Chapter 19, Problem 19.121QP

(a)

Interpretation Introduction

Interpretation:

The first order rate constant for the nuclear decay has to be calculated.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.
• The formula which is used to calculate the half-life has is $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{k}$

Answer to Problem 19.121QP

The first order rate constant for the nuclear decay is $k=0.0247{\text{yr}}^{\text{-1}}$

Explanation of Solution

The rate constant for the given first order nuclear decay is calculated from half-life formula by substituting the given values as follows

$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{k}$

$k=\frac{0.693}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{0.693}{28.1\text{yr}}=0.0247{\text{yr}}^{\text{-1}}$

$k=0.0247{\text{yr}}^{\text{-1}}$

Conclusion

The first order rate constant for the nuclear decay was calculated as $k=0.0247{\text{yr}}^{\text{-1}}$

(b)

Interpretation Introduction

Interpretation:

The fraction of ${}^{\text{90}}\text{Sr}$ that remains after 10 half-lives has to be calculated.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.
• The formula which is used to calculate the half-life has is $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{k}$

Answer to Problem 19.121QP

The fraction of ${}^{\text{90}}\text{Sr}$ that remains after 10 half-lives is $9.8\times {10}^{-4}$

Explanation of Solution

The fraction of ${}^{\text{90}}\text{Sr}$ that remains after 10 half-lives is mathematically calculated as follows

The number of half-lives is n=10

${\left(\frac{1}{2}\right)}^{\text{n}}={\left(\frac{1}{2}\right)}^{10}=9.8\times {10}^{-4}$

Conclusion

The fraction of ${}^{\text{90}}\text{Sr}$ that remains after 10 half-lives was calculated as $9.8\times {10}^{-4}$

(c)

Interpretation Introduction

Interpretation:

The number of years required for 99.0% of ${}^{\text{90}}\text{Sr}$ to disappear has to be calculated.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.
• The formula which is used to calculate the half-life has is $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{k}$

Answer to Problem 19.121QP

The number of years required for 99.0% of ${}^{\text{90}}\text{Sr}$ to disappear is $t=187\text{yr}$

Explanation of Solution

The number of years required for 99.0% of ${}^{\text{90}}\text{Sr}$ to disappear is calculated as follows.

If 99.0% of ${}^{\text{90}}\text{Sr}$ have disappeared then we have remains is 1.0%.  So, ratio of $\raisebox{1ex}{$\left[\text{A}\right]$}\!\left/ \!\raisebox{-1ex}{${\left[\text{A}\right]}_0$}\right.$ is $\raisebox{1ex}{$1.0\%$}\!\left/ \!\raisebox{-1ex}{$100\%$}\right.\text{or}\raisebox{1ex}{$0.010$}\!\left/ \!\raisebox{-1ex}{$1.00$}\right.$

Now, substitute these values in first order rate law, then doing some mathematical calculation we can get the time.

$\ln \frac{{\left[\text{A}\right]}_t}{{\left[\text{A}\right]}_0}=-kt$

$\ln \frac{0.010}{1.0}=-(0.0247{\text{yr}}^{\text{-1}})t$

$-4.61=-(0.0247{\text{yr}}^{\text{-1}})t$

$t=187\text{yr}$

Conclusion

The number of years required for 99.0% of ${}^{\text{90}}\text{Sr}$ to disappear was calculated as $t=187\text{yr}$