Chapter 25, Problem 25.97QP

Interpretation Introduction

Interpretation:

The equilibrium constant at ${\text{25}}^{\circ }\text{C}\text{and}{\text{100}}^{\circ }\text{C}$ for the formation of carbon monoxide from coke has to be calculated.

Concept introduction:

Equilibrium constant: In a reversible chemical reaction the relationship between the amounts of starting materials and products present at equilibrium in a given temperature is given by the equilibrium constant.

To calculate: the equilibrium constant for the formation of carbon monoxide from coke at given temperatures.

Answer to Problem 25.97QP

• The equilibrium constant at ${\text{25}}^{\circ }\text{C}\text{is}\text{9}\text{.61}\times {\text{10}}^{-\text{22}}$.
• The equilibrium constant at ${\text{100}}^{\circ }\text{C}\text{is}\text{1}\text{.2}\times {\text{10}}^{-15}$.

Explanation of Solution

The relationship between equilibrium constant and Gibb’s free energy is given by

$\begin{array}{l}{\text{ΔG}}^{\circ }=-\text{RTlnK}\\ {\text{ΔG}}^{\circ }-\text{change}\text{in}\text{Gibb's}\text{free}\text{enrgy}\\ \text{R}-\text{Gas}\text{constant}\\ \text{K}-\text{equilibrium}\text{constant}\text{and}\text{T}-\text{temperature}\end{array}$

The relationship between Gibb’s free energy, enthalpy and entropy is given by

$\begin{array}{l}{\text{ΔG}}^{\circ }={\text{ΔH}}^{\circ }-{\text{TΔS}}^{\circ }\\ {\text{ΔG}}^{\circ }-\text{change}\text{in}\text{Gibb's}\text{free}\text{enrgy}\\ {\text{ΔH}}^{\circ }-\text{change}\text{in}\text{enthalpy}\\ {\text{ΔS}}^{\circ }-\text{change}\text{in}\text{entropy}\text{and}\text{T}-\text{temperature}\end{array}$

First calculate $\text{ΔG}$ and then calculate the equilibrium constant by the equation given above.

For the formation of carbon monoxide from coke, the change in enthalpy is calculated as follows

$\begin{array}{l}{\text{ΔH° = 2ΔH°}}_{\text{f}}{\text{ [CO}}_{\text{(g)}}\text{] }\text{-}{\text{{ΔH°}}_{\text{f}}{\text{ [C}}_{\text{(s)}}{\text{] + ΔH°}}_{\text{f}}{\text{ [CO}}_{\text{2 (g)}}\text{]}}\\ \text{ΔH°}\text{=}\text{(2)(-110}\text{.5 kJ/mol) - (0 + -393}\text{.5 kJ/mol) }\\ \text{= 172}\text{.5 kJ/mol}\end{array}$

For the formation of carbon monoxide from coke, the change in entropy is calculated as follows

$\begin{array}{l}{\text{ΔS° = 2ΔS° [CO}}_{\text{(g)}}\text{]}\text{-}{\text{{S° [C}}_{\text{(s)}}{\text{] + S°[CO}}_{\text{2 (g)}}\text{]}}\\ \text{ΔS° = (2)(197}\text{.9 J/K×mol) - (5}\text{.69 J/K×mol + 213}\text{.6 J/K}\text{.mol)}\\ \text{ = 176}\text{.5 J/K×mol}\end{array}$

For the formation of carbon monoxide from coke, the change in enthalpy and change in entropy was calculated.

Calculate the change in Gibb’s free energy for the formation of carbon monoxide at ${\text{25}}^{\circ }\text{C}$

$\begin{array}{l}\text{ΔG° = ΔH° - TΔS°}\\ \text{= (172}{\text{.5×10}}^{\text{3}}\text{ J /mol) - (298 K)(176}\text{.5 J /K×mol)}\\ \text{= 1}{\text{.199×10}}^{\text{5}}\text{ J /mol}\end{array}$

Calculate the change in Gibb’s free energy for the formation of carbon monoxide at ${\text{100}}^{\text{o}}\text{C}$

$\begin{array}{l}\text{ΔG° = ΔH° - TΔS°}\\ \text{= (172}{\text{.5×10}}^{\text{3}}\text{ J /mol) - (373 K)(176}\text{.5 J /K×mol)}\\ \text{= 1}{\text{.07×10}}^{\text{5}}\text{ J /mol}\end{array}$

For the formation of carbon monoxide from coke, the change in Gibb’s free energy was calculated at ${\text{25}}^{\circ }{\text{C and 100}}^{\circ }\text{C}$.

The relationship between equilibrium constant and Gibb’s free energy is given by

${\text{ΔG}}^{\circ }=-\text{RTlnK}$

The equilibrium constant at ${\text{25}}^{\circ }\text{C}$

$\begin{array}{l}\text{K}\text{=}{\text{e}}^{\text{-}\frac{{\text{ΔG}}^{\text{o}}}{\text{RT}}}\\ \text{=}{\text{e}}^{\frac{\text{-}\left(\text{1}\text{.99}\text{×}{\text{10}}^{\text{5}}\text{J/mol}\right)}{\left(\text{8}\text{.314}\text{J/mol}\right)\left(\text{298}\text{K}\right)}}\\ \text{=}\text{9}\text{.61}\text{×}{\text{10}}^{\text{-22}}\end{array}$

The equilibrium constant at ${\text{100}}^{\circ }\text{C}$

$\begin{array}{l}\text{K}\text{=}{\text{e}}^{\text{-}\frac{{\text{ΔG}}^{\text{o}}}{\text{RT}}}\\ \text{=}{\text{e}}^{\frac{\text{-}\left(\text{1}\text{.07}\text{×}{\text{10}}^{\text{5}}\text{J/mol}\right)}{\left(\text{8}\text{.314}\text{J/mol}\right)\left(\text{373}\text{K}\right)}}\\ \text{=}\text{1}\text{.2}\text{×}{\text{10}}^{\text{-15}}\end{array}$

For the formation of carbon monoxide from coke, the equilibrium constant was calculated at ${\text{25}}^{\circ }{\text{C and 100}}^{\circ }\text{C}$.

Conclusion

The equilibrium constant for the formation of carbon monoxide from coke at ${\text{25}}^{\circ }\text{C}\text{and}{\text{100}}^{\circ }\text{C}$ was calculated.