91Ó°ÊÓ

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

(a)

Considering the given information:

The following is the xenon hexafluoride formation reaction:

${\text{Xe(g)+3F}}_{\text{2}}\text{(g)}⇌{\text{XeF}}_{\text{6}}\text{(g)}$

Record the enthalpy and free energy values,

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=-402kJ/mol}\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-280.0kJ/mol}\end{array}$

The ${\text{Δ³§}}_{\text{rxn}}^{\text{o}}$ of a given reaction is calculated as follows:

The standard free energy change equation is as follows:

$\begin{array}{l}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}{\text{=Δ±á}}_{\text{rxn}}^{\text{o}}{\text{-TΔ³§}}_{\text{rxn}}^{\text{o}}\\ {\text{Δ³§}}^{\text{o}}\text{=}\frac{{\text{Δ±á}}^{\text{n}}{\text{-Δ³Ò}}^{\text{o}}}{\text{T}}\end{array}$

The values of enthalpy and free energy are plugging into the equation above.

$\begin{array}{c}{\text{Δ³§}}^{\text{°}}\text{=}\frac{\text{-402kJ/mol-(-280.0kJ/mol)}}{\text{298K}}\\ \text{=}\frac{\text{-122.0kJ/mol}}{\text{298K}}\\ \text{=-0.40939597°ì´³/³¾´Ç±ô×°­}\\ \text{=-0.409°ì´³/³¾´Ç±ô×°­}\end{array}$

Therefore, the entropy value is ${\text{Δ³§}}_{\text{rxn}}^{\text{o}}\text{=-0.409°ì´³/³¾´Ç±ô×°­}.$

(b)

Considering the given information:

Thereactionthatproducesxenonhexafluorideisasfollows:

${\text{Xe(g)+3F}}_{\text{2}}\text{(g)}⇌{\text{XeF}}_{\text{6}}\text{(g)}$

Calculate the change in free energy ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}$ next.

Standard Free energy change equation is,

${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}{\text{=Δ±á}}_{\text{rxn}}^{\text{o}}{\text{-TΔ³§}}_{\text{rxn}}^{\text{o}}$

Free energy change ${\text{Δ³Ò}}_{\text{f}}^{\text{o}}$

The values of enthalpy and entropy are,

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=-402kJ/mol}\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-280.0kJ/mol}\end{array}$

These figures are used to fill in the blanks in the standard free energy equation.

$\begin{array}{c}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=(-402°ì´³/³¾´Ç±ô)-°Ú(500°­)(-0.40939597°ì´³/³¾´Ç±ô×°­)±Õ}\\ {\text{=-197.302kJ/molΔ³Ò}}_{\text{rxn}}^{\text{o}}\\ \text{=-197kJ/mol}\end{array}$

Therefore, the given reaction standard free energy value is $\text{-197kJ/mol}$.