Q20.63 P - Chemistry: Molecular Nature Of Matter And Change

Question

The U.S. government requires automobile fuels to contain a renewable component. Fermentation of glucose from corn yields ethanol, which is added to gasoline to fulfil this requirement:

 ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}{\mathbf{\text{O}}}_{\mathbf{\text{6}}}\mathbf{\text{(s)}}\mathbf{\to }{\mathbf{\text{2C}}}_{\mathbf{\text{2}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}{\mathbf{\text{OH(l)+2CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

Calculate ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}{\mathbf{\text{,ΔS}}}^{\mathbf{\text{°}}}$, and ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$ for the reaction at ${\mathbf{\text{25}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}$ . Is the spontaneity of this reaction dependent on T? Explain

Step-by-Step Solution

Answer

The given reaction's standard free energy value is  $\text{-229.1kJ/mol.}$ As ${\text{ΔG}}_{\text{rxn}}^{\text{°}}$  is negative the reaction is spontaneous.

 

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: Calculate ΔH ° ,ΔS ° , and ΔG °

Considering the given information:

The fermentation reaction looks like this:

 ${\text{C}}_{\text{6}}{\text{ H}}_{\text{12}}{\text{ O}}_{\text{6}}\text{(s) }\to {\text{2\hspace{0.17em}C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{\hspace{0.17em}OH(l)+2CO}}_{\text{2}}\text{(g)}$

The typical enthalpy change is,

The formation of values,  ${\text{ΔH}}^{\text{o}}$

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

The reaction's enthalpy change is calculated as follows:

 $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{\text{ΔH}}_{\text{f}}^{\text{°}}\text{P reducts) -}\sum \text{n}{\text{ΔH}}_{\text{f( reactants )}}^{\text{°}}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{2 molC}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\left({\text{ΔH}}_{\text{f}}^{\text{o}}{\text{ of C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\text{+}\left({\text{2 molCOCO}}_{\text{2}}\right)\left({\text{ΔH}}_{\text{f}}^{\text{o}}{\text{ of CO}}_{\text{2}}\right)]\\ \text{-}[\left({\text{1molC}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)\left({\text{ΔH}}_{\text{f}}^{\text{°}}{\text{ of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)]\end{array}$

$\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{2 molC}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\text{(-277.63 kJ/mol)+}\left({\text{2 molCO}}_{\text{2}}\right)\text{(-393.5 kJ/mol)}\right]\\ \left[\left({\text{1 molC}}_{\text{6}}{\text{H}}_{\text{l2}}{\text{O}}_{\text{6}}\right)\text{(-1273.3 kJ/mol)}\right]\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=-68.96 kJ=-69.0kJ}\end{array}$

 

The change in enthalpy is negative.

As a result, the enthalpy $\left({\text{ΔH}}_{\text{rxn}}^{\text{°}}\right)\text{changes is-69.0kJ}$ 

Change in entropy  ${\text{ΔS}}_{\text{system }}^{\text{°}}$

Calculate the entropy change for this reaction as follows:

 ${\text{DS}}_{\text{rxn}}^{\text{°}}\text{=}\sum {\text{mS}}_{{\text{P}}_{\text{roducts }}}^{\text{°}}\text{-}\sum {\text{nS}}_{\text{reactants }}^{\text{°}}$

Here, m, n are moles of individual species, which are given by individual species coefficients in the balanced chemical equation.

 $\begin{array}{c}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{2 molC}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\left({\text{S}}^{\text{o}}{\text{ of C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\text{+}\left({\text{2 molCOCO}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ of CO}}_{\text{2}}\right)]\\ \text{-}\left[\left({\text{1 molC}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)\left({\text{S}}^{\text{o}}\text{ of }\left({\text{1 molC}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)\right)\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}[\left({\text{2 molC}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)\text{(161 J/mol×K)}\text{+}\left({\text{2 molCOCO}}_{\text{2}}\right)\text{(213.7 J/mol×K)}]\\ \text{-}\left[\left({\text{1 molC}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)\text{(212.1 J/mol×K)}\right]\\ \text{=537.3 J/K}\end{array}$

 ${\text{ΔS}}_{\text{rxn}}^{\text{0}}\text{=537J/K}$

Hence, the $({\text{ΔS}}_{\text{rxn}}^{\text{0}}$ ) of the reaction is $\underset{\_}{\text{537 J/K}}$  (or) $\text{0.5373 kJ/K}$ .

Determine the free energy change  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$.

The standard free energy change equation is as follows:

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