Chapter 20, Problem 20.91P

(a)

Interpretation Introduction

Interpretation:

For magnesite decomposition the balanced equation has to be written.

Concept introduction:

Balanced equation: A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

The equation for a reaction, which has same number of atoms and charge of the ions in both reactants and product sides, is known as balanced equation.

Reactant: In a chemical reaction the species that present at left is denoted as reactant which undergoes chemical change and result to given new species called product.

Product: In a chemical reaction the species that present in right side is denoted as product that results from the reactant.

Answer to Problem 20.91P

Balance chemical eqaution is,

  $MgC{O}_3\left(s\right)\underset{}{\rightleftharpoons }MgO\left(s\right)+C{O}_2\left(g\right)$

Explanation of Solution

Balance chemical eqaution is,

  $MgC{O}_3\left(s\right)\underset{}{\rightleftharpoons }MgO\left(s\right)+C{O}_2\left(g\right)$

The $MgC{O}_3$ decomposed at ${1200}^{o}C$ to forms Magnesia and carbon dioxide the balanced equation is shows above. The above reaction is balanced one since both reactant and the product side contains same type and same number of elements.

(b)

Interpretation Introduction

Interpretation:

For the given decomposition reaction the free energy ${\text{ΔG}}^{\text{o}}$ value has to be calculated at $298K$ using the enthalpy and entropy values.

Concept introduction:

Free energy (or) entropy change 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 G.  All spontaneous processes are associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ${\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}$

Where,

  ${\text{ΔG}}^{\text{o}}$ is the standard change in free energy of the system

  $\Delta {{\rm H}}^{\text{o}}$ is the standard change in enthalpy of the system

  $\text{T}$ is the absolute value of the temperature

  $\Delta {\text{S}}^{\text{o}}$ is the change in entropy in the system

Answer to Problem 20.91P

For given decomposition reaction standard free energy values is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=\underset{\_}{\text{65}\text{.2}\text{kJ}}$

Explanation of Solution

Given decomposition reaction is,

  $MgC{O}_3\left(s\right)\underset{}{\rightleftharpoons }MgO\left(s\right)+C{O}_2\left(g\right)$

Calculate the change in Gibb’s free energy at $298K$ as follows,

Standard enthalpy change is,

$\Delta H^o$ Formation of values are shown as follows,

  $\begin{array}{c}MgO\left(g\right)=-601.2kJ/mol\\ C{O}_2\left(g\right)=-393.5kJ/mol\\ MgC{O}_3\left(g\right)=-1112.0kJ/mol\end{array}$

Enthalpy values referred from Appendix Table. 

The enthalpy change for the reaction is calculated as follows,

  ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}\Delta {\text{H}}^{\text{°}}{}_{\text{f(Products)}}-{\displaystyle \sum \text{n}}\Delta {\text{H}}^{\text{°}}{}_{\text{f(reactants)}}$

  $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}MgO\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofMgO\right)+\left(1\text{mol}C{O}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofC{O}_2\right)\right]\\ -\left[\left(1molMgC{O}_3\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofMgC{O}_3\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}MgO\right)\left(-601.2kJ/mol\right)+\left(1\text{mol}C{O}_2\right)\left(-393.5kJ/mol\right)\right]-\\ \left[\left(1molMgC{O}_3\right)\left(-1112kJ/mol\right)\right]\\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}117.3kJ\end{array}$

The enthalpy change is positive. Hence, the enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ value is $\underset{\_}{117.3kJ}$

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

Calculate the change in entropy for this reaction as follows,

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

Where, $\left(m\right)and\left(n\right)$ are the stoichiometric co-efficient.

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}MgO\right)\left(S^{o}ofMgO\right)+\left(1\text{mol}C{O}_2\right)\left(S^{o}ofC{O}_2\right)\right]\\ -\left[\left(1\text{mol}MgC{O}_3\right)\left(S^{o}ofMgC{O}_3\right)\right]\\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}\right)\left(26.9J/mol\times K\right)+\left(1\text{mol}\right)\left(213.7J/mol\times K\right)\right]\\ -\left[\left(1\text{mol}\right)\left(65.86J/mol\times K\right)\right]\\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}174.74J/K\end{array}$

The ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{174.74J/K}$

The enthalpy and entropy changes are positive sign for ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$ indicates the formation of methanol is favored at given temperature.

Calculate the Free enrgy change $\Delta G_{rxn}^o$

Standared Free energy change equation is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ = }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Calcualted enthalpy and entropy values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=117.3kJ\\ \Delta {\text{S}}_{\text{rxn}}^{\text{o}}=174.74J/K\\ {\text{T}}_{\text{1}}=273\text{+}25\text{=}298K\end{array}$

Plugging these values into above standard free energy equation,

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=117.3\text{kJ}\text{-}\left[\left(298\text{K}\right)\left(174.74\text{J/K}\right)\left(1kJ/{10}^3\right)\right]\\ =65.22748kJ\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=65.2kJ\end{array}$

Therefore, for the given decomposition reaction standard free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=\underset{\_}{65.2kJ}$.

(c)

Interpretation Introduction

Interpretation:

The minimum temperature at which the given reaction is spontaneous has to be identified.

Concept introduction:

Free energy (or) entropy change 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 G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ${\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}$

Where,

  ${\text{ΔG}}^{\text{o}}$ is the standard change in free energy of the system

  $\Delta {{\rm H}}^{\text{o}}$ is the standard change in enthalpy of the system

  $\text{T}$ is the absolute value of the temperature

  $\Delta {\text{S}}^{\text{o}}$ is the change in entropy in the system

Answer to Problem 20.91P

The required minimum temperature is $T=671K$.

Explanation of Solution

Given decomposition reaction is,

  $MgC{O}_3\left(s\right)\underset{}{\rightleftharpoons }MgO\left(s\right)+C{O}_2\left(g\right)$

Calculate the change in Gibb’s free energy at $298K$ as follows,

The reaction becomes spontaneous below the temprature where ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=0$

Consider the follwing free energy equation,

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ =}\text{0}\text{= }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}------\left[1\right]\\ \Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{=T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}-------\left[2\right]\end{array}$

Rearrange equation (2) to calculate temprature T,

$\text{T}=\frac{\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}}{\Delta {\text{S}}_{\text{rxn}}^{\text{o}}}$

Hence,

  $\begin{array}{c}T=\frac{117.3kJ}{0.17474kJ/K}=671.283\\ T=671K\end{array}$

At temprature above $671K$, this reaction is spontaneous. Because both enthalpy $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}$ and entropy $\Delta S_{\text{rxn}}^{\text{o}}$ values are positve hence the reaction becomes spontaneous above this temprature. 

(d)

Interpretation Introduction

Interpretation:

For the $MgC{O}_3$ decomposition reaction, the equilibrium constant $P_{C{O}_2}$ value has to be calculated at $298K$.

Concept introduction:

Free energy (or) entropy change 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 G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ${\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}$

Where,

  ${\text{ΔG}}^{\text{o}}$ is the standard change in free energy of the system

  $\Delta {{\rm H}}^{\text{o}}$ is the standard change in enthalpy of the system

  $\text{T}$ is the absolute value of the temperature

  $\Delta {\text{S}}^{\text{o}}$ is the change in entropy in the system

Free energy change$\text{ΔG}$: change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $\text{K}$

  $\begin{array}{l}\text{ΔG =}{\text{ΔG}}^{\text{0}}\text{+}\text{RT}\ln \left(\text{K}\right)\\ {\text{ΔG}}^{\text{0}}\text{=}{\text{ΔH}}^{\text{0}}-{\text{TΔS}}^{\text{0}}\end{array}$

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

  ${\text{ΔG}}^{\text{0}}$, ${\text{ΔH}}^{\text{0}}$ and ${\text{ΔS}}^{\text{0}}$ is standard free energy, enthalpy and entropy values.

Answer to Problem 20.91P

For given reaction, the equilbrium pressure $P_{C{O}_2}$ value is $\underset{\_}{K=3.68\times 10^{-12}atm}$.

Explanation of Solution

Given,

Free energy equation is,

  $MgC{O}_3\left(s\right)\underset{}{\rightleftharpoons }MgO\left(s\right)+C{O}_2\left(g\right)$

Solving for $P_{C{O}_2}$ using following free energy equation,

  $\begin{array}{l}\Delta G=\Delta G^o-RT\ln \left(K\right)------\left[1\right]\\ \left(or\right)\\ \Delta G^o=-RT\ln \left(K\right)\end{array}$

Rearrange the above equation as shown below,

  $InK=-\frac{\Delta G^o}{RT}------\left[2\right]$

Then,

  $\begin{array}{l}InK=\frac{\Delta G^o}{-RT}=-\frac{65.22748kJ/mol}{\left(8.314\text{J/mol}\times \text{K}\right)\left(298K\right)}\left(\frac{1000J}{1kJ}\right)\\ InK=-26.327178\\ K=e^{-26.327178}=3.6834253\times {10}^{-12}\\ K=3.68\times 10^{-12}atmof{P}_{CO}{}_{{}_2}\end{array}$

Therefore, the given equilbrium pressure $P_{C{O}_2}$ value is $\underset{\_}{K=3.68\times 10^{-12}atm}$

(e)

Interpretation Introduction

Interpretation:

For the following $MgC{O}_3$ decomposition reaction, the equilibrium constant $P_{C{O}_2}$ value has to be calculated at $1200K$.

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.

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

Free energy change$\text{ΔG}$: change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $\text{K}$

  $\begin{array}{l}\text{ΔG =}{\text{ΔG}}^{\text{0}}\text{+}\text{RT}\ln \left(\text{K}\right)\\ {\text{ΔG}}^{\text{0}}\text{=}{\text{ΔH}}^{\text{0}}-{\text{TΔS}}^{\text{0}}\end{array}$

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

  ${\text{ΔG}}^{\text{0}}$, ${\text{ΔH}}^{\text{0}}$ and ${\text{ΔS}}^{\text{0}}$ is standard free energy, enthalpy and entropy values.

Answer to Problem 20.91P

For given reaction the equilbrium pressure $P_{C{O}_2}$ value is $\underset{\_}{K=1.0512\times 10^{4}atm}$

Explanation of Solution

Given,

Free energy equation is,

  $MgC{O}_3\left(s\right)\underset{}{\rightleftharpoons }MgO\left(s\right)+C{O}_2\left(g\right)$

Calculate the Free enrgy change  $\Delta G_{rxn}^o$

Standared Free energy change equation iss,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{ = }\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}\text{- T}\Delta {\text{S}}_{\text{rxn}}^{\text{o}}$

Calcualted enthalpy and entropy  values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=117.3kJ\\ \Delta {\text{S}}_{\text{rxn}}^{\text{o}}=174.74J/K\\ {\text{T}}_{\text{1}}=273\text{+}25\text{=}298K\end{array}$

These values are plugging above standard free energy equation,

  $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=117.3\text{kJ}\text{-}\left[\left(1200\text{K}\right)\left(174.74\text{J/K}\right)\left(1kJ/{10}^3\right)\right]\\ =-92.388kJ\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=-92.38kJ\end{array}$

Next, we solving for $P_{C{O}_2}$ using following free energy equation at $1200K$

  $\begin{array}{l}\Delta G=\Delta G^o-RT\ln \left(K\right)------\left[1\right]\\ \left(or\right)\\ \Delta G^o=-RT\ln \left(K\right)\end{array}$

Rearrange the above equation,

  $InK=-\frac{\Delta G^o}{RT}------\left[2\right]$

Then,

  $\begin{array}{l}InK=\frac{\Delta G^o}{-RT}=\frac{-92.388kJ/mol}{-\left(8.314\text{J/mol}\times \text{K}\right)\left(1200K\right)}\left(\frac{1000J}{1kJ}\right)\\ InK=9.26028\\ K=e^{9.26028}=1.0512\times {10}^4\\ K=1.0512\times 10^{4}atmof{P}_{CO}{}_{{}_2}\end{array}$

Therefore, the given equilbrium pressure $P_{C{O}_2}$ value is $\underset{\_}{K=1.0512\times 10^{4}atm}$.