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 .

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

An expression for the root mean square velocity, vrms, of a gas was derived. Using Maxwell’s velocity distribution, one can also calculate the mean velocity and the most probable velocity (mp) of a collection of molecules. The equations used for these two quantities are vmean=(8RT/πM)1/2 and vmp=(2RT/M)1/2 These values have a fixed relationship to each other.(a) Arrange these three quantities in order of increasing magnitude.(b) Show that the relative magnitudes are independent of the molar mass of the gas.(c) Use the smallest velocity as a reference for establishing the order of magnitude and determine the relationship between the larger and smaller values.

The reaction of solid dimethylhydrazine, (CH3)2N2H2, and liquefied dinitrogen tetroxide, N2O4, has been investigated for use as rocket fuel. The reaction produces the gases carbon dioxide (CO2), nitrogen (N2), and water vapor (H2O), which are ejected in the exhaust gases. In a controlled experiment, solid dimethylhydrazine was reacted with excess dinitrogen tetroxide, and the gases were collected in a closed balloon until a pressure of 2.50 atm and a temperature of 400.0 K were reached.(a) What are the partial pressures of CO2, N2, and H2O?(b) When the CO2 is removed by chemical reaction, what are the partial pressures of the remaining gases?

One liter of chlorine gas at 1 atm and 298 K reacts completely with 1.00 L of nitrogen gas and 2.00 L of oxygen gas at the same temperature and pressure. A single gaseous product is formed, which fills a 2.00 L flask at 1.00 atm and 298 K. Use this information to determine the following characteristics of the product:(a) its empirical formula;(b) its molecular formula;(c) the most favorable Lewis formula based on formal charge arguments (the central atom is N);(d) the shape of the molecule.

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Ch. 7.1 - Practice Problem ATTEMPT What element(s) would you...Ch. 7.1 - Practice Problem BUILD Arrange the following...Ch. 7.1 - Practice ProblemCONCEPTUALIZE Three different...Ch. 7.1 - 7.1.1 Which of the following elements would you...Ch. 7.1 - Prob. 2CP Ch. 7.1 - Prob. 3CP Ch. 7.2 - Practice Problem ATTEMPT Without using a periodic...Ch. 7.2 - Practice ProblemBUILD Identify the elements...Ch. 7.2 - Prob. 1PPC Ch. 7.2 - 7.2.1 Which electron configuration is correct for...

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.