Question

Calculate${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$ for each reaction using${\mathbf{\text{ΔG}}}_{\mathbf{\text{f}}}^{\mathbf{\text{°}}}$ values:

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

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

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

Short answer

1. The Gibbs free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{2.6kJ}}$.
2. The Gibbs free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{-48.}}\text{3kJ}$.
3. The Gibbs free energy value is${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=92kJ}$.

Step-by-step solution

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  ΔG°

(a)

Considering the given 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 $\text{ΔG}$is decreasing.

So, their$\Delta G_f^o$values are zero the solid is less than the gas.

Free energy change $\Delta G_f^o$

${\text{ΔG}}^{\text{o}}$Formation of values,

$\begin{array}{l}{\text{H}}_{\text{2}}\text{(g)=0kJ/mol}\\ {\text{I}}_{\text{2}}\text{(g)=0kJ/mol}\\ \text{HI(g)=1.3kJ/mol}\end{array}$

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

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\sum {\text{m}}_{\text{f}}{\text{G}}_{\text{f}}^{\text{°}}\text{(Products)-}\sum \text{n}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(Reactants)}\\ {\text{ΔG}}^{\text{°}}\text{=}\left[\text{(2molHI)}\left({\text{ΔG}}_{\text{f}}^{\text{o}}\text{ofHI}\right)\right]\text{-}\left[\left({\text{1molH}}_{\text{2}}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ofH}}_{\text{2}}\right)\text{+}\left({\text{1molI}}_{\text{2}}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ofI}}_{\text{2}}\right)\right]\\ {\text{ΔG}}^{\text{°}}\text{=[(2molHI)(1.3kJ/mol)]-}[\text{(1molH2)(0kJ/mol)+}\left({\text{1molO}}_{\text{2}}\right)\text{(0kJ/mol)}\\ {\text{ΔG}}^{\text{o}}\text{=2.6kJ}\end{array}$

Therefore, the Gibbs free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{2.6kJ}}$.

Calculate  ΔG°

(b)

Considering the given reaction:

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

Free energy change $\Delta G_f^o$

${\text{ΔG}}^{\text{o}}$Formation of values,

$\begin{array}{l}{\text{MnO}}_{\text{2}}\text{(s)=-466.1kJ/mol}\\ \text{CO(g)=-137.2kJ/mol}\\ \text{Mn(s)=0kJ/mol}\\ {\text{CO}}_{\text{2}}\text{(g)=-394.4kJ/mol}\end{array}$

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

$\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\sum {\text{m}}_{\text{f}}{\text{G}}_{\text{f}}^{\text{°}}\text{(Products)-}\sum {\text{n}}_{\text{f}}{\text{G}}_{\text{f}}^{\text{°}}\text{(Reactants)}\\ {\text{ΔG}}^{\text{°}}\text{=}[\text{(1molMn)}\left({\text{ΔG}}_{\text{f}}^{\text{o}}\text{ofMn}\right)\text{+}\left({\text{2molCO}}_{\text{2}}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ofCO}}_{\text{2}}\right)\\ \text{-}\left[\left({\text{1molMnOm}}_{\text{2}}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ofMnO}}_{\text{2}}\right)\text{+(2molCO)}\left({\text{ΔG}}_{\text{f}}^{\text{o}}\text{ofCO}\right)\right]\\ {\text{ΔG}}^{\text{°}}=\left[{\text{(1molMn)(0kJ/mol)+(2molCO)}}_{\text{2}}\right)\text{(-394.4kJ/mol)}]\\ \text{-(1molMnO2)(-466.1kJ/mol)}\text{+}\left({\text{2COmolO}}_{\text{2}}\right)\text{(-137.2kJ/mol)}]\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{-48.3kJ}}\end{array}$

Therefore, the Gibbs free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{-48.3kJ}}$.

Calculate  ΔG°

(c)

Considering the given reaction:

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

Free energy change $\Delta G_f^o$

${\text{ΔG}}^{\text{o}}$Formation of values,

$\begin{array}{l}{\text{NH}}_{\text{4}}\text{Cl(s)=-203.0kJ/mol}\\ {\text{NH}}_{\text{3}}\text{(g)=-16.0kJ/mol}\\ \text{HCl(g)=-95.30kJ/mol}\end{array}$

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

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{{\text{ΔG}}_{\text{f}}}^{\text{°}}\text{(Products)-}\sum \text{n}{{\text{ΔG}}_{\text{f}}}^{\text{°}}\text{(Reactants)}\\ {\text{ΔG}}^{\text{°}}\text{=}[\left({\text{1molNH}}_{\text{3}}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ofNH}}_{\text{3}}\right)\text{+(1molHCl)}\left({\text{ΔG}}_{\text{f}}^{\text{o}}\text{ofHCl}\right)]\text{-}\left[\left({\text{1molNH}}_{\text{4}}\text{Cl}\right)\left({\text{ΔG}}_{\text{f}}^{\text{o}}{\text{ofNH}}_{\text{4}}\text{Cl}\right)\right]\\ {\text{ΔG}}^{\text{°}}\text{=}\left[{\text{(1mol)(-16kJ/mol)+(1molCO)}}_{\text{2}}\right)\text{(-95.30kJ/mol)}]\text{-[(1mol)(-203.0kJ/mol)]}\\ {\text{ΔG}}^{\text{°}}\text{=91.7kJÞ92kJ}\end{array}$

Therefore, the Gibbs free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{92}}\text{kJ}$.