Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
Question
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Chapter 24, Problem 24.127P

(a)

Interpretation Introduction

Interpretation:

Kinetic Energy of an H atom has to be determined using the given data.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  Nucleus + nuclear binding energy  nucleons

Kinetic energy of a particle can be using the below mentioned formula,

  KE=12mv2= (32)RTNA R= Gas constantT = TemperatureNA = Avogadro's number

(a)

Expert Solution
Check Mark

Explanation of Solution

Given data is shown below:

Temperature, T = 1.00×106 K

Kinetic Energy of an H atom is calculated as shown below,

  Kinetic Energy = (32)RTNA(32)(8.314 J/mol.K)(1.00×106K)(6.022×1023 atom/mol)= 2.07×1017 J/atom 11H

Kinetic Energy of an H atom is 2.07×10-17 J/atom.

(b)

Interpretation Introduction

Interpretation:

Number of H atoms heated from the energy produced by the annihilation of one H and one antiH has to be determined using the given data.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  Nucleus + nuclear binding energy  nucleons

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

  E = Δmc2where,Δm = Mass Differencec= Speed of light

(b)

Expert Solution
Check Mark

Explanation of Solution

Given data is shown below:

Kinetic Energy of an H atom is 2.07×10-17 J/atom

A kilogram of antiH gets annihilated by a kilogram of 1H. Hence 2 kg is converted to energy and it can be determined as follows,

  Energy = mc2 = (2.00 kg)(2.998×108 ms1)2(J/kg.m2s2)= 1.7975048×1017 J per one kg H

Energy produced by the annihilation of one H and one antiH is 1.7975048×1017 J

Number of H atoms heated from the energy produced by the annihilation of one H and one antiH is calculated as shown below,

  No. of atoms = [(1.7975048×1017 J1 kg H)(1 kg103 g)(1.0078 g H1 mol H)(1 mol H6.022×1023 atoms H)(1 H atom2.07×1017 J/atom)]= 1.45×107 H atoms

Number of H atoms is 1.45×107 H atoms.

(c)

Interpretation Introduction

Interpretation:

Energy released during the fusion of 4 Hydrogen atoms forming Helium-4 has to be determined using the given data.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  Nucleus + nuclear binding energy  nucleons

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

  E = Δmc2where,Δm = Mass Differencec= Speed of light

Change in mass of a given reaction can be determined as given,

  Δm = Mass of reactant Mass of products

(c)

Expert Solution
Check Mark

Explanation of Solution

Given data is shown below:

  Reaction: 411 24He + 2+10βNumber of H atoms = 1.45×107Mass  of 11H= 1.007825 amuMass  of 24He= 4.00260 amuMass of β+ = Mass of an electron = 5.48580×104 amu

  • Calculate the mass difference for the formation of Helium-4:

Mass difference of the reaction can be calculated as given,

  Δm = Mass of reactant Mass of products[4×Mass of 11H] [Mass of 24He + (2×Mass of β+)][4×1.007825 amu][(4.00260 amu)+(2×5.48580×104 amu)]=0.02760 amu per 24He formed= 0.02760 g per mol 24He formed

Mass difference during the fusion of 4 Hydrogen atoms is 0.02760 g per mol.

  • Calculate total mass difference in kilogram:

Mass difference during the fusion of 1.45×107 atoms is calculated as follows,

Δm = (0.02760 amumol4He)(1 kg103 g)(1 4He4 H atoms)(1 mol H6.022×1023 H atoms)(1.45×107 H atoms)= 1.6615×1022 kg

  • Calculate energy released:

Energy released is calculated as given below,

  Energy = Δmc2 (1.6615×1022 kg)(2.998×108 ms1)= 1.49×105 J

Energy released during the fusion of 4 Hydrogen atoms forming Helium-4 is 1.49×10-5 J.

(d)

Interpretation Introduction

Interpretation:

Increase in energy produce by the fusion of part (c) comparing with HantiH collision of part (b) has to be determined using the given data.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  Nucleus + nuclear binding energy  nucleons

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

  E = Δmc2where,Δm = Mass Differencec= Speed of light

Change in mass of a given reaction can be determined as given,

  Δm = Mass of reactant Mass of products

(d)

Expert Solution
Check Mark

Explanation of Solution

In part (b), Energy produced by the annihilation of one H and one antiH is 1.7975048×1017 J 

Therefore, energy generated by HantiH collision is determined as below,

  Energy = (1.7975048×1017 J 1 kg H)(1 kg103 g)(1.0078 g H1 mol H)(1 mol H6.022×1023 atoms H) = 3.00×1010 J

In part (c), Energy released during the fusion of 4 Hydrogen atoms forming Helium-4 is 1.49×10-5 J.

Increase in energy is determined as follows,

Energy increase = (1.49×105 J)(3.00×1010 J)= 1.49×105 J

Energy increase is 1.49×10-5 J.

(e)

Interpretation Introduction

Interpretation:

Energy released when 1.00 kg each of antimatter and matter annihilate with each other has to be determined using the given data.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  Nucleus + nuclear binding energy  nucleons

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

  E = Δmc2where,Δm = Mass Differencec= Speed of light

Change in mass of a given reaction can be determined as given,

  Δm = Mass of reactant Mass of products

(e)

Expert Solution
Check Mark

Explanation of Solution

Given data is shown below:

  Reaction: 311 23He + 1+10βNumber of H atoms = 1.45×107Mass  of 11H= 1.007825 amuMass  of 23He= 3.01603 amuMass of β+ = Mass of an electron = 5.48580×104 amu

  • Calculate the mass difference for the formation of Helium-4:

Mass difference of the reaction can be calculated as given,

  Δm = Mass of reactant Mass of products[3×Mass of 11H] [Mass of 23He + Mass of β+][3×1.007825 amu][3.01603+(5.48580×104 amu)]=0.006896 amu per 23He formed= 0.006896 g/mol 23He formed

Mass difference during the fusion of 3 Hydrogen atoms is 0.006896 g/mol.

  • Calculate total mass difference in kilogram:

Mass difference during the fusion of 1.45×107 atoms is calculated as follows,

Δm = (0.006896 amumol4He)(1 kg103 g)(1 3He3 H atoms)(1 mol H6.022×1023 H atoms)(1.45×107 H atoms)= 5.5348×1023 kg

  • Calculate energy released:

Energy released is calculated as given below,

  Energy = Δmc2 (5.5348×1023 kg)(2.998×108 ms1)= 4.97×106 J

Energy released during the fusion of 3 Hydrogen atoms forming Helium-3 is 4.97×10-6 J.

Energy released during the fusion of 4 Hydrogen atoms forming Helium-4 is 1.49×10-5 J. It is higher comparing to the energy produced during the formation of Helium-3.

Therefore, Chief Engineer should advise Captain not to change the technology and to keep the current technology.

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Chapter 24 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 24.2 - Prob. 24.6AFPCh. 24.2 - Prob. 24.6BFPCh. 24.6 - Prob. 24.7AFPCh. 24.6 - Prob. 24.7BFPCh. 24.7 - Prob. B24.1PCh. 24.7 - Prob. B24.2PCh. 24.7 - Prob. B24.3PCh. 24.7 - Prob. B24.4PCh. 24 - Prob. 24.1PCh. 24 - Prob. 24.2PCh. 24 - Prob. 24.3PCh. 24 - Prob. 24.4PCh. 24 - Prob. 24.5PCh. 24 - Prob. 24.6PCh. 24 - Prob. 24.7PCh. 24 - Prob. 24.8PCh. 24 - Prob. 24.9PCh. 24 - Prob. 24.10PCh. 24 - Prob. 24.11PCh. 24 - Prob. 24.12PCh. 24 - Prob. 24.13PCh. 24 - Prob. 24.14PCh. 24 - Prob. 24.15PCh. 24 - Prob. 24.16PCh. 24 - Prob. 24.17PCh. 24 - Prob. 24.18PCh. 24 - Prob. 24.19PCh. 24 - Prob. 24.20PCh. 24 - Prob. 24.21PCh. 24 - Prob. 24.22PCh. 24 - Prob. 24.23PCh. 24 - Prob. 24.24PCh. 24 - Prob. 24.25PCh. 24 - Prob. 24.26PCh. 24 - Prob. 24.27PCh. 24 - Prob. 24.28PCh. 24 - Prob. 24.29PCh. 24 - Prob. 24.30PCh. 24 - Prob. 24.31PCh. 24 - Prob. 24.32PCh. 24 - Prob. 24.33PCh. 24 - Prob. 24.34PCh. 24 - Prob. 24.35PCh. 24 - Prob. 24.36PCh. 24 - Prob. 24.37PCh. 24 - Prob. 24.38PCh. 24 - Prob. 24.39PCh. 24 - Prob. 24.40PCh. 24 - Prob. 24.41PCh. 24 - Prob. 24.42PCh. 24 - Prob. 24.43PCh. 24 - Prob. 24.44PCh. 24 - Prob. 24.45PCh. 24 - Prob. 24.46PCh. 24 - Prob. 24.47PCh. 24 - Prob. 24.48PCh. 24 - Prob. 24.49PCh. 24 - Prob. 24.50PCh. 24 - Prob. 24.51PCh. 24 - Prob. 24.52PCh. 24 - Prob. 24.53PCh. 24 - Prob. 24.54PCh. 24 - Prob. 24.55PCh. 24 - Prob. 24.56PCh. 24 - Prob. 24.57PCh. 24 - Prob. 24.58PCh. 24 - Prob. 24.59PCh. 24 - Prob. 24.60PCh. 24 - Prob. 24.61PCh. 24 - Prob. 24.62PCh. 24 - Prob. 24.63PCh. 24 - Prob. 24.64PCh. 24 - Prob. 24.65PCh. 24 - Prob. 24.66PCh. 24 - Prob. 24.67PCh. 24 - Prob. 24.68PCh. 24 - Prob. 24.69PCh. 24 - Prob. 24.70PCh. 24 - Prob. 24.71PCh. 24 - Prob. 24.72PCh. 24 - Prob. 24.73PCh. 24 - Prob. 24.74PCh. 24 - Prob. 24.75PCh. 24 - Prob. 24.76PCh. 24 - Prob. 24.77PCh. 24 - Prob. 24.78PCh. 24 - Prob. 24.79PCh. 24 - Prob. 24.80PCh. 24 - Prob. 24.81PCh. 24 - Prob. 24.82PCh. 24 - Prob. 24.83PCh. 24 - Prob. 24.84PCh. 24 - Prob. 24.85PCh. 24 - Prob. 24.86PCh. 24 - Prob. 24.87PCh. 24 - Prob. 24.88PCh. 24 - Prob. 24.89PCh. 24 - Prob. 24.90PCh. 24 - Prob. 24.91PCh. 24 - Prob. 24.92PCh. 24 - Prob. 24.93PCh. 24 - Prob. 24.94PCh. 24 - Prob. 24.95PCh. 24 - Prob. 24.96PCh. 24 - Prob. 24.97PCh. 24 - Prob. 24.98PCh. 24 - Prob. 24.99PCh. 24 - Prob. 24.100PCh. 24 - Prob. 24.101PCh. 24 - Prob. 24.102PCh. 24 - Prob. 24.103PCh. 24 - Prob. 24.104PCh. 24 - Prob. 24.105PCh. 24 - Prob. 24.106PCh. 24 - Prob. 24.107PCh. 24 - Prob. 24.108PCh. 24 - Prob. 24.109PCh. 24 - Prob. 24.110PCh. 24 - Prob. 24.111PCh. 24 - Prob. 24.112PCh. 24 - Prob. 24.113PCh. 24 - Prob. 24.114PCh. 24 - Prob. 24.115PCh. 24 - Prob. 24.116PCh. 24 - Prob. 24.117PCh. 24 - Prob. 24.118PCh. 24 - Prob. 24.119PCh. 24 - Prob. 24.120PCh. 24 - Prob. 24.121PCh. 24 - Prob. 24.122PCh. 24 - Prob. 24.123PCh. 24 - Prob. 24.124PCh. 24 - Prob. 24.125PCh. 24 - Prob. 24.126PCh. 24 - Prob. 24.127PCh. 24 - Prob. 24.128PCh. 24 - Prob. 24.129PCh. 24 - Prob. 24.130PCh. 24 - Prob. 24.131PCh. 24 - Prob. 24.132PCh. 24 - Prob. 24.133PCh. 24 - Prob. 24.134PCh. 24 - Prob. 24.135PCh. 24 - Prob. 24.136PCh. 24 - Prob. 24.137PCh. 24 - Prob. 24.138PCh. 24 - Prob. 24.139PCh. 24 - Prob. 24.140PCh. 24 - Prob. 24.141PCh. 24 - Prob. 24.142PCh. 24 - Prob. 24.143PCh. 24 - Prob. 24.144P
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