Q24.77P - Chemistry: Molecular Nature Of Matter And Change

Question

How many joules are released when $\mathbf{1}\mathbf{.}\mathbf{5}$ mol of  $\mathbf{239}$ Pu decays, if each nucleus releases $\mathbf{5}\mathbf{.}\mathbf{243}$ MeV?

Step-by-Step Solution

Answer

The joules released are ${\text{7.587×10}}^{\text{11}}\text{ J}$.

Step 1: To find how many joules are released

-${\text{1.5 mol of }}^{\text{239}}\text{Pu}$ decays

- each nucleus releases $\text{5.243 MeV}$

First, we need to calculate the number of nuclei in $1.5 \text{mol of }{}^{\text{239}}\text{Pu}$

$\text{Number of nuclei}=1.50 \text{mol}\times \frac{{\displaystyle {\text{6.022×10}}^{\text{23}}\text{ nuclei}}}{1 \text{mol}}=\text{9.033×}{10}^{\text{23}}\text{ nuclei}$

Total energy released in $\text{MeV}$ is

$\text{Total energy}={\text{9.033×10}}^{\text{23}}\text{ nuclei×5.243 }\frac{\text{MeV}}{\text{ nuclei }}={\text{4.736×10}}^{\text{24}}\text{ MeV}$

-${\text{1 MeV = 1.602×10}}^{\text{ - 13}}\text{ J}$

Therefore, the number of joules released is

${\text{4.736×10}}^{\text{24}}\text{ MeV}={\text{4.736×10}}^{\text{24}}\text{ MeV×}\frac{{\displaystyle {\text{1.602×10}}^{\text{ - 13}}\text{ J}}}{{\displaystyle \text{1 MeV}}}={\text{7.587×10}}^{\text{11}}\text{ J}$

Hence the joules released are ${\text{7.587×10}}^{\text{11}}\text{ J}$