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 decomposition reaction:

${\text{CH}}_{\text{3}}\text{OH(g)}⇌{\text{CO(g)+2H}}_{\text{2}}\text{(g)}$

Calculate the change in Gibb's free energy at 298 K using the following formula:

Standard enthalpy change is,

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

$\begin{array}{l}\text{CO(g)=-110.5kJ/mol}\\ {\text{H}}_{\text{2}}\text{(g)=0kJ/mol}\\ {\text{CH}}_{\text{3}}\text{OH(g)=-201.2kJ/mol}\end{array}$

The reaction's enthalpy change is calculated as follows:

$\begin{array}{c}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}∑\text{m}{\text{Δ±á}}_{\text{f(Products)}}^{\text{°}}\text{-}∑\text{n}{\text{Δ±á}}_{\text{f((reactants)}}^{\text{°}}\text{-}[\left({\text{1molCH}}_{\text{3}}\text{OH}\right)({\text{Δ±á}}_{\text{f}}^{\text{°}}\text{of}{\text{CH}}_{\text{3}}\text{OH})]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{(1molCO)(-110.5kJ/mol)+}\left({\text{2molHH}}_{\text{2}}\right)\text{(0kJ/mol)}\right]\\ \text{-}\left[\left({\text{1molCHCOH}}_{\text{3}}\right)\text{(-201.2kJ/mol)}\right]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=90.7kJ}\end{array}$

The reaction's enthalpy change is positive.

Hence, the enthalpy $\left({\text{Δ±á}}_{\text{rxn}}^{\text{°}}\right)$ value is $\text{90.7kJ}$.

Entropy change ${\text{Δ³§}}_{\text{system}}^{\text{°}}$

Calculate the entropy change in this reaction as follows:

${\text{DS}}_{\text{rxn}}^{\text{0}}\text{=}∑\text{m}{\text{S}}_{{\text{P}}_{\text{roducts}}^{\text{0}}}^{\text{0}}\text{-}∑\text{n}{\text{S}}_{\text{reactants}}^{\text{0}}$

The stoichiometric co-efficient are (m) and (n), respectively.

$\begin{array}{c}{\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}[\text{(1molCO)}({\text{S}}^{\text{o}}\text{of}\text{CO}){\text{+(2molH)}}_{\text{2}})({\text{S}}^{\text{o}}\text{of}{\text{H}}_{\text{2}})]\\ \text{-}[\left({\text{1molCH}}_{\text{3}}\text{OH}\right)({\text{S}}^{\text{o}}\text{of}{\text{CH}}_{\text{3}}\text{OH})]\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=°Ú(1³¾´Ç±ô)(197.5´³/³¾´Ç±ô×°­)+(2³¾´Ç±ô)(130.6´³/³¾´Ç±ô×°­)}\\ \text{-°Ú(1³¾´Ç±ô)(238.0´³/³¾´Ç±ô×°­)±Õ}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=220.7=221J/K}\end{array}$

The $\left({\text{Δ³§}}_{\text{rxn}}^{\text{0}}\right)$ of the reaction is $\underset{\_}{\text{221J/K}}$.

For ${\text{Δ³§}}_{\text{rxn}}^{\text{°}}$, the enthalpy and entropy changes are positive, indicating that the formation of methanol is favored at the given temperature.

(b)

Considering the given decomposition reaction:

${\text{CH}}_{\text{3}}\text{OH(g)}⇌{\text{CO(g)+2H}}_{\text{2}}\text{(g)}$

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

Standard Free energy change equation is,

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

Temperature$\left({\text{T}}_{\text{1}}\right)$:

The enthalpy and entropy values calculated are,

$\begin{array}{c}{\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=90.7k.J}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{o}}\text{=220.7J/K}\\ {\text{T}}_{\text{1}}\text{=28+273=301K}\end{array}$

These numbers are plugging into the standard free energy equation,

$\begin{array}{c}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=90.7kJ-}\left[\text{(301K)(220.7J/K)}\left({\text{1kJ/10}}^{\text{3}}\right)\right]\\ \text{=24.2693kJ}\\ {\text{Δ³Ò}}_{\text{roo}}^{\text{o}}\text{=24.3kJ}\end{array}$

Temperature $\left({\text{T}}_{\text{2}}\right)$ :

The enthalpy and entropy values calculated are,

$\begin{array}{c}{\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=90.7kJ}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{o}}\text{=220.7J/K}\\ {\text{T}}_{\text{1}}\text{=128+273=401K}\end{array}$

These figures are used to fill in the blanks in the standard free energy equation.

$\begin{array}{c}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=90.7kJ-}\left[\text{(401K)(220.7J/K)}\left({\text{1kJ/10}}^{\text{3}}\right)\right]\\ {\text{=2.1993kJΔ³Ò}}_{\text{rxn}}^{\text{o}}\\ \text{=2.2kJ}\end{array}$

Temperature$\left({\text{T}}_{\text{3}}\right)$:

The enthalpy and entropy values calculated are,

$\begin{array}{c}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=90.7kJ}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{o}}\text{=220.7J/K}\\ {\text{T}}_{\text{1}}\text{=228+273=501K}\end{array}$

These values are plugging above standard free energy equation,

$\begin{array}{c}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=90.7kJ-}\left[\text{(501K)(220.7J/K)}\left({\text{1kJ/10}}^{\text{3}}\right)\right]\\ \text{=-19.8707kJ}\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-19.9kJ}\end{array}$

Independent standard free energy values are ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{24.3kJ,2.2kJand-19.9kJ}$.

(c)

Considering the given decomposition reaction:

${\text{CH}}_{\text{3}}\text{OH(g)}⇌{\text{CO(g)+2H}}_{\text{2}}\text{(g)}$

The enthalpy and entropy value calculated are:

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=90.7kJ}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=220.7J/K}\end{array}$

At ${\text{28}}^{\text{°}}\text{C}$, the reaction is non-spontaneous, near equilibrium at ${\text{128}}^{\text{°}}\text{C}$, and spontaneous at ${\text{228}}^{\text{°}}\text{C}$ for the substances in their standard states.

At high temperatures, reactions involving positive ${\text{Δ±á}}_{\text{rxn}}^{\text{°}}$ and ${\text{Δ³§}}_{\text{rxn}}^{\text{°}}$ become spontaneous.