Chapter 20, Problem 20.54E

Interpretation Introduction

(a)

Interpretation:

The expression for time at which the amount of ${}^{\text{210}}\text{Po}$ is maximum is to be derived using the ${\left[\text{B}\right]}_t$ expression.

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 20.54E

The expression for time at which the amount of ${}^{\text{210}}\text{Po}$ is maximum is $\frac{\ln k_2/k_1}{\left(k_2-k_1\right)}$.

Explanation of Solution

The ${\left[\text{B}\right]}_t$ expression in equation 20.47 is given as follows:

${\left[\text{B}\right]}_t=\frac{k_1\cdot {\left[\text{A}\right]}_0}{k_2-k_1}\cdot \left(e^{-k_{1}t}-e^{-k_{2}t}\right)$

Where,

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

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

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

The derivation of the above expression with respect to time is taken as follows.

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

For the maximum amount of ${}^{\text{210}}\text{Po}$, the above expression is equated to zero as shown below.

$\begin{array}{rllll}\frac{d{\left[\text{B}\right]}_t}{dt}& =& \frac{k_1\cdot {\left[\text{A}\right]}_0}{k_2-k_1}\left(-k_1{e}^{-k_{1}t}+k_2{e}^{-k_{2}t}\right)& =& 0\\ -k_1{e}^{-k_{1}t}+k_2{e}^{-k_{2}t}& =& 0\\ k_1{e}^{-k_{1}t}& =& k_2{e}^{-k_{2}t}\end{array}$

The natural logarithm is taken on both sides as shown below.

$\begin{array}{rll}\ln \left(k_1{e}^{-k_{1}t}\right)& =& \ln \left(k_2{e}^{-k_{2}t}\right)\\ \ln k_1-k_{1}t& =& \ln k_2-k_{2}t\\ t\left(k_2-k_1\right)& =& \ln \frac{k_2}{k_1}\\ t& =& \frac{\ln \frac{k_2}{k_1}}{\left(k_2-k_1\right)}\end{array}$

Therefore, the expression for time at which the amount of ${}^{\text{210}}\text{Po}$ is maximum is $\frac{\ln k_2/k_1}{\left(k_2-k_1\right)}$.

Conclusion

The expression for time at which the amount of ${}^{\text{210}}\text{Po}$ is maximum is $\frac{\ln k_2/k_1}{\left(k_2-k_1\right)}$.

Interpretation Introduction

(b)

Interpretation:

The specific amounts of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ when the amount of ${}^{\text{210}}\text{Po}$ is at a maximum is to be determined using the value for time 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 20.54E

The value of time is $\text{2}\text{.15}\times {\text{10}}^6\text{s}$. The specific amounts of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ at $t=\text{2}\text{.15}\times {\text{10}}^6\text{s}$ are $0.02725\text{M}$, $0.7505\text{M}$ and $0.0728\text{M}$ 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 expression for time is given as follows:

$t=\frac{\ln k_2/k_1}{\left(k_2-k_1\right)}$

Substitute the value of $k_1$ and $k_2$ in above expression.

$\begin{array}{rll}t& =& \frac{\ln \left(5.79\times {10}^{-8}{\text{s}}^{-1}/1.60\times {10}^{-6}{\text{s}}^{-1}\right)}{\left[\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\right]}\\ & =& \frac{-3.3190}{-1.5421\times {10}^{-6}}\text{s}\\ & =& \text{2}\text{.15}\times {10}^6\text{s}\end{array}$

The value of $t$ at which the amount of ${}^{\text{210}}\text{Po}$ is maximum is $\text{2}\text{.15}\times {10}^6\text{s}$.

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 $t=\text{2}\text{.15}\times {10}^6\text{s}$ 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)

From the graph in example 20.7, the ${\left[{}^{\text{210}}\text{Bi}\right]}_0$ is taken as $0.85\text{M}$ at which the amount of ${}^{\text{210}}\text{Po}$ is at a maximum.

Substitute $t=\text{2}\text{.15}\times {10}^6\text{s}$, $k_1=1.60\times {10}^{-6}{\text{s}}^{-1}$ and ${\left[{}^{\text{210}}\text{Bi}\right]}_0=0.85\text{M}$ in first expression of equation (3).

$\begin{array}{rll}{\left[{}^{\text{210}}\text{Bi}\right]}_t& =& \left(\text{0}\text{.85}\text{M}\right)\cdot e^{-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\left(\text{2}\text{.15}\times {\text{10}}^6\text{s}\right)}\\ & =& 0.02725\text{M}\end{array}$

Substitute $t=\text{2}\text{.15}\times {10}^6\text{s}$, $k_1=1.60\times {10}^{-6}{\text{s}}^{-1}$, $k_2=5.79\times {10}^{-8}{\text{s}}^{-1}$ and ${\left[{}^{\text{210}}\text{Bi}\right]}_0=0.85\text{M}$ in second expression of equation (3).

$\begin{array}{rll}{\left[{}^{\text{210}}\text{Po}\right]}_t& =& \frac{\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\cdot \left(0.85\text{M}\right)}{\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)}\cdot \left(e^{-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\left(\text{2}\text{.15}\times {\text{10}}^6\text{s}\right)}-e^{-\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)\left(\text{2}\text{.15}\times {\text{10}}^6\text{s}\right)}\right)\\ & =& \left(-0.8819\text{M}\right)\left(0.032-0.883\right)\\ & =& \left(-0.8819\text{M}\right)\left(-0.851\right)\\ & =& 0.7505\text{M}\end{array}$

Substitute $t=\text{2}\text{.15}\times {10}^6\text{s}$, $k_1=1.60\times {10}^{-6}{\text{s}}^{-1}$, $k_2=5.79\times {10}^{-8}{\text{s}}^{-1}$ and ${\left[{}^{\text{210}}\text{Bi}\right]}_0=0.85\text{M}$ in third expression of equation (3).

$\begin{array}{rll}{\left[{}^{\text{206}}\text{Pb}\right]}_t& =& \left(\text{0}\text{.85}\text{M}\right)\left[\begin{array}{l}1+\frac{1}{\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)-\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)}\\ \left(\begin{array}{l}\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)\cdot e^{-\left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\left(\text{2}\text{.15}\times {\text{10}}^6\text{s}\right)}-\\ \left(1.60\times {10}^{-6}{\text{s}}^{-1}\right)\cdot e^{-\left(5.79\times {10}^{-8}{\text{s}}^{-1}\right)\left(\text{2}\text{.15}\times {\text{10}}^6\text{s}\right)}\end{array}\right)\end{array}\right]\\ & =& \left(\text{0}\text{.85}\text{M}\right)\left[1+\left(0.648\times {10}^6\text{s}\right)\left(0.1853\times {10}^{-8}{\text{s}}^{-1}-1.4128\times {10}^{-6}{\text{s}}^{-1}\right)\right]\\ & =& \left(\text{0}\text{.85}\text{M}\right)\left[1+\left(0.648\times {10}^6\text{s}\right)\left(-1.411\times {10}^{-6}{\text{s}}^{-1}\right)\right]\\ & =& 0.0728\text{M}\end{array}$

Therefore, the specific amounts of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ at $t=\text{2}\text{.15}\times {\text{10}}^6\text{s}$ are $0.02725\text{M}$, $0.7505\text{M}$ and $0.0728\text{M}$ respectively.

Conclusion

The value of time is $\text{2}\text{.15}\times {\text{10}}^6\text{s}$. The specific amounts of ${}^{\text{210}}\text{Bi}$, ${}^{\text{210}}\text{Po}$, and ${}^{\text{206}}\text{Pb}$ at $t=\text{2}\text{.15}\times {\text{10}}^6\text{s}$ are $0.02725\text{M}$, $0.7505\text{M}$ and $0.0728\text{M}$ respectively.