Chapter 6, Problem 6B.5P

Interpretation Introduction

Interpretation:

The values of $K,{\Delta }_r{H}^{\Theta },{\Delta }_r{S}^{\Theta }$ and ${\Delta }_{\text{r}}{G}^{\Theta }$ at $1443\text{K}$ for the dissociation of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ into $\text{CO}\left(\text{g}\right)$ and ${\text{O}}_{\text{2}}\left(\text{g}\right)$ has to be calculated.

Concept introduction:

The Gibbs free energy of the reaction is the key factor of the identification of the direction of spontaneity of the reaction.  For the reaction to be spontaneous, the Gibbs free energy of the reaction must be negative.

Answer to Problem 6B.5P

The values of $K,{\Delta }_r{H}^{\Theta },{\Delta }_r{S}^{\Theta }$ and ${\Delta }_{\text{r}}{G}^{\Theta }$ at $1443\text{K}$ for the dissociation of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ into $\text{CO}\left(\text{g}\right)$ and ${\text{O}}_{\text{2}}\left(\text{g}\right)$ are calculated as $\underset{\_}{2.795\times 10^{-6}}$, $\underset{\_}{289kJmo{l}^{-1}}$, $\underset{\_}{94.24Jmo{l}^{-1}{K}^{-1}}$ and $\underset{\_}{153.4kJmo{l}^{-1}}$ respectively.

Explanation of Solution

The given reaction is shown below.

    ${\text{CO}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)$

Initially total number of moles is considered as one.  After dissociation the total number of moles become $1+\frac{\alpha }{2}$.

The ICE table for the given reaction is shown below.

  $\begin{array}{l}{\text{ CO}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)\text{ + }\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)\\ \text{Initial}\text{amounts }\left(\text{1}-\alpha \right)\alpha \frac{\alpha }{2}\\ \text{Mole}\text{fractions }\frac{\text{1}-\alpha }{1+\frac{\alpha }{2}}\frac{\alpha }{1+\frac{\alpha }{2}}\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}\\ \text{Partial}\text{pressure }\frac{\left(\text{1}-\alpha \right)p}{1+\frac{\alpha }{2}}\frac{\alpha p}{1+\frac{\alpha }{2}}\frac{\frac{\alpha }{2}p}{1+\frac{\alpha }{2}}\end{array}$

The equilibrium constant $K$ can be calculated using the equation shown below.    $K=\frac{\left(\frac{p\text{CO}\left(\text{g}\right)}{p°}\right)\left(\frac{p{\text{O}}_{\text{2}}\left(\text{g}\right)}{p°}\right)}{\left(\frac{p{\text{CO}}_2\left(\text{g}\right)}{p°}\right)}$                                                                          (1)

Where,

• $p\text{CO}\left(\text{g}\right)$ is the partial pressure of carbon monoxide.
• $p{\text{O}}_{\text{2}}\left(\text{g}\right)$ is the partial pressure of oxygen.
• $p{\text{CO}}_2$ is the partial pressure of carbon dioxide.
• $p°$ is the standard pressure.

Substitute the values of partial pressure of carbon monoxide, oxygen and carbon dioxide in equation (1) as shown below.

    $\begin{array}{c}K=\frac{\left(\frac{p\text{CO}\left(\text{g}\right)}{p°}\right)\left(\frac{p{\text{O}}_{\text{2}}\left(\text{g}\right)}{p°}\right)}{\left(\frac{p{\text{CO}}_2\left(\text{g}\right)}{p°}\right)}\\ ={\frac{\left(\frac{\left(\frac{\alpha }{1+\frac{\alpha }{2}}\right)p}{p°}\right)\left(\frac{\left(\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}\right)p}{p°}\right)}{\left(\frac{\left(\frac{1-\alpha }{1+\frac{\alpha }{2}}\right)p}{p°}\right)}}^{1/2}\\ =\frac{\left(\frac{\alpha }{1+\frac{\alpha }{2}}\right)\left(\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}\right){\left(\frac{p}{p°}\right)}^{1/2}}{\left(\frac{1-\alpha }{1+\frac{\alpha }{2}}\right)}\end{array}$

The value of dissociation constant $\left(\alpha \right)$ is less than one at all temperatures, therefore, the above equation can be written as shown below.

    $\begin{array}{c}K=\frac{\left(\frac{\alpha }{1}\right)\left(\frac{\frac{\alpha }{2}}{1}\right){\left(\frac{p}{p°}\right)}^{1/2}}{\left(\frac{1}{1}\right)}\\ =\alpha {\left(\frac{\alpha }{2}\right)}^{1/2}\\ =\frac{{\alpha }^{3/2}}{\sqrt{2}}\end{array}$                                                                              (2)

The value of $\alpha$ at $1443\text{K}$ is given as $2.50\times {10}^{-4}$.  Substitute the value of $\alpha$ in above equation as shown below.

    $\begin{array}{c}K=\frac{{\alpha }^{3/2}}{\sqrt{2}}\\ =\frac{{\left(2.50\times {10}^{-4}\right)}^{3/2}}{\sqrt{2}}\\ =\underset{\_}{2.795\times 10^{-6}}\end{array}$

The value of ${\Delta }_{\text{r}}{G}^{\Theta }$ for the given reaction is calculated by the following formula.

    ${\Delta }_{\text{r}}{G}^{\Theta }=-RT\ln K$                                                                                           (3)

Where,

• ${\Delta }_{\text{r}}{G}^{\Theta }$ is the Gibbs free energy of the given reaction.
• $R$ is the gas constant.
• $K$ is the equilibrium constant.
• $T$ is the temperature.

The value of gas constant is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

The value of equilibrium constant for the given reaction is calculated as $2.795\times {10}^{-6}$.

It is given that the temperature is $1443\text{K}$.

Substitute the values of temperature, equilibrium constant and gas constant in equation (3).

    $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\Theta }=-8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 1443\text{K}\times \text{ln}\left(2.795\times {10}^{-6}\right)\\ =153415.08\text{J}{\text{mol}}^{\text{-1}}\\ =\underset{\_}{153.4kJmo{l}^{-1}}\end{array}$

The relation between standard enthalpy and equilibrium constant is given by van’t Hoff equation as shown below.

    $\ln \left(\frac{K_2}{K_1}\right)=-\frac{{\Delta }_r{H}^{\Theta }}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$                                                                          (4)

Where,

• $K_2,K_1$ are the equilibrium constants.
• $T_1,T_2$ are the initial and final temperatures.
• $R$ is gas constant.
• ${\Delta }_r{H}^{\Theta }$ is the standard enthalpy.

The value of $\alpha$ at $\text{1395}\text{K}$ is given as $1.44\times {10}^{-4}$.

The value of equilibrium constant at $\text{1395}\text{K}$ can be calculated using equation (2)  as shown below.

    $\begin{array}{c}{K}_1=\frac{{\alpha }^{3/2}}{\sqrt{2}}\\ =\frac{{\left(1.44\times {10}^{-4}\right)}^{3/2}}{\sqrt{2}}\\ =1.22\times {10}^{-6}\end{array}$

Substitute $T_1=1395\text{K}$, $T_2=1443\text{K}$, $K_1=1.22\times {10}^{-6}$ and $K_2=2.795\times {10}^{-6}$ in equation (4) as shown below.

    $\begin{array}{c}\ln \left(\frac{K_2}{K_1}\right)=-\frac{{\Delta }_r{H}^{\Theta }}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]\\ \ln \left(\frac{2.795\times {10}^{-6}}{1.22\times {10}^{-6}}\right)=-\frac{{\Delta }_r{H}^{\Theta }}{8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}}\left[\frac{1}{1443\text{K}}-\frac{1}{1395\text{K}}\right]\\ 0.8289=-\frac{{\Delta }_r{H}^{\Theta }}{8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}}\times \left[\frac{\left(1395\text{K}-1443\text{K}\right)}{1443\text{K}\times 1395\text{K}}\right]\\ {\Delta }_r{H}^{\Theta }=-\left(0.8289\times 8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times \left[\frac{1443\text{K}\times 1395\text{K}}{\left(1395\text{K}-1443\text{K}\right)}\right]\right)\end{array}$

The above equation on solving gives the value of standard enthalpy as ${\Delta }_r{H}^{\Theta }=\underset{\_}{289kJmo{l}^{-1}}$.

The ${\Delta }_r{S}^{\Theta }$ is calculated by the formula given below.

    ${\Delta }_r{S}^{\Theta }=\frac{{\Delta }_r{H}^{\Theta }-{\Delta }_{\text{r}}{G}^{\Theta }}{T}$

Substitute the values of ${\Delta }_r{H}^{\Theta }$, ${\Delta }_{\text{r}}{G}^{\Theta }$ and $T$ in above equation as shown below.

    $\begin{array}{c}{\Delta }_r{S}^{\Theta }=\frac{{\Delta }_r{H}^{\Theta }-{\Delta }_{\text{r}}{G}^{\Theta }}{T}\\ =\frac{289\text{kJ}{\text{mol}}^{-1}-153\text{kJ}{\text{mol}}^{-1}}{1443\text{K}}\\ =0.0942\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\\ =\underset{\_}{94.24Jmo{l}^{-1}{K}^{-1}}\end{array}$

Therefore, the values of $K,{\Delta }_r{H}^{\Theta },{\Delta }_r{S}^{\Theta }$ and ${\Delta }_{\text{r}}{G}^{\Theta }$ at $1443\text{K}$ for the dissociation of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ into $\text{CO}\left(\text{g}\right)$ and ${\text{O}}_{\text{2}}\left(\text{g}\right)$ are calculated as $\underset{\_}{2.795\times 10^{-6}}$, $\underset{\_}{289kJmo{l}^{-1}}$, $\underset{\_}{94.24Jmo{l}^{-1}{K}^{-1}}$ and $\underset{\_}{153.4kJmo{l}^{-1}}$ respectively.