What is the change in entropy when 0.200ml of potassium freezes at $\mathbf{63}{\mathbf{.7}}^{\mathbf{°}}\mathbf{\text{C}}\mathbf{(}\mathbf{\Delta }{\mathbf{H}}_{\mathbf{\text{fus}}}\mathbf{=}\mathbf{2}\mathbf{.39}\mathbf{}\mathbf{\text{kJ}}\mathbf{/}\mathbf{\text{mol}}\mathbf{)}$?

It is used in chemistry to indicate a change in enthalpy as well as the addition of heat to the reaction. A triangle symbol is used to represent it.

The following is a description of the reaction:

$K\left(g\right)\to K\left(l\right)$

Freezing is an exothermic process that occurs spontaneously, and the enthalpy of freezing is:

$\begin{array}{l}\Delta H_{rxn}^{°}=-\Delta \\ H_{fus}^{°}=-2.39\text{kJ}/\text{mol}\end{array}$

By definition, enthalpy is the amount of heat gained or lost during a reaction. As a result, we can assume that at constant pressure.

$\begin{array}{c}Q=\Delta H_{rxn}^{°}\\ =-2.39\text{kJ}/\text{mol}\end{array}$

The reaction's entropy can be computed as follows:

$\begin{array}{c}\Delta S_{rxn}^{°}=\frac{Q}{T}\Delta \\ S_{rxn}^{°}=\frac{-\Delta H_{fus}^{°}}{T}\end{array}$

${63.7}^{°}\text{C}$, which is equal to 336.7K, is the specified temperature.

$\begin{array}{l}\Delta S_{rxn}^{°}=\frac{-2.39\cdot {10}^3\text{J}/\text{mol}}{336.7\text{K}}\\ \Delta S_{rxn}^{°}=-7.098\frac{\text{J}}{\text{mol}\cdot \text{K}}\end{array}$

Since only 0.2 mol of potassium is freezing,

$\begin{array}{l}\Delta S_{rxn}^{°}=-7.098\frac{\text{J}}{\text{mol}\cdot \text{K}}\cdot 0.2\text{mol}\\ \Delta S_{rxn}^{°}=-1.4197\text{J}/\text{K}=-1.42\text{J}/\text{K}\end{array}$

Therefore, the entropy of freezing 0.200 mol of potassium is -1.42 J/K.