Chapter 14, Problem 14.34P

(a)

Interpretation Introduction

Interpretation:

Whether there is any advantage in processing the reactions at pressure above $1\text{ bar}$ or not needs to be determined.

Concept introduction:

For a multiple reaction system, the final moles for each of the components present in the products can be estimated by the equation:

  $n_i=n_{i0}+{\displaystyle \sum _j{v}_{ij}{\xi }_j}$ ...... (1)

Here, $n_i$ is the final moles of the component $i$ , $n_{i0}$ is the initial moles of the component $i$ , $v_{ij}$ is the stoichiometric coefficient of the component $i$ in the reaction $j$ and ${\xi }_j$ is the extent of the reaction $j$ . $v_{ij}$ 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.

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 14.34P

There is no advantage in processing the reactions at pressure above $1\text{ bar}$ as increase in the pressure decreases the equilibrium constant for the reaction (i).

Explanation of Solution

Given information:

The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii ):

  $\begin{array}{c}{\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}i\text{)}\\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}ii\text{)}\end{array}$

The equilibrium conditions given for these reactions are $1300\text{ K and 1 bar}$ .

For the given reaction to produce synthesis gas, let the extent of the reaction ( i ) be ${\xi }_1$ and for reaction ( ii ) be ${\xi }_2$ .

  $\begin{array}{l}{\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\\ 1\text{ 1 0 0}\\ 1-\xi 1-\xi \xi \text{ 3}\xi \\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\\ 1\text{1}\text{ 0 0}\\ 1-\xi \text{ 1}-\xi \xi \xi \end{array}$

The overall stoichiometric coefficients for both the reactions are

  $\begin{array}{c}{\nu }_1=1+3-\left(1+1\right)=2\\ {\nu }_2=1+1-\left(1+1\right)=0\end{array}$

The individual stoichiometric coefficients for all the components in both the reactions are

  $\begin{array}{c}{\nu }_1{}_{\left({\text{CH}}_4\right)}=-1\\ {\nu }_1{}_{\left({\text{H}}_2\text{O}\right)}=-1\\ {\nu }_{1\left(\text{CO}\right)}=+1\\ {\nu }_{1\left({\text{H}}_2\right)}=+3\\ {\nu }_2{}_{\left(\text{CO}\right)}=-1\\ {\nu }_2{}_{\left({\text{H}}_2\text{O}\right)}=-1\\ {\nu }_{2\left({\text{CO}}_2\right)}=+1\\ {\nu }_{2\left({\text{H}}_2\right)}=+1\end{array}$

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 both the reactions and simplify the expressions as

  $\begin{array}{c}{K}_1=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K_1={\left(y_{{\text{H}}_2}\right)}^3{\left(y_{\text{CO}}\right)}^1{\left(y_{{\text{H}}_2\text{O}}\right)}^{-1}{\left(y_{{\text{CH}}_4}\right)}^{-1}{\left(\frac{P^0}{P}\right)}^{-\left(2\right)}\\ K_1=\frac{{\left(y_{{\text{H}}_2}\right)}^3{y}_{{\text{CO}}_2}}{y_{{\text{H}}_2\text{O}}{y}_{{\text{CH}}_4}}{\left(\frac{P}{P^0}\right)}^{\left(2\right)}\\ K_2=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K_2={\left(y_{{\text{H}}_2}\right)}^1{\left(y_{{\text{CO}}_2}\right)}^1{\left(y_{{\text{H}}_2\text{O}}\right)}^{-1}{\left(y_{\text{CO}}\right)}^{-1}{\left(\frac{P^0}{P}\right)}^{-\left(0\right)}\\ K_2=\frac{y_{{\text{H}}_2}{y}_{{\text{CO}}_2}}{y_{{\text{H}}_2\text{O}}{y}_{\text{CO}}}\end{array}$

The expression for the equilibrium constant for reaction ( i ) is dependent on pressure of the system. An increase in the pressure increases the equilibrium constant of the reaction. Thus, 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 is more of reactants present at the equilibrium providing no advantage of carrying the reactions at pressure above $1\text{ bar}$ .

(b)

Interpretation Introduction

Interpretation:

Whether there is any advantage in processing the reactions at temperatures below $1300\text{ K}$ or not needs 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 14.34P

The decrease in the equilibrium temperature of the reaction ( i ) does not provide any advantage as the equilibrium shifts towards the reactants with decreasing temperature for an endothermic reaction.

Explanation of Solution

Given information:

The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )

  $\begin{array}{c}{\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}i\text{)}\\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}ii\text{)}\end{array}$

The equilibrium conditions given for these reactions are $1300\text{ K and 1 bar}$ .

From Table C.4 the standard heat of reaction and Gibb’s free energy for both the reactions are

  $\begin{array}{l}{\left(\Delta G_{298}^0\right)}_1=141863\text{J/mol}\\ {\left(\Delta H_{298}^0\right)}_1=205813\text{ J/mol}\\ {\left(\Delta G_{298}^0\right)}_2=-28618\text{J/mol}\\ {\left(\Delta H_{298}^0\right)}_2=-41166\text{ J/mol}\end{array}$

Since, the reaction ( i ) is endothermic, the heat required during the reaction acts as one of the reactants of the reaction.

Decreasing the equilibrium temperature of the reaction below $1300\text{ K}$ increases the heat required for the reaction to proceed. So, according to the Le’ Chatelier’s principle, to undo the effect of the decrease in the reactants, the reaction shifts to the left towards the reactants. Thus, there is more reactants present at the equilibrium providing no advantage of carrying the reactions at temperature below $1300\text{ K}$ .

(c)

Interpretation Introduction

Interpretation:

In the gas synthesis, the ratio of number of moles of hydrogen to carbon monoxide needs to be determined.

Concept introduction:

For a multiple reaction system, the final moles for each of the components present in the products can be estimated by the equation:

  $n_i=n_{i0}+{\displaystyle \sum _j{v}_{ij}{\xi }_j}$ ...... (1)

Here, $n_i$ is the final moles of the component $i$ , $n_{i0}$ is the initial moles of the component $i$ , $v_{ij}$ is the stoichiometric coefficient of the component $i$ in the reaction $j$ and ${\xi }_j$ is the extent of the reaction $j$ . $v_{ij}$ 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 14.34P

The molar ratio of hydrogen to carbon monoxide in the synthesis gas $3$ .

Explanation of Solution

Given information:

The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )

  $\begin{array}{c}{\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}i\text{)}\\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}ii\text{)}\end{array}$

The equilibrium conditions given for these reactions are $1300\text{ K and 1 bar}$ .

Equimolar mixture of steam and methane is fed to the reactor.

From Table C.4 the standard heat of reaction and Gibb’s free energy for reaction ( i ) is

  $\begin{array}{l}\Delta G_{298}^0=141863\text{J/mol}\\ \Delta H_{298}^0=205813\text{ J/mol}\end{array}$

From Table C.1 the coefficients for the heat capacity of the component gases in reaction ( i ) are given as

Substance	$A$	$B$	$C$	$D$
${\text{CH}}_4$	$1.702$	$9.081\times {10}^{-3}$	$-2.164\times {10}^{-6}$	$\_\_$
${\text{H}}_2\text{O}$	$3.470$	$1.450\times {10}^{-3}$	$\_\_$	$0.121\times {10}^5$
$\text{CO}$	$3.376$	$0.557\times {10}^{-3}$	$\_\_$	$-0.031\times {10}^5$
${\text{H}}_2$	$3.249$	$0.422\times {10}^{-3}$	$\_\_$	$0.083\times {10}^5$

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

  $\begin{array}{c}\Delta A=3A_{{\text{H}}_2}+A_{\text{CO}}-A_{{\text{CH}}_4}-A_{{\text{H}}_2\text{O}}\\ =\left(3\times 3.249\right)+3.376-1.702-3.470\\ =7.951\\ \Delta B=3B_{{\text{H}}_2}+B_{\text{CO}}-B_{{\text{CH}}_4}-B_{{\text{H}}_2\text{O}}\\ =\left(\left(3\times 0.422\right)+0.557-9.081-1.450\right)\times {10}^{-3}\\ =-8.708\times {10}^{-3}\\ \Delta C=3C_{{\text{H}}_2}+C_{\text{CO}}-C_{{\text{CH}}_4}-C_{{\text{H}}_2\text{O}}\\ =\left(\left(3\times 0\right)+0-\left(-2.164\right)-0\right)\times {10}^{-6}\\ =2.164\times {10}^{-6}\\ \Delta D=3D_{{\text{H}}_2}+D_{\text{CO}}-D_{{\text{CH}}_4}-D_{{\text{H}}_2\text{O}}\\ =\left(\left(3\times 0.083\right)+\left(-0.031\right)-0-0.121\right)\times {10}^5\\ =9.7\times {10}^3\end{array}$

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 $1300\text{ K}$ as

  $\tau =\frac{T}{T_0}=\frac{1300\text{ K}}{298.15\text{ K}}=4.360$

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

  $=\left(\begin{array}{c}\left(7.951\right)\left(298.15\right)\left(4.360-1\right)+\frac{\left(-8.708\times {10}^{-3}\right)}{2}{\left(298.15\right)}^2\left({\left(4.360\right)}^2-1\right)\\ +\frac{\left(2.164\times {10}^{-6}\right)}{3}{\left(298.15\right)}^3\left({\left(4.360\right)}^3-1\right)+\frac{\left(9.7\times {10}^3\right)}{298.15}\left(\frac{4.360-1}{4.360}\right)\end{array}\right)$

  $=2585.19$

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

  $=\left(7.951\right)\ln \left(4.360\right)+\left(4.360-1\right)\left[\left(-8.708\times {10}^{-3}\right)\left(298.15\right)+\left(\frac{4.360+1}{2}\right)\left(\begin{array}{l}\left(2.164\times {10}^{-6}\right){\left(298.15\right)}^2\\ +\frac{9.7\times {10}^3}{{\left(4.360\right)}^2{\left(298.15\right)}^2}\end{array}\right)\right]$

  $=4.768$

Now, use equation (5) along with the above calculated values to get the value of $\Delta G_1^0$ for reaction ( i ) as

  $\begin{array}{c}\Delta G_1^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ =\left(\left(205813\right)-\left(4.360\right)\left(205813-141863\right)+\left(8.314\right)\left(2585.19\right)-\left(8.314\right)\left(1300\right)\left(4.768\right)\right)\\ =-1.031\times {10}^5\text{ J/mol}\end{array}$

Use this calculated value of $\Delta G_1^0$ to calculate the equilibrium constant of the reaction at $1300\text{ K}$ using equation (4) as

  $\begin{array}{c}\ln K_1=-\frac{\Delta G_1^0}{RT}\\ K_1=\exp \left(-\frac{\left(-1.031\times {10}^5\right)}{\left(8.314\right)\left(1300\right)}\right)\\ K_1=13891.8\end{array}$

From Table C.4 the standard heat of reaction and Gibb’s free energy for reaction ( ii ) 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 in reaction ( ii ) 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

  $\begin{array}{c}\Delta A=A_{{\text{H}}_2}+A_{{\text{CO}}_2}-A_{\text{CO}}-A_{{\text{H}}_2\text{O}}\\ =3.249+5.457-3.376-3.470\\ =1.86\\ \Delta B=B_{{\text{H}}_2}+B_{{\text{CO}}_2}-B_{\text{CO}}-B_{{\text{H}}_2\text{O}}\\ =\left(0.422+1.045-0.557-1.450\right)\times {10}^{-3}\\ =-0.54\times {10}^{-3}\\ \Delta C=C_{{\text{H}}_2}+C_{{\text{CO}}_2}-C_{\text{CO}}-C_{{\text{H}}_2\text{O}}\\ =\left(0+0-0-0\right)\times {10}^{-6}\\ =0\\ \Delta D=D_{{\text{H}}_2}+D_{\text{CO}}{}_{{}_2}-D_{\text{CO}}-D_{{\text{H}}_2\text{O}}\\ =\left(0.083+\left(-1.157\right)-\left(-0.031\right)-0.121\right)\times {10}^5\\ =-1.164\times {10}^5\end{array}$

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 $1300\text{ K}$ as

  $\tau =\frac{T}{T_0}=\frac{1300\text{ K}}{298.15\text{ K}}=4.360$

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

  $=\left(\begin{array}{c}\left(1.86\right)\left(298.15\right)\left(4.360-1\right)+\frac{\left(-0.54\times {10}^{-3}\right)}{2}{\left(298.15\right)}^2\left({\left(4.360\right)}^2-1\right)\\ +\frac{\left(0\right)}{3}{\left(298.15\right)}^3\left({\left(4.360\right)}^3-1\right)+\frac{\left(-1.164\times {10}^5\right)}{298.15}\left(\frac{4.360-1}{4.360}\right)\end{array}\right)$

  $=1130.20$

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

  $=\left(1.86\right)\ln \left(4.360\right)+\left(4.360-1\right)\left[\left(-0.54\times {10}^{-3}\right)\left(298.15\right)+\left(\frac{4.360+1}{2}\right)\left(\begin{array}{l}\left(0\right){\left(298.15\right)}^2\\ +\frac{-1.164\times {10}^5}{{\left(4.360\right)}^2{\left(298.15\right)}^2}\end{array}\right)\right]$

  $=1.578$

Now, use equation (5) along with the above calculated values to get the value of $\Delta G_2^0$ for reaction ( ii ) as

  $\begin{array}{c}\Delta G_2^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ =\left(\left(-41166\right)-\left(4.360\right)\left(\left(-41166\right)-\left(-28618\right)\right)+\left(8.314\right)\left(1130.20\right)-\left(8.314\right)\left(1300\right)\left(1.578\right)\right)\\ =5.884\times {10}^3\text{ J/mol}\end{array}$

Use this calculated value of $\Delta G_2^0$ to calculate the equilibrium constant of the reaction at $1300\text{ K}$ using equation (4) as

  $\begin{array}{c}\ln K_2=-\frac{\Delta G_2^0}{RT}\\ K_2=\exp \left(-\frac{\left(5.884\times {10}^3\right)}{\left(8.314\right)\left(1300\right)}\right)\\ K_2=0.5802\end{array}$

The value of $K_1$ is very large as compared to the value of $K_2$ so, practically reaction ( i ) is considered to go to completion.

Since, there is stoichiometric amount of ${\text{CH}}_{\text{4}}{\text{ and H}}_{\text{2}}\text{O}$ is fed to the reactor, all of ${\text{H}}_2\text{O}$ is consumed in reaction ( i ) and none is left for reaction ( ii ). Also, the stoichiometric amount of ${\text{CO and H}}_2$ is produced.

Accordingt o the stoichiometry of the reaction ( i ), $1{\text{ mol CH}}_4$ reacts with $1{\text{ mol H}}_2\text{O}$ to form $1\text{ mol CO}$ and $3{\text{ mol H}}_2$ .

Therefore, the ratio of moles of ${\text{H}}_2\text{ to CO}$ will be

  $\frac{n_{{\text{H}}_2}}{n_{\text{CO}}}=\frac{3}{1}=3$

(d)

Interpretation Introduction

Interpretation:

In the gas synthesis, the molar ratio of hydrogen to carbon monoxide needs to be determined for the feed ratio of steam-to-methane equals to $2$ .

Concept introduction:

For a multiple reaction system, the final moles for each of the components present in the products can be estimated by the equation,

  $n_i=n_{i0}+{\displaystyle \sum _j{v}_{ij}{\xi }_j}$ ...... (1)

Here, $n_i$ is the final moles of the component $i$ , $n_{i0}$ is the initial moles of the component $i$ , $v_{ij}$ is the stoichiometric coefficient of the component $i$ in the reaction $j$ and ${\xi }_j$ is the extent of the reaction $j$ . $v_{ij}$ 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 14.34P

The molar ratio of hydrogen to carbon monoxide in the synthesis gas $3.638$ .

Explanation of Solution

Given information:

The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )

  $\begin{array}{c}{\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}i\text{)}\\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}ii\text{)}\end{array}$

The equilibrium conditions given for these reactions are $1300\text{ K and 1 bar}$ .

Molar ratio of steam and methane fed to the reactor is $2$ .

From part (c), the value of $K_1\text{ and }{K}_2$ calculated for both the reactions are

  $\begin{array}{c}{K}_1=13891.8\\ K_2=0.5802\end{array}$

Moles of ${\text{H}}_2\text{O}$ fed is $2\text{ mol}$ and moles of ${\text{CH}}_4$ fed is $1\text{ mol}$ . Practically reaction ( i ) is considered to go to completion as the value of $K_1$ is very large. So, after stoichiometrically reacting with ${\text{CH}}_4$ in reaction ( i ), $1{\text{ mol of H}}_2\text{O}$ is left to react by reaction ( ii ). Also, after the completion of reaction ( i ), $1{\text{ mol CO and 3 mol H}}_2$ are present in the reacting mixture.

For reaction ( ii ), initially $n_{0\left(2\right)}=5\text{ mol}$ is present.

Let the extent of the reaction be $\xi$ .

  $\begin{array}{l}{\text{H}}_2\text{O}\left(g\right)+\text{CO}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{CO}}_2\left(g\right)\\ 1\text{1}\text{ 3 0}\\ 1-\xi \text{ 1}-\xi 3+\xi \xi \end{array}$

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 is

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

Using equation (1), write the expressions for final moles of all the components present in the products as gases in reaction ( ii )

  $\begin{array}{c}{n}_{{\text{H}}_2\text{O}}=n_{0\left({\text{H}}_2\text{O}\right)}+v_{{\text{H}}_2\text{O}}\xi =1-\xi \\ n_{\text{CO}}=n_{0\left(\text{CO}\right)}+v_{\text{CO}}\xi =1-\xi \\ n_{{\text{H}}_2}=n_{0\left({\text{H}}_2\right)}+v_{{\text{H}}_2}\xi =3+\xi \\ n_{{\text{CO}}_2}=n_{0\left({\text{CO}}_2\right)}+v_{{\text{CO}}_2}\xi =\xi \end{array}$

Total moles of the products will be

  $\begin{array}{c}n=n_{{\text{H}}_2\text{O}}+n_{\text{CO}}+n_{{\text{H}}_2}+n_{{\text{CO}}_2}\\ =1-\xi +1-\xi +3+\xi +\xi \\ =5\end{array}$

Using equation (2) to write the mole fraction of all the species as

  $\begin{array}{c}{y}_{{\text{H}}_2\text{O}}=\frac{1-\xi }{5}\\ y_{\text{CO}}=\frac{1-\xi }{5}\\ y_{{\text{H}}_2}=\frac{3+\xi }{5}\\ y_{{\text{CO}}_2}=\frac{\xi }{5}\end{array}$

Now, use equation (3) and the calculated value of the equilibrium constant and calculate $\xi$ . The standard pressure is taken as $1\text{ bar}$ .

  $\begin{array}{c}K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K_2={\left(y_{{\text{H}}_2}\right)}^1{\left(y_{{\text{CO}}_2}\right)}^1{\left(y_{{\text{H}}_2\text{O}}\right)}^{-1}{\left(y_{\text{CO}}\right)}^{-1}{\left(\frac{1}{1}\right)}^{-\left(0\right)}\\ 0.5802=\frac{\left(\frac{3+\xi }{5}\right)\left(\frac{\xi }{5}\right)}{\left(\frac{1-\xi }{5}\right)\left(\frac{1-\xi }{5}\right)}\\ \xi =0.1375\end{array}$

Now, the ratio of ${\text{H}}_2\text{ to CO}$ is calculated as

  $\begin{array}{c}\frac{n_{{\text{H}}_2}}{n_{\text{CO}}}=\frac{3+\xi }{1-\xi }\\ =\frac{3+0.1375}{1-0.1375}\\ =3.638\end{array}$

(e)

Interpretation Introduction

Interpretation:

The way to alter the feed composition to yield lower ${\text{H}}_2\text{ to CO}$ ratio as calculated in part (c) needs 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 14.34P

Addition of ${\text{CO}}_2$ in the feed mixture will lead to lower in the ${\text{H}}_2\text{ to CO}$ ratio in the synthesis gas.

Explanation of Solution

Given information:

The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )

  $\begin{array}{c}{\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}i\text{)}\\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}ii\text{)}\end{array}$

The equilibrium conditions given for these reactions are $1300\text{ K and 1 bar}$ .

By adding ${\text{CO}}_2$ in the feed mixture along with ${\text{CH}}_4{\text{ and H}}_{\text{2}}\text{O}$ leads to the shift in the reaction ( ii ) towards left according to the Le’ Chatelier’s principle, as equilibrium is disturbed by this addition of ${\text{CO}}_2$ . To undo this change, the reaction ( ii ) shifts to its left and produces more of $\text{CO}$ while doing so. Therefore, ${\text{H}}_2$ gets consumed due to its reaction with ${\text{CO}}_2$ .

This leads to the increase in the amount of $\text{CO}$ and decrease in the amount of ${\text{H}}_2$ . Therefore, the ratio of ${\text{H}}_2\text{ to CO}$ will be lower than the value obtained in part (c).

(f)

Interpretation Introduction

Interpretation:

Whether there is any danger in the formation of solid carbon by the given side reaction at equilibrium conditions or not needs to be determined for conditions given in part (c) and part (d). Also, a way to alter the feed conditions to prevent this deposition needs to be explained.

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, $n_i$ 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 14.34P

In part (c), there is danger of formation of solid carbon by the given side reaction at equilibrium conditions.

Addition of ${\text{CO}}_2$ in the feed mixture will eliminate this danger.

In part (d), there is no danger of formation of solid carbon by the given side reaction at equilibrium conditions.

Explanation of Solution

Given information:

The reaction for the production of “synthesis gas” by catalytic re-forming of methane with steam is given as ( i ) and the side reaction accompanying this is given as reaction ( ii )

  $\begin{array}{c}{\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}i\text{)}\\ \text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)\mathrm{......}\text{ (}ii\text{)}\end{array}$

The equilibrium conditions given for these reactions are $1300\text{ K and 1 bar}$ .

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 in carbon formation reaction 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 $1300\text{ K}$ as

  $\tau =\frac{T}{T_0}=\frac{1300\text{ K}}{298.15\text{ K}}=4.360$

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

  $=\left(\begin{array}{c}\left(0.476\right)\left(298.15\right)\left(4.360-1\right)+\frac{\left(0.702\times {10}^{-3}\right)}{2}{\left(298.15\right)}^2\left({\left(4.360\right)}^2-1\right)\\ +\frac{\left(0\right)}{3}{\left(298.15\right)}^3\left({\left(4.360\right)}^3-1\right)+\frac{\left(-1.962\times {10}^5\right)}{298.15}\left(\frac{4.360-1}{4.360}\right)\end{array}\right)$

  $=531.65$

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

  $=\left(0.476\right)\ln \left(4.360\right)+\left(4.360-1\right)\left[\left(0.702\times {10}^{-3}\right)\left(298.15\right)+\left(\frac{4.360+1}{2}\right)\left(\begin{array}{l}\left(0\right){\left(298.15\right)}^2\\ +\frac{-1.962\times {10}^5}{{\left(4.360\right)}^2{\left(298.15\right)}^2}\end{array}\right)\right]$

  $=0.359$

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

  $\begin{array}{c}\Delta G^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ =\left(\left(-172459\right)-\left(4.360\right)\left(-172459-\left(-120021\right)\right)+\left(8.314\right)\left(531.65\right)-\left(8.314\right)\left(800\right)\left(0.359\right)\right)\\ =5.82\times {10}^4\text{ J/mol}\end{array}$

Use this calculated value of $\Delta G^0$ to calculate the equilibrium constant of the reaction at $1300\text{ K}$ using equation (4) as

  $\begin{array}{c}\ln K_{\text{eq}}=-\frac{\Delta G^0}{RT}\\ K_{\text{eq}}=\exp \left(-\frac{\left(5.82\times {10}^4\right)}{\left(8.314\right)\left(1300\right)}\right)\\ K_{\text{eq}}=4.586\times {10}^{-3}\end{array}$

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 reduce 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 is:

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

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:

  $\begin{array}{c}{n}_{\text{CO}}=1\\ n_{{\text{CO}}_2}=0\end{array}$

Calculate the ratio of the actual constant as:

  $\begin{array}{c}{K}_{\text{actual}}=\frac{n_{{\text{CO}}_2}}{{\left(n_{\text{CO}}\right)}^2}\\ =\frac{0}{{\left(1\right)}^2}\\ =0\end{array}$

As the actual value of $K$ is lower than the equilibrium value, there will be the danger of carbon deposition.

Addition of ${\text{CO}}_2$ in the feed mixture will eliminate this danger by increasing the value of actual $K$ .

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

  $\begin{array}{c}{n}_{\text{CO}}=1-\xi =1-0.1375=0.8625\\ n_{{\text{CO}}_2}=\xi =0.1375\end{array}$

Calculate the ratio of the actual constant as:

  $\begin{array}{c}{K}_{\text{actual}}=\frac{n_{{\text{CO}}_2}}{{\left(n_{\text{CO}}\right)}^2}\\ =\frac{0.8625}{{\left(0.1375\right)}^2}\\ =45.62\end{array}$

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.