Chapter 24, Problem 24.93P

(a)

Interpretation Introduction

Interpretation:

Mass difference of the given reaction has to be calculated

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Change in mass of a given reaction can be determined as given,

  $\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}Given{\text{ reaction: }}_{96}^{243}Cm\underset{}{\to }{}_{94}^{239}Pu+{}_2^{4}He\\ \\ {\text{Mass of }}_{96}^{243}Cm=\text{ 243}\text{.0614 amu}\\ \\ {\text{Mass of }}_{94}^{239}Pu=\text{ 239}\text{.0522 amu}\\ \\ {\text{Mass of }}_2^{4}He=\text{ 4}\text{.0026 amu}\end{array}$

• Calculate the mass of products:

Mass of products is determined as shown below,

  $\begin{array}{c}Mass{\text{ of products = Mass of }}_2^{4}He{\text{ + Mass of }}_{94}^{239}Pu\\ \\ =\left(\text{4}\text{.0026 amu}\right)+\left(\text{239}\text{.0522 amu}\right)\\ \\ =\text{ 243}\text{.0548 amu}\end{array}$

• Calculate the mass difference:

Mass of reactants $\left({}_{96}^{243}Cm\right)$ is $\text{243}\text{.0614 amu}$

Mass difference of the reaction can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}\\ \\ \text{= }\left(\text{243}\text{.0614 amu}\right)-\left(\text{243}\text{.0548 amu}\right)\\ \\ =\text{ 0}\text{.0066 amu}\end{array}$

Mass difference is $\text{0}\text{.0066 amu}$.

• Convert unit of mass difference to $Kg$:

Mass difference for the reaction is converted from $amu\text{ to kg}$ as follows,

  $\begin{array}{c}\Delta m\text{ = }\left(\text{0}\text{.0066 amu}\right)\left(\frac{\text{1}\text{.661}\times {\text{10}}^{-24}\text{ g}}{1\text{ amu}}\right)\left(\frac{1\text{ kg}}{{10}^{-3}\text{ kg}}\right)\\ \\ =\text{ 1}\text{.1}\times {\text{10}}^{-29}\text{ kg}\end{array}$

Mass difference for the reaction is $\text{1}\text{.1}\times {\text{10}}^{-29}\text{ kg}$.

(b)

Interpretation Introduction

Interpretation:

Energy released (in $J$)during the given reaction has to be determined.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

It can be calculated using the given formula,

  $\begin{array}{l}\Delta \text{E = }\Delta {\text{mc}}^2\\ \\ where,\\ \Delta \text{m = Mass Difference}\\ c=\text{ Speed of light}\end{array}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}Given{\text{ reaction: }}_{96}^{243}Cm\underset{}{\to }{}_{94}^{239}Pu+{}_2^{4}He\\ \\ {\text{Mass of }}_{96}^{243}Cm=\text{ 243}\text{.0614 amu}\\ \\ {\text{Mass of }}_{94}^{239}Pu=\text{ 239}\text{.0522 amu}\\ \\ {\text{Mass of }}_2^{4}He=\text{ 4}\text{.0026 amu}\end{array}$

• Calculate the mass difference:

Mass of reactants $\left({}_{96}^{243}Cm\right)$ is $\text{243}\text{.0614 amu}$

Mass difference of the reaction can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}\\ \\ \text{= }\left(\text{243}\text{.0614 amu}\right)-\left(\text{243}\text{.0548 amu}\right)\\ \\ =\left(\text{0}\text{.0066 amu}\right)\left(\frac{\text{1}\text{.661}\times {\text{10}}^{-24}\text{ g}}{1\text{ amu}}\right)\left(\frac{1\text{ kg}}{{10}^{-3}\text{ kg}}\right)\\ \\ =\text{ 1}\text{.1}\times {\text{10}}^{-29}\text{ kg}\end{array}$

Mass difference for the reaction is $\text{1}\text{.1}\times {\text{10}}^{-29}\text{ kg}$.

• Calculate the energy released:

Energy released is calculated as follows for the reaction,

  $\begin{array}{c}Energyreleased{\text{, E = mc}}^2\\ \\ =\left(\text{1}\text{.1}\times {\text{10}}^{-29}\text{ kg}\right){\left(2.998\times {10}^8\text{ m/s}\right)}^2\\ \\ =9.8527\times {10}^{-13}\text{ J}\end{array}$

Energy released during the given reaction is $9.8527\times 10^{-13}J.$

(c)

Interpretation Introduction

Interpretation:

Energy released (in $\text{kJ/mol}$)during the given reaction has to be determined and difference between this energy values of heat of energy in a chemical reaction has to be commented.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

It can be calculated using the given formula,

  $\begin{array}{l}\Delta \text{E = }\Delta {\text{mc}}^2\\ \\ where,\\ \Delta \text{m = Mass Difference}\\ c=\text{ Speed of light}\end{array}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}Given{\text{ reaction: }}_{96}^{243}Cm\underset{}{\to }{}_{94}^{239}Pu+{}_2^{4}He\\ \\ {\text{Mass of }}_{96}^{243}Cm=\text{ 243}\text{.0614 amu}\\ \\ {\text{Mass of }}_{94}^{239}Pu=\text{ 239}\text{.0522 amu}\\ \\ {\text{Mass of }}_2^{4}He=\text{ 4}\text{.0026 amu}\end{array}$

• Calculate the mass difference:

Mass of reactants $\left({}_{96}^{243}Cm\right)$ is $\text{243}\text{.0614 amu}$

Mass difference of the reaction can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}\\ \\ \text{= }\left(\text{243}\text{.0614 amu}\right)-\left(\text{243}\text{.0548 amu}\right)\\ \\ =\left(\text{0}\text{.0066 amu}\right)\left(\frac{\text{1}\text{.661}\times {\text{10}}^{-24}\text{ g}}{1\text{ amu}}\right)\left(\frac{1\text{ kg}}{{10}^{-3}\text{ kg}}\right)\\ \\ =\text{ 1}\text{.1}\times {\text{10}}^{-29}\text{ kg}\end{array}$

Mass difference for the reaction is $\text{1}\text{.1}\times {\text{10}}^{-29}\text{ kg}$.

• Calculate the energy released:

Energy released for the reaction is calculated as follows,

$\begin{array}{c}Energyreleased{\text{, E = mc}}^2\\ \\ =\left(\text{1}\text{.1}\times {\text{10}}^{-29}\text{ kg}\right){\left(2.998\times {10}^8\text{ m/s}\right)}^2\\ \\ =9.8527\times {10}^{-13}\text{ J}\end{array}$

Energy released during the given reaction is $\text{9}{\text{.8527×10}}^{-\text{13}}\text{ J}\text{.}$

• Convert the unit of energy released to $\text{kJ/mol}$ :

Unit of energy released is converted as follows,

$\begin{array}{c}\text{Binding energy, E in kJ/mol = }\left(\frac{9.8527\times {10}^{-13}\text{ J}}{\text{reactions}}\right)\left(\frac{6.022\times {10}^{23}\text{ reactions}}{1Emol}\right)\left(\frac{1\text{ kJ}}{{10}^3\text{ J}}\right)\\ \\ =\text{ 5}\text{.9}\times {\text{10}}^8\text{ kJ/mol}\end{array}$

Energy released during the given reaction is $5.9\times 10^{8}kJ/mol$.

Energy released during the given reaction is $\text{5}\text{.9}\times {\text{10}}^8\text{ kJ/mol}$ whereas the energy values of heat of energy in a chemical reaction are in few hundreds $\text{kJ/mol}$.

This value is about one million times more than a typical heat of energy in a chemical reaction.