Question

Find ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$for the reactions in Problem 20.51 using ${\mathbf{\text{ΔH}}}_{\mathbf{\text{f}}}^{\mathbf{\text{°}}}$andSvalues.

Short answer

Reaction-A the value of standard free energy is .${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=2.4kJ}$

Reaction-B the value of standard free energy is .${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-48.4kJ}$

Reaction-C the value of standard free energy is${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{91.2kJ}}$

Step-by-step solution

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:Find   ΔG°for the reactions A

Reaction-A

Considering the given Chemical reaction:

${\text{H}}_{\text{2}}{\text{(g)+I}}_{\text{2}}\text{(g)}\to \text{2HI(g)}$

The number of particles decreases as well, indicating that entropy is decreasing.

As a result, the ${\text{ΔG}}_{\text{f}}^{\text{o}}$values are zero, indicating that the solid is less than the gas.

Standard enthalpy change is,

The reaction's enthalpy change is calculated as follows:

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=\hspace{0.17em}}\sum \text{m}{\text{ΔH}}_{\text{f(Products)}}^{\text{°}}\text{-}\sum \text{n}{\text{ΔH}}_{\text{f(reactants)}}^{\text{°}}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=\hspace{0.17em}}[\text{(2molHI)}({{\text{ΔH}}_{\text{f}}^{\text{°}}}_{\text{of}}\text{of}\text{)}]\text{-}[\left({\text{1molH}}_{\text{2}}\right)({\text{ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{H}}_{\text{2}})\text{+}\left({\text{1molI}}_{\text{2}}\right)({{\text{ΔH}}_{\text{f}}^{\text{°}}}_{\text{f}}\text{of}{\text{I}}_{\text{2}})]\\ {\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=\hspace{0.17em}[(2molHI)(25.9kJ/mol)]-}[\left({\text{1molH}}_{\text{2}}\right)\text{(0kJ/mol)+}\left({\text{1molI}}_{\text{2}}\right)\text{(0kJ/mol)}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=\hspace{0.17em}-51.8kJ}\end{array}$

The value of reaction's enthalpy change is negative.

Hence, the enthalpy $\left({\text{ΔH}}_{\text{rxn}}^{\text{°}}\right)$ value is $\text{-51.8kJ}$

Entropy change ${\text{ΔS}}_{\text{system}}^{\text{°}}$

The standard equation for entropy change is:

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

Where, (m) and (n) are the stoichiometric co-efficient.

$\begin{array}{l}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=[(2molHI)(206.33J/mol×K)]-[(2molH}\text{\hspace{1em}+}\left({\text{130.6J/mol×molI}}_{\text{2}}\right)\text{(116.14J/mol×K)}]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=165.92J/K}\end{array}$

Therefore, the $\left({\text{ΔS}}_{\text{rxn}}^{\text{0}}\right)$ of the reaction is $\text{165.92J/K}$

Calculate the change in free energy ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ next.

Standard Free energy change equation is,

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

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

The enthalpy and entropy values calculated are

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=-51.8kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=165.92J/K}\end{array}$

These figures are used to fill in the blanks in the standard free energy equation.

${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=51.8kJ-}\left[\text{(298K)(165.92J/K)}\left({\text{1kJ/10}}^{\text{3}}\text{J}\right)\right]$

Therefore, the standard free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=2.3558kJ}$.

 Find ΔG°  for the reactions-B

Reaction-B

Considering the given Chemical reaction:

${\text{MnO}}_{\text{2}}\text{(s)+2CO(g)}\to {\text{Mn(s)+4CO}}_{\text{2}}\text{(g)}$

The number of particles decreases as well, indicating that entropy is decreasing.

The standard enthalpy change formula is:

The reaction's enthalpy change is calculated as follows:

$\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{\text{ΔH}}_{\text{f(Products)}}^{\text{°}}\text{-}\sum \text{n}{\text{ΔH}}_{\text{f(reactants)}}^{\text{°}}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{(1molMn)(0kJ/mol)+}\left({\text{2molCO}}_{\text{2}}\right)\text{(-393.5kJ/mol)}\right]\\ [\left({\text{1molMnO}}_{\text{2}}\right)\text{(-520.9kJ/mol)}\text{+(2molCO)(-110.5kJ/mol)]}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=-45.1kJ}\end{array}$

The value of reaction's enthalpy change is negative.

Hence, the enthalpy$\left({\text{ΔH}}_{\text{rxn}}^{\text{°}}\right)$ value is $\text{-45.1kJ}$

Entropy change ${\text{ΔS}}_{\text{system}}^{\text{°}}$.

Standard entropy change equation is,

Where, (m) and (n) are the stoichiometric co-efficient.

$\begin{array}{c}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{(1molMn)(31.8J/mol×K)+(2molCO)}}_{\text{2}}\right)\text{(213.7J/mol×K)}]\\ \text{-[(1molMnO2)(53.1J/mol×K)+(2molCO)(197.5J/mol×K)]}\\ \text{=11.1J/k}\end{array}$

Therefore, the $\left({\text{ΔS}}_{\text{rxn}}^{\text{0}}\right)$of the reaction is $\text{11.1J/k}$

Next calculate the Free energy change ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$

Standard Free energy change equation is,

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

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

The values of calculated enthalpy and entropy are,

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=-45.1kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=-91.28J/K}\end{array}$

These figures are used to fill in the blanks in the standard free energy equation.

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-45.1kJ-}\left[\text{(298K)(-11.1J/K)}\left({\text{1kJ/10}}^{\text{3}}\text{J}\right)\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-48.4078kJ}\end{array}$

Therefore, the standard free energy value is $\text{-48.4078kJ}$.

 Find ΔG°  for the reaction- C

Reaction-C

Considering the given Chemical reaction:

${\text{NH}}_{\text{4}}\text{Cl(s)}\to {\text{NH}}_{\text{3}}\text{(g)+HCl(g)}$

The number of particles decreases as well, indicating that entropy is decreasing.

The standard enthalpy change formula is:

The reaction's enthalpy change is calculated as follows:

$\begin{array}{c}{\text{DH}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{\text{DH}}_{\text{f(Products)}}^{\text{°}}\text{-}\sum \text{n}{\text{DH}}_{\text{f(reactants)}}^{\text{°}}\\ {\text{DH}}_{\text{rxn}}^{\text{o}}\text{=}\left[\left({\text{1molNH}}_{\text{3}}\right)\text{(-45.9kJ/mol)+(1molHCl)(-92.3kJ/mol)}\right]\\ \left[\left({\text{1molNH}}_{\text{4}}\text{Cl}\right)\text{(-314.4kJ/mol)}\right]\\ {\text{DH}}_{\text{r×n}}^{\text{o}}\text{=176.2kJ}\end{array}$

The enthalpy change is positive.

Hence, the enthalpy $\left({\text{ΔH}}_{\text{rxn}}^{\text{°}}\right)$ value is $\text{176.2kJ}$

Entropy change ${\text{ΔS}}_{\text{system}}^{\text{°}}$

Standard entropy change equation is,

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

Where, (m) and (n) are the stoichiometric co-efficient.

$\begin{array}{c}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{1molNHH}}_{\text{3}}\right)\text{(193J/mol×K)+(1molHCl)(186.79J/mol×K)}\right]\\ \left[\left({\text{1molNH}}_{\text{4}}\text{Cl}\right)\text{(94.6J/mol×K)}\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=285.19J/K}\end{array}$

Therefore,$\left({\text{ΔS}}_{\text{rxn}}^{\text{0}}\right)$the of the reaction is $\text{285.19J/K}$

Finally calculate the Free energy change ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$

Standard Free energy change equation is,

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

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

Calculated enthalpy and entropy values are

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=176.2kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=285.19J/K}\end{array}$

These values are plugging above standard free energy equation,

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=176.2kJ-}\left[\text{(298K)(285.19J/K)}\left({\text{1kJ/10}}^{\text{3}}\text{J}\right)\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=91.213kJ}\end{array}$

Therefore, the standard free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=91.213kJ}$.