Chapter 15, Problem 96AP

About 75 percent of hydrogen for industrial use is produced by the steam-reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at $\text{800°C}$ to give hydrogen and carbon monoxide:

${\text{CH}}_{\text{4}}\left(g\right){\text{ H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftarrows }\text{ CO}\left(g\right){\text{ + 3H}}_{\text{2}}\left(g\right)\Delta H°\text{ = 206 kJ/mol}$

The secondary stage is carried out at about $\text{1000°C}$. in the presence of air. to convert the remaining methane to hydrogen:

$\text{CH4(}g\text{)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(}g\text{)}\underset{}{\rightleftarrows }\text{CO(}g{\text{)+2H}}_{\text{2}}\text{(}g\text{) Δ}H°\text{=35}\text{.7 kJ/mol}$

(a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondany stages? (b) The equilibrium constant K_c for the primary stage is 18 at $\text{800°C}$. (i) Calculate $K_p$ for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

Interpretation Introduction

Interpretation:

The favorable conditions for the formation of products, the partial pressure constant of thereaction, and the pressures of all the gases at equilibrium are to be determined.

Concept introduction:

When the enthalpy of a reaction is less than zero, it is called an exothermic reaction (enthalpy is negative), and when it is more than zero, it is an endothermic reaction (enthalpy is positive).

Equilibrium constant is the ratio of the concentration of reactants and products present in the chemical reaction.

For a general reaction: $a\text{A}\text{+}b\text{B}\rightleftharpoons c\text{C}\text{+}d\text{D}$

The general formula for writing equilibrium expression for the reaction is given as:

$K_C=\frac{{\left[\text{C}\right]}^c{}_{eqm}{\left[\text{D}\right]}^d{}_{eqm}}{{\left[\text{A}\right]}^a{}_{eqm}{\left[\text{B}\right]}^b{}_{eqm}}$

Here, $K_C$ is the equilibrium constant and $C$ in $K_C$ stands for the concentration. $\left[\text{A}\right]$, $\left[\text{B}\right]$, $\left[\text{C}\right]$, and $\left[\text{D}\right]$ are the equilibrium concentration of reactants A and B and product C and D, respectively.

A and B are reactants, C is products, and $x,y$, and $z$ are their respective stoichiometric coefficients.

For the general reaction: $a\text{A}\text{+}b\text{B}\rightleftharpoons c\text{C}\text{+}d\text{D}$

The general formula for writing equilibrium expression for the reaction is given as:

$K_p=\frac{P_C^c{P}_D^d}{P_A^a{P}_B^b}$

Here, $p$ in $K_p$ stands for the pressure. $P_A,P_B,P_C,\text{and}{P}_D$ are the equilibrium partial pressure of reactants A and B and products C and D.

A and B are reactants, C and D are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

Equilibrium constants of gas phase reaction are written in terms of partial pressures because concentration of gases is directly proportional to partial pressures.

The relationship between $K_P$ and $K_C$ is given as:

$\begin{array}{c}{K}_P=K_C{\left(RT\right)}^{\Delta n}\\ K_c=\frac{K_P}{{\left(RT\right)}^{\Delta n}}\end{array}$

Answer to Problem 96AP

Solution:

(a)

A high temperature favors product formation.

(b)

i) $1.4\times {10}^5$

ii) $\begin{array}{c}{P}_{{\text{CH}}_{\text{4}}}=2\text{atm}\\ P_{{\text{H}}_{\text{2}}\text{O}}=2\text{atm}\\ P_{\text{CO}}=13\text{atm}\\ P_{{\text{H}}_{\text{2}}}=39\text{atm}\end{array}$

Explanation of Solution

Given information:

The temperature at the first stage is ${800}^{\circ }C$.

The temperature at thesecond stage is ${1000}^{\circ }C$.

$\Delta H^{\circ }=2.6\text{kJ}/\text{mol}$

$\Delta H^{\circ }=35.7\text{kJ}/\text{mol}$

a)The conditions of temperature and pressure that would favor that formation of products both in primary and secondary stage.

The optimum condition for pressure is $200-500$ atmospheres.

Standard conditions for pressure in air is,

$1\text{ atm}=760\text{ mmHg}$.

At a temperature of ${25}^{\circ }\text{C}$, the pressure is $101.325\text{ kPa}$.

$\begin{array}{c}1\text{Pa}={10}^{-6}\text{N}/{\text{mm}}^2\\ ={10}^{-5}\text{bar}\end{array}$

Based on the given reactions, the enthalpy is positive. So, the forward reaction is an endothermic reaction. According to LeChatelier’s principle, at high temperatures, more products are formed. In the steam-reforming process, alow temperature is favored, according to Le Chatelier’s principle.

An interesting fact to note is that the plants use natural gases, such as methane, for heating and generating hydrogen. To maintain the high temperatures, one-third of the methane gas is burned.

The first stage of the reaction is as follows:

${\text{CH}}_4\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\Delta H^{\circ }=206\text{kJ}/\text{mol}$

The second stage of the reaction is as follows:

${\text{CH}}_4\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\rightleftharpoons \text{CO}\left(g\right)+3{\text{H}}_2\left(g\right)\Delta H^{\circ }=35.7\text{kJ}/\text{mol}$

In both the primary and the secondary reaction, a high temperature is favorable for product formation because both are endothermic reactions.

The sum of the number of moles of the product is more than that of the reactants in both the reactions. So, a decrease in pressure favors the forward reaction to form more products.

b)The equilibrium constant is $18$.

i) $K_p$ for the reaction

The value of $K_p$ for the reaction is as follows:

The relation between $K_p$ and $K_c$ is:

$K_p=K_c{\left(RT\right)}^{\Delta n}$

Here, $K_p$ is the equilibrium constant in terms of pressure, $K_c$ is the equilibrium constant, $R$ is the gas constant, $T$ is the temperature, and $\Delta n$ is the difference between the number of moles of the reactants and the products.

The number of moles of the reaction is calculated as follows:

$\begin{array}{c}\Delta n=\text{the sum of number of moles of gaseous products}-\text{the sum of number of moles of gaseous reactants}\\ =4-2\\ =2\end{array}$

Substitute the values of $K_c$, $R$, $T$, and $\Delta n$ in the above equation,

$\begin{array}{c}{K}_p=18{\left(0.0821\times 1073\right)}^2\\ =18\left(7760.429\right)\\ =139687.74\\ =1.4\times {10}^5\end{array}$

The value of $K_p$ for the reaction is $1.4\times {10}^5$.

ii) The pressure of all gases at equilibrium

The partial pressures of methane and steam start at $15atm$. The pressures of all the gases at equilibrium is as follows:

The initial change equilibrium table for thereaction is as follows:

$\begin{array}{ccccc}& {\text{CH}}_{\text{4}}& {\text{H}}_{\text{2}}\text{O}& \text{CO}& {\text{3H}}_{\text{2}}\\ \text{Initial}\left(\text{atm}\right)& 15& 15& 0& 0\\ \text{Change}\left(\text{atm}\right)& -x& -x& +x& +x\\ \text{Equilibrium}\left(\text{atm}\right)& 15-x& 15-x& x& 3x\end{array}$

The equilibrium constant is:

$K_p=\frac{P_C^c{P}_D^d}{P_A^a{P}_B^b}$

$K_p=\frac{P_{\text{CO}}{P}_{{\text{H}}_{\text{2}}}^3}{P_{{\text{CH}}_{\text{4}}}{P}_{{\text{H}}_{\text{2}}\text{O}}}$

Here, $K_p$ is the equilibrium constant in terms ofpressure, $P_{CO}$ is the partial pressure of $\text{CO}$, $P_{{\text{H}}_2}^3$ is the partial pressure of ${\text{H}}_2$, $P_{{\text{CH}}_{\text{4}}}$ is the partial pressure of ${\text{CH}}_4$, and $P_{{\text{H}}_{\text{2}}\text{O}}$ is the partial pressure of ${\text{H}}_2\text{O}$.

Substitute the value of $K_p$ to calculate the value of $x$,

$\begin{array}{c}1.4\times {10}^5=\frac{\left(x\right){\left(3x\right)}^3}{{\left(15-x\right)}^2}\\ =\frac{27x^4}{{\left(15-x\right)}^2}\end{array}$

Take the square root on both sides,

$\begin{array}{c}3.7\times {10}^2=\frac{5.2x^2}{\left(15-x\right)}\\ 5.2x^2+\left(3.7\times {10}^{2}x\right)-\left(5.6\times {10}^3\right)=0\\ x_1=13\\ x_2=-0.61\end{array}$

From the quadratic expression, $x_2$ is negative. So, $x_1=13$ is possible.

The partial pressure of ${\text{CH}}_4$ at equilibrium is as follows:

$P_{{\text{CH}}_{\text{4}}}=\left(15-13\right)=2\text{atm}$

The partial pressure of ${\text{H}}_2\text{O}$ at equilibrium is as follows:

$P_{{\text{H}}_{\text{2}}\text{O}}=\left(15-13\right)=2\text{atm}$

The partial pressure of $\text{CO}$ at equilibrium is as follows:

$P_{\text{CO}}=13\text{atm}$

The partial pressure of ${\text{H}}_2$ at equilibrium is as follows:

$P_{{\text{H}}_{\text{2}}}=3\left(13\right)=39\text{atm}$