Calculate K at$\mathbf{\text{298K}}$ for each reaction:

(a)${\mathbf{\text{SrSO}}}_{\mathbf{\text{4}}}\mathbf{\text{(s)}}\mathbf{\rightleftharpoons }{\mathbf{\text{Sr}}}^{\mathbf{\text{2+}}}{{\mathbf{\text{(aq)+SO}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2-}}}\mathbf{\text{(aq)}}$

(b)${\mathbf{\text{2NO(g)+Cl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\rightleftharpoons }\mathbf{\text{2NOCl(g)}}$

(c)${\mathbf{\text{Cu}}}_{\mathbf{\text{2}}}{\mathbf{\text{S(s)+O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\rightleftharpoons }{\mathbf{\text{2Cu(s)+SO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

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.

(a)

Consider the reaction given below,

${\text{SrSO}}_{\text{4}}\text{(s)}\rightleftharpoons {\text{Sr}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

To solve for K, we can use the equation below.

${\text{ΔG}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for using the Gibb's free energy constants found in Appendix

B.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=}\sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(product)-}\sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}\left({\text{Sr}}^{\text{2+}}\text{(aq)}\right){\text{+nΔG}}_{\text{f}}^{\text{°}}\left({\text{SO}}_{\text{4}}^{\text{2-}}\text{(aq)}\right)\right]\text{-}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}\left({\text{SrSO}}_{\text{4}}\text{(s)}\right)\right]\\ \text{=([1(-557.3)+1(-741.99)]-[1(-1334)])kJ/mol}\\ {\text{ΔG}}^{\text{°}}\text{=34.71kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{ΔC}}^{\text{°}}}{\text{HI}}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{34.71kJ/mol×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314J/mol×K×298K}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=-14.01}\\ {\text{K=e}}^{\text{-14.01}}\\ {\text{K=8.2×10}}^{\text{-7}}\end{array}$

Therefore, the required value is ${\text{8.2×10}}^{\text{-7}}$.

(b)

Consider the reaction given below,

${\text{2NO(g)+Cl}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2NOCl(g)}$

To solve for K,

we can use the equation below.

${\text{ΔG}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for ${\text{ΔG}}^{\text{°}}$using the Gibb's free energy constants found in Appendix

B.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=}\sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(product)-}\sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\text{[}{\text{nΔG}}_{\text{f}}^{\text{°}}\text{(NOCl(g))}\text{]}\text{-}\text{[}{\text{nΔG}}_{\text{f}}^{\text{°}}{\text{(NO(g))+nΔG}}_{\text{f}}^{\text{°}}\text{(}{\text{Cl}}_{\text{2}}\text{(g)}\text{)}\text{]}\\ \text{=([2(66.07)]-[2(86.60)+1(0)])kJ/mol}\\ {\text{ΔG}}^{\text{°}}\text{=-41.06kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{ΔC}}^{\text{°}}}{\text{RT}}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-41.06kJ/mol×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314J/mol×K×298K}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=16.57}\\ {\text{K=e}}^{\text{16.57}}\\ {\text{K=1.6×10}}^{\text{7}}\end{array}$

Therefore, the required value is ${\text{1.6×10}}^{\text{7}}$.

(c)

Consider the reaction given below,

${\text{Cu}}_{\text{2}}{\text{S(s)+O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2Cu(s)+SO}}_{\text{2}}\text{(g)}$

To solve for K, we can use the equation below.

${\text{ΔG}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for ${\text{ΔG}}^{\text{°}}$using the Gibb's free energy constants found in Appendix

B.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=}\sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(product)-}\sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}{\text{(Cu(s))+nΔG}}_{\text{f}}^{\text{°}}\left({\text{SO}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}\left({\text{Cu}}_{\text{2}}\text{S(s)}\right){\text{+nΔG}}_{\text{f}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\\ \text{=([2(0)+1(-300.2)]-[1(-86.2)+1(0)])kJ/mol}\\ {\text{ΔG}}^{\text{°}}\text{=-214kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{ΔC}}^{\text{°}}}{\text{KT}}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-214kJ/mol×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314J/mol×K×298K}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=86.37}\\ {\text{K=e}}^{\text{86.37}}\\ {\text{K=3.3×10}}^{\text{37}}\end{array}$

Therefore, the required value is ${\text{3.3×10}}^{\text{37}}$.