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)

The reaction of ${\text{H}}_{\text{2}}$ with ${\text{O}}_{\text{2}}$ to give ${\text{H}}_{\text{2}}\text{O}$ formation is,

${\text{H}}_{\text{2}}\text{(g)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}→{\text{H}}_{\text{2}}\text{O(g)}$

The coefficient are written this way instead of

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

The specific thermodynamic values (per${\text{1molH}}_{\text{2}}$) not per ${\text{2molH}}_{\text{2}}$.

Standard enthalpy change is

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

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

The enthalpy change for the reaction is calculated as follows,

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}∑\text{m}{\text{Δ±á}}_{\text{f(Products)}}^{\text{°}}\text{-}∑\text{n}{\text{Δ±á}}^{\text{°}}{\text{f}}_{\text{(reactants)}}\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1mol}}_{\text{2}}\text{O}\right)\left({{\text{Δ±á}}_{\text{f}}^{\text{°}}}_{\text{of}}{\text{H}}_{\text{2}}\text{O}\right)\right]\end{array}$

$\begin{array}{c}\text{-}\left[\left({\text{1molH}}_{\text{2}}\right)\left({{\text{Δ±á}}_{\text{f}}^{\text{°}}}_{\text{f}}{\text{ofH}}_{\text{2}}\right)\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\left({\text{Δ±á}}_{\text{f}}^{\text{°}}{\text{ofO}}_{\text{2}}\right)\right]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1molH}}_{\text{2}}\text{O}\right)\text{(-241.826kJ/mol)}\right]\\ \text{-}[\left({\text{1molH}}_{\text{2}}\right)\text{(0kJ/mol)}\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\text{(0kJ/mol)}]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=-241.826kJ}\end{array}$

The enthalpy change is negative.

Hence, the enthalpy $\left({\text{Δ±á}}_{\text{rxn}}^{\text{°}}\right)\text{changesis-241.826kJ}$.

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

Calculate the change in entropy for this reaction as follows,

$\begin{array}{c}{\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}∑{\text{mS}}_{\text{Products}}^{\text{0}}\text{-}∑{\text{nS}}_{\text{reactants}}^{\text{0}}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1molH}}_{\text{2}}\text{O}\right)\left({\text{S}}^{\text{°}}{\text{ofH}}_{\text{2}}\text{O}\right)\right]\\ \text{-}\left[\text{(1molH2)}\left({\text{S}}^{\text{o}}{\text{ofN}}_{\text{2}}\right)\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ofO}}_{\text{2}}\right)\right]\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1molH}}_{\text{2}}\text{O}\right)\text{(188.72´³/³¾´Ç±ô×°­)}\right]\\ \text{-}\left[\text{(1³¾´Ç±ô±á2)(130.6´³/³¾´Ç±ô×°­)+}\left({\text{1/2molO}}_{\text{2}}\right)\text{(205.0´³/³¾´Ç±ô×°­)}\right]\\ {{\text{Δ³§}}^{\text{°}}}_{\text{rxn}}\text{=-44.38}\\ \text{=-44.4J/K}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=-0.0444kJ/K}\end{array}$

Hence, the $\left({\text{Δ³§}}_{\text{rxn}}^{\text{0}}\right)$of the reaction is $\text{-44.4J/K}$

Calculate the Free energy change ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}$ is,

Standard Free energy change equation is,

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

Free energy change ${\text{Δ³Ò}}_{\text{f}}^{\text{o}}$

Enthalpy and entropy values are

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=-241.826}\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-0.0444kJ/K}\end{array}$

These values are plugging above standard free energy equation,

$\begin{array}{c}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=(-241.826kJ)-[(298K)(-0.04438kJ/K)]}\\ \text{=-228.6008kJ/mol}\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-228.6kJ}\end{array}$

Therefore, the given reaction standard free energy value is $\text{-228.6kJ}$.

(b)

Given reaction,

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

This reaction is does not dependent on temperature T, because enthalpy and values are negative.

This reaction will become non-spontaneous at higher temperature because the positive$\text{(-TΔ³§)}$ term becomes larger than negative $\text{Δ±á}$ term. Further in this reaction.

(c)

Given reaction,

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

The reaction becomes spontaneous below the temperature where ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=0}$

Consider the following free energy equation,

$\begin{array}{l}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}{\text{=0=Δ±á}}_{\text{rxn}}^{\text{o}}{\text{-TΔ³§}}_{\text{rxn}}^{\text{o}}\text{-----[1]}\\ {\text{Δ±á}}_{\text{rxn}}^{\text{o}}{\text{=TΔ³§}}_{\text{rxn}}^{\text{°}}\text{------[2]}\end{array}$

Rearrange equation (2) to calculate temperature T,

$\text{T=}\frac{{\text{Δ±á}}_{\text{rnn}}^{\text{o}}}{{\text{Δ³§}}_{\text{ran}}^{\text{o}}}$

Hence,

$\begin{array}{c}\text{T=}\frac{\text{-241.826kJ}}{\text{-0.04438kJ/K}}\\ \text{=5448.986}\\ {\text{°Õ=5.45×10}}^{\text{3}}\text{K}\end{array}$

At temperature become above $\text{7281.45K}$, this reaction is non-spontaneous. Because both enthalpy ${\text{Δ±á}}_{\text{rxn}}^{\text{o}}$and entropy ${\text{Δ³§}}_{\text{rxn}}^{\text{o}}$values are negative, the reaction becomes non-spontaneous calculated temperature T.