Chapter 9, Problem 78QAP

Trichloroethane, C₂H₃Cl₃, is the active ingredient in aerosols that claim to stain-proof men's ties. Trichloroethane has a vapor pressure of 100.0 mm Hg at 20.0°C and boils at 74.1°C. An uncovered cup $\left(\frac{1}{2}\text{p}\mathrm{int}\right)$ of trichloroethane $\left(d=1.325\text{g}/\text{mL}\right)$ is kept in an 18-ft³ refrigerator at 39°F. What percentage (by mass) of the trichloroethane is left as a liquid when equilibrium is established?

Interpretation Introduction

Interpretation:

The percentage (by mass) of the trichloroethane that remains as a liquid when equilibrium is established should be calculated.

Concept introduction:

The relation between vapor pressure ( P), temperature ( T) and molar heat of vaporization $\left(\Delta H_{\text{vap}}\right)$ is as follows:

$\ln P=\left(\frac{-\Delta H_{\text{vap}}}{R}\right)\frac{1}{T}+A$

Here,  A is the y-intercept, which is a constant for any given line.

R is the gas constant.

Answer to Problem 78QAP

The amount of trichloroethane is left as a liquid is $44.25\%$.

Explanation of Solution

Given:

Vapor pressure of trichloroethane = 100.0 mm Hg at 20 °C

Boiling point = 74.1 °C

Volume of cup = ½ pint

Density = 1.325 g/mL

18 ft³ refrigerator at 39 °F

The relation between vapor pressure ( P), temperature ( T) and molar heat of vaporization $\left(\Delta H_{\text{vap}}\right)$ is as follows:

$\ln P=\left(\frac{-\Delta H_{\text{vap}}}{R}\right)\frac{1}{T}+A$

Here,  A is the y-intercept, which is a constant for any given line.

R is the gas constant. $\left(8.314\times {10}^{-3}\text{kJ/mol}\text{K}\right)$.

For vapor pressure $P_1$ and $P_2$ at temperature $T_1$ and $T_2$ the above equation becomes:

$\ln \frac{P_2}{P_1}=\left(\frac{-\Delta H_{\text{vap}}}{R}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$   ..........(1)

At $T_1=20.0°\text{C}$ vapor pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}$

$\left(P_1\right)$ is $100.0\text{mm}\text{Hg}$

The vapor pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}$ at its normal boiling point $T_2=74.1°\text{C}$, $P_2=760\text{mm}\text{Hg}$

$\begin{array}{c}{T}_1=20.0°\text{C}\\ \text{=}\left(\text{273+20}\text{.0}\right)\text{K}\\ \text{=293}\text{.0}\text{K}\end{array}$

$\begin{array}{c}{T}_2=74.1°\text{C}\\ \text{=}\left(273+74.1\right)\text{K}\\ \text{=347}\text{.1}\text{K}\end{array}$

Substitute the values in equation (1) and calculate $\Delta H_{\text{vap}}$ as follows:

$\begin{array}{c}\ln \frac{760\cancel{\text{mm}\text{Hg}}}{100.0\cancel{\text{mm}\text{Hg}}}=\left(\frac{-\Delta H_{\text{vap}}}{8.314\times {10}^{-3}\text{kJ/mol}\text{K}}\right)\left(\frac{1}{347.1\text{K}}-\frac{1}{293.0\text{K}}\right)\\ \ln \left(760\right)-\ln \left(100.0\right)=\left(\frac{-\Delta H_{\text{vap}}}{8.314\times {10}^{-3}\text{kJ/mol}\text{K}}\right)\left(0.0029{\text{K}}^{-1}-0.0034{\text{K}}^{-1}\right)\\ 2.03=\left(\frac{-\Delta H_{\text{vap}}}{8.314\times {10}^{-3}\text{kJ/mol}\cancel{\text{K}}}\right)\left(-0.0005\cancel{{\text{K}}^{-1}}\right)\end{array}$

$\begin{array}{c}\Delta H_{\text{vap}}=\frac{2.03\times 8.314\times {10}^{-3}\text{kJ/mol}}{0.0005}\\ =33.754\text{kJ/mol}\end{array}$

Molar heat of vaporization of liquid ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}$ is $33.754\text{kJ/mol}$.

Convert the temperature of refrigerator $\left(39°\text{F}\right)$ is as follows:

$\begin{array}{c}T\left(°\text{C}\right)=\frac{T\left(°\text{F}\right)-32}{1.8}\\ =\frac{32°\text{F-32}}{1.8}\\ =3.9°\text{C}\end{array}$

With the values of $T_1$ and $P_1,$ calculate the value of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}$ at $3.9°\text{C}$.

$\begin{array}{l}\ln \frac{P}{100.0\text{mm}\text{Hg}}=\left(-\frac{33.754\text{}\cancel{\text{kJ/mol}}}{8.314\times {10}^{-3}\cancel{\text{kJ/mol}}\text{K}}\right)\left(\frac{1}{\left(3.9+273\right)\text{K}}-\frac{1}{293\text{K}}\right)\\ \ln \frac{P}{100.0\text{mm}\text{Hg}}=\left(-4.06\cancel{\text{K}}\right)\left(0.2\cancel{{\text{K}}^{-1}}\right)\\ P=\left(100.0\text{mm}\text{Hg}\right)\times e^{-0.812}\\ P=\left(100.0\text{mm}\text{Hg}\right)\times 0.4440\\ P=44.40\text{mm}\text{Hg}\end{array}$

Therefore, vapor pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}$ at $3.9°\text{C}$ is $44.40\text{mm}\text{Hg}$.

Ideal gas law is:

$PV=nRT$

Here, P represent pressure, V represents volume, n represents number of moles, R represents gas constant and T represents temperature.

Given, volume of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}$ is $18f{t}^3$.

Calculate the number of moles and the mass of vapor formed in the refrigerator as follows:

$\begin{array}{c}{n}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}\left(g\right)}=\frac{PV}{RT}\\ =\frac{44.40\cancel{\text{mm}\text{Hg}}\times \text{18}\cancel{{\text{ft}}^3}}{0.08206\cancel{\text{L}}\cancel{\text{atm}}\text{/mol}\cancel{\text{K}}\times 276.9\cancel{\text{K}}}\times \frac{\cancel{\text{atm}}}{\text{760}\cancel{\text{mm}\text{Hg}}}\times \frac{28.317\cancel{\text{L}}}{\cancel{{\text{ft}}^3}}\\ =1.31\text{mol}\end{array}$

$\begin{array}{c}{m}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}\left(g\right)}=1.31\cancel{\text{mol}}\times \frac{133.4\text{g}}{\cancel{\text{mol}}}\\ =174.75\text{g}\end{array}$

Calculate the mass of liquid ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}$ placed in the refrigerator, from its volume and density.

$\begin{array}{c}{m}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}\left(l\right)}=\left(\frac{1}{2}\cancel{\text{pint}}\times \frac{473.176\cancel{\text{mL}}}{1\cancel{\text{pint}}}\right)\left(1.325\text{g/}\cancel{\text{mL}}\right)\\ =313.5\text{g}\end{array}$

313.5 g of the liquid, 174.75 is vaporized. Thus, the mass of liquid at left is

$\begin{array}{c}{m}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}\left(l\right)}\left(left\right)=m_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}\left(l\right)}-m_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}\left(g\right)}\\ =313.5\text{g-174}\text{.75}\text{g}\\ \text{=138}\text{.75}\text{g}\end{array}$

Calculate the mass percentage of liquid left at equilibrium to take the ratio of liquid mass left to the initial liquid mass and then multiply by $100$.

$\begin{array}{c}\frac{m_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}\left(l\right)}\left(left\right)}{m_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{Cl}}_{\text{3}}\left(l\right)}}\times 100\%=\frac{138.75}{313.5}\times 100\%\\ =44.25\%\end{array}$

Thus, the amount of trichloroethane is left as a liquid is $44.25\%$.

Conclusion

The amount of trichloroethane is left as a liquid is $44.25\%$.