Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
Question
Book Icon
Chapter 21, Problem 21.69QA
Interpretation Introduction

To calculate:

The energy released in each of the given reactions.

Expert Solution & Answer
Check Mark

Answer to Problem 21.69QA

Solution:

The energy released is for each of the given reactions is as follows:

a. 4.36 ×10-12 J

b. 6.79 ×10-12 J

c. 2.68 ×10-12 J

d. 1.59 ×10-12 J

Explanation of Solution

1) Concept:

The reactions of formation of silicon isotope are given, and we are asked to find the energy released during each of the given reactions in Joules. First we need to find the mass defect between the masses of silicon isotope and given reactant isotopes. Due to this mass defect, energy is released, which can be calculated using the Einstein’s equation  E=mc2. The masses of the particles are in kilograms, which is convenient because the units of energy and mass are related as 1J = 1kg (m/s)2 . Thus, using the conversion factor, we will convert the calculated mass defect from amu to kilogram and hence calculate the energy released.

2) Formulae:

i) m= mproducts- mreactants

ii) E=mc2

3) Given:

i) Mass of Nitrogen =14.00307 amu

ii) Mass of Boron =10.0129 amu

iii) Mass of  Oxygen =15.99491 amu

iv) Mass of Hydrogen =2.0146 amu

v) Mass of Carbon =12.000 amu

vi) Mass of Magnesium =23.98504 amu

vii) Mass of Helium =4.00260 amu

viii) Mass of Silicon =27.97693 amu

ix) Velocity of light c=2.998 ×108 m/s

x) 1 amu=1.66054 ×10-27 kg

4) Calculations:

a. For the reaction of formation of Si 28, the given equation is

N 14+N  14Si 28

First, we calculate the difference in the mass of the particles, that is, mass defect:

m= mSilicon- 2 × mNitrogen

m= 27.97693 amu- 2 × 14.00307 amu

m= 27.97693 amu- 28.00614 amu

 m= -0.02921 amu

The mass difference in kg:

-0.02921 amu × 1.66054 ×10-27kg1 amu=-4.850437 ×10-29kg

Now the energy released during the reaction is

E=(m)c2

E= - 4.85 ×10-29 kg × 2.998 ×108 m/s 2

 E= -4.36 ×10-12 kgms2= -4.36 ×10-12 J

Thus, the energy released during the first given reaction is 4.36 ×10-12J.

b. For the reaction of formation of Si 28, the given equation is

B+ 10O 16+ H 2 Si 28

First, we calculate the difference in the mass of the particles:

m= mSilicon- mBoron+ mOxygen+mHydrogen

m= 27.97693 amu- 10.0129 amu+ 15.99491 amu+ 2.0146 amu 

m= 27.97693 amu- 28.02241 amu

 m= -0.04548 amu

The mass difference in kg:

-0.04548  amu × 1.66054 ×10-27 kg1 amu=-7.552135 ×10-29kg

Now the energy released during the reaction:

E=(m)c2

E= - 7.5591 ×10-29 kg × 2.998 ×108 m/s 2

 E= -6.79 ×10-12 kgms2= -6.79 ×10-12 J

Thus, the energy released during the second given reaction is 6.79 ×10-12J.

c. The third given equation is

O 16+C 12 Si 28

First, we calculate the difference in the mass of the particles:

m= mSilicon- mOxygen+ mcarbon

m= 27.97693 amu-  15.99491 amu+12.000 amu

m= 27.97693 amu- 27.99491 amu

 m= -0.01798 amu

The mass difference in kg:

-0.01798 amu × 1.66054 ×10-27 kg1 amu=2.9856×10-29kg

Now the energy released during the reaction:

E=(m)c2

E= - 2.98 ×10-29 kg × 2.998 ×108 m/s 2

 E= -2.68 ×10-12 kg· m2/s2= -2.68 ×10-12 J

Thus, the energy released during the third given reaction is 2.68 ×10-12 J

d. The fourth equation given as:

Mg 24+He 4 Si 28

First, we calculate the difference in the mass of the particles:

m= mSilicon- mMagnesium+ mHelium

m= 27.97693 amu- 23.98504 amu+4.00260 amu

m= 27.97693 amu- 27.98764 amu

 m= -0.01071 amu

The mass difference in kg:

-0.01071 amu × 1.66054 ×10-27 kg1 amu=1.7784 ×10-29 kg

Now the energy released during the reaction:

E=(m)c2

E= - 1.78 ×10-29 kg × 2.998 ×108 m/s 2

 E= -1.60 ×10-12 kgms2= -1.60 ×10-12 J

Conclusion:

The lost mass during the reaction results in energy release that can be calculated using Einstein’s equation.

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!

Chapter 21 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

Ch. 21 - Prob. 21.11VPCh. 21 - Prob. 21.12VPCh. 21 - Prob. 21.13QACh. 21 - Prob. 21.14QACh. 21 - Prob. 21.15QACh. 21 - Prob. 21.16QACh. 21 - Prob. 21.17QACh. 21 - Prob. 21.18QACh. 21 - Prob. 21.19QACh. 21 - Prob. 21.20QACh. 21 - Prob. 21.21QACh. 21 - Prob. 21.22QACh. 21 - Prob. 21.23QACh. 21 - Prob. 21.24QACh. 21 - Prob. 21.25QACh. 21 - Prob. 21.26QACh. 21 - Prob. 21.27QACh. 21 - Prob. 21.28QACh. 21 - Prob. 21.29QACh. 21 - Prob. 21.30QACh. 21 - Prob. 21.31QACh. 21 - Prob. 21.32QACh. 21 - Prob. 21.33QACh. 21 - Prob. 21.34QACh. 21 - Prob. 21.35QACh. 21 - Prob. 21.36QACh. 21 - Prob. 21.37QACh. 21 - Prob. 21.38QACh. 21 - Prob. 21.39QACh. 21 - Prob. 21.40QACh. 21 - Prob. 21.41QACh. 21 - Prob. 21.42QACh. 21 - Prob. 21.43QACh. 21 - Prob. 21.44QACh. 21 - Prob. 21.45QACh. 21 - Prob. 21.46QACh. 21 - Prob. 21.47QACh. 21 - Prob. 21.48QACh. 21 - Prob. 21.49QACh. 21 - Prob. 21.50QACh. 21 - Prob. 21.51QACh. 21 - Prob. 21.52QACh. 21 - Prob. 21.53QACh. 21 - Prob. 21.54QACh. 21 - Prob. 21.55QACh. 21 - Prob. 21.56QACh. 21 - Prob. 21.57QACh. 21 - Prob. 21.58QACh. 21 - Prob. 21.59QACh. 21 - Prob. 21.60QACh. 21 - Prob. 21.61QACh. 21 - Prob. 21.62QACh. 21 - Prob. 21.63QACh. 21 - Prob. 21.64QACh. 21 - Prob. 21.65QACh. 21 - Prob. 21.66QACh. 21 - Prob. 21.67QACh. 21 - Prob. 21.68QACh. 21 - Prob. 21.69QACh. 21 - Prob. 21.70QACh. 21 - Prob. 21.71QACh. 21 - Prob. 21.72QACh. 21 - Prob. 21.73QACh. 21 - Prob. 21.74QACh. 21 - Prob. 21.75QACh. 21 - Prob. 21.76QACh. 21 - Prob. 21.77QACh. 21 - Prob. 21.78QACh. 21 - Prob. 21.79QACh. 21 - Prob. 21.80QACh. 21 - Prob. 21.81QACh. 21 - Prob. 21.82QACh. 21 - Prob. 21.83QACh. 21 - Prob. 21.84QACh. 21 - Prob. 21.85QACh. 21 - Prob. 21.86QACh. 21 - Prob. 21.87QACh. 21 - Prob. 21.88QACh. 21 - Prob. 21.89QACh. 21 - Prob. 21.90QACh. 21 - Prob. 21.91QACh. 21 - Prob. 21.92QACh. 21 - Prob. 21.93QACh. 21 - Prob. 21.94QACh. 21 - Prob. 21.95QACh. 21 - Prob. 21.96QACh. 21 - Prob. 21.97QACh. 21 - Prob. 21.98QACh. 21 - Prob. 21.99QACh. 21 - Prob. 21.100QACh. 21 - Prob. 21.101QACh. 21 - Prob. 21.102QACh. 21 - Prob. 21.103QACh. 21 - Prob. 21.104QACh. 21 - Prob. 21.105QACh. 21 - Prob. 21.106QACh. 21 - Prob. 21.107QACh. 21 - Prob. 21.108QACh. 21 - Prob. 21.109QACh. 21 - Prob. 21.110QACh. 21 - Prob. 21.111QACh. 21 - Prob. 21.112QACh. 21 - Prob. 21.113QACh. 21 - Prob. 21.114QACh. 21 - Prob. 21.115QACh. 21 - Prob. 21.116QACh. 21 - Prob. 21.117QACh. 21 - Prob. 21.118QACh. 21 - Prob. 21.119QACh. 21 - Prob. 21.120QACh. 21 - Prob. 21.121QACh. 21 - Prob. 21.122QACh. 21 - Prob. 21.123QACh. 21 - Prob. 21.124QACh. 21 - Prob. 21.125QACh. 21 - Prob. 21.126QACh. 21 - Prob. 21.127QACh. 21 - Prob. 21.128QACh. 21 - Prob. 21.129QACh. 21 - Prob. 21.130QA
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY