Q12.125CP - Chemistry: Molecular Nature Of Matter And Change

Question

Mercury (Hg) vapor is toxic and readily absorbed from the lungs. At  $\mathbf{20}\mathbf{°}\mathbf{C}$, mercury ( $\mathbf{\Delta Hvap}$ = 59.1 kJ/mol) has a vapor pressure of  $\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{-3}\mathbf{torr}$, which is high enough to be hazardous. To reduce the danger to workers in processing plants, Hg is cooled to lower its vapor pressure. At what temperature would the vapor pressure of Hg be at the safer level of  $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{-5}\mathbf{torr}$?

Step-by-Step Solution

Answer

The temperature at which pressure is $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{-5}\mathbf{torr}$  is 259.7k.

Step 1: Definition

Enthalpy of vaporization may be defined as the energy required to convert the liquid into the vapor phase. It is also known as Enthalpy of evaporation.

 

Clausius - Clapeyron equation

$\mathbf{ln}\frac{P2}{P1}\mathbf{=}\frac{{\mathbf{\Delta H}}^o\mathbf{vap}}{R}\mathbf{[}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{]}$

P is Pressure

T is Temperature

R is Gas Constant

$\text{ΔH}{°}_{\text{vap.}}^{}$  is Enthalpy of vaporisation.

Step 2: The temperature at which pressure is 5 . 0 × 10 - 5 torr .

Vapour pressure,  $\mathbf{P}\mathbf{1}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{-3}\mathbf{torr}$

Temperature, $\mathbf{T}\mathbf{1}\mathbf{=}{\mathbf{20}}^o\mathbf{C}\mathbf{=}\mathbf{293}\mathbf{k}$

Vapour pressure,  $\mathbf{P}\mathbf{2}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{-5}\mathbf{torr}$

Temperature  ${\mathbf{T}}_2\mathbf{=}?$

The Enthalpy of the vaporisation, $\mathbf{\Delta Hvap}\mathbf{=}\mathbf{59}\mathbf{.}\mathbf{1}\mathbf{kJ}\mathbf{/}\mathbf{mole}$ 

  $\begin{array}{c}\ln \frac{\mathrm{P}1}{\mathrm{P}2}=\frac{{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{}[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}]\\ \ln \frac{1.20\times {10}^{-3}\mathrm{torr}}{5.0\times {10}^{-5}\mathrm{torr}}=\frac{59.1\mathrm{kJ}/\mathrm{mole}}{8.314}[\frac{1}{\mathrm{T}2}-\frac{1}{293}]\\ \ln 24=\frac{59.1\mathrm{kJ}/\mathrm{mole}}{8.314}[\frac{1}{\mathrm{T}2}-\frac{1}{293}]\\ [\frac{1}{\mathrm{T}2}-\frac{1}{293}]=\frac{8.314\times 3.18}{59100\mathrm{J}/\mathrm{mole}}\end{array}$

$\begin{array}{c}\frac{1}{\mathrm{T}2}=0.00045+0.0034\\ \frac{1}{\mathrm{T}2}=0.00385\\ \mathrm{T}2=\frac{}{0.00385}\\ \mathrm{T}2=259.7\mathrm{k}\end{array}$

The temperature is 259.7k.