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.

Considering the given information:

The fermentation reaction looks like this:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}→{\text{2 C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{ O±á(±ô)+2°ä°¿}}_{\text{2}}\text{(g)}$

The typical enthalpy change is,

The formation of values, ${\text{Δ±á}}^{\text{o}}$

$\begin{array}{c}{\text{H}}_{\text{2}}\text{(g)=0kJ/mol}\\ {\text{O}}_{\text{2}}\text{(g)=0kJ/mol}\\ {\text{H}}_{\text{2}}\text{O(g)=-241.826kJ/mol}\end{array}$

The reaction's enthalpy change is calculated as follows:

$\begin{array}{c}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}∑\text{m}{\text{Δ±á}}_{\text{f}}^{\text{°}}\text{Preducts)-}∑\text{n}{\text{Δ±á}}_{\text{f(reactants)}}^{\text{°}}\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{2molC}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\left({\text{Δ±á}}_{\text{f}}^{\text{o}}{\text{ofC}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\text{+}\left({\text{2molCOCO}}_{\text{2}}\right)\left({\text{Δ±á}}_{\text{f}}^{\text{o}}{\text{ofCO}}_{\text{2}}\right)]\\ \text{-}[\left({\text{1molC}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)\left({\text{Δ±á}}_{\text{f}}^{\text{°}}{\text{ofC}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)]\end{array}$

$\begin{array}{c}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{2molC}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\text{(-277.63kJ/mol)+}\left({\text{2molCO}}_{\text{2}}\right)\text{(-393.5kJ/mol)}\right]\\ \left[\left({\text{1molC}}_{\text{6}}{\text{H}}_{\text{l2}}{\text{O}}_{\text{6}}\right)\text{(-1273.3kJ/mol)}\right]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=-68.96kJ=-69.0kJ}\end{array}$

The change in enthalpy is negative.

As a result, the enthalpy $\left({\text{Δ±á}}_{\text{rxn}}^{\text{°}}\right)\text{changesis-69.0kJ}$

Change in entropy ${\text{Δ³§}}_{\text{system}}^{\text{°}}$

Calculate the entropy change for this reaction as follows:

${\text{DS}}_{\text{rxn}}^{\text{°}}\text{=}∑{\text{mS}}_{{\text{P}}_{\text{roducts}}}^{\text{°}}\text{-}∑{\text{nS}}_{\text{reactants}}^{\text{°}}$

Here, m, n are moles of individual species, which are given by individual species coefficients in the balanced chemical equation.

${\text{Δ³§}}_{\text{rxn}}^{\text{0}}\text{=537J/K}$

Hence, the$({\text{Δ³§}}_{\text{rxn}}^{\text{0}}$) of the reaction is$\underset{\_}{\text{537J/K}}$(or)$\text{0.5373kJ/K}$.

Determine the free energy change ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}$.

The standard free energy change equation is as follows:

${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}{\text{=Δ±á}}_{\text{rxn}}^{\text{o}}{\text{-TΔ³§}}_{\text{rxn}}^{\text{o}}$