Q20.53 P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

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 1: 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:

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

The value of reaction's enthalpy change is negative. 

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

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.

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

 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

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

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

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

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

Step 3: 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:

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

The value of reaction's enthalpy change is negative. 

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

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

Standard entropy change equation is,

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

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

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

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,

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

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

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

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

 

Step 4: 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:

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

The enthalpy change is positive. 

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

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.

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

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

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

These values are plugging above standard free energy equation,

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

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