Chapter 20, Problem 20.102QP

(a)

Interpretation Introduction

Interpretation:

Tellurium-132 decays by emission of beta particle to produce highly unstable intermediate which is decays promptly to give stable product nuclei. The half-life of Tellurium-132 is 3.26d.

The balanced equation for decay of Tellurium-132 has to be written and the product has to be identified.

Answer to Problem 20.102QP

The balanced equation for decay of Tellurium-132 is,

${}_{\text{52}}^{\text{132}}\text{Te}\stackrel{}{\to }{}_{\text{54}}^{\text{132}}{\text{Xe + 2}}_{\text{-1}}^{\text{ 0}}\text{β}$

Explanation of Solution

When Tellurium-132 undergoes beta ( ${}_{\text{-1}}^{\text{0}}\text{β}$) decay it gives Xenon-132. This balanced equation is given as,

${}_{\text{52}}^{\text{132}}\text{Te}\stackrel{}{\to }{}_{\text{54}}^{\text{132}}{\text{Xe + 2}}_{\text{-1}}^{\text{ 0}}\text{β}$

(b)

Interpretation Introduction

Interpretation:

Tellurium-132 decays by emission of beta particle to produce highly unstable intermediate which is decays promptly to give stable product nuclei. The half-life of Tellurium-132 is 3.26d.

The balanced equation for production of stable product has to be written.

Answer to Problem 20.102QP

The balanced equation for stable product formed in the given nuclear reaction is,

${\text{H}}_{\text{2}}{}_{\text{52}}^{\text{132}}{\text{Te}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{+ }}_{\text{54}}^{\text{132}}{\text{Xe +2}}_{\text{-1}}^{\text{ 0}}\text{β}$

Explanation of Solution

Already given that ${\text{H}}_{\text{2}}\text{Te}$ made with Te-132 isotope.  Thus, Tellurium isotope undergoes beta emission to gives ${\text{H}}_{\text{2}}$ and eventual decay product of Tellurium-132.  Hence, the equation is given,

${\text{2H}}_{\text{35}}^{\text{82}}{\text{Br}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{+ 2}}_{\text{36}}^{\text{82}}{\text{Kr +}}_{\text{-1}}^{\text{ 0}}\text{β}$

(c)

Interpretation Introduction

Interpretation:

Tellurium-132 decays by emission of beta particle to produce highly unstable intermediate which is decays promptly to give stable product nuclei. The half-life of Tellurium-132 is 3.26d.

The quantity of ${\text{H}}_{\text{2}}\text{Te}$ remains after 93h has to be calculated and the pressure in the flask at $\text{25}°\text{C}$ should be identified.

Answer to Problem 20.102QP

The quantity of ${\text{H}}_{\text{2}}\text{Te}$ remains after 93h is $\text{0}\text{.005484 mol}$ .

At $\text{25}°\text{C}$ the pressure in the flask is $\text{0}\text{.318 atm}$ .

Explanation of Solution

Use the rate law to identify the fraction after 93hh. The half-life is 3.26d.

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{o}}}\text{= -kt =}\frac{\text{-0}\text{.693t}}{{\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}}\text{=}\frac{\text{-0}\text{.693(93}\text{.0d)}}{\text{3}\text{.26d}}\text{×}\frac{\text{1d}}{\text{24h}}\text{= -0}\text{.8237}$

Taking antilog on both sides gives,

$\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{o}}}\text{=}\frac{{\text{A}}_{\text{t}}}{{\text{A}}_{\text{o}}}{\text{=e}}^{\text{-0}\text{.8237}}\text{= 0}\text{.4387}$

So, the number of moles remains after 93.0h is

$\begin{array}{l}{\text{A}}_{\text{t}}\text{ = 0}\text{.0125 mol × 0}\text{.4387}\\ \\ \text{= 0}\text{.005484 mol}\end{array}$

The reaction can be summarized as follows,

$\begin{array}{l}{\text{H}}_{\text{2}}\text{Te }\underset{}{\to }{\text{ H}}_{\text{2}}\text{ + Xe}\\ \text{0}\text{.0125-y}\text{y}\text{y}\end{array}$

$\begin{array}{l}\text{From this,we can calculate y,}\\ \\ \text{0}\text{.0125mol-y = 0}\text{.005484 mol}\\ \\ \text{y = 0}\text{.007015 mol}\end{array}$

To determine the pressure in the flask, we required the total moles of gas.  This is,

$\begin{array}{l}{\text{Total mass = mol H}}_2^{}{\text{Te + mol H}}_{\text{2}}\text{+ mol Xe}\\ \\ \text{= (0}\text{.0125-y) + y + y}\\ \\ \text{= 0}\text{.0125 + y = 0}\text{.0125 + 0}\text{.007015 = 0}\text{.01951mol}\end{array}$

Therefore, the pressure in the flask is,

$\begin{array}{l}\text{P =}\frac{\text{nRT}}{\text{V}}\text{=}\frac{\text{0}\text{.01951 mol×0}\text{.0821L•atm/K•mol×298K}}{\text{1}\text{.50L}}\\ \\ \text{=0}\text{.318 atm}\end{array}$