Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 21, Problem 21.20QA
Interpretation Introduction

To calculate:

Calculate the energy released in each of the given reactions.

Expert Solution & Answer
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Answer to Problem 21.20QA

Solution:

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

a. 2.65 ×10-12 J

b. 1.11 ×10-12 J

c. 6.88 ×10-12 J

d. 2.71 ×10-12 J

Explanation of Solution

1) Concept:

The reaction of formation of the sulfur isotope is 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 sulfur isotope and the given reactant isotopes. Due to this mass defect, the energy released 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=m)c2

3) Given:

i) Mass of Nitrogen =14.00307 amu

ii) Mass of Lithium =6.01512 amu

iii) Mass of  Oxygen =15.99491 amu

iv) Mass of Sulphur =31.97207 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. Calculating mass defect and corresponding energy lost during the first given reaction:

The first given reaction is

O 16+O  16S 32

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

m= mSulphur- 2 × mOxygen

m= 31.97207 amu- 2 ×15.99491 amu

m= 31.97207 amu- 31.98982 amu

 m= -0.01775 amu

The mass difference in kg:

-0.01775 amu × 1.66054 ×10-27 kg1 amu=-2.9474 ×10-29kg

Now, energy released during the reaction:

E=(m)c2

E= - 2.9474 ×10-29 kg × 2.998 ×108 m/s 2

 E= -2.65 ×10-12 kg·ms2= -2.65 ×10-12 J

Thus, energy released during the first given reaction is  2.65 ×10-12 J.

b. Calculating mass defect and corresponding energy lost during the second given reaction:

The given reaction is

Si 28+ He 4 S 32

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

m= mSulphur- mSilicon+ mHelium

m= 31.97207 amu-  27.97693 amu+4.00260  amu

m= 31.97207 amu- 31.97953 amu

 m= -0.00746 amu

The mass difference in kg:

-0.00746 amu ×1.66054 ×10-27 kg1 amu=-1.23876 ×10-29kg

Now, energy released during the reaction:

E=(m)c2

E= -1.23876 ×10-29kg × 2.998 ×108 m/s 2

 E= -1.11 ×10-12 kg·ms2= -1.11 ×10-12 J

Thus, energy released during the second reaction is 1.11 ×10-12 J.

c. Calculating mass defect and corresponding energy lost during the third given reaction:

The given reaction is

N 14+C 12+ Li 6 S 32

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

m= mSulphur- mNitrogen+ mcarbon+ mLithium

m= 31.97207 amu-   14.00307 amu+12.000 amu+ 6.01512 amu

m= 31.97207 amu- 32.01819 amu

 m= -0.04612 amu

The mass difference in kg:

-0.04612 amu× 1.66054 ×10-27 kg1 amu= -7.6584 ×10-29 kg

Now, energy released during the reaction:

E=(m)c2

E= -7.6584 ×10-29 kg × 2.998 ×108 m/s 2

 E= -6.88×10-12 kg·ms2= -6.88 ×10-12 J

Thus, energy released during the reaction is 6.88 ×10-12 J.

d. Calculating mass defect and corresponding energy lost during the fourth given reaction:

The given reaction is

Mg 24+2 He 4 S 32

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

m= mSulphur- mMagnesium+2× mHelium

m= 31.97207 amu- 23.98504 amu+2 ×4.00260 amu

m= 31.97207 amu- 31.99024 amu

 m= -0.01817 amu

The mass difference in kg:

-0.01817 amu × 1.66054 ×10-27 kg1 amu=-3.0172 ×10-29kg

Now, energy released during the reaction:

E=(m)c2

E= - 3.0172 ×10-29kg  × 2.998 ×108 m/s 2

 E= -2.71×10-12 kg·ms2= -2.71 ×10-12 J

Thus, energy released during the reaction is 2.71 ×10-12 J.

Conclusion:

The mass defect corresponds to the amount of energy released during the reaction.

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Chapter 21 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 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. 21.22QACh. 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. 21.99QA
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