Chapter 15, Problem 117AP

The vapor pressure of mercury is 0.0020 mmHg at $\text{26°C}$. (a) Calculate $K_c$ and $K_p$ for the process $\text{Hg(}l\text{)}\underset{}{\rightleftarrows }\text{Hg(}g\text{)}$. (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring 6.1 m long.

5.3 m wide, and 3.1 m high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor $\left(\text{in mg/}{\text{m}}^{\text{3}}\right)$. Does this concentration exceed the safety limit of $\text{0}\text{.05 mg/}{\text{m}}^{\text{3}}\text{?}$ (Ignore the volume of furniture and other objects in the laboratory.)

Interpretation Introduction

Interpretation:

The equilibrium constants $K_p$ and $K_c$ for the process is to be calculated and the mass of mercury in the laboratory, its concentration, and whether it exceeds the safety limit or not are to be determined.

Concept introduction:

According to Dalton’s law, the total pressure of the mixture of a nonreactive gas is equal to the partial pressure of individual gases. The pressure exerted by the individual gas is called its partial pressure.

The ratio of the concentration of products and the concentration of reactants at equilibrium state in the chemical reaction is known as equilibrium constant.

Equilibrium constant is the ratio of the concentration of reactants and products present in the chemical reaction.

For a general reaction: $\text{xA}\text{+}\text{yB}\rightleftarrows \text{zC}$

The general formula for writing equilibrium expression for the reaction is given as:

$K_C=\frac{{\left[\text{C}\right]}^Z{}_{eqm}}{{\left[\text{A}\right]}^X{}_{eqm}{\left[\text{B}\right]}^Y}$

Here, $K_C$ is the equilibrium constant and $C$ in $K_C$ stands for the concentration. $\left[\text{A}\right]$, $\left[\text{B}\right]$, and $\left[\text{C}\right]$ are the equilibrium concentration of reactants A and B and product C.

A and B are reactants, C is products, and $x,y$, and $z$ are their respective stoichiometric coefficients.

For the general reaction: $a\text{W}\text{+}b\text{X}\to c\text{Y}\text{+}d\text{Z}$

The general formula for writing equilibrium expression for the reaction is given as:

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

Here, $p$ in $K_p$ stands for the pressure. $P_W,P_X,P_Y,\text{and}{P}_Z$ are the equilibrium partial pressure of reactants W and X and products Y and Z.

W and X are reactants, Y and Z are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

Equilibrium constants of gas phase reaction are written in terms of partial pressures because concentration of gases is directly proportional to partial pressures.

The relationship between $K_P$ and $K_C$ is given as:

$\begin{array}{c}{K}_P=K_C{\left(RT\right)}^{\Delta n}\\ K_c=\frac{K_P}{{\left(RT\right)}^{\Delta n}}\end{array}$

$\text{Volume}\text{=}\text{length×}\text{breadth}\text{×}\text{height}$.

The relationship between $\text{L}$

and ${\text{cm}}^3$

can be expressed as: $1\text{L}\text{=}{\text{1000cm}}^3$.

For converting ${\text{cm}}^3$ to $\text{L}$, conversion factor is $\frac{1\text{L}}{1000c{m}^3}$.

The relationship between meter and centimeter can be expressed as: $1\text{m}\text{=}\text{100cm}$.

For converting meter to centimeter, conversion factor is $\frac{100cm}{1m}$.

The relationship between $\text{atm}$

and $\text{mm Hg}$ can be expressed as: $1\text{atm}\text{=}76\text{0mm Hg}$.

For converting $\text{mm Hg}$

to $\text{atm}$, conversion factor is $\frac{1\text{atm}}{760mmHg}$.

Answer to Problem 117AP

Solution:

a)

$K_p=2.6\times {10}^{-6}$ and $K_c=1.059\times {10}^{-7}$

b)

$m=2.2g,Concentration=22mg/m^3$, and the concentration of mercury exceeds the safety limit of $0.05mg/m^3$.

Explanation of Solution

Given information:

The vapor pressure of mercury is $0.0020mmHg$.

a) $K_c$ and $K_p$ for the process $\text{Hg}(l)\stackrel{}{\rightleftarrows }\text{Hg}(g)$.

The vapor pressure of mercury is equivalent to the partial pressure of mercury.

The conversion of partial pressure of mercury into $\text{atm}$ units is given as follows:

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

$\begin{array}{c}{K}_p=P_{Hg}=0.0020mmHg\\ =\frac{0.0020\cancel{mmHg}}{760.0\cancel{mmHg}/atm}\\ =2.6\times {10}^{-6}atm\end{array}$

Equilibrium constant is expressed without units, so the equilibrium pressure constant is $2.6\times {10}^{-6}$.

The equilibrium constant $K_c$ is calculated by the expression given as follows:

$K_P=K_C{\left(RT\right)}^{\Delta n}$

$K_c=\frac{K_p}{{\left(RT\right)}^{\Delta n}}$

Here, $K_c$ is the equilibrium constant in terms of concentration, $K_p$ is the equilibrium constant in terms of partial pressure, $\text{R}$ is the gas constant, $T$ is the temperature, and $\Delta n$ is the change in number of moles.

Substitute the values of $K_p,R,T$, and $\Delta n$ in the above equation,

$\begin{array}{c}{K}_c=\frac{2.6\times {10}^{-6}}{{\left(0.0821\times 299\right)}^1}\\ =\frac{2.6\times {10}^{-6}}{24.548}\\ =1.1\times {10}^{-7}\end{array}$

Hence, $K_p=2.6\times {10}^{-6}$ and $K_c=1.1\times {10}^{-7}$.

b) The mass of mercury (in grams) vaporized at equilibrium and concentration of mercury vapor (in ${\text{mg/m}}^{\text{3}}$

)

The laboratory measurements are given as:

Length= $6.1m$

Breadth= $5.3m$

Height= $3.1m$

The volume of the lab is calculated as follows:

$\begin{array}{c}V=\left(6.1m\right)\left(5.3m\right)\left(3.1m\right)\\ =100m^3\end{array}$

The total mass of mercury vapor is calculated as follows:

$\begin{array}{c}m=\frac{1.1\times {10}^{-7}mol}{1L}\left(\frac{200.6g}{1mol}\right)\left(\frac{1\text{L}}{1000c{m}^3}\right){\left(\frac{1cm}{0.01m}\right)}^3\left(100m^3\right)\\ =2.2g\end{array}$

The concentration of mercury vapor in the room is calculated as follows:

$\begin{array}{c}Concentrationofmercury=\frac{2.2g}{100m^3}\\ =0.022g/m^3\\ =22mg/m^3\end{array}$

Therefore, the concentration of mercury exceeds the safety limit of $0.05mg/m^3$.