Chapter 20.3, Problem 20.8BFP

Interpretation Introduction

Interpretation:

The temperature at which the following reaction becomes spontaneous has to be calculated using the Appendix B values.

    $CaO\left(s\right)+C{O}_2\left(g\right)\stackrel{}{\to }CaC{O}_3\left(s\right)$

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

  ${\text{ΔH}}_{\text{rxn}}^{\text{o}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

  ${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

  ${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Entropy is the measure of randomness in the system.  Standard entropy change in a reaction is the difference in entropy of the products and reactants  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ can be calculated by the following equation.

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

Where,

  ${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

  ${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Standard 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 $G^o$.  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.

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

Where,

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

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

  $\text{T}$ is the absolute temperature

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

Answer to Problem 20.8BFP

The reaction becomes a spontaneous at temperature $\underset{\_}{<1121K}$.

Explanation of Solution

Given,

Standared free energy equation is,

  $CaO\left(s\right)+C{O}_2\left(g\right)\stackrel{}{\to }CaC{O}_3\left(s\right)$

Calculation for enthalpy ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}$ value 

Standard enthalpy change equation is,

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

  $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}CaC{O}_3\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofCaC{O}_3\right)\right]\\ -\left[\left(2molCaO\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofCaO\right)+\left(1molC{O}_2\right)\left({\text{ΔH}}^{\text{°}}{}_{f}ofC{O}_2\right)\right]\\ \\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}CaC{O}_3\right)\left(-1206.9kJ/mol\right)\right]-\\ \left[\left(1molCaO\right)\left(-635.1kJ/mol\right)+\left(1molC{O}_2\right)\left(-393.5kJ/mol\right)\right]\\ \\ {\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ =}-178.3kJ\end{array}$

Therefore, the calculated enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ value is $\underset{\_}{-178.3kJ}$ 

Calculation for entropy ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$ value

Standard entropy change equation is,

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

  $\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}CaC{O}_3\right)\left(S^{o}ofCaC{O}_3\right)\right]\\ -\left[\left(2molCaO\right)\left(S^{0}ofCaO\right)+\left(1molC{O}_2\right)\left(S^{o}ofC{O}_2\right)\right]\\ \\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}\left[\left(1\text{mol}CaC{O}_3\right)\left(92.9J/mol\times K\right)\right]-\\ \left[\left(2molCaO\right)\left(38.2J/mol\times K\right)+\left(1molC{O}_2\right)\left(213.7J/mol\times K\right)\right]\\ \\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}-159.0\text{J/K}\left(or\right)\\ \\ {\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ =}-0.159\text{kJ/K}\end{array}$

The entropy change is expected to be negative because in gaseous is reactant.

Therefore, the entropy ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ value is $\underset{\_}{-0.159kJ/K}$ 

Calculation of standard entropy value $\Delta {\text{G}}_{\text{rxn}}^{\text{o}}$

The standard free energy equation is,

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

Calculated enthalpy and entropy value are plugging equation (1),

  $\begin{array}{l}\text{T=}\frac{\Delta {{\rm H}}^{\text{o}}}{\Delta {\text{S}}^{\text{o}}}\\ \\ \text{T=}\frac{-178.3\text{kJ}}{-0.1590\text{kJ/K}}=1121.384\\ \\ \text{T=}1121K\end{array}$

Therefore, the reaction becomes spontaneous at temperature $\underset{\_}{<1121K}$.