Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 18, Problem 43E
Interpretation Introduction

Interpretation: The atomic masses of neutron, proton, electron 232Pu and 231Pa are given. Using these values, the change in energy when 1 mole of 232Pu nuclei and 1 mole of 231Pa nuclei are formed from their respective number of protons and neutrons is to be calculated.

Concept introduction: The difference between the mass of an atom and the mass of its constituent particles is known as mass defect. This mass can be converted into energy by Einstein’s mass-energy equation as,

ΔE=Δmc2

To determine: The change in energy when 1 mole of 232Pu nuclei and 1 mole of 231Pa nuclei are formed from their respective number of protons and neutrons.

Expert Solution & Answer
Check Mark

Answer to Problem 43E

Solution

The change in energy when 1 mole of 232Pu nuclei formed is 1.6702×1014J/mol_

The change in energy when 1 mole of 231Pa nuclei formed is 1.6704×1014J/mol_

Explanation of Solution

Explanation

Given

The atomic mass of 232Pu is 3.85285×1022g .

The mass of a neutron is 1.67493×1024g .

The mass of a proton is 1.67262×1024g .

The atomic number of 232Pu is 94 . Therefore, the number of protons in 232Pu is 94.

The number of neutrons in 232Pu is,

Number of neutrons =23294=138

The mass defect is calculated by the formula,

Δm=Atomicmass of 232Pu[Numberofprotons×massofprotonNumberofneutrons×massofneutron]

Where,

  • Δm is the mass defect.

Substitute the value of the atomic mass of 232Pu , the number of protons and neutrons and their respective masses in the above equation.

Δm=3.85285×1022g[(94×1.67262×1024g)+(138×1.67493×1024g)]=0.0308162×1022g

To calculate the mass defect for one mole of 232Pu , multiply this value by Avogadro’s number (6.022×1023permol) .

Δm'=0.0308162×1022g×6.022×1023permol=1.85575g/mol_

The change in energy of 232Pu is. 1.6702×1014J/mol_ .

Explanation

The mass defect is 1.85575g/mol .

The conversion of gram (g) into kilogram (kg) is done as,

1g=103kg

Hence, the conversion of 1.85575g/mol into kilogram/mol is,

1.85575g/mol=1.85575×103kg/mol

The energy change is calculated by Einstein’s mass energy equation,

ΔE=Δm'c2

Where,

  • ΔE is the change in energy.
  • Δm' is the mass defect.
  • c is the speed of light (3.0×108m/s) .

Substitute the values of Δm' and c in the above equation.

ΔE=Δm'c2=(1.85575×103kg/mol)(3.0×108m/s)2=1.6702×1014J/mol_

The mass defect of 231Pa nucleus is 1.85635g/mol_ .

Explanation

Given

The atomic mass of 231Pa is 3.83616×1022g .

The mass of a neutron is 1.67493×1024g .

The mass of a proton is 1.67262×1024g .

The atomic number of 231Pa is 91 . Therefore, the number of protons in 231Pa is 91 .

The number of neutrons in 232Pu is,

Number of neutrons =23191=140

The mass defect is calculated by the formula,

Δm=Atomicmass of 231Pa[Numberofprotons×massofprotonNumberofneutrons×massofneutron]

Substitute the value of the atomic mass of 231Pa , the number of protons and neutrons and their respective masses in the above equation.

Δm=3.83616×1022g[(91×1.67262×1024g)+(140×1.67493×1024g)]=0.0308262×1022g

To calculate the mass defect for one mole of 231Pa , multiply this value by Avogadro’s number (6.022×1023permol) .

Δm'=0.0308262×1022g×6.022×1023permol=1.85635g/mol_

The binding energy of 231Pa is. 1.6707×1014J/mol_ .

Explanation

The mass defect is 1.85635g/mol .

The conversion of gram (g) into kilogram (kg) is done as,

1g=103kg

Hence, the conversion of 1.85635g/mol into kilogram/mol is,

1.85635g/mol=1.85635×103kg/mol

The energy change is calculated by Einstein’s mass energy equation,

ΔE=Δm'c2

Where,

  • ΔE is the change in energy.
  • Δm' is the mass defect.
  • c is the speed of light (3.0×108m/s) .

Substitute the values of Δm' and c in the above equation.

ΔE=Δm'c2=(1.85635×103kg/mol)(3.0×108m/s)2=1.6707×1014J/mol_

Conclusion

Conclusion

The change in energy when 1 mole of 232Pu nuclei formed is 1.6702×1014J/mol_ .

The change in energy when 1 mole of 231Pa nuclei formed is 1.6707×1014J/mol_ .

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Chapter 18 Solutions

Chemistry: An Atoms First Approach

Ch. 18 - Prob. 1QCh. 18 - Prob. 2QCh. 18 - Prob. 3QCh. 18 - Prob. 4QCh. 18 - Prob. 5QCh. 18 - Prob. 6QCh. 18 - Prob. 7QCh. 18 - Prob. 8QCh. 18 - Prob. 9QCh. 18 - Prob. 10QCh. 18 - Prob. 11ECh. 18 - Prob. 12ECh. 18 - Prob. 13ECh. 18 - Prob. 14ECh. 18 - Prob. 15ECh. 18 - Prob. 16ECh. 18 - Prob. 17ECh. 18 - Prob. 18ECh. 18 - Prob. 19ECh. 18 - Prob. 20ECh. 18 - Prob. 21ECh. 18 - Prob. 22ECh. 18 - Prob. 23ECh. 18 - Prob. 24ECh. 18 - Prob. 25ECh. 18 - Prob. 26ECh. 18 - Prob. 27ECh. 18 - Prob. 28ECh. 18 - Prob. 29ECh. 18 - Prob. 30ECh. 18 - Prob. 31ECh. 18 - Prob. 32ECh. 18 - Prob. 33ECh. 18 - Prob. 34ECh. 18 - Prob. 35ECh. 18 - Prob. 36ECh. 18 - Prob. 37ECh. 18 - Prob. 38ECh. 18 - Prob. 39ECh. 18 - Prob. 40ECh. 18 - Prob. 41ECh. 18 - Prob. 42ECh. 18 - Prob. 43ECh. 18 - Prob. 44ECh. 18 - Prob. 45ECh. 18 - Prob. 46ECh. 18 - Prob. 47ECh. 18 - Prob. 48ECh. 18 - Prob. 49ECh. 18 - Prob. 50ECh. 18 - Prob. 51ECh. 18 - Prob. 52ECh. 18 - Prob. 53ECh. 18 - A chemist studied the reaction mechanism for the...Ch. 18 - Prob. 55ECh. 18 - Prob. 56ECh. 18 - Prob. 57ECh. 18 - Prob. 58ECh. 18 - Prob. 59AECh. 18 - Prob. 60AECh. 18 - Prob. 61AECh. 18 - Prob. 62AECh. 18 - Prob. 63AECh. 18 - Prob. 64AECh. 18 - Prob. 65AECh. 18 - Prob. 66AECh. 18 - Prob. 67AECh. 18 - Prob. 68AECh. 18 - Prob. 69AECh. 18 - Prob. 70AECh. 18 - Prob. 71AECh. 18 - Prob. 72AECh. 18 - Prob. 73CWPCh. 18 - Prob. 74CWPCh. 18 - Prob. 75CWPCh. 18 - Prob. 76CWPCh. 18 - Prob. 77CWPCh. 18 - Prob. 78CWPCh. 18 - Prob. 79CPCh. 18 - Prob. 80CPCh. 18 - Prob. 81CPCh. 18 - Prob. 82CPCh. 18 - Prob. 83CPCh. 18 - Prob. 84CPCh. 18 - Prob. 85CPCh. 18 - Prob. 86CPCh. 18 - Prob. 87IPCh. 18 - Prob. 88IP
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