Question

Question: Tungsten melts at$\mathbf{3410}\mathbf{°}\mathbf{C}$and has an enthalpy change of fusion of. Calculate the entropy of $\mathbf{35}\mathbf{.}\mathbf{4}\mathbf{kJ}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$fusion of tungsten.

Short answer

The entropy of fusion of tungsten is $9.61{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\mathrm{TT}$.

Step-by-step solution

Given information

The melting point of tungsten is ${\mathrm{T}}_{\mathrm{f}}=3410°\mathrm{C}$.

Given enthalpy change of fusion of tungsten$H_{fus}=35.4kJmo{l}^{-1}$

Concept of entropy of fusion

The term "entropy of fusion" refers to the change in entropy that occurs when one mole of a solid material melts into a liquid.

For instance, as the ice evaporates,

$\mathbf{∆}{\mathbf{S}}_{\mathbf{fusion}\mathbf{}}\mathbf{=}{\mathbf{S}}_{\mathbf{water}}\mathbf{}\mathbf{-}{\mathbf{S}}_{\mathbf{ice}}\mathbf{}\mathbf{=}\frac{\mathbf{∆}{\mathbf{H}}_{\mathbf{fusion}}}{{\mathbf{T}}_{\mathbf{f}}}$

where, $∆{\mathrm{H}}_{\mathrm{fusion}}$ is the enthalpy of fusion and T_f is the fusion temperature.

Convert unit of temperature

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

The melting point of tungsten is${\mathrm{T}}_{\mathrm{f}}=3410°\mathrm{C}$.

Calculate entropy of fusion

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

Put the value of $H_{fus}$and T_f in above equation.
$$\begin{gathered} S_{\mathrm{fus}}=35.4k{\mathrm{Jmol}}^{-1}/3683.15K \\ {S}_{\mathrm{fus}}=9.61{\mathrm{JK}}^{-1}\mathrm{mol} \end{gathered}$$

Thus, the entropy of fusion of tungsten is .