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 butane from its constituent elements is shown below.

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{(g)+}\frac{\text{13}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}→{\text{4 C°¿}}_{\text{2}}{\text{(g)+5H}}_{\text{2}}\text{O(g)}$

One mole of butane reacted with oxygen to give eight moles of ${\text{CO}}_{\text{2}}$and ${\text{H}}_{\text{2}}\text{O}$

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

In this reaction, $\text{1,}\frac{\text{13}}{\text{2}}$moles of gaseous reactants are converted into 9 moles of gaseous product.

$\begin{array}{l}{\text{Δ³§}}_{\text{rxn}}^{\text{0}}{\text{=S}}_{\text{Products}}^{\text{0}}{\text{-S}}_{\begin{array}{l}\text{reactants}\\ \end{array}}^{\text{0}}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=9-2}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=7}\end{array}$

An increase in the number of moles of gas in this reaction should result in a positive ${\text{Δ³§}}^{\text{°}}$ value.

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

Butane combustion results in the release of energy or a negative enthalpy ${\text{Δ±á}}^{\text{o}}$ value

(b)

Considering the given information:

The chemical formula for producing butane from its constituent elements is shown below.

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{(g)+}\frac{\text{13}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}→{\text{4 C°¿}}_{\text{2}}{\text{(g)+5H}}_{\text{2}}\text{O(g)}$

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:

$\begin{array}{c}{\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{4molCO}}_{\text{2}}\right)\left({\text{Δ³Ò}}_{\text{f}}^{\text{o}}{\text{ofCO}}_{\text{2}}\right)\text{+}\left({\text{5molH}}_{\text{2}}\text{O}\right)\left({\text{Δ³Ò}}_{\text{f}}^{\text{o}}{\text{ofH}}_{\text{2}}\text{O}\right)]\\ \text{-}[\left({\text{1molC}}_{\text{4}}{\text{H}}_{\text{10}}\right)\left({\text{Δ³Ò}}_{\text{f}}^{\text{o}}{\text{ofC}}_{\text{4}}{\text{H}}_{\text{10}}\right)\text{+(12/2molO}\\ {\text{Δ³Ò}}_{\text{°ùײÔ}}\text{=[(4mol)(-394.4kJ/mol)+(5mol)(-228.60kJ/mol)]}\\ \text{-}[\left({\text{1molCC}}_{\text{4}}{\text{H}}_{\text{10}}\right)\text{(-16.7kJ/mol)}\text{+}\left({\text{13/2molO}}_{\text{2}}\right)\text{(0kJ/mol)}]\end{array}$

${\text{Δ³Ò}}_{\text{rxn}}^{\text{0}}\text{=-2703.9kJ}$

As a result, the standard free energy value is ${\text{Δ³Ò}}_{\text{rxn}}^{\text{0}}\text{=-2703.9kJ}$.

2nd Method

Given the reaction,

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{(g)+}\frac{\text{13}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}→{\text{4 C°¿}}_{\text{2}}{\text{(g)+5H}}_{\text{2}}\text{O(g)}$

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 change in enthalpy for the reaction is calculated as follows:

$\begin{array}{c}{\text{Δ±á}}_{\text{rxn}}^{\text{°}}\text{=}∑\text{m}{\text{Δ±á}}_{\text{f(Products)}}^{\text{°}}\text{-}∑\text{n}{\text{Δ±á}}_{\text{f(reactants)}}^{\text{°}}\\ {\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=}\left[\left({\text{4molCO}}_{\text{2}}\right){\left({\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{ofCO}\right)}_{\text{2}}\right)\text{+}\left({\text{5molH}}_{\text{2}}\text{O}\right)({\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{of}{\text{H}}_{\text{2}}\text{O})]\\ \text{-}[\left({\text{1molC}}_{\text{4}}{\text{H}}_{\text{8}}\right)({\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}})\text{+}\left({\text{13/2molO}}_{\text{2}}\right)({\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{of}{\text{O}}_{\text{2}})\\ {\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=}\left[\left({\text{1molCO}}_{\text{2}}\right)\text{(-393.5kJ/mol)+}\left({\text{5molH}}_{\text{2}}\text{O}\right)\text{(-241.826kJ/mol)}\right]\\ [\left({\text{1molC}}_{\text{4}}{\text{H}}_{\text{8}}\right)\text{(-126kJ/mol)}\text{+(13/2molO2)(0kJ/mol)]}\\ {\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=-2657.13kJ}\end{array}$

The enthalpy change is negative.

As a result, the enthalpy${\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=-2657.13kJ}$

Entropy change

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).

$\begin{array}{c}{\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{4molCO}}_{\text{2}}\right)({\text{S}}^{\text{o}}\text{of}{\text{CO}}_{\text{2}})\text{+(5molH2O)}({\text{S}}^{\text{o}}\text{of}{\text{H}}_{\text{2}}\text{O})]\\ \text{-}[\left({\text{1molC4H}}_{\text{8}}\right)({\text{S}}^{\text{o}}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}})\text{+}\left({\text{13/2molO}}_{\text{2}}\right)({\text{S}}^{\text{o}}\text{of}{\text{O}}_{\text{2}})]\\ {\text{Δ³§}}_{\text{rxn}}^{\text{o}}\text{=}\left[\left({\text{4molCO}}_{\text{2}}\right)\text{(213.7´³/³¾´Ç±ô×°­)+}\left({\text{5molH}}_{\text{2}}\text{O}\right)\text{(188.72´³/³¾´Ç±ô×°­)}\right]\\ [\left({\text{1molC4H}}_{\text{8}}\right)\text{(310´³/³¾´Ç±ô×°­)}\text{+(1/2³¾´Ç±ô°¿°¿2)(205.0´³/³¾´Ç±ô×°­)±Õ}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{°}}\text{=155.9J/K}\end{array}$

Therefore, the $\left({\text{Δ³§}}_{\text{rxn}}^{\text{°}}\right)$of the reaction is $\text{155.9J/K}$

Finally, 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

$\begin{array}{l}{\text{Δ±á}}_{\text{rxn}}^{\text{o}}\text{=-2657.13kJ}\\ {\text{Δ³§}}_{\text{rxn}}^{\text{o}}\text{=155.9J/K}\end{array}$

These values are entered into the standard free energy equation,

$\begin{array}{l}{\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-2657.13kJ-}\left[\text{(298K)(155.9J/K)}\left({\text{1kJ/10}}^{\text{3}}\text{J}\right)\right]\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-2703.588kJ}\end{array}$

Therefore, the standard free energy value is ${\text{Δ³Ò}}_{\text{rxn}}^{\text{o}}\text{=-2704kJ}$.