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 information:

The chemical equation for the formation of ${\text{CO}}_{\text{2}}\text{(g)}$ from its elements is shown below:

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

According to the above equation, one mole of ${\text{CO}}_{\text{2}}$is formed from its elements ${\text{CO}}_{\text{2}}$ in their standard states.

Predicting entropy ${{\text{Δ³§}}^{\text{°}}}_{\text{system}}$:

$\text{1,}\frac{\text{1}}{\text{2}}$moles of gaseous reactants are converted into one mole of gaseous product in this reaction.

$\begin{array}{c}{\text{Δ³§}}_{\text{rxn}}^{\text{0}}{\text{=S}}_{\text{Products}}^{\text{0}}{\text{-S}}_{\text{reactants}}^{\text{0}}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=1-2}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=-1}\end{array}$

Because the number of moles of gas decreases from reactants to products, the reaction entropy ${\text{Δ³§}}^{\text{°}}$is negative.

Enthalpy as a predictor of ${\text{Δ±á}}_{\text{rxn}}^{\text{o}}$:

The oxidation of carbon monoxide CO requires an initial energy input to start the reaction but then releases energy that is exothermic and has a negative enthalpy ${\text{Δ±á}}_{\text{rxn}}^{\text{o}}$.

(b)

Considering the given information:

The chemical equation for the formation of ${\text{CO}}_{\text{2}}\text{(g)}$from its elements is shown below:

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

1st method:

Calculate ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}$ from ${\text{Δ³Ò}}_{\text{f}}^{\text{o}}$values of products and reactants,

The Gibbs free energy equation is as follows:

${\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=}∑\text{m}{{\text{Δ³Ò}}_{\text{f}}}^{\text{°}}\text{(Products)-}∑\text{n}{{\text{Δ³Ò}}_{\text{f}}}^{\text{°}}\text{(Reactants)}$

The reaction's free energy change is calculated as follows:

$\begin{array}{l}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=}[\left({\text{1molCO}}_{\text{2}}\right)\text{ }({\text{Δ³Ò}}_{\text{f}}^{\text{o}}\text{of}{\text{CO}}_{\text{2}})]\text{-}[\text{(1molCO)}({\text{Δ³Ò}}_{\text{f}}^{\text{o}}\text{ofCO)+}\left({\text{1/2molO}}_{\text{2}}\right)({\text{Δ³Ò}}_{\text{f}}^{\text{o}}\text{of}{\text{O}}_{\text{2}})]\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=[(2mol)(-394.4kJ/mol)]-[(2mol)(-137.2kJ/mol)}\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\text{(0kJ/mol)}]\end{array}$

${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-257.2kJ}$

As a result, the standard free energy value is ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{-257.2}}\text{kJ}$.

2nd Method

Given the reaction,

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

Calculate ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}$from ${\text{Δ±á}}_{\text{rxn}}^{\text{o}}$and ${\text{Δ³§}}_{\text{rxn}}^{\text{o}}$ at $\text{298K}$

The change in standard enthalpy is,

The enthalpy change for the reaction is calculated as follows:

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}∑\text{m}{\text{Δ±á}}_{\text{f(Products)}}^{\text{°}}\text{-}∑{\text{nΔ±á}}_{\text{f(reactants)}}^{\text{°}}\\ {\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=}\left[\left({\text{1molCOO}}_{\text{2}}\right)\left({\text{Δ±á}}_{\text{rxn}}^{\text{o}}{\text{ofCO}}_{\text{2}}\right)\right]\text{-}\left[\text{(1molCO)}\left({\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{ofCO}\right)\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\left({\text{Δ±á}}_{\text{rxn}}^{\text{o}}{\text{ofO}}_{\text{2}}\right)\right]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=}\left[{\text{(1molCO)}}_{\text{2}}\right)\text{(-393.5kJ/mol)}]\text{-[(1molCO)(-110.5kJ/mol)}\text{+}\left({\text{1/2molOO}}_{\text{2}}\right)\text{(0kJ/mol)}]\\ {\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=-283.0kJ}\end{array}$

The enthalpy change is negative.

As a result, the enthalpy $\left({\text{Δ±á}}_{\text{rxn}}^{\text{°}}\right)\text{valueis-283.0kJ}$

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

The standard entropy change equation is,

${\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}∑{\text{mS}}_{\text{Products}}^{\text{°}}\text{-}∑\text{n}{\text{S}}_{\text{reactants}}^{\text{°}}$

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

$\begin{array}{l}{\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{2molCO}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ofCO}}_{\text{2}}\right)\right]\text{-}\left[\text{(1molCO)}\left({\text{S}}^{\text{o}}\text{ofCO}\right)\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ofO}}_{\text{2}}\right)\right]\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=°Ú(1³¾´Ç±ô°ä°¿2)(213.7´³/³¾´Ç±ô×°­)±Õ-°Ú(1³¾´Ç±ô°ä°¿)(197.5´³/³¾´Ç±ô×°­)}\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\text{(205.0´³/³¾´Ç±ô×°­)}]\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=-86.3J/K}\end{array}$

Therefore, the $\left({\text{Δ³§}}_{\text{rxn}}^{\text{°}}\right)$of the reaction is $\text{-86.3JJ/K}$

Next, compute the change in free energy ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}$

Standard The equation for free energy change 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 calculated are

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=-283.0kJ}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=-283.0kJ}\end{array}$

These values are entered into the standard free energy equation,

$\begin{array}{l}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-283.0kJ-}\left[\text{(298K)(-86.3J/K)}\left({\text{1kJ/10}}^{\text{3}}\text{J}\right)\right]\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-257.2826kJ}\end{array}$

Therefore, the standard free energy value is ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{-257.3kJ}}$.