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,

${\text{MgCO}}_{\text{3}}\text{(s)}⇌{\text{Mg}}^{\text{2+}}{\text{(aq)+CO}}_{\text{3}}^{\text{2-}}\text{(aq)}$

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.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{Mg}}^{\text{2+}}\text{(aq)}\right){\text{+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{CO}}_{\text{3}}^{\text{2-}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{MgCO}}_{\text{3}}\text{(s)}\right)\right]\\ \text{=([1(-456.01)+1(-528.10)]-[1(-1028)])kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=43.89kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ°ä}}^{\text{°}}}{\text{kT}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{43.89°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=-17.71}\\ {\text{K=e}}^{\text{-17.71}}\\ {\text{°­=2.0×10}}^{\text{-8}}\end{array}$

Therefore, the required value is ${\text{2.0×10}}^{\text{-8}}$.

(b)

Consider the reaction given below,

${\text{2HCl(g)+Br}}_{\text{2}}\text{(l)}⇌{\text{2HBr(g)+Cl}}_{\text{2}}\text{(g)}$

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.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}{\text{(HBr(g))+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{Cl}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}{\text{(HCl(g))+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{Br}}_{\text{2}}\text{(l)}\right)\right]\\ \text{=([2(-53.5)+1(0)]-[2(-95.30)+1(0)])kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=83.6kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{\text{Δ³¦}}{\text{kT}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{83.6°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=-33.74}\\ {\text{K=e}}^{\text{-33.74}}\\ {\text{°­=2.2×10}}^{\text{-15}}\end{array}$

Therefore, the required value is ${\text{2.2×10}}^{\text{-15}}$.

(c)

Consider the reaction given below,

${\text{H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}⇌{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(l)}$

To solve for K,

we can use the equation below.

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

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

B.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=}∑{\text{n}}_{\text{p}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(product)-}∑{\text{n}}_{\text{r}}{\text{Δ³Ò}}_{\text{f}}^{\text{°}}\text{(reactant)}\\ \text{=}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(l)}\right)\right]\text{-}\left[{\text{nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{(g)}\right){\text{+nΔ³Ò}}_{\text{f}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\\ \text{=([1(-120.4)]-[1(0)+1(0)])kJ/mol}\\ {\text{Δ³Ò}}^{\text{°}}\text{=-120.4kJ/mol}\end{array}$

Then solve for K.

$\begin{array}{c}{\text{Δ³Ò}}^{\text{°}}\text{=-RTlnK}\\ {\text{K=e}}^{\frac{{\text{Δ°ä}}^{\text{°}}}{\text{KI}}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-120.4°ì´³/³¾´Ç±ô×}\frac{\text{1000J}}{\text{1kJ}}}{\text{-8.314´³/³¾´Ç±ô×°­Ã—298°­}}\\ \frac{{\text{Δ³Ò}}^{\text{°}}}{\text{-RT}}\text{=48.60}\\ {\text{K=e}}^{\text{48.60}}\\ {\text{°­=1.3×10}}^{\text{21}}\end{array}$

Therefore, the required value is ${\text{1.3×10}}^{\text{21}}$.