Q20.55 P - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the combustion of butane gas:

 ${\mathbf{\text{C}}}_{\mathbf{\text{4}}}{\mathbf{\text{H}}}_{\mathbf{\text{10}}}\mathbf{\text{(g)+}}\frac{\mathbf{\text{13}}}{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{4CO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g)+5H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}}$

 (a) Predict the signs of  ${\mathbf{\text{ΔS}}}^{\mathbf{\text{°}}}{\mathbf{\text{ and ΔH}}}^{\mathbf{\text{°}}}$. Explain.

(b) Calculate ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$ by two different methods.

Step-by-Step Solution

Answer

1. For a given reaction the entropy   $\Delta {\text{S}}^{°}$sign is positive and enthalpy  ${\text{ΔH}}^{\text{°}}$ value is negative.
2. The standard free energy value is  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-2704kJJ}$.

Step 1: Definition of Concept

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.

Step 2: Predict the signs of ΔS ° , ΔH ° and Explain

(a)

Considering the given information:

The chemical equation for the formation of butane from its constituent elements is shown below.

${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{(g)+}\frac{\text{13}}{\text{2}}{\text{ O}}_{\text{2}}\text{(g)}\to {\text{4\hspace{0.17em}CO}}_{\text{2}}{\text{(g)+5H}}_{\text{2}}\text{O(g)}$ 

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 ${{\text{ΔS}}^{\text{°}}}_{\text{system }}$ :

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

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

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

Enthalpy as a predictor of ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ :

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

Step 3: Calculate ΔG ° by two different methods

(b)

Considering the given information:

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

 ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{(g)+}\frac{\text{13}}{\text{2}}{\text{ O}}_{\text{2}}\text{(g)}\to {\text{4\hspace{0.17em}CO}}_{\text{2}}{\text{(g)+5H}}_{\text{2}}\text{O(g)}$

1st method:

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

The Gibbs free energy equation is as follows:

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

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

 $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{4 molCO}}_{\text{2}}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ of CO}}_{\text{2}}\right)\text{+}\left({\text{5 molH}}_{\text{2}}\text{O}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ of H}}_{\text{2}}\text{O}\right)]\\ \text{-}[\left({\text{1 molC}}_{\text{4}}{\text{H}}_{\text{10}}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ of C}}_{\text{4}}{\text{H}}_{\text{10}}\right)\text{+(12/2molO}\\ {\text{ΔG}}_{\text{r×n}}\text{=[(4 mol)(-394.4 kJ/mol)+(5 mol)(-228.60 kJ/mol)]}\\ \text{-}[\left({\text{1 molCC}}_{\text{4}}{\text{H}}_{\text{10}}\right)\text{(-16.7 kJ/mol)}\text{+}\left({\text{13/2molO}}_{\text{2}}\right)\text{(0 kJ/mol)}]\end{array}$

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

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

2nd Method

Given the reaction,

 ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{(g)+}\frac{\text{13}}{\text{2}}{\text{ O}}_{\text{2}}\text{(g)}\to {\text{4\hspace{0.17em}CO}}_{\text{2}}{\text{(g)+5H}}_{\text{2}}\text{O(g)}$

Calculate${\text{ΔG}}_{\text{rxn}}^{\text{o}}$   from  ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ and  ${\text{ΔS}}_{\text{rxn}}^{\text{o}}$ at  $\text{298K}$

The change in standard enthalpy is,

The change in enthalpy for the reaction is calculated as follows: 

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

The enthalpy change is negative.

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

Entropy change  

Standard entropy change equation is,

 ${\text{ΔS}}_{\text{rxn }}^{\text{°}}\text{=}\sum {\text{mS}}_{\text{Products }}^{\text{°}}\text{-}\sum \text{n}{\text{S}}_{\text{reactants }}^{\text{°}}$

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

 $\begin{array}{c}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{4 molCO}}_{\text{2}}\right)({\text{S}}^{\text{o}}\text{of}{\text{CO}}_{\text{2}})\text{+(5 molH2O)}({\text{S}}^{\text{o}}\text{of}{\text{H}}_{\text{2}}\text{O})]\\ \text{-}[\left({\text{1 molC4H}}_{\text{8}}\right)({\text{S}}^{\text{o}}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}})\text{+}\left({\text{13/2molO}}_{\text{2}}\right)({\text{S}}^{\text{o}}\text{of}{\text{O}}_{\text{2}})]\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=}\left[\left({\text{4 molCO}}_{\text{2}}\right)\text{(213.7 J/mol×K)+}\left({\text{5 molH}}_{\text{2}}\text{O}\right)\text{(188.72 J/mol×K)}\right]\\ [\left({\text{1 molC4H}}_{\text{8}}\right)\text{(310 J/mol×K)}\text{+(1/2molOO2)(205.0 J/mol×K)]}\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=155.9 J/K}\end{array}$

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

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

Standard The equation for free energy change is,

 ${\text{ΔG}}_{\text{rxn}}^{\text{o}}{\text{=ΔH}}_{\text{rxn}}^{\text{o}}{\text{-TΔS}}_{\text{rxn}}^{\text{o}}$

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

Enthalpy and entropy values calculated are

 $\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=-2657.13 kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=155.9 J/K}\end{array}$

These values are entered into the standard free energy equation,

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

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