Chapter 14, Problem 14.1KSP

Using $\Delta G_{\text{f}}°$ values from Appendix 2, calculate the standard free-energy change $(\Delta G{°}_{\text{rxn}})$ of the following reaction at 25.0°C.

${\text{Mg(OH)}}_{\text{2}}(s)\to \text{MgO(}s{\text{) + H}}_{\text{2}}\text{O(}l\text{)}$

(a) − 35.6 kJ/mol

(b) − 1166.2 kJ/mol

(c) + 35.6 kJ/mol

(d) + 1166.2 kJ/mol

(e) + 27.0 kJ/mol

Interpretation Introduction

Interpretation:

The standard free energy change ${\text{(ΔG}}_{\text{rxn}}{}^{\text{°}}\text{)}$ in the given reaction has to be calculated.

Concept introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

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

Where $\text{'n'}$ is the number of moles of products and reactants.

Answer to Problem 14.1KSP

The standard free energy change $\left({\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\right)$ of the given reaction $\text{-800}{\text{.8kJmol}}^{\text{-1}}$

Hence, option (e) is found to be correct option.

Explanation of Solution

Given,

$\begin{array}{l}{\text{ΔG}}_{\text{f}}{}^{\text{°}}\left(\text{MgO}(s)\right)=\text{-569}{\text{.6kJmol}}^{\text{-1}}\\ \\ {\text{ΔG}}_{\text{f}}{}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(l)}\right)=\text{-237}{\text{.2kJmol}}^{\text{-1}}\\ \\ {\text{ΔG}}_{\text{f}}{}^{\text{°}}\left({\text{Mg(OH)}}_{\text{2}}\text{(s}\right)=\text{-833}{\text{.75kJmol}}^{\text{-1}}\end{array}$

To find the ${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}$of the given reaction

${\text{Mg(OH)}}_{\text{2}}\text{(s)}\to \text{MgO(s)}+{\text{H}}_{\text{2}}\text{O(l)}$

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}$ the given reaction is calculated by plugging in the values of standard free energy of the reactants and products in the given equation.

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

$\begin{array}{l}{\text{= ΔG}}_{\text{f}}{}^{\text{°}}\left(\text{MgO}(s)\right){\text{+ΔG}}_{\text{f}}{}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(l)}\right)\text{-}\left[{\text{ΔG}}_{\text{f}}{}^{\text{°}}\left({\text{Mg(OH)}}_{\text{2}}\text{(s}\right)\right]\\ \text{=}\text{(1)(-569}{\text{.6kJmol}}^{\text{-1}}\text{)+(1)(-237}{\text{.2kJmol}}^{\text{-1}}\text{)-}\left[\text{(1)(-833}{\text{.75kJmol}}^{\text{-1}})\right]\\ \text{=}27{\text{kJmol}}^{\text{-1}}\end{array}$

Hence, option (e) is found to be the correct option.

Conclusion

The standard free energy change ${\text{(ΔG}}_{\text{rxn}}{}^{\text{°}}\text{)}$ in the given reaction has been calculated.