Q20.56 P - Chemistry: Molecular Nature Of Matter And Change

Question

For the gaseous reaction of xenon and fluorine to form xenon hexafluoride:

(a) Calculate ${\mathbf{\text{ΔS}}}^{\mathbf{\text{°}}}$ at  $\mathbf{\text{298 K}}\mathbf{(}{\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}\mathbf{\text{=-402 kJ/mol}}$ and  ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}\mathbf{\text{=-280.kJ/mol}}$ ).

(b) Assuming that ${\mathbf{\text{ΔS}}}^{\mathbf{\text{°}}}$ and  ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}$change little with temperature, calculate   ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$at 500. K.

Step-by-Step Solution

Answer

1. The entropy  $\left({\text{ΔS}}_{\text{rxn }}^{\text{0}}\right)$ value is  $\text{-0.409kJ/mol×K}$.
2. The standard free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-197 kJ/mol}$ .

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: Calculate ΔS °

 (a)

Considering the given information:

The following is the xenon hexafluoride formation reaction:

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

Record the enthalpy and free energy values,

 $\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=-402 kJ/mol}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-280.0 kJ/mol}\end{array}$

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

The standard free energy change equation is as follows:

 $\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}{\text{=ΔH}}_{\text{rxn}}^{\text{o}}{\text{-TΔS}}_{\text{rxn}}^{\text{o}}\\ {\text{ΔS}}^{\text{o}}\text{=}\frac{{\text{ΔH}}^{\text{n}}{\text{-ΔG}}^{\text{o}}}{\text{T}}\end{array}$

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

 $\begin{array}{c}{\text{ΔS}}^{\text{°}}\text{=}\frac{\text{-402 kJ/mol-(-280.0 kJ/mol)}}{\text{298 K}}\\ \text{ =}\frac{\text{-122.0 kJ/mol}}{\text{298 K}}\\ \text{ =-0.40939597 kJ/mol×K}\\ \text{ =-0.409 kJ/mol×K}\end{array}$

Therefore, the entropy value is  ${\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=-0.409 kJ/mol×K}.$

 

Step 3: Calculate ΔG °

(b)

Considering the given information:

The reaction that produces xenon hexafluoride is as follows:

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

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

Standard Free energy change equation is,

 ${\text{ΔG}}_{\text{rxn}}^{\text{o}}{\text{=ΔH}}_{\text{rxn}}^{\text{o}}{\text{-TΔS}}_{\text{rxn}}^{\text{o}}$

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

The values of enthalpy and entropy are,

 $\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=-402 kJ/mol}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-280.0 kJ/mol}\end{array}$

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

 $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=(-402 kJ/mol)-[(500 K)(-0.40939597 kJ/mol×K)]}\\ {\text{ =-197.302 kJ/molΔG}}_{\text{rxn}}^{\text{o}}\\ \text{ =-197 kJ/mol}\end{array}$

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