Question

(a) The normal boiling point of carbon tetrachloride $\left({\text{CCl}}_4\right)$is$\mathbf{76}\mathbf{.}\mathbf{5}\mathbf{°}\mathit{C}$. A student looks up the standard enthalpies of formation of${\mathbf{\text{CCl}}}_{\mathbf{4}}\mathbf{\left(}\mathit{\ell }\mathbf{\right)}$and of${\mathbf{\text{CCl}}}_{\mathbf{4}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$in Appendix D. They are listed as$\mathbf{-}\mathbf{135}\mathbf{.}\mathbf{44}$and$\mathbf{-}\mathbf{102}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{\text{kJ}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$, respectively. By subtracting the first from the second, she computes$\mathbf{∆}\mathit{H}\mathbf{°}$for the vaporization of$\mathbf{\left(}{\mathbf{\text{CCl}}}_{\mathbf{4}}\mathbf{\right)}$to be$\mathbf{32}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{kJ}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$. But Tablestates that the$\mathbf{∆}{\mathit{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}}^{\mathbf{°}}$of$\mathbf{\left(}{\mathbf{\text{CCl}}}_{\mathbf{4}}\mathbf{\right)}$is$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{kJ}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$. Explain the discrepancy.

(b) Calculate the molar entropy change of vaporizationof$\mathbf{\left(}{\mathbf{\text{CCl}}}_{\mathbf{4}}\mathbf{\right)}$at.

Short answer

The molar entropy change of vaporization is $85.84\text{J}{\text{K}}^{-1}$and the explanation of the discrepancy is stated below

Step-by-step solution

Given information

There is vaporization of carbon tetrachlorideand$\left({\text{CCl}}_4\right)$the boiling point is:

$T_B=76.5°C$

The standard formationenthalpies are:

$$\begin{gathered} \Delta H_f^{°}\left({\text{CCl}}_4,l\right)=-135.44\text{kJ}\Delta H_f^{°}\left({\text{CCl}}_4,g\right)=-102.9\text{kJ}\Delta H_{vap}^{°}\left(\text{calculated}\right)=32.5\text{kJ} \\ \Delta H_{vap}^{°}\left(\text{experimental}\right)=30.0\text{kJ} \end{gathered}$$

Concept of enthalpy of vaporization

A certain amount of energy is involved when a material changes phase from solid to liquid or from liquid to gas. This amount of energy is known as the enthalpy of vaporization, (symbol$\mathit{\Delta }{\mathit{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}}^{\mathbf{°}}$; unit: J), also known as the (latent) heat of vaporization or heat of evaporation, in the case of a liquid to gas phase change.

Explanation of the discrepancy

The enthalpy of vaporization discovered empirically is higher in value. This means that breaking apart the bonds of liquid requires greater energy.

This is owing to the possibility of liquid-gaseous bonding during the reaction, resulting in a higher number of intermolecular forces that must be broken apart

Calculate the entropy of vaporization

Calculate the entropy of vaporization of${\text{CCl}}_4$

$$\begin{gathered} \Delta S_{vap}^{°}=\frac{\Delta H_{vap}^{°}}{T_B} \\ \Delta S_{vap}^{°}=\frac{30.0\times {10}^3}{76.5+273} \\ \Delta S_{vap}^{°}=85.84\text{J}{\text{K}}^{-1} \end{gathered}$$