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)

Considering the given reaction:

The free energy equation is as follows:

$2\mathrm{Mg}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)→2\mathrm{MgO}\left(\mathrm{s}\right)$

Two moles of solid and one mole of gas react to produce two moles of solid product in the above reaction.

The number of particles decreases as well, indicating that entropy is decreasing.

So, there ${\text{Δ³Ò}}_f^o$values are zero the solid is less than gas.

Free energy change ${\text{Δ³Ò}}_f^o$

According to the source,

$\begin{array}{c}\mathrm{MgO}\left(\mathrm{s}\right)=-569.0\mathrm{kJ}/\mathrm{mol}\\ \mathrm{Mg}\left(\mathrm{s}\right)=0\mathrm{kJ}/\mathrm{mol}\\ {\mathrm{O}}_2\left(\mathrm{g}\right)=0\mathrm{kJ}/\mathrm{mol}\end{array}$

The free energy change equation is as follows:

role="math" localid="1663371711802" $\begin{array}{c}{\mathrm{Δ³Ò}}_{\mathrm{rxn}}^{\mathrm{o}}=∑\mathrm{m}{\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Products}\right)-∑\mathrm{n}{\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Reactants}\right)\\ {\mathrm{Δ³Ò}}^{∘}=\left[\left(2\mathrm{molMgO}\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{ofMgO}\right)\right]-\left[\left(2\mathrm{molMg}\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{ofMg}\right)+\left(1{\mathrm{molO}}_2\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}{\mathrm{ofO}}_2\right)\right]\\ {\mathrm{Δ³Ò}}^{∘}=\left[\right(2\mathrm{molMgO}\left)\right(-569.0\mathrm{kJ}/\mathrm{mol}\left)\right]-\left[\left(2\mathrm{molMg}\right)(0\mathrm{kJ}/\mathrm{mol})+\left(1{\mathrm{molO}}_2\right)(0\mathrm{kJ}/\mathrm{mol})\right.\\ {\mathrm{Δ³Ò}}^{∘}=-1138.0\mathrm{kJ}\end{array}$

Therefore, the standard free energy value is ${\mathrm{Δ³Ò}}_{\mathrm{rxn}}^{\mathrm{o}}=-1138.0\mathrm{kJ}$.

Considering the given reaction:

$2{\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{g}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)→2{\mathrm{CO}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

A five-mole gaseous molecule reacts to produce a six-mole gaseous product in this reaction. As a result, the ${\text{Δ³Ò}}_f^o$ values are zero.

Free energy change ${\text{Δ³Ò}}_f^o$

According to the source,

$\begin{array}{c}{\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{g}\right)=-161.9\mathrm{kJ}/\mathrm{mol}\\ {\mathrm{O}}_2=0\mathrm{kJ}/\mathrm{mol}\\ {\mathrm{CO}}_2\left(\mathrm{g}\right)=-394.4\mathrm{kJ}/\mathrm{mol}\\ {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=-228.6\mathrm{kJ}/\mathrm{mol}\end{array}$

The free energy change equation is as follows:

role="math" localid="1663372070295" $\begin{array}{c}{\mathrm{Δ³Ò}}_{\mathrm{rxn}}^{\mathrm{o}}=∑{\mathrm{m}}_{\mathrm{f}}{\mathrm{G}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Products}\right)-∑\mathrm{n}{\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Reactants}\right)\\ {\mathrm{Δ³Ò}}^{\mathrm{o}}=\left[\left(2{\mathrm{molCO}}_2\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}{\mathrm{ofCO}}_2\right)\right.\left.+\left(4{\mathrm{molH}}_2\mathrm{O}\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}{\mathrm{ofH}}_2\mathrm{O}\right)\right]\\ -\left[\left(2{\mathrm{molCH}}_3\mathrm{OH}\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}{\mathrm{ofCH}}_3\mathrm{OH}\right)+\left(3{\mathrm{molO}}_2\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}{\mathrm{ofO}}_2\right)\right]\\ {\mathrm{Δ³Ò}}^{\mathrm{o}}=\left[\left(2{\mathrm{molCO}}_2\right)(-394.4\mathrm{kJ}/\mathrm{mol})+\left(4{\mathrm{molH}}_2\mathrm{O}\right)(-288.60\mathrm{kJ}/\mathrm{mol})\right]\\ \left[\left(2{\mathrm{molCH}}_3\mathrm{OH}\right)(-161.9\mathrm{kJ}/\mathrm{mol})\right.\left.+\left(3{\mathrm{molO}}_2\right)(0\mathrm{kJ}/\mathrm{mol})\right]\\ {\mathrm{Δ³Ò}}^{\mathrm{o}}=-1379.4\mathrm{kJ}\end{array}$

Therefore, the standard free energy value is ${\mathrm{Δ³Ò}}^{\mathrm{o}}=-1379.4\mathrm{kJ}$.

(c)

Considering the given reaction:

$\mathrm{BaO}\left(\mathrm{s}\right)+\mathrm{C}{\mathrm{O}}_2\left(\mathrm{g}\right)→{\mathrm{BaCO}}_3\left(\mathrm{s}\right)$

The number of particles decreases as well, indicating that entropy is decreasing.

As a result, the ${\text{Δ³Ò}}_f^o$values are zero, indicating that the solid is less than the gas.

Free energy change${\text{Δ³Ò}}_f^o$

According to the source,

$\begin{array}{c}\mathrm{BaO}\left(\mathrm{s}\right)=-520.4\mathrm{kJ}/\mathrm{mol}\\ {\mathrm{CO}}_2\left(\mathrm{g}\right)=-394.4\mathrm{kJ}/\mathrm{mol}\\ {\mathrm{BaCO}}_3\left(\mathrm{g}\right)=-1139\mathrm{kJ}/\mathrm{mol}\end{array}$

The free energy change equation is as follows:

$\begin{array}{c}{\mathrm{Δ³Ò}}_{\mathrm{rxn}}^{\mathrm{o}}=∑\mathrm{m}{\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Products}\right)-∑\mathrm{n}{\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Reactants}\right)\\ {\mathrm{Δ³Ò}}^0=\left[\left(1{\mathrm{molBaCO}}_3\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}{\mathrm{ofBaCO}}_3\right)\right]-\left[\left(1\mathrm{molBaO}\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{ofBaO}\right)+\left(1{\mathrm{molCO}}_2\right)\left({\mathrm{Δ³Ò}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{ofCO}2\right)\right]\\ {\mathrm{Δ³Ò}}^{\mathrm{o}}=\left[\left(1{\mathrm{molBaCO}}_3\right)(-1139\mathrm{kJ}/\mathrm{mol})\right]-\left[\right(1\mathrm{molBaO}\left)\right(-520.4\mathrm{kJ}/\mathrm{mol})\left.+\left(1{\mathrm{molCO}}_2\right)(-394.4\mathrm{kJ}/\mathrm{mol})\right]\\ {\mathrm{Δ³Ò}}^{\mathrm{o}}=-224.2\mathrm{kJ}\end{array}$

Therefore, the standard free energy value is ${\mathrm{Δ³Ò}}^{\mathrm{o}}=-224.2\mathrm{kJ}$.