Chapter 19, Problem 19.126QP

Interpretation Introduction

Interpretation:

The rate constant and half-life for the given radium decay and the activity after 500yr have to be determined.

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

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Answer to Problem 19.126QP

The rate constant and half-life for the given radium decay and the activity after 500yr is determined.

The rate constant of the given reaction is $\text{k =1}{\text{.4×10}}^{\text{-11}}\text{/s}$

The half-life of the given reaction is $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=5.0\times {10}^{10}\text{s}$

The activity after 500yr is $\text{activity}=3.1\times {10}^{10}\text{disintigarastions/s}$

Explanation of Solution

First of all, calculate the number of radium nuclei in 1.0g

$\text{1}\text{.0g×}\frac{\text{1}\text{mol}\text{Ra}}{\text{226}\text{.03}\text{g}\text{Ra}}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{Ra}\text{nuclei}}{\text{1 mol Ra}}=2.7\times {10}^{21}\text{Ra}\text{nuclei}$

From the given information and the above value we can easily calculate the rate constant and then from the rate constant we can calculate the half-life of the given reaction as follows.

The rate constant is:

$\text{activity = kN}$

$\text{k =}\frac{\text{ activity}}{\text{N}}\text{=}\frac{\text{3}{\text{.70×10}}^{\text{10}}\text{nuclear}\text{disitegrations/s}}{\text{2}{\text{.7×10}}^{\text{21}}\text{Ra}\text{nuclei}}\text{=1}{\text{.4×10}}^{\text{-11}}\text{/s}$

$\text{k =1}{\text{.4×10}}^{\text{-11}}\text{/s}$

The half-life is:

$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{k}=\frac{0.693}{\text{1}{\text{.4×10}}^{\text{-11}}\text{/s}}=5.0\times {10}^{10}\text{s}$

$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=5.0\times {10}^{10}\text{s}$

The activity after 500yr is:

Let us convert 500 years into seconds.  Then we can calculate the number of nuclei remaining after 500 years.

$\text{500yr×}\frac{\text{365days}}{\text{1yr}}\text{×}\frac{\text{24h}}{\text{1day}}\text{×}\frac{\text{3600s}}{\text{1h}}\text{=1}{\text{.58×10}}^{\text{10}}\text{s}$

Now, use the first order rate law for the calculation of the number of nuclei remaining after 500 years.

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{ = -kt}$

$\text{ln}\left(\frac{{\text{N}}_{\text{t}}}{\text{2}{\text{.7×10}}^{\text{21}}}\right)\text{= -}\left(\text{1}{\text{.4×10}}^{\text{-1}}\text{/s}\right)\left(\text{1}{\text{.58×10}}^{\text{10}}\text{s}\right)$

$\frac{{\text{N}}_{\text{t}}}{\text{2}{\text{.7×10}}^{\text{21}}}=e^{-0.22}$

${\text{N}}_{\text{t}}\text{=2}{\text{.2×10}}^{\text{21}}\text{Ra}\text{nuclei}$

Finally, after 500 years the number of nuclei remains and the rate constants, we can determine the activity as follows.

$\text{activity = kN}$

$\text{activity = }\left(\text{1}{\text{.4×10}}^{\text{-11}}\text{/s}\right)\left(\text{2}{\text{.2×10}}^{\text{21}}\text{Ra}\text{nuclei}\right)=3.1\times {10}^{10}\text{disintigarastions/s}$

$\text{activity}=3.1\times {10}^{10}\text{disintigarastions/s}$

Conclusion

The rate constant and half-life for the given radium decay and the activity after 500yr were determined.

The rate constant of the given reaction was $\text{k =1}{\text{.4×10}}^{\text{-11}}\text{/s}$

The half-life of the given reaction was $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=5.0\times {10}^{10}\text{s}$

The activity after 500yr was $\text{activity}=3.1\times {10}^{10}\text{disintigarastions/s}$