91Ó°ÊÓ

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:

$K\left(g\right)→K\left(l\right)$

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

$\begin{array}{l}\mathrm{Δ}{H}_{rxn}^{°}=−\mathrm{Δ}\\ H_{fus}^{°}=−2.39\text{kJ}/\text{mol}\end{array}$

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

$\begin{array}{c}Q=\mathrm{Δ}{H}_{rxn}^{°}\\ =−2.39\text{kJ}/\text{mol}\end{array}$

The reaction's entropy can be computed as follows:

$\begin{array}{c}\mathrm{Δ}{S}_{rxn}^{°}=\frac{Q}{T}\mathrm{Δ}\\ S_{rxn}^{°}=\frac{−\mathrm{Δ}{H}_{fus}^{°}}{T}\end{array}$

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

$\begin{array}{l}\mathrm{Δ}{S}_{rxn}^{°}=\frac{−2.39â‹\dots {10}^3\text{J}/\text{mol}}{336.7\text{K}}\\ \mathrm{Δ}{S}_{rxn}^{°}=−7.098\frac{\text{J}}{\text{mol}â‹\dots \text{K}}\end{array}$

Since only 0.2 mol of potassium is freezing,

$\begin{array}{l}\mathrm{Δ}{S}_{rxn}^{°}=−7.098\frac{\text{J}}{\text{mol}â‹\dots \text{K}}â‹\dots 0.2\text{mol}\\ \mathrm{Δ}{S}_{rxn}^{°}=−1.4197\text{J}/\text{K}=−1.42\text{J}/\text{K}\end{array}$

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