Chapter 21, Problem 21.69QA

Interpretation Introduction

To calculate:

The energy released in each of the given reactions.

Answer to Problem 21.69QA

Solution:

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

a. $4.36 \times {10}^{-12} J$

b. $6.79 \times {10}^{-12} J$

c. $2.68 \times {10}^{-12} J$

d. $1.59 \times {10}^{-12} J$

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 $E=\left(∆m\right)c^2$. 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 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 $c=2.998 \times {10}^8 m/s$

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

4) Calculations:

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

$N+N \to Si$

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

$∆m= \left(m_{Silicon}\right)- \left(2 \times m_{Nitrogen}\right)$

$∆m= \left(27.97693 amu\right)- \left(2 \times 14.00307 amu\right)$

$∆m= 27.97693 amu- \left(28.00614 amu\right)$

$\therefore ∆m= -0.02921 amu$

The mass difference in $kg$:

$-0.02921 amu \times \frac{1.66054 \times {10}^{-27}kg}{1 amu}=-4.850437 \times {10}^{-29}kg$

Now the energy released during the reaction is

$E=(∆m)c^2$

$E= \left(- 4.85 \times {10}^{-29} kg\right) \times {\left(2.998 \times {10}^8 m/s \right)}^2$

$\therefore E= -4.36 \times {10}^{-12} kg{\left(\frac{m}{s}\right)}^2= -4.36 \times {10}^{-12} J$

Thus, the energy released during the first given reaction is $4.36 \times {10}^{-12}J$.

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

$B+O+ H \to Si$

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

$∆m= \left(m_{Silicon}\right)- \left(m_{Boron}+ m_{Oxygen}+m_{Hydrogen}\right)$

$∆m= \left(27.97693 amu\right)- \left(10.0129 amu+ 15.99491 amu+ 2.0146 amu \right)$

$∆m= 27.97693 amu- \left(28.02241 amu\right)$

$\therefore ∆m= -0.04548 amu$

The mass difference in $kg$:

$-0.04548 amu \times \frac{1.66054 \times {10}^{-27} kg}{1 amu}=-7.552135 \times {10}^{-29}kg$

Now the energy released during the reaction:

$E=(∆m)c^2$

$E= \left(- 7.5591 \times {10}^{-29} kg\right) \times {\left(2.998 \times {10}^8 m/s \right)}^2$

$\therefore E= -6.79 \times {10}^{-12} kg{\left(\frac{m}{s}\right)}^2= -6.79 \times {10}^{-12} J$

Thus, the energy released during the second given reaction is $6.79 \times {10}^{-12}J$.

c. The third given equation is

$O+C \to Si$

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

$∆m= \left(m_{Silicon}\right)- \left(m_{Oxygen}+ m_{carbon}\right)$

$∆m= \left(27.97693 amu\right)- \left( 15.99491 amu+12.000 amu\right)$

$∆m= 27.97693 amu- \left(27.99491 amu\right)$

$\therefore ∆m= -0.01798 amu$

The mass difference in $kg$:

$-0.01798 amu \times \frac{1.66054 \times {10}^{-27} kg}{1 amu}=2.9856\times {10}^{-29}kg$

Now the energy released during the reaction:

$E=(∆m)c^2$

$E= \left(- 2.98 \times {10}^{-29} kg\right) \times {\left(2.998 \times {10}^8 m/s \right)}^2$

$\therefore E= -2.68 \times {10}^{-12} kg· m^2/s^2= -2.68 \times {10}^{-12} J$

Thus, the energy released during the third given reaction is $2.68 \times {10}^{-12} J$

d. The fourth equation given as:

$Mg+He \to Si$

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

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

$∆m= \left(27.97693 amu\right)- \left(23.98504 amu+4.00260 amu\right)$

$∆m= 27.97693 amu- \left(27.98764 amu\right)$

$\therefore ∆m= -0.01071 amu$

The mass difference in $kg$:

$-0.01071 amu \times \frac{1.66054 \times {10}^{-27} kg}{1 amu}=1.7784 \times {10}^{-29} kg$

Now the energy released during the reaction:

$E=(∆m)c^2$

$E= \left(- 1.78 \times {10}^{-29} kg\right) \times {\left(2.998 \times {10}^8 m/s \right)}^2$

$\therefore E= -1.60 \times {10}^{-12} kg{\left(\frac{m}{s}\right)}^2= -1.60 \times {10}^{-12} J$

Conclusion:

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