24.142 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Determine the age of a rock containing 0.065 g of uranium$\mathbf{-}\mathbf{238}\mathbf{(}{\mathit{t}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathit{y}\mathit{r}\mathbf{)}$ and 0.23 g of lead-206 . (Assume all the lead-206 came from U-238 decay.)

Step-by-Step Solution

Answer

The age of the rock is  $2.2271x{10}^{9}years$.

Step 1: Definition of Half-life

The half-life of a chemical reaction is the time it takes for a particular reactant's concentration to reach   of its initial concentration (i.e., the time taken for the reactant concentration to reach half of its initial value). It is commonly represented in seconds and is denoted by the sign ' t_1/2'.

Step 2: The amount of U-238

In this problem, we need to determine the amount of ²³⁸U first.

 $$\begin{gathered} Mass=0.023g^{206} Pb\left[\frac{1mo{l}^{206} Pb}{206g^{206} Pb}\right]\left[\frac{1mo{l}^{238}U}{1mo{l}^{206} Pb}\right]\left[\frac{238g^{238}U}{1mo{l}^{238}U}\right] \\ Mass=0.0266g^{238}U \end{gathered}$$

Original Amount of ²³⁸U :

 $(0.065g+0.0266g{)}^{238}U=0.0916g^{238}U$

We need to identify the constant (k) using the formula for half-life.

 $$\begin{gathered} t_{\frac{1}{2}}=\frac{\ln 2}{k} \\ k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ =\frac{\ln 2}{4.5x{10}^{9}yr} \\ =\frac{1.540327x{10}^{-10}}{yr} \end{gathered}$$

Step 3: Calculation for the determination of the age of lead.

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

$\ln \frac{N_0}{N_t}=kt$ 

 $\ln \frac{0.0916g}{0.065g}=\frac{1.540327\times {10}^{-10}\times t}{yr}$

 $t=2.2271x{10}^{9}years$