Chapter 22, Problem 22.104QP

As we saw in Section 21.2, the reduction of iron oxides is accomplished by using carbon monoxide as a reducing agent. Starting with coke in a blast furnace, the following equilibrium plays a key role in the extraction of iron:

$\text{C(}s\text{)}+{\text{CO}}_2(g)\stackrel{}{\rightleftharpoons }2\text{CO}(g)$

Use the data in Appendix 2 to calculate the equilibrium constant at 25°C and 1000°C. Assume ΔH° and ΔS° to be independent of temperature.

Appendix 2

Thermodynamic data at 1 atm and 25°C*

Inorganic Substances

Interpretation Introduction

Interpretation:

For the given reaction, the equilibrium constant at ${\text{25}}^{\text{o}}\text{C}$ and ${\text{1000}}^{\text{o}}\text{C}$ has to be calculated.

Concept Introduction:

Equilibrium constant:

Equilibrium constant is the ratio of the concentration of product and the concentration of the reactant at equilibrium.

It can be calculated using the relationship between Gibb’s free energy and rate constant:

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=}\text{-RTlnK}\\ \text{K}=\exp \left(-\frac{{\text{ΔG}}^{\text{0}}}{\text{RT}}\right)\end{array}$

Where,

$\begin{array}{l}{\text{ΔG}}^{\text{0}}=\text{standard}\text{gibb's}\text{free}\text{energy}\text{change}\text{.}\\ \text{R}=\text{Universal}\text{gas}\text{constant}\text{=}\text{8}\text{.314}{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\\ \text{K}=\text{Equilibrium}\text{constant}\text{.}\\ \text{T}\text{=}\text{Temperature}\text{.}\end{array}$

Gibbs free energy: Gibbs free is a thermodynamic parameter, which can be defined as the amount of energy available with the system to perform a useful work.

Gibbs free energy Equation:

$\text{ΔG°=ΔH°-TΔS°}$

Where,

$\text{ΔG°}$- is the change in standard Gibbs free energy.

$\text{ΔH°}$-is the change in standard enthalpy.

$\text{ΔS°}$- is the change in standard entropy.

T– is Temperature in K.

The change in standard entropy can be calculated as follows:

${\text{ΔS}}_{{}^{\text{reaction}}}^{\text{o}}\text{ = }{\displaystyle \sum {\text{mΔS}}^{\text{o}}\left(\text{products}\right)}\text{- }{\displaystyle \sum {\text{nΔS}}^{\text{o}}}\left(\text{reactants}\right)$

$\begin{array}{l}\text{Where,}\\ {\displaystyle \sum {\text{mΔS}}^{\text{o}}\left(\text{products}\right)}\text{is}\text{the}\text{total}\text{sum}\text{of}\text{standard entropy}\text{change}\text{of}\text{products}\text{.}\\ {\displaystyle \sum {\text{nΔS}}^{\text{o}}\left(\text{reactants}\right)}\text{is}\text{the}\text{total}\text{sum}\text{of}\text{standard entropy}\text{change}\text{of}\text{reactants}\text{. }\\ {\text{ΔS}}^{\text{o}}\text{is}\text{the}\text{change}\text{in}\text{the}\text{standard entropy}\text{.}\end{array}$

${\text{ΔS}}_{{}^{\text{reaction}}}^{\text{o}}\text{is}\text{the}\text{change}\text{in}\text{the}\text{standard entropy}\text{of}\text{the}\text{reaction}\text{.}$

$\begin{array}{l}\text{m}\text{is the number of moles of the products}\text{.}\\ \text{n}\text{is the number of moles of the reactants}\text{.}\end{array}$

The change in standard enthalpy can be calculated as follows:

${\text{ΔH}}_{{}^{\text{reaction}}}^{\text{o}}\text{ = }{\displaystyle \sum {\text{mΔH}}_{\text{f}}^{\text{o}}\left(\text{products}\right)}\text{- }{\displaystyle \sum {\text{nΔH}}_{\text{f}}^{\text{o}}}\left(\text{reactants}\right)$

$\begin{array}{l}\text{Where,}\\ {\displaystyle \sum {\text{mΔH}}_{\text{f}}^{\text{o}}\left(\text{products}\right)}\text{is}\text{the}\text{total}\text{sum}\text{of}\text{standard enthalpy}\text{change}\text{of}\text{products}\text{.}\\ {\displaystyle \sum {\text{nΔH}}_{\text{f}}^{\text{o}}\left(\text{reactants}\right)}\text{is}\text{the}\text{total}\text{sum}\text{of}\text{standard enthalpy}\text{change}\text{of}\text{reactants}\text{. }\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\text{is}\text{the}\text{change}\text{in}\text{the}\text{standard enthalpy}\text{of}\text{formation}\text{.}\end{array}$

${\text{ΔH}}_{{}^{\text{reaction}}}^{\text{o}}\text{is}\text{the}\text{change}\text{in}\text{the}\text{standard enthalpy}\text{of}\text{the}\text{reaction}\text{.}$

$\begin{array}{l}\text{m}\text{is the number of moles of the products}\text{.}\\ \text{n}\text{is the number of moles of the reactants}\text{.}\end{array}$

Answer to Problem 22.104QP

For the given reaction, $\text{C}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\rightleftharpoons }\text{2CO}\left(\text{g}\right)$:

The equilibrium constant at $298\text{K}$ $\left({\text{25}}^{\text{o}}\text{C}\right)$ is $\text{K}=9.61\times {10}^{-22}$

The equilibrium constant at $1273\text{K}$ $\left({\text{1000}}^{\text{o}}\text{C}\right)$ is $\text{K}=138.46$

Explanation of Solution

The given reaction is:

$\text{C}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)\stackrel{}{\rightleftharpoons }\text{2CO}\left(\text{g}\right)$

The following table gives the data of standard enthalpy of formation and standard entropy values for the reactants and products which are extracted from the appendix 2.

Reactant/Product	Substance	${\text{ΔH}}_{\text{f}}^{\text{o}}\left(\text{kJ/mol}\right)$	${\text{S}}^{\text{o}}\left(\text{J/K}\text{.mol}\right)$
Reactant	$\text{C}\left(\text{s}\right)$	0$\text{kJ/mol}$	5.69$\text{J/K}\text{.mol}$
Reactant	${\text{CO}}_{\text{2}}\left(\text{g}\right)$	-393.5$\text{kJ/mol}$	213.6$\text{J/K}\text{.mol}$
Product	$\text{CO}\left(\text{g}\right)$	-110.5$\text{kJ/mol}$	197.9$\text{J/K}\text{.mol}$

Converting the temperature from ${}^{\text{o}}\text{C}$ into $\text{K}$-Kelvin :

$\begin{array}{c}{\text{25}}^{\text{o}}\text{C}={\text{25}}^{\text{o}}\text{C}\text{+}\text{273}\\ \text{=298}\text{K}\end{array}$

The change in standard enthalpy can be calculated as follows:

${\text{ΔH}}_{{}^{\text{reaction}}}^{\text{o}}\text{ = }{\displaystyle \sum {\text{mΔH}}_{\text{f}}^{\text{o}}\left(\text{products}\right)}\text{- }{\displaystyle \sum {\text{nΔH}}_{\text{f}}^{\text{o}}}\left(\text{reactants}\right)$

$\begin{array}{c}=\left(2\times {\text{ΔH}}^{\text{o}}{}_{\text{f}}\left(\text{CO}\left(\text{g}\right)\right)\right)\text{-}\left(1\times {\text{ΔH}}^{\text{o}}{}_{\text{f}}\left(\text{C}\left(\text{s}\right)\right)\text{+}1\times {\text{ΔH}}^{\text{o}}{}_{\text{f}}\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\\ =\left(2\times -110.5\text{kJ/mol}\right)\text{-}\left(0-393.5\text{kJ/mol}\right)\\ \text{=}-\text{221 kJ/mol}\text{+}393.5\text{kJ/mol}\\ \text{= }172.5\text{kJ/mol}\end{array}$

Converting it into $\text{J/mol}$:

$\begin{array}{c}\text{1}\text{kJ}=\text{1000}\text{J}\\ 172.5\text{kJ/mol}=\frac{\text{1000}\text{J}}{\text{1}\text{kJ}}\times 172.5\text{kJ}\\ =172500\text{J/mol}\end{array}$

The change in standard entropy can be calculated as follows:

${\text{ΔS}}_{{}^{\text{reaction}}}^{\text{o}}\text{ = }{\displaystyle \sum {\text{mΔS}}^{\text{o}}\left(\text{products}\right)}\text{- }{\displaystyle \sum {\text{nΔS}}^{\text{o}}}\left(\text{reactants}\right)$

$\begin{array}{c}=\left(2\times {\text{ΔS}}^{\text{o}}\left(\text{CO}\left(\text{g}\right)\right)\right)\text{-}\left(1\times {\text{ΔS}}^{\text{o}}\left(\text{C}\left(\text{s}\right)\right)\text{+}1\times {\text{ΔS}}^{\text{o}}\left({\text{CO}}_{\text{2}}\left(\text{g}\right)\right)\right)\\ =\left(2\times 197.9\text{J/K}\text{.mol}\right)\text{-}\left(5.69\text{+}213.6\text{J/K}\text{.mol}\right)\\ \text{=}395.8\text{J/K}\text{.mol}\text{-}219.29\text{J/K}\text{.mol}\\ \text{= }176.51\text{J/K}\text{.mol}\end{array}$

Calculating the Gibb’s free energy at $298K$ $\left(25^{o}C\right)$:

$\begin{array}{c}\text{ΔG°=ΔH°-TΔS°}\\ \text{=}172.5\text{kJ/mol}-\left(298\text{K}\right)\times 176.51\text{J/K}\text{.mol}\\ =172500\text{J/mol}-52599.98\text{J/mol}\\ \text{=}\text{119900}\text{.02}\text{J/mol}\end{array}$

Calculating the equilibrium constant for the given reaction at $298K$ $\left(25^{o}C\right)$:

$\begin{array}{c}\text{K}=\exp \left(-\frac{{\text{ΔG}}^{\text{0}}}{\text{RT}}\right)\\ =\exp \left(-\frac{\text{119900}\text{.02}\text{J/mol}}{\text{8}\text{.314}{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\times \text{298}\text{K}}\right)\\ =9.61\times {10}^{-22}\end{array}$

Thus, the equilibrium constant at $298\text{K}$ $\left({\text{25}}^{\text{o}}\text{C}\right)$ is $\text{K}=9.61\times {10}^{-22}$

Calculating the Gibb’s free energy at $1273K$ $\left(1000^{o}C\right)$:

$\begin{array}{c}\text{ΔG°=ΔH°-TΔS°}\\ \text{=}172.5\text{kJ/mol}-\left(\text{1273}\text{K}\right)\times 176.51\text{J/K}\text{.mol}\\ =172500\text{J/mol}-224684.5\text{J/mol}\\ \text{=}-\text{52184}\text{.5}\text{J/mol}\end{array}$

Calculating the equilibrium constant for the given reaction at $1273K$ $\left(1000^{o}C\right)$

$\begin{array}{c}\text{K}=\exp \left(-\frac{{\text{ΔG}}^{\text{0}}}{\text{RT}}\right)\\ =\exp \left(-\frac{-\text{52184}\text{.5}\text{J/mol}}{\text{8}\text{.314}{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\times \text{1273}\text{K}}\right)\\ =\exp \left(+\frac{\text{52184}\text{.5}\text{J/mol}}{\text{8}\text{.314}{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\times \text{1273}\text{K}}\right)\\ =138.46\end{array}$

Thus, the equilibrium constant at $1273\text{K}$ $\left({\text{1000}}^{\text{o}}\text{C}\right)$ is $\text{K}=138.46$

Conclusion

For the given reaction, the equilibrium constants at ${\text{25}}^{\text{o}}\text{C}$ and ${\text{1000}}^{\text{o}}\text{C}$ have been calculated.