Chapter 21, Problem 21.203QP

Interpretation Introduction

Interpretation:

• Chemical equation for given reaction has to be written.
• At $\text{2000°C}$ the equilibrium constant for this reaction has to be calculated by using thermodynamic data and respective approximation.

Answer to Problem 21.203QP

• Chemical equation for given reaction is

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

• Equilibrium constant for this reaction at $\text{2000°C}$ is $\text{1}\text{.6}\times {\text{10}}^{\text{5}}$

Explanation of Solution

To write: Chemical equation for given reaction

Carbon burns in a air it gives Carbon dioxide and carbon monoxide also presented because the product the hot Carbon is react with Carbon dioxide.

Thermodynamic data and the reaction gives the equation as follows,

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

To determine: constant for this reaction at $\text{2000°C}$

$\begin{array}{l}{\text{ CO}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{ + C}}_{\left(\text{graphite}\right)}\stackrel{}{\to }{\text{2CO}}_{\left(\text{g}\right)}\\ \\ {\text{ΔH}}_{\text{f}}^{\text{0}}\text{-393}\text{.5 }\text{ 0}\text{-110kJ/mol}\\ \\ \text{ S°}\text{213}\text{.7}\text{5}\text{.740}\text{197}\text{.5J/mol}\text{.K}\end{array}$

Standard enthalpy change and standard entropy change is calculated.

$\begin{array}{l}{\text{ ΔH}}_{\text{rxn}}\text{°= }\left[\text{2}\left(\text{-110}\text{.5}\right)\text{ - }\left(\text{-393}\text{.5}\right)\right]\text{ kJ = 172}\text{.5 kJ }\\ \\ {\text{ ΔS}}_{\text{rxn}}\text{°= }\left[\text{2}\left(\text{197}\text{.5}\right)\text{ - }\left(\text{213}\text{.7}\right)\text{ - }\left(\text{5}\text{.740}\right)\right]\text{ J/K = 175}\text{.56 J/K = 0}\text{.17556 kJ/K}\end{array}$

Assume that change in enthalpy and entropy they do not vary with temperature and use them to determine $\text{ΔG}$ at greater temperature.

${\text{ ΔG}}_{\text{rxn}}{\text{=ΔH}}_{\text{rxn}}{\text{-TΔS}}_{\text{rxn}}\text{= }\left(\text{172}\text{.5 kJ}\right)\text{ - }\left(\text{2273 K}\right)\left(\text{0}\text{.17556 kJ/K}\right)\text{ = -226}\text{.54 kJ }$

Write the free-energy change in terms of equilibrium constant

$\begin{array}{l}{\text{ ΔG}}_{\text{rxn}}\text{= -RTlnK= -226}\text{.54 kJ/mol}\\ \\ \text{Rearrange and solve for K}\text{. }\\ \\ \text{lnK= }\frac{\text{-ΔG}}{\text{RT}}\text{= }\frac{\text{-(-226}\text{.54kJ/mol)}}{\text{(8}\text{.314×10-3kJ/mol•K)(2273K)}}\text{= 11}\text{.988 }\\ \\ {\text{K=e}}^{\text{11}\text{.988}}\text{= 1}\text{.60}\times {\text{10}}^{\text{5}}\text{= 1}\text{.6}\times {\text{10}}^{\text{5}}\end{array}$

Conclusion

• Chemical equation for given reaction was given as,

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

• Equilibrium constant for this reaction at $\text{2000°C}$ was calculated as $\text{1}\text{.6}\times {\text{10}}^{\text{5}}$