Chapter 20, Problem 20.56E

Interpretation Introduction

Interpretation:

The concentration of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ at $t=\text{1}\text{.00}\times {\text{10}}^6\text{s}$ are to be determined using the half-lives from the reaction ${}_{83}^{\text{210}}\text{Bi}\underset{k_1}{\overset{t_{1/2,1}}{\to }}{}_{84}^{\text{210}}\text{Po}\underset{k_2}{\overset{t_{1/2,2}}{\to }}{}_{82}^{\text{206}}\text{Pb}$ and equations 20.47.

Concept introduction:

The rate of a reaction is defined as the speed by which the reaction is proceeding. The rate of reaction depends on several factors such as the concentration of reactant and temperature.

Answer to Problem 20.56E

The concentration of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ at $t=\text{1}\text{.00}\times {\text{10}}^6\text{s}$ are $0.201897\text{M}$, $0.76970\text{M}$ and $0.02994\text{M}$ respectively.

Explanation of Solution

The reaction in Example 20.7 is shown below.

${}_{83}^{\text{210}}\text{Bi}\underset{k_1}{\overset{t_{1/2,1}}{\to }}{}_{84}^{\text{210}}\text{Po}\underset{k_2}{\overset{t_{1/2,2}}{\to }}{}_{82}^{\text{206}}\text{Pb}$

The half-lives $t_{1/2,1}$ and $t_{1/2,2}$ are $5.01$ days and $138.4$ days, respectively.

Conversion of half-lives from days to seconds is shown below.

$\text{1}\text{day}=\text{60}\times \text{60}\times \text{24}\text{seconds}$…(1)

Substitute $t_{1/2,1}$ in equation (1).

$\begin{array}{rll}\text{5}\text{.01}\text{days}& =& \text{5}\text{.01}\times \text{60}\times \text{60}\times \text{24}\text{seconds}\\ & =& \text{4}\text{.33}\times {10}^5\text{seconds}\end{array}$

Substitute $t_{1/2,2}$ in equation (1).

$\begin{array}{rll}138.4\text{days}& =& 138.4\times \text{60}\times \text{60}\times \text{24}\text{seconds}\\ & =& \text{1}\text{.196}\times {10}^7\text{seconds}\end{array}$

The half-lives, $t_{1/2,1}$ and $t_{1/2,2}$ are $\text{4}\text{.33}\times {10}^5\text{s}$ and $\text{1}\text{.196}\times {10}^7\text{s}$, respectively.

The rate constant for first order reaction is given by the formula shown below.

$k=\frac{0.693}{t_{1/2}}$…(2)

Where,

• $t_{1/2}$ is the half-life.

Substitute $t_{1/2,1}$ in equation (2).

$\begin{array}{rll}{k}_1& =& \frac{0.693}{\text{4}\text{.33}\times {10}^5\text{s}}\\ & =& 1.60\times {10}^{-6}{\text{s}}^{-1}\end{array}$

Substitute $t_{1/2,2}$ in equation (2).

$\begin{array}{rll}{k}_2& =& \frac{0.693}{\text{1}\text{.196}\times {10}^7\text{s}}\\ & =& 5.79\times {10}^{-8}{\text{s}}^{-1}\end{array}$

The value of $k_1$ and $k_2$ are $1.60\times {10}^{-6}{\text{s}}^{-1}$ and $5.79\times {10}^{-8}{\text{s}}^{-1}$ respectively.

The Equation 20.47 is given as follows:

$\begin{array}{rll}{\left[\text{A}\right]}_t& =& {\left[\text{A}\right]}_0\cdot e^{-k_{1}t}\\ {\left[\text{B}\right]}_t& =& \frac{k_1\cdot {\left[\text{A}\right]}_0}{k_2-k_1}\cdot \left(e^{-k_{1}t}-e^{-k_{2}t}\right)\\ {\left[\text{C}\right]}_t& =& {\left[\text{A}\right]}_0\left[1+\frac{1}{k_1-k_2}\left(k_2\cdot e^{-k_{1}t}-k_1\cdot e^{-k_{2}t}\right)\right]\end{array}$

Where,

• $k_1$ and $k_2$ are the equilibrium constant.

• ${\left[\text{A}\right]}_t$ is the concentration of $\text{A}$ at time $t$.

• ${\left[\text{B}\right]}_t$ is the concentration of $\text{B}$ at time $t$.

• ${\left[\text{C}\right]}_t$ is the concentration of $\text{C}$ at time $t$.

• ${\left[\text{A}\right]}_0$ is the initial concentration of $\text{A}$.

The above expression in terms of concentration of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ at $t=\text{1}\text{.00}\times {\text{10}}^6\text{s}$ is written as follows.

$\begin{array}{rll}{\left[{}^{\text{210}}\text{Bi}\right]}_t& =& {\left[{}^{\text{210}}\text{Bi}\right]}_0\cdot e^{-k_{1}t}\\ {\left[{}^{\text{210}}\text{Po}\right]}_t& =& \frac{k_1\cdot {\left[{}^{\text{210}}\text{Bi}\right]}_0}{k_2-k_1}\cdot \left(e^{-k_{1}t}-e^{-k_{2}t}\right)\\ {\left[{}^{\text{206}}\text{Pb}\right]}_t& =& {\left[{}^{\text{210}}\text{Bi}\right]}_0\left[1+\frac{1}{k_1-k_2}\left(k_2\cdot e^{-k_{1}t}-k_1\cdot e^{-k_{2}t}\right)\right]\end{array}$…(3)

Substitute $t=\text{1}\text{.00}\times {\text{10}}^6\text{s}$, $k_1=1.60\times {10}^{-6}{\text{s}}^{-1}$ and ${\left[{}^{\text{210}}\text{Bi}\right]}_0=\text{1}\text{.000}\text{M}$ in the first expression of equation (3).

$\begin{array}{rll}{\left[{}^{\text{210}}\text{Bi}\right]}_t& =& \left(\text{1}\text{.000}\text{M}\right)\cdot e^{-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\left(\text{1}\text{.00}\times {\text{10}}^6\text{s}\right)}\\ & =& 0.201897\text{M}\end{array}$

Substitute $t=\text{1}\text{.00}\times {\text{10}}^6\text{s}$, $k_1=1.60\times {10}^{-6}{\text{s}}^{-1}$, $k_2=5.79\times {10}^{-8}{\text{s}}^{-1}$ and ${\left[{}^{\text{210}}\text{Bi}\right]}_0=\text{1}\text{.000}\text{M}$ in second expression of equation (3).

$\begin{array}{rll}{\left[{}^{\text{210}}\text{Po}\right]}_t& =& \frac{\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\cdot \left(1.000\text{M}\right)}{\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)}\cdot \left(e^{-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\left(\text{1}\text{.00}\times {\text{10}}^6\text{s}\right)}-e^{-\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)\left(\text{1}\text{.00}\times {\text{10}}^6\text{s}\right)}\right)\\ & =& \left(-1.03755\text{M}\right)\left(0.201897-0.943744\right)\\ & =& \left(-1.03755\text{M}\right)\left(-0.74185\right)\\ & =& 0.76970\text{M}\end{array}$

Substitute $t=\text{1}\text{.00}\times {\text{10}}^6\text{s}$, $k_1=1.60\times {10}^{-6}{\text{s}}^{-1}$, $k_2=5.79\times {10}^{-8}{\text{s}}^{-1}$ and ${\left[{}^{\text{210}}\text{Bi}\right]}_0=\text{1}\text{.000}\text{M}$ in third expression of equation (3).

$\begin{array}{rll}{\left[{}^{\text{206}}\text{Pb}\right]}_t& =& \left(\text{1}\text{.000}\text{M}\right)\left[\begin{array}{l}1+\frac{1}{\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)-\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)}\\ \left(\begin{array}{l}\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)\cdot e^{-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\left(\text{1}\text{.00}\times {\text{10}}^6\text{s}\right)}-\\ \left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\cdot e^{-\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)\left(\text{1}\text{.00}\times {\text{10}}^6\text{s}\right)}\end{array}\right)\end{array}\right]\\ & =& \left(\text{1}\text{.000}\text{M}\right)\left[1+\left(0.648\times {10}^6\text{s}\right)\left(1.169\times {10}^{-8}{\text{s}}^{-1}-1.509\times {10}^{-6}{\text{s}}^{-1}\right)\right]\\ & =& \left(\text{1}\text{.000}\text{M}\right)\left[1+\left(0.648\times {10}^6\text{s}\right)\left(-1.497\times {10}^{-6}{\text{s}}^{-1}\right)\right]\\ & =& 0.02994\text{M}\end{array}$

Therefore, the concentration of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ at $t=\text{1}\text{.00}\times {\text{10}}^6\text{s}$ are $0.201897\text{M}$, $0.76970\text{M}$ and $0.02994\text{M}$ respectively.

Conclusion

The concentration of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ at $t=\text{1}\text{.00}\times {\text{10}}^6\text{s}$ are $0.201897\text{M}$, $0.76970\text{M}$ and $0.02994\text{M}$ respectively.