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.

(a)

Consider the reaction given below,

$\text{NO(g)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}⇌{\text{NO}}_{\text{2}}\text{(g)}$

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

${\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}$

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

B.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{NO}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}{\text{(NO(g))+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\\ \text{=}\left(\text{[1(51)]-}\left[\text{1(86.60)+}\frac{\text{1}}{\text{2}}\text{(0)}\right]\right)\text{kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=-35.6kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ°ä}}^{\text{°}}}{\text{HI}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-35.6°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=14.37}\\ {\text{K=e}}^{\text{14.37}}\\ \text{K=1739148.759}\\ {\text{=1.7×10}}^{\text{6}}\end{array}$

Therefore, the required value is ${\text{1.7×10}}^{\text{6}}$.

(b)

Consider the reaction given below,

$\text{2HCl(g)}⇌{\text{H}}_{\text{2}}{\text{(g)+Cl}}_{\text{2}}\text{(g)}$

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

${\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}$

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

B.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{(g)}\right){\text{+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{Cl}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\text{(HCl(g))}\right]\\ \text{=([1(0)+1(0)]-[2(-95.30)])kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=190.6kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ°ä}}^{\text{°}}}{\text{RT}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{190.6°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=-76.93K}\\ {\text{=e}}^{\text{-76.93}}\\ {\text{°­=3.9×10}}^{\text{-34}}\end{array}$

Therefore, the required value is ${\text{3.9×10}}^{\text{-34}}$.

(c)

Consider the reaction given below,

${\text{2C(graphite)+O}}_{\text{2}}\text{(g)}⇌\text{2CO(g)}$

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

${\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for ${\text{Δ³Ò}}^{\text{°}}$using the Gibb's free energy constants found in Appendix

B.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\text{(CO(g))}\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}{\text{(C(graphite))+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\\ \text{=([2(-137.2)]-[2(0)+1(0)])kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=-274.4kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ³¦}}^{\text{°}}}{\text{HI}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-274.4°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=110.75}\\ {\text{K=e}}^{\text{110.75}}\\ {\text{°­=1.3×10}}^{\text{48}}\end{array}$

Therefore, the required value is ${\text{1.3×10}}^{\text{48}}$.