24.145 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Some nuclear power plants use ²³⁹Pu , which is produced in breeder reactors. The rate-determining step is the second emission. How long does it take to make 1.00 kg of ²³⁹Pu  if the reaction is complete when the product is  90% ²³⁹Pu ?

Step-by-Step Solution

Answer

The time taken to complete 90% of the reaction is 7.80 days . 

Step 1: Nuclear Chemistry

Nuclear chemistry is the study of the chemical and physical properties of elements that are involved in nuclear processes or reactions that occur within the nucleus' structure. From the research of element production in space to the development of radiopharmaceuticals for diagnostic medicine, modern nuclear chemistry, also known as radiochemistry, has become increasingly interdisciplinary in its applications.

Step 2: Calculations for the determination of the time.

To calculate the half-life t_1/2, we set N_t equal to  $\frac{1}{2}{N}_0$ and solve for t_1/2 .

 $\ln \frac{N_0}{\frac{1}{2}{N}_0}=k{t}_{\frac{1}{2}}$

Rearranging the derived formula,

 $t_{\frac{1}{2}}=\frac{\ln 2}{k}$

The expression for finding the number of nuclei remaining, N_t , after a given time t is: $N_t=N_0{e}^{-kt}$ or $\ln \frac{N_0}{N_t}=kt$

 

Step 3: Calculations for the determination of the time.

In this problem, we need to identify the constant (k) using the formula for half-life.

 $$\begin{gathered} t_{\frac{1}{2}}=\frac{\ln 2}{k} \\ k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ =\frac{\ln 2}{2.35d} \\ =\frac{0.294956}{d} \end{gathered}$$

Using the expression for finding the number of nuclei remaining, we can solve for the value of  t.

 $$\begin{gathered} \ln \frac{N_0}{N_t}=kt \\ \ln \left[\frac{1kg}{\frac{1kg}{\left(\frac{0.1}{1}\right)}}\right]=\frac{0.294956\times t}{days} \\ t=7.80 days \end{gathered}$$