Chapter 14, Problem 14.32P

(a)

Interpretation Introduction

Interpretation:

If there is any advantage in processing the reaction at pressure above $1\text{ bar}$ needs to be stated.

Concept introduction:

For a single reaction system, the final moles of each of the present components can be estimated by the equation:

  $n_i=n_{io}+v_i\xi$ ...... (1)

Here, $n_i$ is the final moles of the component $i$ , $n_{io}$ is the initial moles of the component $i$ , $v_i$ is the stoichiometric coefficient of the component $i$ in the reaction and $\xi$ is the extent of the reaction. $v_i$ is taken as negative for reactants and positive for products.

Mole fraction $\left(y_i\right)$ of any component is given by

  $y_i=\frac{n_i}{{\displaystyle \sum n_i}}$ ...... (2)

Here, $n_i$ are the moles of component $i$ and ${\displaystyle \sum n_i}$ is the total moles of all the components present in the mixture.

Equilibrium constant of this reaction from equation 14.28 can be written as:

  $K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }$ ...... (3)

Where, $y_i$ is the mole fraction of component $i$ , $P$ is the pressure of the reaction, and $P^0$ is the standard pressure.

Answer to Problem 14.32P

There is no advantage in processing the reaction at pressure above $1\text{ bar}$ as pressure has no effect on the equilibrium constant for the given reaction.

Explanation of Solution

Given information:

By the reaction of steam with “water gas”, hydrogen gas is produced. The reaction by which the steam is passed over a catalyst to convert ${\text{CO to CO}}_2$ is

  ${\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{CO}}_2\left(g\right)$

The equilibrium conditions given for this reaction is $800\text{ K and 1 bar}$ .

For the given reaction to produce hydrogen gas, let the extent of the reaction be $\xi$ . Water gas is the combination of ${\text{H}}_2\text{ and CO}$ which is fed in equimolar quantity.

  $\begin{array}{l}{\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{CO}}_2\left(g\right)\\ 1\text{ 1}\text{ 1 0}\\ 1-\xi \text{ 1}-\xi 1+\xi \xi \end{array}$

The overall stoichiometric coefficient for this reaction is

  $\nu =1+1-\left(1+1\right)=0$

The individual stoichiometric coefficients for all the components in this reaction are

  $\begin{array}{c}{\nu }_{{\text{H}}_2\text{O}}=-1\\ {\nu }_{\text{CO}}=-1\\ {\nu }_{{\text{H}}_2}=+1\\ {\nu }_{{\text{CO}}_2}=+1\end{array}$

The initial and final pressure of the system is taken as

  $\begin{array}{l}P=P\text{ bar}\\ P^0=1\text{ bar}\end{array}$ .

Now, use equation (3) for the equilibrium constant of this reaction and simplify the expression as

  $\begin{array}{c}K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K={\left(y_{{\text{H}}_2}\right)}^1{\left(y_{{\text{CO}}_2}\right)}^1{\left(y_{{\text{H}}_2\text{O}}\right)}^{-1}{\left(y_{\text{CO}}\right)}^{-1}{\left(\frac{P^0}{P}\right)}^{-\left(0\right)}\\ K=\frac{y_{{\text{H}}_2}{y}_{{\text{CO}}_2}}{y_{{\text{H}}_2\text{O}}{y}_{\text{CO}}}\end{array}$

Since the expression for the equilibrium constant for this reaction does not depend on the pressure of the system, carrying the reaction above $1\text{ bar}$ will not affect the reaction and will not have any advantage on the reaction.

(b)

Interpretation Introduction

Interpretation:

The effect of increase in the equilibrium temperature for the given reaction on the conversion of $\text{CO}$ is to be determined.

Concept introduction:

According to the Le’ Chatelier’s principle, if a reaction is subject to any change at its equilibrium, then the reaction tends to shift its equilibrium in the direction so as to undo the effect of that change on its equilibrium.

Answer to Problem 14.32P

The increase in the equilibrium temperature of the reaction does not increase the conversion of $\text{CO,}$ as the equilibrium constant decreases with increasing temperature for an exothermic reaction.

Explanation of Solution

Given information:

By the reaction of steam with “water gas”, hydrogen gas is produced. The reaction by which the steam is passed over a catalyst to convert ${\text{CO to CO}}_2$ is

  ${\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{CO}}_2\left(g\right)$

The equilibrium conditions given for this reaction is $800\text{ K and 1 bar}$ .

The standard heat of reaction and Gibb’s free energy for this reaction is

  $\begin{array}{l}\Delta G_{298}^0=-28618\text{J/mol}\\ \Delta H_{298}^0=-41166\text{ J/mol}\end{array}$

Since the reaction is exothermic, the heat released during the reaction acts as one of the products of the reaction.

Increasing the equilibrium temperature of the reaction increases the heat released of the system. So, according to the Le’ Chatelier’s principle, to undo the effect of the increase in the products, the reaction shifts to the left toward the reactants. Thus, there are more reactants present at the equilibrium and the conversion of $\text{CO}$ deceases as the equilibrium constant decreases.

(c)

Interpretation Introduction

Interpretation:

The molar ratio of steam to water gas for the given reaction is to be determined.

Concept introduction:

For a single reaction system, the final moles of each of the present components can be estimated by the equation:

  $n_i=n_{io}+v_i\xi$ ...... (1)

Here, $n_i$ is the final moles of the component $i$ , $n_{io}$ is the initial moles of the component $i$ , $v_i$ is the stoichiometric coefficient of the component $i$ in the reaction and $\xi$ is the extent of the reaction. $v_i$ is taken as negative for reactants and positive for products.

Mole fraction $\left(y_i\right)$ of any component is given by:

  $y_i=\frac{n_i}{{\displaystyle \sum n_i}}$ ...... (2)

Here, $n_i$ are the moles of component $i$ and ${\displaystyle \sum n_i}$ is the total moles of all the components present in the mixture.

Equilibrium constant of this reaction from equation 14.28 can be written as:

  $K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }$ ...... (3)

Where, $y_i$ is the mole fraction of component $i$ , $P$ is the pressure of the reaction, and $P^0$ is the standard pressure.

Gibb’s free energy in terms of equilibrium constant is written as:

  $\ln K=-\frac{\Delta G^0}{RT}$ ...... (4)

Also, Gibbs free energy is calculated using heat of reaction from the equation given as $14.11b$ .

  $\Delta G^0=\Delta H_0^0-\frac{T}{T_0}\left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}$ ...... (5)

Here, ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}\text{ and }{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}$ are defines as:

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}=\left(\Delta A\right)T_0\left(\tau -1\right)+\frac{\Delta B}{2}{T}_0^2\left({\tau }^2-1\right)+\frac{\Delta C}{3}{T}_0^3\left({\tau }^3-1\right)+\frac{\Delta D}{T_0}\left(\frac{\tau -1}{\tau }\right)\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}=\Delta A\ln \tau +\left(\tau -1\right)\left[\Delta B{T}_0+\left(\frac{\tau +1}{2}\right)\left(\Delta C{T}_0^2+\frac{\Delta D}{{\tau }^2{T}_0^2}\right)\right]\end{array}$ ...... (6)

Where, $\tau =\frac{T}{T_0}$

Answer to Problem 14.32P

The molar ratio of steam to water gas fed for the given reaction is $4.09$ .

Explanation of Solution

Given information:

By the reaction of steam with “water gas”, hydrogen gas is produced. The reaction by which the steam is passed over a catalyst to convert ${\text{CO to CO}}_2$ is:

  ${\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{CO}}_2\left(g\right)$

The equilibrium conditions given for this reaction is $800\text{ K and 1 bar}$ .

After cooling the products to ${20}^{\circ }\text{C}$ , the gaseous products contains only $2\text{ mol\% CO}$ and all the unreacted ${\text{H}}_2\text{O}$ is virtually condensed.

From Table C.4 the standard heat of reaction and Gibb’s free energy for this reaction is:

  $\begin{array}{l}\Delta G_{298}^0=-28618\text{J/mol}\\ \Delta H_{298}^0=-41166\text{ J/mol}\end{array}$

From Table C.1 the coefficients for the heat capacity of the component gases are given as:

Substance	$A$	$B$	$C$	$D$
${\text{H}}_2\text{O}$	$3.470$	$1.450\times {10}^{-3}$	$\_\_$	$0.121\times {10}^5$
$\text{CO}$	$3.376$	$0.557\times {10}^{-3}$	$\_\_$	$-0.031\times {10}^5$
${\text{H}}_2$	$3.249$	$0.422\times {10}^{-3}$	$\_\_$	$0.083\times {10}^5$
${\text{CO}}_2$	$5.457$	$1.045\times {10}^{-3}$	$\_\_$	$-1.157\times {10}^5$

$\Delta A$ , $\Delta B$ , $\Delta C$ , and $\Delta D$ from above given values can be calculated as:

  $\begin{array}{c}\Delta A=A_{{\text{H}}_2}+A_{{\text{CO}}_2}-\left(1\times A_{\text{CO}}\right)-\left(1\times A_{{\text{H}}_2\text{O}}\right)\\ =3.249+5.457-\left(1\times 3.376\right)-\left(1\times 3.470\right)\\ =1.86\\ \Delta B=B_{{\text{H}}_2}+B_{{\text{CO}}_2}-\left(1\times B_{\text{CO}}\right)-\left(1\times B_{{\text{H}}_2\text{O}}\right)\\ =\left(0.422+1.045-\left(1\times 0.557\right)-\left(1\times 1.450\right)\right)\times {10}^{-3}\\ =-0.54\times {10}^{-3}\\ \Delta C=C_{{\text{H}}_2}+C_{{\text{CO}}_2}-\left(1\times C_{\text{CO}}\right)-\left(1\times C_{{\text{H}}_2\text{O}}\right)\\ =0+0-\left(1\times 0\right)-\left(1\times 0\right)\\ =0\\ \Delta D=D_{{\text{H}}_2}+D_{{\text{CO}}_2}-\left(1\times D_{\text{CO}}\right)-\left(1\times D_{{\text{H}}_2\text{O}}\right)\\ =\left(0.083+\left(-1.157\right)-\left(1\times \left(-0.031\right)\right)-\left(1\times 0.121\right)\right)\times {10}^5\\ =-1.164\times {10}^5\end{array}$

Now, use equations set (6) to evaluate the values of ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}\text{ and }{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}$ at equilibrium temperature of $800\text{ K}$ as:

$\tau =\frac{T}{T_0}=\frac{800\text{ K}}{298.15\text{ K}}=2.683$

${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}=\left(\Delta A\right)T_0\left(\tau -1\right)+\frac{\Delta B}{2}{T}_0^2\left({\tau }^2-1\right)+\frac{\Delta C}{3}{T}_0^3\left({\tau }^3-1\right)+\frac{\Delta D}{T_0}\left(\frac{\tau -1}{\tau }\right)$

$=\left(\begin{array}{c}\left(1.86\right)\left(298.15\right)\left(2.683-1\right)+\frac{\left(-0.54\times {10}^{-3}\right)}{2}{\left(298.15\right)}^2\left({\left(2.683\right)}^2-1\right)\\ +\frac{\left(0\right)}{3}{\left(298.15\right)}^3\left({\left(2.683\right)}^3-1\right)+\frac{\left(-1.164\times {10}^5\right)}{298.15}\left(\frac{2.683-1}{2.683}\right)\end{array}\right)$]$=539.656$

${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}=\Delta A\ln \tau +\left(\tau -1\right)\left[\Delta B{T}_0+\left(\frac{\tau +1}{2}\right)\left(\Delta C{T}_0^2+\frac{\Delta D}{{\tau }^2{T}_0^2}\right)\right]$

$\begin{array}{c}=\left(1.86\right)\ln \left(2.683\right)+\left(2.683-1\right)\left[\left(-0.54\times {10}^{-3}\right)\left(298.15\right)+\left(\frac{2.683+1}{2}\right)\left(\begin{array}{l}\left(0\right){\left(298.15\right)}^2\\ +\frac{-1.164\times {10}^5}{{\left(2.683\right)}^2{\left(298.15\right)}^2}\end{array}\right)\right]\\ =1.001\end{array}$

Now, use equation (5) along with the above calculated values to get the value of $\Delta G^0$ as:

  $\begin{array}{c}\Delta G^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ =\left(\left(-41166\right)-\left(2.683\right)\left(-41166-\left(-28618\right)\right)+\left(8.314\right)\left(539.656\right)-\left(8.314\right)\left(800\right)\left(1.001\right)\right)\\ =-9.67\times {10}^3\text{ J/mol}\end{array}$

Use this calculated value of $\Delta G^0$ to calculate the equilibrium constant of the reaction at $800\text{ K}$ using equation (4) as:

  $\begin{array}{c}\ln K=-\frac{\Delta G^0}{RT}\\ K=\exp \left(-\frac{\left(-9.67\times {10}^3\right)}{\left(8.314\right)\left(800\right)}\right)\\ K=4.279\end{array}$

For the given reaction to produce hydrogen gas, let the extent of the reaction be $\xi$ .

  $\begin{array}{l}{\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{CO}}_2\left(g\right)\\ 1\text{ 1}\text{ 1 0}\\ 1-\xi \text{ 1}-\xi 1+\xi \xi \end{array}$

The overall stoichiometric coefficient for this reaction is,

  $\nu =1+1-\left(1+1\right)=0$

The individual stoichiometric coefficients for all the components in this reaction are:

  $\begin{array}{c}{\nu }_{{\text{H}}_2\text{O}}=-1\\ {\nu }_{\text{CO}}=-1\\ {\nu }_{{\text{H}}_2}=+1\\ {\nu }_{{\text{CO}}_2}=+1\end{array}$

Using equation (1), write the expressions for final moles of all the components present in the products as gases. All of the unreacted ${\text{H}}_2\text{O}$ is virtually condensed.

  $\begin{array}{c}{n}_{\text{CO}}=n_{0\left(\text{CO}\right)}+v_{\text{CO}}\xi =1-\xi \\ n_{{\text{H}}_2}=n_{0\left({\text{H}}_2\right)}+v_{{\text{H}}_2}\xi =1+\xi \\ n_{{\text{CO}}_2}=n_{0\left({\text{CO}}_2\right)}+v_{{\text{CO}}_2}\xi =\xi \end{array}$

According to the given conditions on the product gases,

  $\begin{array}{c}\frac{n_{\text{CO}}}{n_{\text{CO}}+n_{{\text{H}}_2}+n_{{\text{CO}}_2}}=0.02\\ \frac{1-\xi }{1-\xi +1+\xi +\xi }=0.02\\ \xi =0.941\end{array}$

Let the mole of steam $\left({\text{H}}_2\text{O}\right)$ fed be $n_{0\left({\text{H}}_2\text{O}\right)}$ . Also, the ratio of steam to the water gas fed to the reactor be $z$ . So,

  $\begin{array}{c}z=\frac{n_{0\left({\text{H}}_2\text{O}\right)}}{n_{0\left({\text{H}}_2\right)}+n_{0\left(\text{CO}\right)}}\\ z=\frac{n_{0\left({\text{H}}_2\text{O}\right)}}{1+1}\\ z=\frac{n_{0\left({\text{H}}_2\text{O}\right)}}{2}\mathrm{......}\text{ (7)}\end{array}$

Now, the moles of all the species in the products will be

  $\begin{array}{c}{n}_{{\text{H}}_2\text{O}}=n_{0\left({\text{H}}_2\text{O}\right)}-\xi \\ n_{\text{CO}}=1-\xi \\ n_{{\text{H}}_2}=1+\xi \\ n_{{\text{CO}}_2}=\xi \end{array}$

Total moles of the products will be

  $\begin{array}{c}n=2+n_{0\left({\text{H}}_2\text{O}\right)}+\left(0\right)\xi \\ =2+n_{0\left({\text{H}}_2\text{O}\right)}\end{array}$

Using equation (2) and substituting the value of $n_{0\left({\text{H}}_2\text{O}\right)}$ in terms of $z$ from equation (7), mole fraction of all the species will be

  $\begin{array}{c}{y}_{{\text{H}}_2\text{O}}=\frac{n_{0\left({\text{H}}_2\text{O}\right)}-\xi }{2+n_{0\left({\text{H}}_2\text{O}\right)}}=\frac{2z-\xi }{2+2z}\\ y_{\text{CO}}=\frac{1-\xi }{2+n_{0\left({\text{H}}_2\text{O}\right)}}=\frac{1-\xi }{2+2z}\\ y_{{\text{H}}_2}=\frac{1+\xi }{2+n_{0\left({\text{H}}_2\text{O}\right)}}=\frac{1+\xi }{2+2z}\\ y_{{\text{CO}}_2}=\frac{\xi }{2+n_{0\left({\text{H}}_2\text{O}\right)}}=\frac{\xi }{2+2z}\end{array}$

Now, use equation (3) and the calculated value of the equilibrium constant and $\xi$ to get the value of $z$ as

  $\begin{array}{c}K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ 4.279={\left(\frac{1+\xi }{2+2z}\right)}^1{\left(\frac{\xi }{2+2z}\right)}^1{\left(\frac{2z-\xi }{2+2z}\right)}^{-1}{\left(\frac{1-\xi }{2+2z}\right)}^{-1}{\left(\frac{1}{1}\right)}^{-\left(0\right)}\\ 4.279=\frac{\left(\frac{1+0.941}{2+2z}\right)\left(\frac{0.941}{2+2z}\right)}{\left(\frac{2z-0.941}{2+2z}\right)\left(\frac{1-0.941}{2+2z}\right)}\\ z=4.09\end{array}$

Therefore, the molar ratio of steam to water gas fed for the given reaction is $4.09$ .

(d)

Interpretation Introduction

Interpretation:

The danger of formation of solid carbon by the given side reaction at equilibrium conditions is to be determined.

Concept introduction:

For a single reaction system, the final moles of each of the components present, can be estimated by the equation:

  $n_i=n_{io}+v_i\xi$ ...... (1)

Here, $n_i$ is the final moles of the component $i$ , $n_{io}$ is the initial moles of the component $i$ , $v_i$ is the stoichiometric coefficient of the component $i$ in the reaction and $\xi$ is the extent of the reaction. $v_i$ is taken as negative for reactants and positive for products.

Mole fraction $\left(y_i\right)$ of any component is given by

  $y_i=\frac{n_i}{{\displaystyle \sum n_i}}$ ...... (2)

Here, $n_i$ is the moles of component $i$ and ${\displaystyle \sum n_i}$ is the total moles of all the components present in the mixture.

Equilibrium constant of this reaction from equation 14.28 can be written as

  $K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }$ ...... (3)

Where, $y_i$ is the mole fraction of component $i$ , $P$ is the pressure of the reaction, and $P^0$ is the standard pressure.

Gibb’s free energy in terms of equilibrium constant is written as

  $\ln K=-\frac{\Delta G^0}{RT}$ ...... (4)

Also, Gibbs free energy is calculated using heat of reaction from the equation given as $14.11b$ .

  $\Delta G^0=\Delta H_0^0-\frac{T}{T_0}\left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}$ ...... (5)

Here, ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}\text{ and }{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}$ are defines as

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}=\left(\Delta A\right)T_0\left(\tau -1\right)+\frac{\Delta B}{2}{T}_0^2\left({\tau }^2-1\right)+\frac{\Delta C}{3}{T}_0^3\left({\tau }^3-1\right)+\frac{\Delta D}{T_0}\left(\frac{\tau -1}{\tau }\right)\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}=\Delta A\ln \tau +\left(\tau -1\right)\left[\Delta B{T}_0+\left(\frac{\tau +1}{2}\right)\left(\Delta C{T}_0^2+\frac{\Delta D}{{\tau }^2{T}_0^2}\right)\right]\end{array}$ ...... (6)

Where, $\tau =\frac{T}{T_0}$

Answer to Problem 14.32P

There is no danger of formation of solid carbon by the given side reaction at equilibrium conditions.

Explanation of Solution

Given information:

By the reaction of steam with “water gas”, hydrogen gas is produced. The reaction by which the steam is passed over a catalyst to convert ${\text{CO to CO}}_2$ is:

  ${\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{CO}}_2\left(g\right)$

The equilibrium conditions given for this reaction is $800\text{ K and 1 bar}$ .

At this equilibrium condition, the side reaction taking place to form carbon is

  $\text{2CO}\left(g\right)\to {\text{CO}}_2\left(g\right)+\text{c}\left(s\right)$

From Table C.4 the standard heat of reaction and Gibb’s free energy for the formation of carbon is

  $\begin{array}{l}\Delta G_{298}^0=-120021\text{J/mol}\\ \Delta H_{298}^0=-172459\text{ J/mol}\end{array}$

From Table C.1 the coefficients for the heat capacity of the component gases are given as

Substance	$A$	$B$	$C$	$D$
$\text{CO}$	$3.376$	$0.557\times {10}^{-3}$	$\_\_$	$-0.031\times {10}^5$
${\text{CO}}_2$	$5.457$	$1.045\times {10}^{-3}$	$\_\_$	$-1.157\times {10}^5$
$\text{C}$	$1.771$	$0.771\times {10}^{-3}$	$\_\_$	$-0.867\times {10}^5$

$\Delta A$ , $\Delta B$ , $\Delta C$ , and $\Delta D$ from above given values can be calculated as

  $\begin{array}{c}\Delta A=A_{\text{C}}+A_{{\text{CO}}_2}-\left(2\times A_{\text{CO}}\right)\\ =1.771+5.457-\left(2\times 3.376\right)\\ =0.476\\ \Delta B=B_{\text{C}}+B_{{\text{CO}}_2}-\left(2\times B_{\text{CO}}\right)\\ =\left(0.771+1.045-\left(2\times 0.557\right)\right)\times {10}^{-3}\\ =0.702\times {10}^{-3}\\ \Delta C=C_{\text{C}}+C_{{\text{CO}}_2}-\left(2\times C_{\text{CO}}\right)\\ =0+0-\left(1\times 0\right)\\ =0\\ \Delta D=D_{\text{C}}+D_{{\text{CO}}_2}-\left(2\times D_{\text{CO}}\right)\\ =\left(-0.867+\left(-1.157\right)-\left(2\times \left(-0.031\right)\right)\right)\times {10}^5\\ =-1.962\times {10}^5\end{array}$

Now, use equations set (6) to evaluate the values of ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}\text{ and }{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}$ at equilibrium temperature of $800\text{ K}$ as

$\tau =\frac{T}{T_0}=\frac{800\text{ K}}{298.15\text{ K}}=2.683$

${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}=\left(\Delta A\right)T_0\left(\tau -1\right)+\frac{\Delta B}{2}{T}_0^2\left({\tau }^2-1\right)+\frac{\Delta C}{3}{T}_0^3\left({\tau }^3-1\right)+\frac{\Delta D}{T_0}\left(\frac{\tau -1}{\tau }\right)$

$=\left(\begin{array}{c}\left(0.476\right)\left(298.15\right)\left(2.683-1\right)+\frac{\left(0.702\times {10}^{-3}\right)}{2}{\left(298.15\right)}^2\left({\left(2.683\right)}^2-1\right)\\ +\frac{\left(0\right)}{3}{\left(298.15\right)}^3\left({\left(2.683\right)}^3-1\right)+\frac{\left(-1.962\times {10}^5\right)}{298.15}\left(\frac{2.683-1}{2.683}\right)\end{array}\right)$

$=19.465$

${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}=\Delta A\ln \tau +\left(\tau -1\right)\left[\Delta B{T}_0+\left(\frac{\tau +1}{2}\right)\left(\Delta C{T}_0^2+\frac{\Delta D}{{\tau }^2{T}_0^2}\right)\right]$

$\begin{array}{c}=\left(0.476\right)\ln \left(2.683\right)+\left(2.683-1\right)\left[\left(0.702\times {10}^{-3}\right)\left(298.15\right)+\left(\frac{2.683+1}{2}\right)\left(\begin{array}{l}\left(0\right){\left(298.15\right)}^2\\ +\frac{-1.962\times {10}^5}{{\left(2.683\right)}^2{\left(298.15\right)}^2}\end{array}\right)\right]\\ =-0.128\end{array}$

Now, use equation (5) along with the above calculated values to get the value of $\Delta G^0$ of the side reaction as

  $\begin{array}{c}\Delta G^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ =\left(\left(-172459\right)-\left(2.683\right)\left(-172459-\left(-120021\right)\right)+\left(8.314\right)\left(19.465\right)-\left(8.314\right)\left(800\right)\left(-0.128\right)\right)\\ =-3.0754\times {10}^4\text{ J/mol}\end{array}$

Use this calculated value of $\Delta G^0$ to calculate the equilibrium constant of the reaction at $800\text{ K}$ using equation (4) as

  $\begin{array}{c}\ln K_{\text{eq}}=-\frac{\Delta G^0}{RT}\\ K_{\text{eq}}=\exp \left(-\frac{\left(-3.0754\times {10}^4\right)}{\left(8.314\right)\left(800\right)}\right)\\ K_{\text{eq}}=101.883\end{array}$

This is the equilibrium constant for this reaction and if the actual value of this constant is greater than the equilibrium value then the reaction tries to shift to the left and reducing the formation of carbon.

Calculate the actual value of this constant as

For the given reaction to produce carbon, let the extent of the reaction be $\xi$ .

  $\begin{array}{l}2\text{CO}\left(g\right)\to {\text{CO}}_2\left(g\right)+\text{C}\left(s\right)\\ \text{ 2}\text{ 0}\text{ 0}\\ 2-\xi \xi \xi \end{array}$

The overall stoichiometric coefficient for this reaction is (for gases only)

  $\nu =1-\left(2\right)=-1$

The individual stoichiometric coefficients for all the gaseous components in this reaction are

  $\begin{array}{c}{\nu }_{\text{CO}}=-2\\ {\nu }_{{\text{CO}}_2}=+1\end{array}$

Use equation (3) which is applicable only for gaseous species, such that,

  $\begin{array}{c}{K}_{\text{actual}}=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K_{\text{actual}}={\left(y_{{\text{CO}}_2}\right)}^1{\left(y_{\text{CO}}\right)}^{-2}{\left(\frac{1}{1}\right)}^{-\left(-1\right)}\\ K_{\text{actual}}=\frac{y_{{\text{CO}}_2}}{{\left(y_{\text{CO}}\right)}^2}\end{array}$

From part (c), the actual value of the mole fraction of ${\text{CO and CO}}_2$ will be

  $\begin{array}{c}{y}_{\text{CO}}=\frac{1-\xi }{2+n_{0\left({\text{H}}_2\text{O}\right)}}=\frac{1-\xi }{2+2z}=\frac{1-0.941}{2+2\left(4.09\right)}=0.0057\\ y_{{\text{CO}}_2}=\frac{\xi }{2+n_{0\left({\text{H}}_2\text{O}\right)}}=\frac{\xi }{2+2z}=\frac{0.941}{2+2\left(4.09\right)}=0.0924\end{array}$

Calculate the ratio of the actual constant as:

  $\begin{array}{c}{K}_{\text{actual}}=\frac{y_{{\text{CO}}_2}}{{\left(y_{\text{CO}}\right)}^2}\\ =\frac{0.0924}{{\left(0.0057\right)}^2}\\ =2.843\times {10}^3\end{array}$

As the actual value of $K$ is much greater than the equilibrium value, there will not be any danger of formation of carbon as the side reaction of carbon formation tries to remain on the left side of the reaction.