Question

Use ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}$and${\mathbf{\text{ΔS}}}^{\mathbf{\text{°}}}$ values for the following process at$\mathbf{\text{1atm}}$ to find the normal boiling point of${\mathbf{\text{Br}}}_{\mathbf{\text{2}}}$ :

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

Short answer

The value is $\text{332K}$

Step-by-step solution

Definition of Concept

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.

Find the normal boiling point of  Br2

Considering the given reaction:

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

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

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

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

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

Finally, find T at equilibrium. ,

$\begin{array}{c}{\text{ΔG}}^{\text{°}}{\text{=ΔH}}^{\text{°}}{\text{-TΔS}}^{\text{°}}\\ {\text{0=ΔH}}^{\text{°}}{\text{-TΔS}}^{\text{°}}\\ {\text{TΔS}}^{\text{°}}{\text{=ΔH}}^{\text{°}}\\ \text{T=}\frac{{\text{ΔH}}^{\text{°}}}{{\text{ΔS}}^{\text{°}}}\\ \text{=}\frac{\text{30.9kJ×}\frac{\text{1000J}}{\text{1k.J}}}{\text{93.15J/K}}\\ \text{T=332K}\end{array}$

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