Chapter 21, Problem 21.70QA

Interpretation Introduction

To calculate:

Calculate the energy released in each of the given reactions.

Answer to Problem 21.70QA

Solution:

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

a. $2.65 \times {10}^{-12} J$

b. $1.11 \times {10}^{-12} J$

c. $6.88 \times {10}^{-12} J$

d. $2.71 \times {10}^{-12} J$

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 $E=\left(∆m\right)c^2$. 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) $∆m= m_{products}- m_{reactants}$

ii) $E=∆m)c^2$

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 $c=2.998 \times {10}^8 m/s$

x) $1 amu=1.66054 \times {10}^{-27} kg$

4) Calculations:

a. Calculating mass defect and corresponding energy lost during the first given reaction:

The first given reaction is

$O+O \to S$

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

$∆m= \left(m_{Sulphur}\right)- \left(2 \times m_{Oxygen}\right)$

$∆m= \left(31.97207 amu\right)- \left(2 \times 15.99491 amu\right)$

$∆m= 31.97207 amu- \left(31.98982 amu\right)$

$\therefore ∆m= -0.01775 amu$

The mass difference in $kg$:

$-0.01775 amu \times \frac{1.66054 \times {10}^{-27} kg}{1 amu}=-2.9474 \times {10}^{-29}kg$

Now, energy released during the reaction:

$E=(∆m)c^2$

$E= \left(- 2.9474 \times {10}^{-29} kg\right) \times {\left(2.998 \times {10}^8 m/s \right)}^2$

$\therefore E= -2.65 \times {10}^{-12} kg·{\left(\frac{m}{s}\right)}^2= -2.65 \times {10}^{-12} J$

Thus, energy released during the first given reaction is $2.65 \times {10}^{-12} J$.

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

The given reaction is

$Si+ He \to S$

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

$∆m= \left(m_{Sulphur}\right)- \left(m_{Silicon}+ m_{Helium}\right)$

$∆m= \left(31.97207 amu\right)- \left( 27.97693 amu+4.00260 amu\right)$

$∆m= 31.97207 amu- \left(31.97953 amu\right)$

$\therefore ∆m= -0.00746 amu$

The mass difference in $kg$:

$-0.00746 amu \times \frac{1.66054 \times {10}^{-27} kg}{1 amu}=-1.23876 \times {10}^{-29}kg$

Now, energy released during the reaction:

$E=(∆m)c^2$

$E= \left(-1.23876 \times {10}^{-29}kg\right) \times {\left(2.998 \times {10}^8 m/s \right)}^2$

$\therefore E= -1.11 \times {10}^{-12} kg·{\left(\frac{m}{s}\right)}^2= -1.11 \times {10}^{-12} J$

Thus, energy released during the second reaction is $1.11 \times {10}^{-12} J$.

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

The given reaction is

$N+C+ Li \to S$

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

$∆m= \left(m_{Sulphur}\right)- \left(m_{Nitrogen}+ m_{carbon}+ m_{Lithium}\right)$

$∆m= \left(31.97207 amu\right)- \left( 14.00307 amu+12.000 amu+ 6.01512 amu\right)$

$∆m= 31.97207 amu- \left(32.01819 amu\right)$

$\therefore ∆m= -0.04612 amu$

The mass difference in $kg$:

$-0.04612 amu\times \frac{1.66054 \times {10}^{-27} kg}{1 amu}= -7.6584 \times {10}^{-29} kg$

Now, energy released during the reaction:

$E=(∆m)c^2$

$E= \left(-7.6584 \times {10}^{-29} kg\right) \times {\left(2.998 \times {10}^8 m/s \right)}^2$

$\therefore E= -6.88\times {10}^{-12} kg·{\left(\frac{m}{s}\right)}^2= -6.88 \times {10}^{-12} J$

Thus, energy released during the reaction is $6.88 \times {10}^{-12} J$.

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

The given reaction is

$Mg+2 He \to S$

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

$∆m= \left(m_{Sulphur}\right)- \left(m_{Magnesium}+2\times m_{Helium}\right)$

$∆m= \left(31.97207 amu\right)- \left(23.98504 amu+2 \times 4.00260 amu\right)$

$∆m= 31.97207 amu- \left(31.99024 amu\right)$

$\therefore ∆m= -0.01817 amu$

The mass difference in $kg$:

$-0.01817 amu \times \frac{1.66054 \times {10}^{-27} kg}{1 amu}=-3.0172 \times {10}^{-29}kg$

Now, energy released during the reaction:

$E=(∆m)c^2$

$E= \left(- 3.0172 \times {10}^{-29}kg \right) \times {\left(2.998 \times {10}^8 m/s \right)}^2$

$\therefore E= -2.71\times {10}^{-12} kg·{\left(\frac{m}{s}\right)}^2= -2.71 \times {10}^{-12} J$

Thus, energy released during the reaction is $2.71 \times {10}^{-12} J.$

Conclusion:

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