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 information:

The following is the reaction:

${\text{I}}_{\text{2}}{\text{(g)+C}}_{\text{2}}\text{(g)}⇌\text{2ICl(g)}$

Calculate the change in Gibb's free energy at 298k by using the following formula:

Gibbs free energy equation is ${\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=}∑\text{m}{{\text{Δ³Ò}}_{\text{f}}}^{\text{°}}\text{(Products)-}∑\text{n}{{\text{ D±ð±ô³Ù²¹³Ò}}_{\text{f}}}^{\text{°}}\text{(Reactants)}$

${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}$Formation of values,

$\begin{array}{c}{\text{I}}_{\text{2}}\text{(g)=19.37kJ/mol}\\ {\text{Cl}}_{\text{2}}\text{(g)=0kJ/mol}\\ \text{ICl(g)=-6.075kJ/mol}\end{array}$

The reaction's free energy change is calculated as follows:

$\begin{array}{l}{\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{(2molICl)}\left({\text{Δ³Ò}}_{\text{f}}^{\text{o}}\text{ofICl}\right)\right]\text{-}[\left({\text{1molI}}_{\text{2}}\right)\left({\text{Δ³Ò}}_{\text{f}}^{\text{o}}{\text{ofI}}_{\text{2}}\right)\text{+}\left({\text{1molCl}}_{\text{2}}\right)\left({\text{Δ³Ò}}_{\text{f}}^{\text{o}}{\text{ofCl}}_{\text{2}}\right)]\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=[(2molICl)(-6.075kJ/mol)]-}[\left({\text{1molI}}_{\text{2}}\right)\text{(19.38kJ/mol)}\text{+}\left({\text{1molCl}}_{\text{2}}\right)\text{(0kJ/mol)}]\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=-31.53kJ}\end{array}$

The value of ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}$is -31.53kJ and these value of ${\text{Δ³Ò}}_{\text{f}}^{\text{o}}$are referred from the Appendix B.

K, the equilibrium constant, is calculated.

We are aware of the equilibrium equation.

${\text{Δ³Ò=Δ³Ò}}^{\text{o}}\text{+RTln(K)}$

Rearrange the equation above,

$\begin{array}{c}{\text{lnK}}_{\text{p}}\text{=-}\frac{{\text{Δ³Ò}}^{\text{°}}}{\text{RT}}\\ \text{=}\left(\frac{\text{-31.53kJ/mol}}{\text{-(8.314´³/³¾´Ç±ô×°­)(298°­)}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1kJ}}\right)\\ {\text{lnK}}_{\text{p}}\text{=12.726169}\\ {\text{K}}_{\text{p}}{\text{=e}}^{\text{12.726169}}\\ {\text{K}}_{\text{p}}{\text{=3.3643794×10}}^{\text{5}}\text{(or)}\\ {\text{K}}_{\text{p}}{\text{=3.36×10}}^{\text{5}}\end{array}$

Therefore, the standard equilibrium K_p value is$\underset{\_}{{\text{3.36×10}}^{\text{5}}}$