Question

For the reaction ${\mathbf{\text{I}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g)+Cl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\rightleftharpoons }\mathbf{\text{2ICl(g)}}$, calculate K_p at${\mathbf{\text{25}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}\mathbf{[}{\mathbf{\text{ΔG}}}_{\mathbf{\text{f}}}^{\mathbf{\text{°}}}$of$\mathbf{\text{ICl(g)=-6.075kJ/mol}}\mathbf{]}$.

Short answer

The K_p value is$\text{3.36×10}$

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 value of Kp

Considering the given information:

The following is the reaction:

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

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

Gibbs free energy equation is ${\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{{\text{ΔG}}_{\text{f}}}^{\text{°}}\text{(Products)-}\sum \text{n}{{\text{\hspace{0.17em}DeltaG}}_{\text{f}}}^{\text{°}}\text{(Reactants)}$

${\text{ΔG}}_{\text{rxn}}^{\text{o}}$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:

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

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

K, the equilibrium constant, is calculated.

We are aware of the equilibrium equation.

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

Rearrange the equation above,

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

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