Chapter 14, Problem 14.21P

(a)

Interpretation Introduction

Interpretation:

The equilibrium mole fraction of methanol at $300\text{ K and 1 bar}$ is to be determined for the given methanol synthesis reaction.

Concept introduction:

Mole fraction formethanol synthesis reaction is calculated with the help of the extent of reaction which is defined as a measure of the extent to which the reaction proceeds.

  $\begin{array}{l}{\text{CO}}_2\left(g\right)+2{\text{H}}_2\left(g\right)\to {\text{CH}}_3\text{OH}\left(g\right)\\ 120\\ 1-\xi \text{ 2}-2\xi \xi \end{array}$

Where, $\xi$ is the extent of the reaction.

The overall stoichiometric coefficient for this reaction is,

  $\nu =1-\left(1+2\right)=-2$   (1)

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

  $\begin{array}{c}{\nu }_1=-1\\ {\nu }_2=-2\\ {\nu }_3=+1\end{array}$   (2)

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.

Gibbs 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 14.21P

The equilibrium mole fraction of methanol at $300\text{ K and 1 bar}$ is $0.9294$ .

Explanation of Solution

From Table C.4 the standard heat of reaction and Gibbs free energy for methanol formation is:

  $\begin{array}{l}\Delta G_{298}^0=-24791\text{J/mol}\\ \Delta H_{298}^0=-90135\text{ J/mol}\end{array}$

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

Substance	$A$	$B$	$C$	$D$
${\text{CH}}_3\text{OH}$	$2.211$	$12.216\times {10}^{-3}$	$-3.450\times {10}^{-6}$	$\_\_$
$\text{CO}$	$3.376$	$0.557\times {10}^{-3}$	$\_\_$	$-0.031\times {10}^5$
${\text{H}}_2$	$3.249$	$0.422\times {10}^{-3}$	$\_\_$	$0.083\times {10}^5$

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

  $\begin{array}{c}\Delta A=A_{{\text{CH}}_3\text{OH}}-\left(1\times A_{\text{CO}}\right)-\left(2\times A_{{\text{H}}_{\text{2}}}\right)\\ =2.211-\left(1\times 3.376\right)-\left(2\times 3.249\right)\\ =-7.663\\ \Delta B=B_{{\text{CH}}_3\text{OH}}-\left(1\times B_{\text{CO}}\right)-\left(2\times B_{{\text{H}}_{\text{2}}}\right)\\ =\left(12.216\times {10}^{-3}\right)-\left(1\times 0.557\times {10}^{-3}\right)-\left(2\times 0.422\times {10}^{-3}\right)\\ =10.805\times {10}^{-3}\\ \Delta C=C_{{\text{CH}}_3\text{OH}}-\left(1\times C_{\text{CO}}\right)-\left(2\times C_{{\text{H}}_{\text{2}}}\right)\\ =\left(-3.450\times {10}^{-6}\right)-\left(1\times 0\right)-\left(2\times 0\right)\\ =-3.450\times {10}^{-6}\\ \Delta D=D_{{\text{CH}}_3\text{OH}}-\left(1\times D_{\text{CO}}\right)-\left(2\times D_{{\text{H}}_{\text{2}}}\right)\\ =\left(0\right)-\left(1\times \left(-0.031\times {10}^5\right)\right)-\left(2\times \left(0.083\times {10}^5\right)\right)\\ =-1.35\times {10}^4\end{array}$

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

$\tau =\frac{T}{T_0}=\frac{300\text{ K}}{298\text{ K}}=1.0067$

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

$=\left(\begin{array}{c}\left(-7.663\right)\left(298\right)\left(1.0067-1\right)+\frac{\left(10.815\times {10}^{-3}\right)}{2}{\left(298\right)}^2\left({\left(1.0067\right)}^2-1\right)\\ +\frac{\left(-3.450\times {10}^{-3}\right)}{3}{\left(298\right)}^3\left({\left(1.0067\right)}^3-1\right)+\frac{\left(-1.35\times {10}^4\right)}{298}\left(\frac{1.0067-1}{1.0067}\right)\end{array}\right)$

$=-9.76$

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

$=\left(-7.663\right)\ln \left(1.0067\right)+\left(1.0067-1\right)\left[\left(10.815\times {10}^{-3}\right)\left(298\right)+\left(\frac{1.0067+1}{2}\right)\left(\begin{array}{l}\left(-3.450\times {10}^{-6}\right){\left(298\right)}^2\\ +\frac{-1.35\times {10}^4}{{\left(1.0067\right)}^2{\left(298\right)}^2}\end{array}\right)\right]$

$=-0.0326$

Now, use equation (5) to calculate the value of $\Delta G^0$ at $300\text{ K}$ as:

  $\begin{array}{c}\Delta G^0=\Delta H_{298}^0-\frac{T}{T_{298}}\left(\Delta H_{298}^0-\Delta G_{298}^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ =\left(\begin{array}{l}\left(-90135\right)-\left(\frac{300}{298}\right)\left(-90135-\left(-24791\right)\right)\\ +\left(8.314\right)\left(-9.76\right)-\left(8.314\right)\left(300\right)\left(-0.0326\right)\end{array}\right)\\ =-2.44\times {10}^4\text{ J/mol}\end{array}$

The equilibrium constant for the methanol synthesis reaction at 300 K can be calculated as given below:

  $\begin{array}{c}\ln K=-\frac{\Delta G^0}{RT}\\ \ln K=-\frac{\left(-2.44\times {10}^4\right)}{8.314\times 300}\\ K=17724.4\end{array}$

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

The mole fraction of all the species in terms of extent of reaction are:

  $\begin{array}{c}{y}_{{\text{CH}}_3\text{OH}}=\frac{\xi }{3-2\xi }\\ y_{\text{CO}}=\frac{1-\xi }{3-2\xi }\\ y_{{\text{H}}_2}=\frac{2-2\xi }{3-2\xi }\end{array}$

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

  $\begin{array}{c}K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ 17724.4={\left(\frac{\xi }{3-2\xi }\right)}^1{\left(\frac{1-\xi }{3-2\xi }\right)}^{-1}{\left(\frac{2-2\xi }{3-2\xi }\right)}^{-2}{\left(\frac{1}{1}\right)}^{-\left(-2\right)}\\ 17724.4=\frac{\left(\frac{\xi }{3-2\xi }\right)}{\left(\frac{1-\xi }{3-2\xi }\right){\left(\frac{2-2\xi }{3-2\xi }\right)}^2}\\ \xi =0.9753\end{array}$

Use this extent of reaction to calculate the mole fraction of methanol as:

  $\begin{array}{c}{y}_{{\text{CH}}_3\text{OH}}=\frac{\xi }{3-2\xi }\\ =\frac{0.9753}{3-2\left(0.9753\right)}\\ =0.9294\end{array}$

Therefore, the equilibrium mole fraction of methanol at $300\text{ K and 1 bar}$ is $0.9294$ .

(b)

Interpretation Introduction

Interpretation:

The temperature for the given equilibrium mole fraction of methanol at $1\text{ bar}$ is to be determined.

Concept introduction:

Mole fraction for methanol synthesis reaction is calculated with the help of the extent of reaction which is defined as a measure of the extent to which the reaction proceeds.

  $\begin{array}{l}{\text{CO}}_2\left(g\right)+2{\text{H}}_2\left(g\right)\to {\text{CH}}_3\text{OH}\left(g\right)\\ 120\\ 1-\xi \text{ 2}-2\xi \xi \end{array}$

Where, $\xi$ is the extent of the reaction.

The overall stoichiometric coefficient for this reaction is,

  $\nu =1-\left(1+2\right)=-2$   (1)

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

  $\begin{array}{c}{\nu }_1=-1\\ {\nu }_2=-2\\ {\nu }_3=+1\end{array}$   (2)

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.

Gibbs 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 14.21P

The equilibrium mole fraction of methanol at $1\text{ bar}$ is $0.5$ at temperature, $T=372.05\text{ K}$ .

Explanation of Solution

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

The mole fraction of all the species in terms of extent of reaction $\left(\xi \right)$ are:

  $\begin{array}{c}{y}_{{\text{CH}}_3\text{OH}}=\frac{\xi }{3-2\xi }\\ 0.5=\frac{\xi }{3-2\xi }\\ \xi =0.75\\ y_{\text{CO}}=\frac{1-\xi }{3-2\xi }=\frac{1-0.75}{3-2\left(0.75\right)}=0.167\\ y_{{\text{H}}_2}=\frac{2-2\xi }{3-2\xi }=\frac{2-2\left(0.75\right)}{3-2\left(0.75\right)}=0.333\end{array}$

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

  $\begin{array}{c}K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K={\left(0.5\right)}^1{\left(0.167\right)}^{-1}{\left(0.333\right)}^{-2}{\left(\frac{1}{1}\right)}^{-\left(-2\right)}\\ K=27\end{array}$

Use this equilibrium constant to calculate $\Delta G^0$ of the reaction at $300\text{ K}$ using equation (4) as:

  $\begin{array}{c}\ln K=-\frac{\Delta G^0}{RT}\\ \ln \left(27\right)=-\frac{\Delta G^0}{\left(8.314\right)\left(300\right)}\\ \Delta G^0=-8220.48\text{ J/mol}\end{array}$

From Table C.4 the standard heat of reaction and Gibbs free energy for methanol formation is:

  $\begin{array}{l}\Delta G_{298}^0=-24791\text{J/mol}\\ \Delta H_{298}^0=-90135\text{ J/mol}\end{array}$

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

  $\begin{array}{c}\Delta A=-7.663\\ \Delta B=10.805\times {10}^{-3}\\ \Delta C=-3.450\times {10}^{-6}\\ \Delta D=-1.35\times {10}^4\end{array}$

Now, use equations set (6) in equation (5) along with the value of $\Delta G^0$ and calculate the temperature as:

  $\begin{array}{c}\Delta G^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ \Delta G^0=\left(\begin{array}{l}\left(-90135\right)-\tau \left(-90135-\left(-24791\right)\right)+\left(8.314\right)\left(\begin{array}{l}\left(-7.663\right)\left(298\right)\left(\tau -1\right)+\frac{\left(10.805\times {10}^{-3}\right)}{2}{\left(298\right)}^2\left({\tau }^2-1\right)\\ +\frac{\left(-3.450\times {10}^{-6}\right)}{3}{\left(298\right)}^3\left({\tau }^3-1\right)+\frac{\left(-1.35\times {10}^4\right)}{298}\left(\frac{\tau -1}{\tau }\right)\end{array}\right)\\ -\left(8.314\right)T\left(\left(-7.663\right)\ln \tau +\left(\tau -1\right)\left[\left(10.805\times {10}^{-3}\right)\left(298\right)+\left(\frac{\tau +1}{2}\right)\left(\begin{array}{l}\left(-3.450\times {10}^{-6}\right){\left(298\right)}^2\\ +\frac{\left(-1.35\times {10}^4\right)}{{\tau }^2{\left(298\right)}^2}\end{array}\right)\right]\right)\end{array}\right)\end{array}$

Now, by trying different values of $T$ in the above equation in the right hand side, fit a value which gives the left hand side value of $\Delta G^0=-8220.48\text{ J/mol}$ . At $T=372.05\text{ K}$, this condition is fulfilled as shown below:

  $\begin{array}{c}\Delta G^0=\left(\begin{array}{l}\left(-90135\right)-\left(1.2485\right)\left(-90135-\left(-24791\right)\right)+\left(8.314\right)\left(\left(\begin{array}{l}\left(-7.663\right)\left(298\right)\left(1.2485-1\right)+\frac{\left(10.805\times {10}^{-3}\right)}{2}{\left(298\right)}^2\left({\left(1.2485\right)}^2-1\right)+\\ \frac{\left(-3.450\times {10}^{-6}\right)}{3}{\left(298\right)}^3\left({\left(1.2485\right)}^3-1\right)+\frac{\left(-1.35\times {10}^4\right)}{298}\left(\frac{1.2485-1}{1.2485}\right)\end{array}\right)\right)\\ -\left(8.314\right)\left(372.05\right)\left(\left(-7.663\right)\ln \left(1.2485\right)+\left(1.2485-1\right)\left[\left(10.805\times {10}^{-3}\right)\left(298\right)+\left(\frac{1.2485+1}{2}\right)\left(\begin{array}{l}\left(-3.450\times {10}^{-6}\right){\left(298\right)}^2\\ +\frac{\left(-1.35\times {10}^4\right)}{{\left(1.2485\right)}^2{\left(298\right)}^2}\end{array}\right)\right]\right)\end{array}\right)\\ =-8221.92\text{ J/mol}\end{array}$

Thus, the temperature at which the equilibrium mole fraction of methanol is $0.5$ at $1\text{ bar}$ is calculated to be $372.05\text{ K}$ .

(c)

Interpretation Introduction

Interpretation:

The temperature for the given equilibrium mole fraction of methanol at $100\text{ bar}$ is to be determined.

Concept introduction:

Mole fraction for methanol synthesis reaction is calculated with the help of the extent of reaction which is defined as a measure of the extent to which the reaction proceeds.

  $\begin{array}{l}{\text{CO}}_2\left(g\right)+2{\text{H}}_2\left(g\right)\to {\text{CH}}_3\text{OH}\left(g\right)\\ 120\\ 1-\xi \text{ 2}-2\xi \xi \end{array}$

Where, $\xi$ is the extent of the reaction.

The overall stoichiometric coefficient for this reaction is,

  $\nu =1-\left(1+2\right)=-2$   (1)

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

  $\begin{array}{c}{\nu }_1=-1\\ {\nu }_2=-2\\ {\nu }_3=+1\end{array}$   (2)

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.

Gibbs 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 14.21P

The equilibrium mole fraction of methanol at $100\text{ bar}$ is $0.5$ at temperature, $T=468.5\text{ K}$ .

Explanation of Solution

The initial and final pressure of the system are taken as:

  $\begin{array}{c}{P}^0=1\text{ bar}\\ P=100\text{ bar}\end{array}$ .

The mole fraction of all the species in terms of extent of reaction $\left(\xi \right)$ are:

  $\begin{array}{c}{y}_{{\text{CH}}_3\text{OH}}=\frac{\xi }{3-2\xi }\\ 0.5=\frac{\xi }{3-2\xi }\\ \xi =0.75\\ y_{\text{CO}}=\frac{1-\xi }{3-2\xi }=\frac{1-0.75}{3-2\left(0.75\right)}=0.167\\ y_{{\text{H}}_2}=\frac{2-2\xi }{3-2\xi }=\frac{2-2\left(0.75\right)}{3-2\left(0.75\right)}=0.333\end{array}$

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

  $\begin{array}{c}K=\prod _i{\left(y_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K={\left(0.5\right)}^1{\left(0.167\right)}^{-1}{\left(0.333\right)}^{-2}{\left(\frac{1}{100}\right)}^{-\left(-2\right)}\\ K=2.7\times {10}^{-3}\end{array}$

Use this equilibrium constant to calculate $\Delta G^0$ of the reaction at $300\text{ K}$ using equation (4) as:

  $\begin{array}{c}\ln K=-\frac{\Delta G^0}{RT}\\ \ln \left(2.7\times {10}^{-3}\right)=-\frac{\Delta G^0}{\left(8.314\right)\left(300\right)}\\ \Delta G^0=-1.48\times {10}^4\text{ J/mol}\end{array}$

From Table C.4 the standard heat of reaction and Gibbs free energy for methanol formation is:

  $\begin{array}{l}\Delta G_{298}^0=-24791\text{J/mol}\\ \Delta H_{298}^0=-90135\text{ J/mol}\end{array}$

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

  $\begin{array}{c}\Delta A=-7.663\\ \Delta B=10.805\times {10}^{-3}\\ \Delta C=-3.450\times {10}^{-6}\\ \Delta D=-1.35\times {10}^4\end{array}$

Now, use equations set (6) in equation (5) along with the value of $\Delta G^0$ and calculate the temperature as:

  $\begin{array}{c}\Delta G^0=\Delta H_0^0-\tau \left(\Delta H_0^0-\Delta G_0^0\right)+R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}dT}-RT{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P^0}{R}\frac{dT}{T}}\\ \Delta G^0=\left(\begin{array}{l}\left(-90135\right)-\tau \left(-90135-\left(-24791\right)\right)+\left(8.314\right)\left(\begin{array}{l}\left(-7.663\right)\left(298\right)\left(\tau -1\right)+\frac{\left(10.805\times {10}^{-3}\right)}{2}{\left(298\right)}^2\left({\tau }^2-1\right)\\ +\frac{\left(-3.450\times {10}^{-6}\right)}{3}{\left(298\right)}^3\left({\tau }^3-1\right)+\frac{\left(-1.35\times {10}^4\right)}{298}\left(\frac{\tau -1}{\tau }\right)\end{array}\right)\\ -\left(8.314\right)T\left(\left(-7.663\right)\ln \tau +\left(\tau -1\right)\left[\left(10.805\times {10}^{-3}\right)\left(298\right)+\left(\frac{\tau +1}{2}\right)\left(\begin{array}{l}\left(-3.450\times {10}^{-6}\right){\left(298\right)}^2\\ +\frac{\left(-1.35\times {10}^4\right)}{{\tau }^2{\left(298\right)}^2}\end{array}\right)\right]\right)\end{array}\right)\end{array}$

Now, by trying different values of $T$ in the above equation in the right hand side, fit a value which gives the left hand side value of $\Delta G^0=-1.48\times {10}^4\text{ J/mol}$ . At $T=468.5\text{ K}$, this condition is fulfilled as shown below:

  $\begin{array}{c}\Delta G^0=\left(\begin{array}{l}\left(-90135\right)-\left(1.572\right)\left(-90135-\left(-24791\right)\right)+\left(8.314\right)\left(\left(\begin{array}{l}\left(-7.663\right)\left(298\right)\left(1.572-1\right)+\frac{\left(10.805\times {10}^{-3}\right)}{2}{\left(298\right)}^2\left({\left(1.572\right)}^2-1\right)+\\ \frac{\left(-3.450\times {10}^{-6}\right)}{3}{\left(298\right)}^3\left({\left(1.572\right)}^3-1\right)+\frac{\left(-1.35\times {10}^4\right)}{298}\left(\frac{1.572-1}{1.572}\right)\end{array}\right)\right)\\ -\left(8.314\right)\left(468.5\right)\left(\begin{array}{l}\left(-7.663\right)\ln \left(1.572\right)+\left(1.572-1\right)\\ \left[\left(10.805\times {10}^{-3}\right)\left(298\right)+\left(\frac{1.572+1}{2}\right)\left(\left(-3.450\times {10}^{-6}\right){\left(298\right)}^2+\frac{\left(-1.35\times {10}^4\right)}{{\left(1.572\right)}^2{\left(298\right)}^2}\right)\right]\end{array}\right)\end{array}\right)\\ =-1.41\times {10}^4\text{ J/mol}\end{array}$

Thus, the temperature at which the equilibrium mole fraction of methanol is $0.5$ at $100\text{ bar}$ is calculated to be $468.5\text{ K}$ .

(d)

Interpretation Introduction

Interpretation:

The temperature for the given equilibrium mole fraction of methanol at $100\text{ bar}$ for an ideal solution of gases is to be determined.

Concept introduction:

Mole fraction for methanol synthesis reaction is calculated with the help of the extent of reaction which is defined as a measure of the extent to which the reaction proceeds.

  $\begin{array}{l}{\text{CO}}_2\left(g\right)+2{\text{H}}_2\left(g\right)\to {\text{CH}}_3\text{OH}\left(g\right)\\ 120\\ 1-\xi \text{ 2}-2\xi \xi \end{array}$

Where, $\xi$ is the extent of the reaction.

The overall stoichiometric coefficient for this reaction is,

  $\nu =1-\left(1+2\right)=-2$   (1)

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

  $\begin{array}{c}{\nu }_1=-1\\ {\nu }_2=-2\\ {\nu }_3=+1\end{array}$   (2)

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

Equilibrium constant of this reaction for an ideal solution of gases from equation 14.28 can be written as:

  $K=\prod _i{\left(y_i{\varphi }_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }$   (7)

Where, $y_i$ is the mole fraction of component $i$, ${\varphi }_i$ is the fugacity coefficient for the component $i$, $P$ is the pressure of the reaction, and $P^0$ is the standard pressure.

The formula to calculate ${\varphi }_i$ is:

  ${\varphi }_i=\exp \left[\frac{P_r}{T_r}\left(B^0+\omega B^1\right)\right]$   (8)

Here, $P_r\text{ and }{T}_r$ are the reduced pressure and temperature of the system.The $\omega$ is the acentric factor associated with the substance. The formula to calculate $B^0\text{ and }{B}^1$ is:

  $\begin{array}{c}{B}^0=0.083-\frac{0.422}{T_r^{1.6}}\\ B^1=0.139-\frac{0.172}{T_r^{4.2}}\end{array}$   (9)

Answer to Problem 14.21P

The equilibrium mole fraction of methanol at $100\text{ bar}$ for an ideal solution of gases is $0.5$ at temperature, $T=480\text{ K}$ .

Explanation of Solution

The initial and final pressure of the system are taken as:

  $\begin{array}{c}{P}^0=1\text{ bar}\\ P=100\text{ bar}\end{array}$ .

The mole fraction of all the species in terms of extent of reaction $\left(\xi \right)$ are:

  $\begin{array}{c}{y}_{{\text{CH}}_3\text{OH}}=\frac{\xi }{3-2\xi }\\ 0.5=\frac{\xi }{3-2\xi }\\ \xi =0.75\\ y_{\text{CO}}=\frac{1-\xi }{3-2\xi }=\frac{1-0.75}{3-2\left(0.75\right)}=0.167\\ y_{{\text{H}}_2}=\frac{2-2\xi }{3-2\xi }=\frac{2-2\left(0.75\right)}{3-2\left(0.75\right)}=0.333\end{array}$

From Table-B.1 of appendix B, the critical properties and acentric factor of methanol is:

Component	Pc (bar)	Tc (K)	$\omega$
${\text{CH}}_3\text{OH}$	$80.97$	$512.6$	$0.564$
$\text{CO}$	$34.99$	$132.9$	$0.048$
${\text{H}}_2$	$13.13$	$33.19$	$-0.216$

Let the guessed value of the temperature be $480\text{ K}$ . Calculate the reduced temperature and pressure as:

  $\begin{array}{c}{T}_{r\left({\text{CH}}_3\text{OH}\right)}=\frac{T_{{\text{CH}}_3\text{OH}}}{T_{c\left({\text{CH}}_3\text{OH}\right)}}=\frac{480}{512.6}=0.936\\ T_{r\left(\text{CO}\right)}=\frac{T_{\text{CO}}}{T_{c\left(\text{CO}\right)}}=\frac{480}{132.9}=3.612\\ T_{r\left({\text{H}}_2\right)}=\frac{T_{{\text{H}}_2}}{T_{c\left({\text{H}}_2\right)}}=\frac{480}{33.19}=14.462\\ P_{r\left({\text{CH}}_3\text{OH}\right)}=\frac{P_{{\text{CH}}_3\text{OH}}}{P_{c\left({\text{CH}}_3\text{OH}\right)}}=\frac{100}{80.97}=1.235\\ P_{r\left(\text{CO}\right)}=\frac{P_{\text{CO}}}{P_{c\left(\text{CO}\right)}}=\frac{100}{34.99}=2.858\\ P_{r\left({\text{H}}_2\right)}=\frac{P_{{\text{H}}_2}}{P_{c\left({\text{H}}_2\right)}}=\frac{100}{13.13}=7.616\end{array}$

Use equations set (9) to calculate the values of $B^0\text{ and }{B}^1$ for all the components as:

  $\begin{array}{c}{B}_{\left({\text{CH}}_3\text{OH}\right)}^0=0.083-\frac{0.422}{T_{r\left({\text{CH}}_3\text{OH}\right)}^{1.6}}=0.083-\frac{0.422}{{\left(0.936\right)}^{1.6}}=-0.386\\ B_{\left(\text{CO}\right)}^0=0.083-\frac{0.422}{T_{r\left(\text{CO}\right)}^{1.6}}=0.083-\frac{0.422}{{\left(3.612\right)}^{1.6}}=0.0289\\ B_{\left({\text{H}}_2\right)}^0=0.083-\frac{0.422}{T_{r\left({\text{H}}_2\right)}^{1.6}}=0.083-\frac{0.422}{{\left(14.462\right)}^{1.6}}=0.0771\\ B_{\left({\text{CH}}_3\text{OH}\right)}^1=0.139-\frac{0.172}{T_{r\left({\text{CH}}_3\text{OH}\right)}^{4.2}}=0.139-\frac{0.172}{{\left(0.936\right)}^{4.2}}=-0.0881\\ B_{\left(\text{CO}\right)}^1=0.139-\frac{0.172}{T_{r\left(\text{CO}\right)}^{4.2}}=0.139-\frac{0.172}{{\left(3.612\right)}^{4.2}}=0.1382\\ B_{\left({\text{H}}_2\right)}^1=0.139-\frac{0.172}{T_{r\left({\text{H}}_2\right)}^{4.2}}=0.139-\frac{0.172}{{\left(14.462\right)}^{4.2}}=0.139\end{array}$

Use equation (8) to calculate the value of ${\varphi }_i$ as:

  $\begin{array}{c}{\varphi }_{\left({\text{CH}}_3\text{OH}\right)}=\exp \left[\frac{P_{r\left({\text{CH}}_3\text{OH}\right)}}{T_{r\left({\text{CH}}_3\text{OH}\right)}}\left(B_{\left({\text{CH}}_3\text{OH}\right)}^0+\omega B_{\left({\text{CH}}_3\text{OH}\right)}^1\right)\right]=\exp \left[\frac{1.235}{0.936}\left(-0.386+\left(0.564\right)\left(-0.0881\right)\right)\right]=0.563\\ {\varphi }_{\left(\text{CO}\right)}=\exp \left[\frac{P_{r\left(\text{CO}\right)}}{T_{r\left(\text{CO}\right)}}\left(B_{\left(\text{CO}\right)}^0+\omega B_{\left(\text{CO}\right)}^1\right)\right]=\exp \left[\frac{2.858}{3.612}\left(0.0289+\left(0.048\right)\left(0.1382\right)\right)\right]=1.029\\ {\varphi }_{\left({\text{H}}_2\right)}=\exp \left[\frac{P_{r\left({\text{H}}_2\right)}}{T_{r\left({\text{H}}_2\right)}}\left(B_{\left({\text{H}}_2\right)}^0+\omega B_{\left({\text{H}}_2\right)}^1\right)\right]=\exp \left[\frac{7.616}{14.462}\left(0.0771+\left(-0.216\right)\left(0.139\right)\right)\right]=1.025\end{array}$

Now, use equation (7) to calculate the equilibrium constant of this reaction as:

  $\begin{array}{c}K=\prod _i{\left(y_i{\varphi }_i\right)}^{{\nu }_i}{\left(\frac{P^0}{P}\right)}^{-\nu }\\ K={\left(0.5\times 0.563\right)}^1{\left(0.167\times 1.029\right)}^{-1}{\left(0.333\times 1.025\right)}^{-2}{\left(\frac{1}{100}\right)}^{-\left(-2\right)}\\ K=1.41\times {10}^{-3}\end{array}$

Use this equilibrium constant to calculate $\Delta G^0$ of the reaction at $300\text{ K}$ using equation (4) as:

  $\begin{array}{c}\ln K=-\frac{\Delta G^0}{RT}\\ \ln \left(1.41\times {10}^{-3}\right)=-\frac{\Delta G^0}{\left(8.314\right)\left(300\right)}\\ \Delta G^0=-1.64\times {10}^4\text{ J/mol}\end{array}$

From Table C.4 the standard heat of reaction and Gibbs free energy for methanol formation is:

  $\begin{array}{l}\Delta G_{298}^0=-24791\text{J/mol}\\ \Delta H_{298}^0=-90135\text{ J/mol}\end{array}$

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

  $\begin{array}{c}\Delta A=-7.663\\ \Delta B=10.805\times {10}^{-3}\\ \Delta C=-3.450\times {10}^{-6}\\ \Delta D=-1.35\times {10}^4\end{array}$

Now, use equations set (6) in equation (5) and calculate the value of $\Delta G^0$ at the assumed temperatureof $480\text{ K}$ as:

Since, the calculated value of $\Delta G^0$ by equation (5) and by using equilibrium constant is almost same, the assumed temperature, $T=480\text{ K}$ is correct.

Thus, the temperature at which the equilibrium mole fraction of methanol is $0.5$ at $100\text{ bar}$ for an ideal solution of gases is calculated to be $T=480\text{ K}$ .