Chapter 20, Problem 20.101QP

(a)

Interpretation Introduction

Interpretation:

Bromine-82 decays by emission of beta particle. The half-life of Bromine-82 is 1.471d.

The balanced equation for decay of Bromine-82 and decomposition of ${\text{H}}^{\text{82}}\text{Br}$ has to be written.

Answer to Problem 20.101QP

The balanced equation for decay of Bromine-82 is,

${}_{\text{35}}^{\text{82}}\text{Br}\stackrel{}{\to }{}_{\text{36}}^{\text{82}}{\text{Kr +}}_{\text{-1}}^{\text{ 0}}\text{β}$

The balanced equation for decomposition of ${\text{H}}^{\text{82}}\text{Br}$

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

Explanation of Solution

When Bromine-82 undergoes beta ( ${}_{\text{-1}}^{\text{0}}\text{β}$) decay it gives Krypton-82. This balanced equation is given as,

${}_{\text{35}}^{\text{82}}\text{Br}\stackrel{}{\to }{}_{\text{36}}^{\text{82}}{\text{Kr +}}_{\text{-1}}^{\text{ 0}}\text{β}$

Already given that Hydrogen bromide made with Bromine-82 isotope. Above equation shows that ${\text{Br}}^{\text{82}}$ decays to gives $\text{Kr-82}$ and beta particle. Hence Decomposition of ${\text{H}}^{\text{82}}\text{Br}$ is given by,

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

(b)

Interpretation Introduction

Interpretation:

Bromine-82 decays by emission of beta particle. The half-life of Bromine-82 is 1.471d.

The quantity of $\text{HBr}$ remains after 10h has to be calculated and the pressure in the flask at $\text{22}°\text{C}$ should be identified.

Answer to Problem 20.101QP

The quantity of $\text{HBr}$ remains after 10h is $\text{0}\text{.0123 mol}$.

At $\text{22}°\text{C}$ the pressure in the flask is $\text{0}\text{.3957 atm}$ .

Explanation of Solution

Use the rate law to identify the fraction after 10.0h. The half-life is 1.471d.

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{o}}}\text{= -kt =}\frac{\text{-0}\text{.693t}}{{\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}}\text{=}\frac{\text{-0}\text{.693(0}\text{.4167d)}}{\text{1}\text{.471 d}}\text{= -0}\text{.1963}$

Taking antilog on both sides gives,

$\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{o}}}\text{=}\frac{{\text{A}}_{\text{t}}}{{\text{A}}_{\text{o}}}{\text{=e}}^{\text{-0}\text{.1963}}\text{= 0}\text{.8218}$

So, the number of moles of ${\text{H}}^{\text{82}}\text{Br}$ remains after 10h is

$\begin{array}{l}{\text{A}}_{\text{t}}\text{ = 0}\text{.0150 mol × 0}\text{.8218}\\ \\ \text{= 0}\text{.0123 mol}\end{array}$

The reaction can be summarized as follows,

$\begin{array}{l}\text{2HBr }\underset{}{\to }{\text{ H}}_{\text{2}}\text{ + 2Kr}\\ \text{0}\text{.0150-2y}\text{y}2\text{y}\end{array}$

$\begin{array}{l}\text{From this,we can calculate y,}\\ \\ \text{0}\text{.0150mol-2y = 0}\text{.01233 mol}\\ \\ \text{y = 0}\text{.00134mol}\end{array}$

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

$\begin{array}{l}{\text{Total mass = mol H}}_{\text{35}}^{\text{82}}\text{Br + mol Kr}\\ \\ \text{= (0}\text{.0150-2y) + y + 2y}\\ \\ \text{= 0}\text{.0150 + y = 0}\text{.0150 + 0}\text{.00134 = 0}\text{.01634mol}\end{array}$

Therefore, the pressure in the flask is,

$\begin{array}{l}\text{P =}\frac{\text{nRT}}{\text{V}}\text{=}\frac{\text{0}\text{.01634 mol×0}\text{.0821L•atm/K•mol×295K}}{\text{1}\text{.00L}}\\ \\ \text{=0}\text{.396 atm}\end{array}$