Chapter 4, Problem 4.3Q

Example 4-1. Would the example be correct if water were considered an inert? Explain.

Interpretation Introduction

Interpretation:

The value of equilibrium constant is to be calculated for a given elementary reversible gas phase reaction in a flow reactor.

Concept Introduction :

The general rate law is represented as follows:

  $r=k{\left[A\right]}^x$

Here, $\left[A\right]$ is the concentration of reactant A and x is order of the reaction with respect to A.

Also,

  $r=-\frac{d\left[A\right]}{dt}$

The rate law expression becomes equal to zero at equilibrium.

Answer to Problem 4.3Q

The value of equilibrium constant (K_c) is $0.328{\text{dm}}^{\text{3}}/\text{mol}$ .

Explanation of Solution

The elementary reversible reaction which is carried out in a flow reactor is given as

  $2\text{A}\rightleftarrows \text{B}$

The inlet concentration of A $\left(C_{A0}\right)$ = $4\text{mol}/{\text{dm}}^{\text{3}}$

The equilibrium conversion $\left({\text{X}}_{\text{e}}\right)$ = 0.6

The rate expression of the reaction is written as

  $-r_A=k_1{C}_A^2-k_2{C}_B$

The equilibrium rate constant (K_C) = $\frac{k_1}{k_2}$

The above rate expression can be simplified in terms of the equilibrium constant as

  $\begin{array}{l}-r_A=k_1{C}_A^2-k_2{C}_B\\ =k_1\left[C_A^2-\frac{k_2}{k_1}{C}_B\right]\\ =k_1\left[C_A^2-\frac{C_B}{K_C}\right]\end{array}$

At equilibrium, $-r_A=0$

  $\begin{array}{l}-r_A=0\\ 0=k_1\left[C_{Ae}^2-\frac{C_{Be}}{K_C}\right]\\ C_{Ae}^2-\frac{C_{Be}}{K_C}=0\\ C_{Ae}^2=\frac{C_{Be}}{K_C}\end{array}$

  $K_C=\frac{C_{Be}}{C_{Ae}^2}$ ...... (1)

Where $C_{Ae}$ and $C_{Be}$ are the concentrations of A and B at equilibrium respectively.

Since the reaction takes place in the gas phase, the volume expansion factor $\left(\epsilon \right)$ is to be calculated.

  $\begin{array}{l}\epsilon =y_{A0}\delta \\ =y_{A0}\left[\frac{F_{Tf}-F_{T0}}{F_{T0}}\right]\end{array}$

Where $F_{Tf}$ represents the total number of moles at exit

  $F_{T0}$ represents the initial total number of moles

  $y_{A0}$ represents mole fraction of A in the feed

Since the feed is pure A, therefore $y_{A0}=1$

  $\begin{array}{l}\epsilon =y_{A0}\delta \\ =1\times \left[\frac{1-2}{2}\right]\\ \epsilon =-\frac{1}{2}\end{array}$

The values of $C_{Ae}$ and $C_{Be}$ are calculated as follows:

  $C_{Ae}=C_{A0}\left[\frac{1-X_e}{1+\epsilon X_e}\right]$

  $\begin{array}{l}{C}_{Ae}^2=C_{A0}^2{\left[\frac{1-X_e}{1+\epsilon X_e}\right]}^2\\ =4^2{\left[\frac{1-0.6}{1+\left(-0.5\right)0.6}\right]}^2\\ =16{\left[\frac{0.4}{0.7}\right]}^2\\ =5.224\text{mol}/{\text{dm}}^{\text{3}}\end{array}$

And,

  $\begin{array}{l}{C}_{Be}=C_{B0}+\frac{b}{a}{C}_{A0}\left(\frac{X_e}{1+\epsilon X_e}\right)\\ =0+\frac{b}{a}{C}_{A0}\left(\frac{X_e}{1+\epsilon X_e}\right)\\ =\frac{1}{2}{C}_{A0}\left(\frac{X_e}{1+\epsilon X_e}\right)\end{array}$

Thus,

  $\begin{array}{l}{C}_{Be}=\frac{1}{2}\times 4\text{mol}/{\text{dm}}^{\text{3}}\times \left(\frac{0.6}{1+\left(-0.5\right)0.6}\right)\\ =0.714\text{mol}/{\text{dm}}^{\text{3}}\end{array}$

Where a and b are stoichiometric coefficient of A and B respectively.

Plugin the values of $C_{Ae}$ and $C_{Be}$ in equation (1)

  $\begin{array}{l}{K}_C=\frac{1.2}{{\left(2.285\right)}^2}\\ =\frac{1.714\text{mol}/{\text{dm}}^{\text{3}}}{5.221{\left(\text{mol}/{\text{dm}}^{\text{3}}\right)}^2}\\ =0.328{\text{dm}}^{\text{3}}/\text{mol}\end{array}$

Interpretation Introduction

Interpretation:

The value of the equilibrium constant is to be calculated for a given elementary reversible liquid phase reaction in a flow reactor.

Concept Introduction :

The rate law expression becomes equal to zero at equilibrium and for liquid phase reaction, the value of the volume expansion factor becomes equal to zero.

Answer to Problem 4.3Q

The value of equilibrium constant (K_c) is $0.469{\text{dm}}^{\text{3}}/\text{mol}$ .

Explanation of Solution

The expression of equilibrium constant will remain the same which is derived in part a.

For liquid phase reaction volume of the reaction mixture will not change. Therefore, the volume expansion factor becomes equal to zero.

So, the expressions of $C_{Ae}$ becomes as follows:

  $\begin{array}{l}{C}_{Ae}=C_{A0}\left[\frac{1-X_e}{1+\epsilon X_e}\right]\\ =C_{A0}\left[\frac{1-X_e}{1+0\times X_e}\right]\end{array}$

  $C_{Ae}=C_{A0}\left(1-X_e\right)$ ...... (2)

Plug in the values of initial concentration and equilibrium conversion in equation (2)

  $\begin{array}{l}{C}_{Ae}=C_{A0}\left(1-X_e\right)\\ =4\left(1-0.6\right)\\ =1.6\text{mol}/{\text{dm}}^{\text{3}}\end{array}$

Similarly, for $C_{Be}$ the expression becomes as follows:

  $\begin{array}{l}{C}_{Be}=C_{B0}+\frac{b}{a}{C}_{A0}\left(\frac{X_e}{1+\epsilon X_e}\right)\\ =0+\frac{b}{a}{C}_{A0}\left(\frac{X_e}{1+0\times X_e}\right)\end{array}$

  $C_{Be}=\frac{1}{2}{C}_{A0}{X}_e$ ...... (3)

Plug in the values of initial concentration and equilibrium conversion in equation (3)

  $\begin{array}{l}{C}_{Be}=0.5\times 4\text{mol}/{\text{dm}}^{\text{3}}\times 0.6\\ =1.2\text{mol}/{\text{dm}}^{\text{3}}\end{array}$

Plug in the values of $C_{Ae}$ and $C_{Be}$ in equation (1)

  $\begin{array}{l}{K}_C=\frac{1.2\text{mol}/{\text{dm}}^{\text{3}}}{{\left(1.6\right)}^2{\left(\text{mol}/{\text{dm}}^{\text{3}}\right)}^2}\\ =\frac{1.2\text{mol}/{\text{dm}}^{\text{3}}}{2.56{\left(\text{mol}/{\text{dm}}^{\text{3}}\right)}^2}\\ =0.469{\text{dm}}^{\text{3}}/\text{mol}\end{array}$

Interpretation Introduction

Interpretation:

The rate law equation is to be expressed in terms of conversion for a reversible, elementary, gas-phase, isothermal reaction.

Concept Introduction : For an elementary reaction, the rate law equation corresponds to the stoichiometric equation.

Answer to Problem 4.3Q

The rate equation is $-r_A=32{\left(\frac{1-X_A}{1-0.5X_A}\right)}^2-8\left(\frac{X_A}{1-0.5\times X_A}\right)$ .

Explanation of Solution

For a reaction $2\text{A}\rightleftarrows \text{B}$ , the rate equation is written as:

  $-r_A=k_1{C}_A^2-k_2{C}_B$ ...... (4)

The value of rate constant $\left(k_1\right)$ = $2{\text{dm}}^6/\text{mol}$

The value of equilibrium constant $\left(K_C\right)$ = $0.5$

The relationship between $k_1,k_2\text{and}{K}_C$ is as follows:

  $K_C=\frac{k_1}{k_2}$ ....... (5)

Plug in the values in equation (5)

  $\begin{array}{l}0.5=\frac{2{\text{dm}}^6/\text{mol}}{k_2}\\ k_2=\frac{2{\text{dm}}^6/\text{mol}}{0.5}\\ =4{\text{dm}}^6/\text{mol}\end{array}$

The values of $C_A$ and $C_B$ are calculated as follows:

  $C_A=C_{A0}\left[\frac{1-X_A}{1+\epsilon X_A}\right]$

  $C_A=C_{A0}\left[\frac{1-X_A}{1-0.5X_A}\right]$

  $C_A=4\left[\frac{1-X_A}{1-0.5X_A}\right]$

  $C_B=C_{B0}+\frac{b}{a}{C}_{A0}\left(\frac{X_A}{1+\epsilon X_A}\right)$

  $\begin{array}{l}{C}_B=0+\frac{b}{a}{C}_{A0}\left(\frac{X_A}{1+\left(-0.5\right)\times X_A}\right)\\ =\frac{1}{2}\times 4\left(\frac{X_A}{1-0.5\times X_A}\right)\\ =2\left(\frac{X_A}{1-0.5\times X_A}\right)\end{array}$

Plug in the values of $k_1,k_2,C_A\text{and}{C}_B$ in the equation (4)

  $\begin{array}{l}-r_A=2{\left[4\left(\frac{1-X_A}{1-0.5X_A}\right)\right]}^2-4\times 2\left(\frac{X_A}{1-0.5\times X_A}\right)\\ =2\times 16\times {\left(\frac{1-X_A}{1-0.5X_A}\right)}^2-8\left(\frac{X_A}{1-0.5\times X_A}\right)\\ =32{\left(\frac{1-X_A}{1-0.5X_A}\right)}^2-8\left(\frac{X_A}{1-0.5\times X_A}\right)\end{array}$

Interpretation Introduction

Interpretation:

The rate equation is to be expressed in terms of conversion for a constant volume batch reactor.

Concept Introduction: For an elementary reaction, the rate law equation corresponds to the stoichiometric equation and the value of the volume expansion factor becomes equal to zero for a constant volume batch reactor.

Answer to Problem 4.3Q

The rate expression for a constant volume batch reactor is $-r_A=32{\left(1-X_A\right)}^2-8X_A$ .

Explanation of Solution

For constant volume batch reactor, the expressions of $C_A$ and $C_B$ becomes as:

  $\begin{array}{l}{C}_A=C_{A0}\left[\frac{1-X_A}{1+\epsilon X_A}\right]\\ =C_{A0}\left[\frac{1-X_A}{1+0\times X_A}\right]\end{array}$

  $C_A=C_{A0}\left(1-X_A\right)$ ...... (6)

  $\begin{array}{l}{C}_B=C_{B0}+\frac{b}{a}{C}_{A0}\left(\frac{X_A}{1+\epsilon X_A}\right)\\ C_B=0+\frac{b}{a}{C}_{A0}\left(\frac{X_A}{1+0\times X_A}\right)\end{array}$

  $C_B=\frac{b}{a}{C}_{A0}{X}_A$ ...... (7)

Plug in the values of the initial concentration of A and stoichiometric coefficient in equation (6) and (7)

  $C_A=4\left(1-X_A\right)$

  $\begin{array}{l}{C}_B=\frac{1}{2}\times 4X_A\\ =2X_A\end{array}$

The values of rate constants will remain the same.

Plug in the values of $k_1,k_2,C_A\text{and}{C}_B$ in equation in (4)

  $\begin{array}{l}-r_A=2{\left[4\left(1-X_A\right)\right]}^2-4\times 2X_A\\ =2\times 16{\left(1-X_A\right)}^2-8X_A\\ =32{\left(1-X_A\right)}^2-8X_A\end{array}$