Chapter 14, Problem 14.16P

(a)

Interpretation Introduction

Interpretation:

The fractional conversion of propane for the given reaction is to be calculated at $625\text{ K}$ .

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 14.16P

The fractional conversion of propane for the given reaction at $625\text{ K}$ is $0.777$ .

Explanation of Solution

Given information:

The cracking reaction of propane is:

  ${\text{C}}_3{\text{H}}_8\left(g\right)\to {\text{C}}_2{\text{H}}_4\left(g\right)+{\text{CH}}_4\left(g\right)$

The equilibrium pressure given for this reaction is $\text{1 bar}$ .

The equilibrium conversion at $300\text{ K}$ is negligible but at temperatures above $500\text{ K}$, it becomes appreciable.

The equilibrium temperature for this reaction is taken as $625\text{ K}$ .

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

  $\begin{array}{l}\Delta G_{298}^0=42290\text{J/mol}\\ \Delta H_{298}^0=82670\text{ J/mol}\end{array}$

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

Substance	$A$	$B$	$C$	$D$
${\text{C}}_3{\text{H}}_8$	$1.213$	$28.785\times {10}^{-3}$	$-8.824\times {10}^{-6}$	$\_\_$
${\text{C}}_2{\text{H}}_4$	$1.424$	$14.394\times {10}^{-3}$	$-4.392\times {10}^{-6}$	$\_\_$
${\text{CH}}_4$	$1.702$	$9.081\times {10}^{-3}$	$-2.164\times {10}^{-6}$	$\_\_$

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

  $\begin{array}{c}\Delta A=A_{{\text{C}}_2{\text{H}}_4}+A_{{\text{CH}}_4}-A_{{\text{C}}_3{\text{H}}_8}\\ =1.424+1.702-1.213\\ =1.913\\ \Delta B=B_{{\text{C}}_2{\text{H}}_4}+B_{{\text{CH}}_4}-B_{{\text{C}}_3{\text{H}}_8}\\ =\left(14.394+9.081-28.785\right)\times {10}^{-3}\\ =-5.31\times {10}^{-3}\\ \Delta C=C_{{\text{C}}_2{\text{H}}_4}+C_{{\text{CH}}_4}-C_{{\text{C}}_3{\text{H}}_8}\\ =\left(-4.392+\left(-2.164\right)-\left(-8.824\right)\right)\times {10}^{-6}\\ =2.268\times {10}^{-6}\\ \Delta D=D_{{\text{C}}_2{\text{H}}_4}+D_{{\text{CH}}_4}-D_{{\text{C}}_3{\text{H}}_8}\\ =0+0-0\\ =0\end{array}$

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 $625\text{ K}$ as:

  $\tau =\frac{T}{T_0}=\frac{625\text{ K}}{298.15\text{ K}}=2.096$

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

  $=\left(\begin{array}{c}\left(1.913\right)\left(298.15\right)\left(2.096-1\right)+\frac{\left(-5.31\times {10}^{-3}\right)}{2}{\left(298.15\right)}^2\left({\left(2.096\right)}^2-1\right)\\ +\frac{\left(2.268\times {10}^{-6}\right)}{3}{\left(298.15\right)}^3\left({\left(2.096\right)}^3-1\right)+\frac{\left(0\right)}{298.15}\left(\frac{2.096-1}{2.096}\right)\end{array}\right)$

  $=-11.259$

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

  $=\left(1.913\right)\ln \left(2.096\right)+\left(2.096-1\right)\left[\left(-5.31\times {10}^{-3}\right)\left(298.15\right)+\left(\frac{2.096+1}{2}\right)\left(\begin{array}{l}\left(2.268\times {10}^{-6}\right){\left(298.15\right)}^2\\ +\frac{0}{{\left(2.096\right)}^2{\left(298.15\right)}^2}\end{array}\right)\right]$

  $=0.0226$

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

  $\begin{array}{c}\Delta G^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ =\left(82670\right)-\left(2.096\right)\left(82670-42290\right)+\left(8.314\right)\left(-11.259\right)-\left(8.314\right)\left(625\right)\left(0.0226\right)\\ =-2177.5\text{ J/mol}\end{array}$

Use this calculated value of $\Delta G^0$ to calculate the equilibrium constant of the reaction at $625\text{ K}$ using equation (4) as:

  $\begin{array}{c}\ln K=-\frac{\Delta G^0}{RT}\\ K=\exp \left(-\frac{\left(-2177.5\right)}{\left(8.314\right)\left(625\right)}\right)\\ K=1.521\end{array}$

For the given cracking reaction, let the extent of the reaction be $\xi$ .

  $\begin{array}{l}{\text{C}}_3{\text{H}}_8\left(g\right)\to {\text{C}}_2{\text{H}}_4\left(g\right)+{\text{CH}}_4\left(g\right)\\ 1\text{ 0 0}\\ 1-\xi \xi \xi \end{array}$

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

  $\begin{array}{c}{\nu }_{{\text{C}}_3{\text{H}}_8}=-1\\ {\nu }_{{\text{C}}_2{\text{H}}_4}=+1\\ {\nu }_{{\text{CH}}_4}=+1\end{array}$

The overall stoichiometric coefficient for this reaction is,

  $\nu =1+1-\left(1\right)=1$

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

  $\begin{array}{c}{n}_{{\text{C}}_3{\text{H}}_8}=n_{0\left({\text{C}}_3{\text{H}}_8\right)}+v_{{\text{C}}_3{\text{H}}_8}\xi =1-\xi \\ n_{{\text{C}}_2{\text{H}}_4}=n_{0\left({\text{C}}_2{\text{H}}_4\right)}+v_{{\text{C}}_2{\text{H}}_4}\xi =\xi \\ n_{{\text{CH}}_4}=n_{0\left({\text{CH}}_4\right)}+v_{{\text{CH}}_4}\xi =\xi \end{array}$

Total moles of the products will be:

  $\begin{array}{c}n=n_{{\text{C}}_3{\text{H}}_8}+n_{{\text{C}}_2{\text{H}}_4}+n_{{\text{CH}}_4}\\ =1-\xi +\xi +\xi \\ =1+\xi \end{array}$

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

  $\begin{array}{c}{y}_{{\text{C}}_3{\text{H}}_8}=\frac{1-\xi }{1+\xi }\\ y_{{\text{C}}_2{\text{H}}_4}=\frac{\xi }{1+\xi }\\ y_{{\text{CH}}_4}=\frac{\xi }{1+\xi }\end{array}$

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

  $\begin{array}{c}K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ 1.521={\left(y_{{\text{C}}_2{\text{H}}_4}\right)}^1{\left(y_{{\text{CH}}_4}\right)}^1{\left(y_{{\text{C}}_3{\text{H}}_8}\right)}^{-1}{\left(\frac{1}{1}\right)}^{-\left(1\right)}\\ 1.521=\frac{\left(\frac{\xi }{1+\xi }\right)\left(\frac{\xi }{1+\xi }\right)}{\left(\frac{1-\xi }{1+\xi }\right)}\\ \xi =0.777\end{array}$

Therefore, the fractional conversion of propane is $0.777$ .

(b)

Interpretation Introduction

Interpretation:

The temperature for the given fractional conversion of propane at $1\text{ bar}$ 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 14.16P

The temperature for $85\%$ fractional conversion of propane at $1\text{ bar}$ is $647\text{ K}$ .

Explanation of Solution

Given information:

The cracking reaction of propane is:

  ${\text{C}}_3{\text{H}}_8\left(g\right)\to {\text{C}}_2{\text{H}}_4\left(g\right)+{\text{CH}}_4\left(g\right)$

The equilibrium pressure given for this reaction is $\text{1 bar}$ .

The equilibrium conversion at $300\text{ K}$ is negligible but at temperatures above $500\text{ K}$, it becomes appreciable.

The fractional conversion at equilibrium is $85\%$ .

The initial and final pressure of the system is kept at $1\text{ bar}$, so, $P=P^0=1\text{ bar}$ .

Fractional conversion of propane is given as $85\%$, so, $\xi =0.85$ for the given reaction.

The final mole fraction of all the species will be:

  $\begin{array}{c}{y}_{{\text{C}}_3{\text{H}}_8}=\frac{1-\xi }{1+\xi }=\frac{1-0.85}{1+0.85}=0.0811\\ y_{{\text{C}}_2{\text{H}}_4}=\frac{\xi }{1+\xi }=\frac{0.85}{1+0.85}=0.4595\\ y_{{\text{CH}}_4}=\frac{\xi }{1+\xi }=\frac{0.85}{1+0.85}=0.4595\end{array}$

Now, use equation (3) to calculate the equilibrium constant of this reaction as:

  $\begin{array}{c}K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K={\left(y_{{\text{C}}_2{\text{H}}_4}\right)}^1{\left(y_{{\text{CH}}_4}\right)}^1{\left(y_{{\text{C}}_3{\text{H}}_8}\right)}^{-1}{\left(\frac{1}{1}\right)}^{-\left(1\right)}\\ K=\frac{\left(0.4595\right)\left(0.4595\right)}{\left(0.0811\right)}\\ K=2.604\end{array}$

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

  $\begin{array}{l}\Delta G_{298}^0=42290\text{J/mol}\\ \Delta H_{298}^0=82670\text{ J/mol}\end{array}$

From part (a), the calculated values of $\Delta A$, $\Delta B$, $\Delta C$, and $\Delta D$ are:

  $\begin{array}{c}\Delta A=1.913\\ \Delta B=-5.31\times {10}^{-3}\\ \Delta C=2.268\times {10}^{-6}\\ \Delta D=0\end{array}$

Make an initial guess for the equilibrium temperature at $85\%$ conversion. Basing the guess on t he given information and the result of part (s), let the equilibrium temperature be $647\text{K}$ .

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 this equilibrium temperature of $647\text{ K}$ as:

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

  $\begin{array}{c}\Delta G^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ =\left(82670\right)-\left(2.170\right)\left(82670-42290\right)+\left(8.314\right)\left(-23.318\right)-\left(8.314\right)\left(625\right)\left(0.0036\right)\\ =-5167.2\text{ J/mol}\end{array}$

Use this calculated value of $\Delta G^0$ to calculate the equilibrium constant of the reaction at $647\text{ K}$ using equation (4) as:

  $\begin{array}{c}\ln K=-\frac{\Delta G^0}{RT}\\ K=\exp \left(-\frac{-5167.2}{\left(8.314\right)\left(647\right)}\right)\\ K=2.613\end{array}$

Since, the calculated value of $K$ and the value calculated using the guessed value of $T$ are almost same, the temperature initially guessed is the equilibrium temperature for $85\%$ conversion of propane.