Q20.61 P - Chemistry: Molecular Nature Of Matter And Change

Question

Use  ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}$and  ${\mathbf{\text{ΔS}}}^{\mathbf{\text{°}}}$values to find the temperature at which these sulfur allotropes reach equilibrium at  $\mathbf{\text{1 atm}}$:

 $\mathbf{\text{S(rhombic)}}\mathbf{\rightleftharpoons }\mathbf{\text{S(monoclinic)}}$

Step-by-Step Solution

Answer

The value is $\text{429K}$ .

Step 1: 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.

Step 2: Find the temperature at which these sulfur allotropes reach equilibrium

Considering the given reaction:

 $\text{S(rhombic)}\rightleftharpoons \text{S(monoclinic)}$

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

 $\begin{array}{c}{\text{ΔH}}^{\text{°}}\text{=}\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{°}}\text{( product )-}\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{°}}\text{( reactant )}\\ \text{ =}\left[{\text{nΔH}}_{\text{f}}^{\text{°}}\text{(S( monoclinic ))}\right]\text{-}\left[{\text{nΔH}}_{\text{f}}^{\text{°}}\text{(S( rhombic ))}\right]\\ {\text{ =([1(0.3)]-[1(0)])kJΔH}}^{\text{°}}\\ \text{ =0.3kJ}\end{array}$

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

 $\begin{array}{c}{\text{ΔS}}^{\text{°}}\text{=}\sum {\text{n}}_{\text{p}}{\text{S}}^{\text{°}}\text{( product )-}\sum {\text{n}}_{\text{r}}{\text{S}}^{\text{°}}\text{( reactant )}\\ \text{ =}\left[{\text{nS}}^{\text{°}}\text{(S( monoclinic ))}\right]\text{-}\left[{\text{nS}}^{\text{°}}\text{(S( rhombic ))}\right]\\ {\text{ =([1(32.6)]-[1(31.9)])J/KΔS}}^{\text{°}}\\ \text{ =0.7J/K}\end{array}$

Finally, find T at equilibrium.${\text{ΔG}}^{\text{°}}\text{=0}$  ,

 $\begin{array}{c}{\text{ΔG}}^{\text{°}}{\text{=ΔH}}^{\text{°}}{\text{-TΔS}}^{\text{°}}\\ {\text{ 0=ΔH}}^{\text{°}}{\text{-TΔS}}^{\text{°}}\\ {\text{TΔS}}^{\text{°}}{\text{=ΔH}}^{\text{°}}\\ \text{ T=}\frac{{\text{ΔH}}^{\text{°}}}{{\text{ΔS}}^{\text{°}}}\\ \text{ =}\frac{\text{0.3kJ×}\frac{\text{1000J}}{\text{1kJ}}}{\text{0.7J/K}}\\ \text{ T=429K}\end{array}$

Therefore, the T at equilibrium of sulfur allotropes is at $\text{429 K}$ .