=−nban+1 (I)

Here, the constants are n, b, and inter-ionic distance is a.

Write the formula for net potential energy (Enet).

Enet=Fattraction+Frepulsion=z1z2e24πεoa+ban (II)

Here, the electrons removed from Ba is z1, the electrons added to S is z2, the electron charge is e, the permittivity of space is εo and the inter-ionic distance is a.

When, the bond is formed, the net potential energy become zero.

0=Fattraction+FrepulsionFrepulsion=−Fattraction

Write the expression for inter-ionic distance.

a=r1+r2 (III)

Here the ionic distance of Ba2+ is r1 and the ionic radius of S2- is r2.

Conclusion:

Substitute 0.143 nm for r1 and 0.174 nm for r2 in equation (II).

a=0.143 nm+0.174 nm=(0.317 nm)(10−9 m1 nm)=3.17×10−10 m

Refer problem 2.58. The force of attraction between Ba2+ and S2- ions is,

Fattraction=9.1628×10−9 N

Here,

Frepulsion=−Fattraction=−(9.1628×10−9 N)=−9.1628×10−9 N

Substitute −9.1628×10−9 N for Frepulsion, 10.5 for n, and 3.17×10−10 m for a in

Equation (I).

−9.1628×10−9 N=−(10.5)b(3.17×10−10 m)10.5+1b=(3.17×10−10 m)11.5(9.1628×10−9 N)10.5b=5.2992×10−11810.5b=5.0468×10−119 N⋅m11.5

Substitute 2 for z1, −2 for z2, 1.6×10−19 C for e, 8.85×10−12 C2/N⋅m2 for εo, 3.17×10−10 m for a, 5.0468×10−119 N⋅m11.5 for b, and 10.5 for n in Equation (III).

Enet=(2)(−2)(1.6×10−19 C)24π(8.85×10−12 C2/N⋅m2)(3.17×10−10 m)+5.0468×10−119 N⋅m11.5(3.17×10−10 m)10.5=−(2.9046×10−18 N⋅m)+(2.7663×10−19N⋅m)=−2.6279×10−18 N⋅m=−2.6279×10−18J

Thus, the net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Answer 7

Chapter 2.9, Problem 59AAP

To determine

The net potential energy for a Ba2+S2- ion pair.

Answer 8

Expert Solution & Answer

Answer to Problem 59AAP

The net potential energy for a Ba2+S2- ion pair is −2.6279×10−18J.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The net force acting between two atoms is,

Fnet=Fattraction−Frepulsion

Refer Equation (2.8),

Write the formula for repulsion force (Frepulsion).

Frepulsion=−nban+1 (I)

Here, the constants are n, b, and inter-ionic distance is a.

Write the formula for net potential energy (Enet).

Enet=Fattraction+Frepulsion=z1z2e24πεoa+ban (II)

Here, the electrons removed from Ba is z1, the electrons added to S is z2, the electron charge is e, the permittivity of space is εo and the inter-ionic distance is a.

When, the bond is formed, the net potential energy become zero.

0=Fattraction+FrepulsionFrepulsion=−Fattraction

Write the expression for inter-ionic distance.

a=r1+r2 (III)

Here the ionic distance of Ba2+ is r1 and the ionic radius of S2- is r2.

Conclusion:

Substitute 0.143 nm for r1 and 0.174 nm for r2 in equation (II).

a=0.143 nm+0.174 nm=(0.317 nm)(10−9 m1 nm)=3.17×10−10 m

Refer problem 2.58. The force of attraction between Ba2+ and S2- ions is,

Fattraction=9.1628×10−9 N

Here,

Frepulsion=−Fattraction=−(9.1628×10−9 N)=−9.1628×10−9 N

Substitute −9.1628×10−9 N for Frepulsion, 10.5 for n, and 3.17×10−10 m for a in

Equation (I).

−9.1628×10−9 N=−(10.5)b(3.17×10−10 m)10.5+1b=(3.17×10−10 m)11.5(9.1628×10−9 N)10.5b=5.2992×10−11810.5b=5.0468×10−119 N⋅m11.5

Substitute 2 for z1, −2 for z2, 1.6×10−19 C for e, 8.85×10−12 C2/N⋅m2 for εo, 3.17×10−10 m for a, 5.0468×10−119 N⋅m11.5 for b, and 10.5 for n in Equation (III).

Enet=(2)(−2)(1.6×10−19 C)24π(8.85×10−12 C2/N⋅m2)(3.17×10−10 m)+5.0468×10−119 N⋅m11.5(3.17×10−10 m)10.5=−(2.9046×10−18 N⋅m)+(2.7663×10−19N⋅m)=−2.6279×10−18 N⋅m=−2.6279×10−18J