The second ionization energy of lithium is to be calculated. Concept introduction: Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state. The ionization energy is in kJ / mol .

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Answer 1

Question

Chapter 7, Problem 104AP

Interpretation Introduction

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in kJ/mol.

Expert Solution & Answer

Answer to Problem 104AP

Solution: 7.28× 103 kJ/mol

Explanation of Solution

Given information: I1=520 kJ/mol

En=−(2.18 × 10−18 J) Z2(1n2)

E=1.96× 104 kJ/mol

The total energy required to remove all the three electrons from lithium is given as: E=I1+I2+I3 …… (1)

For lithium, the value of effective nuclear charge (Z) is 3 and the value of principle quantum number (n) is 1 because the third electron will come from the orbital of 1s. The ionization energy is the difference between the final state and the initial state. So, the third ionization energy is calculated as follows:

I3=ΔE = E∞ − E1

Or

I3= E∞ − E1=−(2.18 × 10−18 J) Z2(1∞2)−−(2.18 × 10−18 J) Z2(112)

Substitute 3 for Z in the above equation as follows:

I3=−(2.18 × 10−18 J) (3)2(1∞2)−[−(2.18 × 10−18 J)(3)2(112)]=+1.96×10−17J

In one mole, there are 6.022 × 1023 atoms present.

So, the third ionization energy per mole is given as follows:

I3=(1.96 × 10−17 J) × (6.022 × 10231 mol )= 1.18 × 107 J/mol

Since 1 J=10−3 kJ,

Therefore,

1.18 × 104 J/mol=(1.18 × 104 Jmol)(10−3 kJJ)= 1.18 × 104 kJ/mol

Now, substitute 1.18 × 104 kJ/mol for I3, 520 kJ/mol for I1 and 1.96× 104 kJ/mol for E in equation (1) as follows:

1.96× 104 kJ/mol=(520 kJ/mol)+I2+(1.18 × 104 kJ/mol)I2=7.28× 103 kJ/mol

Conclusion

The second ionization energy of lithium is 7.28× 103 kJ/mol.

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Students have asked these similar questions

Which of the following reactions represents an electron gain energy that would be positive (i.e. endothermic). S(g) → S+(g) + e- Cl(g) → Cl+(g) + e- Kr(g) + e- → Kr-(g) S(g) + e- → S-(g) Kr(g) → Kr+(g)+ e- Cl(g) + e- →Cl-(g)

Use the following data and the Born-Haber cycle to calculate the first ionization energy (ΔH IE1) of K(g) K(s) → K(g) 89 kJ mol-1 Cl(g) + e– → Cl–(g) -349 kJ mol-1 K(s) + ½ Cl2(g) → KCl(s) -437 kJ mol-1 K(g) → K+(g) + e– ΔH IE1 Cl2(g) → 2Cl(g) 244 kJ mol-1 K+(g) + Cl–(g) → KCl(s) -717 kJ mol-1 Use the bond energies listed in the table below to estimate the standard enthalpy of the reaction: CCl3CHCl2(g) + 2HF(g) → CCl3CHF2(g) + 2HCl(g) C – H 412 kJ mol-1 C – C 348 kJ mol-1 C – Cl 338 kJ mol-1 H – Cl 431 kJ mol-1 C – F 484 kJ mol-1 H – F 565 kJ mol-1

Use the following data and the Born-Haber cycle to calculate the first ionization energy (ΔH IE1) of K(g) K(s) → K(g) 89 kJ mol-1 Cl(g) + e– → Cl–(g) -349 kJ mol-1 K(s) + ½ Cl2(g) → KCl(s) -437 kJ mol-1 K(g) → K+(g) + e– ΔH IE1 Cl2(g) → 2Cl(g) 244 kJ mol-1 K+(g) + Cl–(g) → KCl(s) -717 kJ mol-1

Answer 2

Question

Chapter 7, Problem 104AP

Interpretation Introduction

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in kJ/mol.

Expert Solution & Answer

Answer to Problem 104AP

Solution: 7.28× 103 kJ/mol

Explanation of Solution

Given information: I1=520 kJ/mol

En=−(2.18 × 10−18 J) Z2(1n2)

E=1.96× 104 kJ/mol

The total energy required to remove all the three electrons from lithium is given as: E=I1+I2+I3 …… (1)

For lithium, the value of effective nuclear charge (Z) is 3 and the value of principle quantum number (n) is 1 because the third electron will come from the orbital of 1s. The ionization energy is the difference between the final state and the initial state. So, the third ionization energy is calculated as follows:

I3=ΔE = E∞ − E1

Or

I3= E∞ − E1=−(2.18 × 10−18 J) Z2(1∞2)−−(2.18 × 10−18 J) Z2(112)

Substitute 3 for Z in the above equation as follows:

I3=−(2.18 × 10−18 J) (3)2(1∞2)−[−(2.18 × 10−18 J)(3)2(112)]=+1.96×10−17J

In one mole, there are 6.022 × 1023 atoms present.

So, the third ionization energy per mole is given as follows:

I3=(1.96 × 10−17 J) × (6.022 × 10231 mol )= 1.18 × 107 J/mol

Since 1 J=10−3 kJ,

Therefore,

1.18 × 104 J/mol=(1.18 × 104 Jmol)(10−3 kJJ)= 1.18 × 104 kJ/mol

Now, substitute 1.18 × 104 kJ/mol for I3, 520 kJ/mol for I1 and 1.96× 104 kJ/mol for E in equation (1) as follows:

1.96× 104 kJ/mol=(520 kJ/mol)+I2+(1.18 × 104 kJ/mol)I2=7.28× 103 kJ/mol

Conclusion

The second ionization energy of lithium is 7.28× 103 kJ/mol.

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Answer 3

Question

Chapter 7, Problem 104AP

Interpretation Introduction

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in kJ/mol.

Expert Solution & Answer

Answer to Problem 104AP

Solution: 7.28× 103 kJ/mol

Explanation of Solution

Given information: I1=520 kJ/mol

En=−(2.18 × 10−18 J) Z2(1n2)

E=1.96× 104 kJ/mol

The total energy required to remove all the three electrons from lithium is given as: E=I1+I2+I3 …… (1)

For lithium, the value of effective nuclear charge (Z) is 3 and the value of principle quantum number (n) is 1 because the third electron will come from the orbital of 1s. The ionization energy is the difference between the final state and the initial state. So, the third ionization energy is calculated as follows:

I3=ΔE = E∞ − E1

Or

I3= E∞ − E1=−(2.18 × 10−18 J) Z2(1∞2)−−(2.18 × 10−18 J) Z2(112)

Substitute 3 for Z in the above equation as follows:

I3=−(2.18 × 10−18 J) (3)2(1∞2)−[−(2.18 × 10−18 J)(3)2(112)]=+1.96×10−17J

In one mole, there are 6.022 × 1023 atoms present.

So, the third ionization energy per mole is given as follows:

I3=(1.96 × 10−17 J) × (6.022 × 10231 mol )= 1.18 × 107 J/mol

Since 1 J=10−3 kJ,

Therefore,

1.18 × 104 J/mol=(1.18 × 104 Jmol)(10−3 kJJ)= 1.18 × 104 kJ/mol

Now, substitute 1.18 × 104 kJ/mol for I3, 520 kJ/mol for I1 and 1.96× 104 kJ/mol for E in equation (1) as follows:

1.96× 104 kJ/mol=(520 kJ/mol)+I2+(1.18 × 104 kJ/mol)I2=7.28× 103 kJ/mol

Conclusion

The second ionization energy of lithium is 7.28× 103 kJ/mol.

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Answer 4

Question

Chapter 7, Problem 104AP

Interpretation Introduction

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in kJ/mol.

Expert Solution & Answer

Answer to Problem 104AP

Solution: 7.28× 103 kJ/mol

Explanation of Solution

Given information: I1=520 kJ/mol

En=−(2.18 × 10−18 J) Z2(1n2)

E=1.96× 104 kJ/mol

The total energy required to remove all the three electrons from lithium is given as: E=I1+I2+I3 …… (1)

For lithium, the value of effective nuclear charge (Z) is 3 and the value of principle quantum number (n) is 1 because the third electron will come from the orbital of 1s. The ionization energy is the difference between the final state and the initial state. So, the third ionization energy is calculated as follows:

I3=ΔE = E∞ − E1

Or

I3= E∞ − E1=−(2.18 × 10−18 J) Z2(1∞2)−−(2.18 × 10−18 J) Z2(112)

Substitute 3 for Z in the above equation as follows:

I3=−(2.18 × 10−18 J) (3)2(1∞2)−[−(2.18 × 10−18 J)(3)2(112)]=+1.96×10−17J

In one mole, there are 6.022 × 1023 atoms present.

So, the third ionization energy per mole is given as follows:

I3=(1.96 × 10−17 J) × (6.022 × 10231 mol )= 1.18 × 107 J/mol

Since 1 J=10−3 kJ,

Therefore,

1.18 × 104 J/mol=(1.18 × 104 Jmol)(10−3 kJJ)= 1.18 × 104 kJ/mol

Now, substitute 1.18 × 104 kJ/mol for I3, 520 kJ/mol for I1 and 1.96× 104 kJ/mol for E in equation (1) as follows:

1.96× 104 kJ/mol=(520 kJ/mol)+I2+(1.18 × 104 kJ/mol)I2=7.28× 103 kJ/mol

Conclusion

The second ionization energy of lithium is 7.28× 103 kJ/mol.

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Answer 5

Chapter 7, Problem 104AP

Interpretation Introduction

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in kJ/mol.

Expert Solution & Answer

Answer to Problem 104AP

Solution: 7.28× 103 kJ/mol

Explanation of Solution

Given information: I1=520 kJ/mol

En=−(2.18 × 10−18 J) Z2(1n2)

E=1.96× 104 kJ/mol

The total energy required to remove all the three electrons from lithium is given as: E=I1+I2+I3 …… (1)

For lithium, the value of effective nuclear charge (Z) is 3 and the value of principle quantum number (n) is 1 because the third electron will come from the orbital of 1s. The ionization energy is the difference between the final state and the initial state. So, the third ionization energy is calculated as follows:

I3=ΔE = E∞ − E1

Or

I3= E∞ − E1=−(2.18 × 10−18 J) Z2(1∞2)−−(2.18 × 10−18 J) Z2(112)

Substitute 3 for Z in the above equation as follows:

I3=−(2.18 × 10−18 J) (3)2(1∞2)−[−(2.18 × 10−18 J)(3)2(112)]=+1.96×10−17J

In one mole, there are 6.022 × 1023 atoms present.

So, the third ionization energy per mole is given as follows:

I3=(1.96 × 10−17 J) × (6.022 × 10231 mol )= 1.18 × 107 J/mol

Since 1 J=10−3 kJ,

Therefore,

1.18 × 104 J/mol=(1.18 × 104 Jmol)(10−3 kJJ)= 1.18 × 104 kJ/mol

Now, substitute 1.18 × 104 kJ/mol for I3, 520 kJ/mol for I1 and 1.96× 104 kJ/mol for E in equation (1) as follows:

1.96× 104 kJ/mol=(520 kJ/mol)+I2+(1.18 × 104 kJ/mol)I2=7.28× 103 kJ/mol

Conclusion

The second ionization energy of lithium is 7.28× 103 kJ/mol.

Answer 6

Chapter 7, Problem 104AP

Interpretation Introduction

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in kJ/mol.

Expert Solution & Answer

Answer to Problem 104AP

Solution: 7.28× 103 kJ/mol

Explanation of Solution

Given information: I1=520 kJ/mol

En=−(2.18 × 10−18 J) Z2(1n2)

E=1.96× 104 kJ/mol

The total energy required to remove all the three electrons from lithium is given as: E=I1+I2+I3 …… (1)

For lithium, the value of effective nuclear charge (Z) is 3 and the value of principle quantum number (n) is 1 because the third electron will come from the orbital of 1s. The ionization energy is the difference between the final state and the initial state. So, the third ionization energy is calculated as follows:

I3=ΔE = E∞ − E1

Or

I3= E∞ − E1=−(2.18 × 10−18 J) Z2(1∞2)−−(2.18 × 10−18 J) Z2(112)

Substitute 3 for Z in the above equation as follows:

I3=−(2.18 × 10−18 J) (3)2(1∞2)−[−(2.18 × 10−18 J)(3)2(112)]=+1.96×10−17J

In one mole, there are 6.022 × 1023 atoms present.

So, the third ionization energy per mole is given as follows:

I3=(1.96 × 10−17 J) × (6.022 × 10231 mol )= 1.18 × 107 J/mol

Since 1 J=10−3 kJ,

Therefore,

1.18 × 104 J/mol=(1.18 × 104 Jmol)(10−3 kJJ)= 1.18 × 104 kJ/mol

Now, substitute 1.18 × 104 kJ/mol for I3, 520 kJ/mol for I1 and 1.96× 104 kJ/mol for E in equation (1) as follows:

1.96× 104 kJ/mol=(520 kJ/mol)+I2+(1.18 × 104 kJ/mol)I2=7.28× 103 kJ/mol

Conclusion

The second ionization energy of lithium is 7.28× 103 kJ/mol.

Answer 7

Chapter 7, Problem 104AP

Interpretation Introduction

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in kJ/mol.

Answer 8

Expert Solution & Answer

Answer to Problem 104AP

Solution: 7.28× 103 kJ/mol

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information: I1=520 kJ/mol

En=−(2.18 × 10−18 J) Z2(1n2)

E=1.96× 104 kJ/mol

The total energy required to remove all the three electrons from lithium is given as: E=I1+I2+I3 …… (1)

For lithium, the value of effective nuclear charge (Z) is 3 and the value of principle quantum number (n) is 1 because the third electron will come from the orbital of 1s. The ionization energy is the difference between the final state and the initial state. So, the third ionization energy is calculated as follows:

I3=ΔE = E∞ − E1

Or

I3= E∞ − E1=−(2.18 × 10−18 J) Z2(1∞2)−−(2.18 × 10−18 J) Z2(112)

Substitute 3 for Z in the above equation as follows:

I3=−(2.18 × 10−18 J) (3)2(1∞2)−[−(2.18 × 10−18 J)(3)2(112)]=+1.96×10−17J

In one mole, there are 6.022 × 1023 atoms present.

So, the third ionization energy per mole is given as follows:

I3=(1.96 × 10−17 J) × (6.022 × 10231 mol )= 1.18 × 107 J/mol

Since 1 J=10−3 kJ,

Therefore,

1.18 × 104 J/mol=(1.18 × 104 Jmol)(10−3 kJJ)= 1.18 × 104 kJ/mol

Now, substitute 1.18 × 104 kJ/mol for I3, 520 kJ/mol for I1 and 1.96× 104 kJ/mol for E in equation (1) as follows:

1.96× 104 kJ/mol=(520 kJ/mol)+I2+(1.18 × 104 kJ/mol)I2=7.28× 103 kJ/mol

Answer 12

Conclusion

The second ionization energy of lithium is 7.28× 103 kJ/mol.

Answer 13

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The second ionization energy of lithium is to be calculated. Concept introduction: Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state. The ionization energy is in kJ / mol .

Interpretation: The second ionization energy of lithium is to be calculated. Concept introduction: Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state. The ionization energy is in kJ/mol.

Answer to Problem 104AP

Explanation of Solution

Want to see more full solutions like this?

Chapter 7 Solutions

Interpretation: The second ionization energy of lithium is to be calculated.

Concept introduction:

Ionization energy is defined as the minimum energy needed to remove avalence electron from a neutral atom in the gaseous state.

The ionization energy is in kJ/mol.