Question

Question: Tetraphenyl germane, ${\mathbf{\left(}{\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}\mathbf{\right)}}_{\mathbf{4}}\mathbf{\text{Ge}}$,has a melting point of $\mathbf{232}\mathbf{.}\mathbf{5}\mathbf{°}\mathbf{C}$, and its enthalpy increasesby $\mathbf{106}\mathbf{.}\mathbf{7}\mathbf{}{\mathbf{Jg}}^{\mathbf{-}\mathbf{1}}$during fusion. Calculate the molar enthalpy of fusion and molar entropy of fusion of tetraphenyl germane.

Short answer

The molar enthalpy of fusion and molar entropy of fusion of tetraphenyl germane is$80.39{\text{JK}}^{-1}\text{mol}$ .

Step-by-step solution

Given information

The melting point of tetraphenyl germane ${\left({\mathrm{C}}_6{\mathrm{H}}_5\right)}_4\mathrm{Ge},{\mathrm{T}}_{\mathrm{f}}=232.5°\mathrm{C}$and the enthalpy changes of fusion $H_{fus}=106.7g^{-1}$.

Concept of molar entropy and enthalpy of fusion

The molar heat of fusion is divided by the melting point to get the entropy of fusion. The energy required to convert one mole of a substance from the solid to the liquid phase at constant temperature and pressure is known as the molar enthalpy of fusion.

Convert unit of temperature

The unit of temperature is converted from degree Celsius to kelvin as:

The melting point of tetraphenyl germane ${\left({\mathrm{C}}_6{\mathrm{H}}_5\right)}_4\mathrm{Ge},{\mathrm{T}}_{\mathrm{f}}=232.5°\mathrm{C}$

localid="1663817125450" $$\begin{gathered} T_f=232.5+273.15 \\ {T}_f=505.65\text{K} \end{gathered}$$

Calculate molar enthalpy of tetraphenyl germane

The molar mass of tetraphenyl germane${\left(C_6{H}_5\right)}_{4}Ge:=4^{*}\left(\right(612.01)+(51\left)\right)+72.5$.

The molar mass of tetraphenyl germane ${\left(C_6{H}_5\right)}_{4}Ge:=381.0{\text{gmol}}^{-1}$.

Thus, molar enthalpy changes of fusion of tetraphenyl germane $=106.7^{*}381.0$

The molar enthalpy changes of fusion of tetraphenyl germane$=4.{065}^{*}{10}^4{\text{Jmol}}^{-1}$

Calculate molar entropy of fusion

The formula of Entropy of fusion is ${\mathrm{S}}_{\mathrm{fus}}={\mathrm{H}}_{\mathrm{fus}}/{\mathrm{T}}_{\mathrm{f}}$.

Put the value of $H_{fus}$and$T_f$ in above equation.
$$\begin{gathered} S_{fus}=4.{065}^{*}{10}^{4}Jmol-1/505.65K \\ {S}_{fus}=80.39J{K}^{-1}mol \end{gathered}$$