Chapter 6, Problem 41Q

(a)

Interpretation Introduction

Interpretation:

For the given reaction, the difference in mass between reactants and products has to be calculated.

Explanation of Solution

Given reaction and information,

$\begin{array}{l}{}_{\text{1}}^{\text{2}}\text{H}\text{+}{}_{\text{2}}^{\text{3}}\text{He}\stackrel{}{\to }\left[{}_{\text{3}}^{\text{5}}\text{Li}\right]\stackrel{}{\to }{}_{\text{2}}^{\text{4}}\text{He}\text{+}{}_{\text{1}}^{\text{1}}\text{H}\\ \text{2}\text{.01345}\text{g}\text{3}\text{.01493}\text{g}\text{4}\text{.00150}\text{g}\text{1}\text{.00728}\text{g}\end{array}$

Calculate the mass difference between the reactants and products,

$\begin{array}{l}{}_{\text{1}}^{\text{2}}\text{H}\text{+}{}_{\text{2}}^{\text{3}}\text{He}\stackrel{}{\to }\left[{}_{\text{3}}^{\text{5}}\text{Li}\right]\stackrel{}{\to }{}_{\text{2}}^{\text{4}}\text{He}\text{+}{}_{\text{1}}^{\text{1}}\text{H}\\ \text{2}\text{.01345}\text{g}\text{3}\text{.01493}\text{g}\text{4}\text{.00150}\text{g}\text{1}\text{.00728}\text{g}\\ \\ \text{total}\text{mass}\text{of}\text{reactants}\text{=}\text{5}\text{.02838}\text{g}\text{;}\text{total}\text{mass}\text{of}\text{reactants}\text{=}\text{5}\text{.00878}\text{g}\\ \text{mass}\text{difference}\text{=}\text{0}\text{.0196}\text{g}\end{array}$

Therefore, the mass difference is $\text{0}\text{.0196 g}$.

(b)

Interpretation Introduction

Interpretation:

Energy released from one of reactants has to be calculated.

Explanation of Solution

Given reaction and information,

$\begin{array}{l}{}_{\text{1}}^{\text{2}}\text{H}\text{+}{}_{\text{2}}^{\text{3}}\text{He}\stackrel{}{\to }\left[{}_{\text{3}}^{\text{5}}\text{Li}\right]\stackrel{}{\to }{}_{\text{2}}^{\text{4}}\text{He}\text{+}{}_{\text{1}}^{\text{1}}\text{H}\\ \text{2}\text{.01345}\text{g}\text{3}\text{.01493}\text{g}\text{4}\text{.00150}\text{g}\text{1}\text{.00728}\text{g}\end{array}$

Calculate the mass difference between the reactants and products,

$\begin{array}{l}{}_{\text{1}}^{\text{2}}\text{H}\text{+}{}_{\text{2}}^{\text{3}}\text{He}\stackrel{}{\to }\left[{}_{\text{3}}^{\text{5}}\text{Li}\right]\stackrel{}{\to }{}_{\text{2}}^{\text{4}}\text{He}\text{+}{}_{\text{1}}^{\text{1}}\text{H}\\ \text{2}\text{.01345}\text{g}\text{3}\text{.01493}\text{g}\text{4}\text{.00150}\text{g}\text{1}\text{.00728}\text{g}\\ \\ \text{total}\text{mass}\text{of}\text{reactants}\text{=}\text{5}\text{.02838}\text{g}\text{;}\text{total}\text{mass}\text{of}\text{reactants}\text{=}\text{5}\text{.00878}\text{g}\\ \text{mass}\text{difference}\text{=}\text{0}\text{.0196}\text{g}\end{array}$

Therefore, the mass difference is $\text{0}\text{.0196 g}$.

Einstein’s mass energy relationship is given by,

$\begin{array}{l}\text{E}\text{=}\text{m}{\text{c}}^{\text{2}}\\ \text{where,}\\ \text{E}\text{-}\text{energy}\\ \text{m}\text{-}\text{mass}\text{and}\text{c}\text{-}\text{speed}\text{of}\text{light}\end{array}$

Calculate the energy released by using the above equation,

$\begin{array}{l}\text{E}\text{=}\text{0}\text{.0196}\text{g}\text{×}\frac{\text{1}\text{kg}}{{\text{10}}^{\text{3}}\text{g}}\text{×}{\left(\frac{\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m}}{\text{s}}\right)}^{{}^{\text{2}}}\text{×}\left(\frac{\text{1}\text{J}}{\text{kg}{\text{.m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right)\\ \\ \text{E}\text{=}\frac{\text{0}\text{.0196}\text{×}\text{9}\text{×}{\text{10}}^{\text{16}}}{{\text{10}}^{\text{3}}}\text{J}\text{=}\text{0}\text{.1764}\text{×}{\text{10}}^{\text{13}}\text{J}\\ \text{E}\text{=}\text{1}\text{.76}\text{×}{\text{10}}^{\text{12}}\text{J}\end{array}$