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.

Considering the given reaction:

${\text{Br}}_{\text{2}}\text{(l)}⇌{\text{Br}}_{\text{2}}\text{(g)}$

First, use the enthalpy constants from Appendix B to solve for ${\text{Δ±á}}^{\text{°}}$.

$\begin{array}{c}{\text{Δ±á}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ±á}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ±á}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ±á}}_{\text{f}}^{\text{°}}\text{(Br(g))}\right]\text{-}\left[{\text{nΔ±á}}_{\text{f}}^{\text{°}}\left({\text{Br}}_{\text{2}}\text{(l)}\right)\right]\\ \text{=([1(30.91)]-[1(0)])kJ}\\ {\text{Δ±á}}^{\text{°}}\text{=30.91kJ}\end{array}$

Then, using the entropy constants in Appendix B, solve for ${\text{Δ³§}}^{\text{°}}$.

$\begin{array}{c}{\text{Δ³§}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{S}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{S}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nS}}^{\text{°}}\left({\text{Br}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nS}}^{\text{°}}\left({\text{Br}}_{\text{2}}\text{(l)}\right)\right]\\ \text{=([1(245.38)]-[1(152.23)])J/K}\\ {\text{Δ³§}}^{\text{°}}\text{=93.15J/K}\end{array}$

Finally, find T at equilibrium. ,

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}{\text{=Δ±á}}^{\text{°}}{\text{-TΔ³§}}^{\text{°}}\\ {\text{0=Δ±á}}^{\text{°}}{\text{-TΔ³§}}^{\text{°}}\\ {\text{TΔ³§}}^{\text{°}}{\text{=Δ±á}}^{\text{°}}\\ \text{T=}\frac{{\text{Δ±á}}^{\text{°}}}{{\text{Δ³§}}^{\text{°}}}\\ \text{=}\frac{\text{30.9°ì´³Ã—}\frac{\text{1000J}}{\text{1k.J}}}{\text{93.15J/K}}\\ \text{T=332K}\end{array}$

Therefore, the boiling point of ${\text{Br}}_{\text{2}}\text{isat332K.}$