24.95 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Curium-243 undergoes   decay to plutonium-239:

(a) Find the change in mass, $\Delta m$  (in kg). (Masses: ${}^{243}Cm=$${\text{243.0614amu;}}^{\text{239}}{\text{Pu = 239.0522amu;}}^{\text{4}}\text{He = 4.0026amu}$  ; $1amu=1.661\times {10}^{-24} g$.) 

(b) Find the energy released in joules.

(c) Find the energy released in  $\text{kJ/mol}$ of reaction, and comment on the difference between this value and a typical heat of reaction for a chemical change, which is a few hundred $\text{kJ/mol}$ .

Step-by-Step Solution

Answer

1. The value of change in mass is ${\text{1.0963x10}}^{\text{ - 29}}$.
2. The value of energy released in joules is  $9.8530x{10}^{-13} J$ .
3. Energy released during the given reaction is  $593347660\frac{k.J}{ mol}$ .

Step 1: Definition of Concept

Nuclear reaction: A nuclear reaction occurs when a lighter nucleus fuses together to form new stable nuclei, or when a heavier nucleus splits into stable daughter nuclei, releasing a large amount of energy.

Step 2: Find the change in mass

(a)

Considering the given information:

 ${}_{\text{96}}{}^{\text{243}}\text{Cm}{\to }_{\text{94}}^{\text{239}}{\text{Pu + }}_{\text{2}}^{\text{4}}\text{He}$

The mass difference between reactants and products is  :

 $$\begin{gathered} \text{4m} \text{ = } {\text{m}}_{\text{products }}{\text{ - m}}_{\text{reactants }} \\ \text{4m} \text{ = } \text{[243.0614amu - (4.0026 + 239.0522)amu]}\left[\frac{{\text{1.661x10}}^{\text{ - 24}}}{\text{1amu}}\right]\left[\frac{\text{1kg}}{\text{1000g}}\right] \\ \text{ = } {\text{1.0963x10}}^{\text{ - 29}} \end{gathered}$$

Therefore, the required value is ${\text{1.0963x10}}^{\text{ - 29}}$  .

Step 3: Find the energy released in joules

Considering the given information:

 ${}_{\text{96}}{}^{\text{243}}\text{Cm}{\to }_{\text{94}}^{\text{239}}{\text{Pu + }}_{\text{2}}^{\text{4}}\text{He}$

Using Einstein's equation to calculate the energy release:

 $$\begin{gathered} E= 4m{c}^2 \\ = \left(1.0963x{10}^{-}29 kg\right){\left(2.99792x{10}^8\frac{m}{s}\right)}^2\left(\frac{1.J}{\frac{kg\times m^2}{ s^2}}\right) \\ = 9.8530x{10}^{-13} J \end{gathered}$$

Therefore, the required value is $9.8530x{10}^{-13} J$ .

Step 4: Find the energy released in kJ/mol of reaction

Considering the given information:

  

We must use the Avogadro's number to solve for the energy released in  .

 $\left(9.8530x{10}^{-13}J\right)\left(\frac{6.022x{10}^{23}}{mol}\right)\left(\frac{1k\times J}{1000J}\right)=593347660\frac{k.J}{ mol}$

This is higher than a typical heat reaction.