Question

Carbon tetrachloride (${\mathbf{CCl}}_4$ ) has a vapor pressure of 213 torr at $\mathbf{40}\mathbf{°}\mathbf{C}$and 836 torr at$\mathbf{80}\mathbf{}\mathbf{°}\mathbf{C}$ . What is the normal boiling point of ${\mathbf{CCl}}_4$?

Short answer

The normal boiling of ${\mathrm{CCl}}_4$is $77°\mathrm{C}$.

Step-by-step solution

Calculating enthalpy of vaporization

It is given that the vapor pressure at 40°C is 213 torr(P_vapT1)and at 80°C is 836 torr. Substitute the values to calculate the enthalpy of vaporization (∆H_vap)

One kilojoule is equal to 1000 joules.

$\begin{array}{c}\mathrm{n}\left(\frac{213\mathrm{torr}}{836\mathrm{torr}}\right)=\frac{{\mathrm{\Delta H}}_{\mathrm{vap}}}{8.314\mathrm{J}/\mathrm{mol}.\mathrm{K}}\left(\frac{1}{353\mathrm{K}}-\frac{1}{313\mathrm{K}}\right)\\ \ln \left(0.255\right)=\frac{{\mathrm{\Delta H}}_{\mathrm{vap}}}{8.314\mathrm{J}/\mathrm{mol}.\mathrm{K}}(0.002833-0.003195)\\ -1.366=-\frac{{\mathrm{\Delta H}}_{\mathrm{vap}}}{8.314\mathrm{J}/\mathrm{mol}.\mathrm{K}}\left(0.000362\right)\\ {\mathrm{\Delta H}}_{\mathrm{vap}}=31.373\mathrm{J}\end{array}$

Boiling point of CCl4

The boiling point is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure. So, the vapor pressure of mercury is equal to the atmospheric pressure, which is 760 torr.

$\begin{array}{c}\ln \left(\frac{213\mathrm{torr}}{760\mathrm{torr}}\right)=\frac{31373\mathrm{J}/\mathrm{mol}}{8.314\mathrm{J}/\mathrm{mol}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{313\mathrm{K}}\right)\\ \ln \left(0.280\right)=3774\left(\frac{1}{{\mathrm{T}}_2}-0.003195\right)\\ -1.273=-\frac{3774}{{\mathrm{T}}_2}-12.058\\ \frac{3774}{{\mathrm{T}}_2}=10.785\\ {\mathrm{T}}_2=\frac{3774}{10.785}\\ {\mathrm{T}}_2=350\mathrm{K}\end{array}$

The boiling point of carbon tetrachloride is $77°\mathrm{C}$.