Chapter 14, Problem 14.101QP

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 800°C to give hydrogen and carbon monoxide:

$\begin{array}{l}{\text{CH}}_4(g)+{\text{H}}_2\text{O}(g)\rightleftharpoons \text{CO}(g)+3{\text{H}}_2(g)\hfill \\ \Delta H°=260\text{kJ/mol}\hfill \end{array}$

The secondary stage is carried out at about 1000°C, in the presence of air, to convert the remaining methane to hydrogen:

$\begin{array}{l}{\text{CH}}_4(g)+\frac{1}{2}{\text{O}}_2(g)\rightleftharpoons \text{CO}(g)+2{\text{H}}_2(g)\hfill \\ \Delta H°=35.7\text{kJ/mol}\hfill \end{array}$

(a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (b) The equilibrium constant K_c for the primary stage is 18 at 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 conditions of temperature and pressure which would favour the formation of products in both the primary and secondary stage has to be calculated using the data.

Concept Introduction:

Endothermic reaction:

In an endothermic reaction, heat will be a reactant. Therefore, increasing the temperature will shift the reaction from reactant side to the product side and the value of equilibrium constant will increases.

Exothermic reaction:

In an exothermic reaction, heat will be a product. Therefore, increasing the temperature will shift the reaction from product side to the reactant side and the value of equilibrium constant will decreases.

Explanation of Solution

Given data:

  $\begin{array}{l}{\text{ΔH}}_{\text{1}}{}^{\text{°}}\text{=}\text{260}\text{kJ}/\text{mol}\\ {\text{ΔH}}_{\text{2}}{}^{\text{°}}\text{=}\text{35}\text{.7}\text{kJ}/\text{mol}\end{array}$

The reactions at primary stage and at the secondary stage are given below:

  $\begin{array}{l}C{H}_4\left(\text{g}\right)+H_{2}O\left(g\right)\underset{}{\rightleftharpoons }CO\left(\text{g}\right)+3H_2\left(g\right)-{\text{ΔH}}_{\text{1}}{}^{\text{°}}\text{=}\text{260}\text{kJ}/\text{mol}\\ \\ C{H}_4\left(\text{g}\right)+\frac{1}{2}{O}_2\left(g\right)\underset{}{\rightleftharpoons }CO\left(\text{g}\right)+2H_2\left(g\right)-{\text{ΔH}}_{\text{2}}{}^{\text{°}}\text{=}\text{35}\text{.7}\text{kJ}/\text{mol}\end{array}$

The reaction at primary stage has a temperature of ${\text{800}}^{\text{°}}\text{C}$and the reaction at secondary stage has a temperature of ${\text{1000}}^{\text{°}}\text{C}$

The both reactions are endothermic

If the reactions are endothermic, heat will be the reactant and high temperatures will favour the product side.

Therefore, high temperature is maintained at steam-reforming process.

Examining the reactions, it can be seen that number of moles of products is greater than the number of moles of reactants. Therefore, the expectation is that the products will be favoured at low pressures.

The reality is that the reactions are carried out at high pressures. It is for the production of higher yields of ammonia by the hydrogen gas produced which requires a high pressure.

Interpretation Introduction

Interpretation:

The equilibrium constant ${\text{K}}_p$ for the reaction and the pressure of all the gases at equilibrium has to be calculated.

Concept Introduction

Equilibrium constant at constant pressure:$K_p$

It is used to express the relationship between product pressures and reactant pressures.

  For a general reaction, $\text{aA+bB}\underset{}{\rightleftharpoons }\text{cC+dD}$

$\text{Equilibrium}\text{constant}{\text{K}}_p\text{=}\frac{P_C{}^{\text{c}}{P}_D{}^{\text{d}}}{P_A{}^{\text{a}}{P}_B{}^{\text{b}}}$

${\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}$

$\text{Δn}$- Change in number of moles

R- Gas constant

T- Temperature

Explanation of Solution

Given data:

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\text{18}\\ {\text{P}}_{{\text{CH}}_{\text{4}}}\text{=}\text{15}\text{atm}\\ {\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\text{15}\text{atm}\end{array}$

The reaction is given below:

$C{H}_4\left(\text{g}\right)+H_{2}O\left(g\right)\rightleftarrows CO\left(\text{g}\right)+3H_2\left(g\right)$

Equilibrium constant $K_p$can be calculated as shown below:

  $\begin{array}{c}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}\\ {\text{K}}_{\text{p}}\text{=}\text{18}{\left(\text{0}\text{.0821×1073}\right)}^{\text{2}}\\ \text{=}\text{1}{\text{.4×10}}^{\text{5}}\end{array}$

The $K_p$ is calculated to be $\text{1}{\text{.4×10}}^{\text{5}}$.

The pressure of all the gases at equilibrium has to be calculated.

The amount of ${\text{CH}}_{\text{4}}\text{and}{\text{H}}_{\text{2}}\text{O}$ reacted can be taken as x.

The ICE table for the reaction is given below

  $A\left(g\right)\text{+}B\left(g\right)\underset{}{\rightleftharpoons }C\left(g\right)$

$\begin{array}{l}\begin{array}{cccccc}& C{H}_4\left(g\right)& +& H_{2}O\left(g\right)& \underset{}{\rightleftharpoons }& CO\left(g\right)+3H_2\\ Initial& 15& & 15& & 0\\ Change& -x& & -x& & +x+3x\end{array}\\ \underset{¯}{}\\ \begin{array}{cccccc}Equilibirium& 15-x& & & 15-x& x3x\end{array}\end{array}$

Substituting the values in the equilibrium constant equation,

  $\begin{array}{c}{\text{K}}_{\text{P}}\text{=}\frac{{\text{P}}_{\text{CO}}{\text{P}}^{\text{3}}{}_{\text{H}{\text{2}}_{}}}{{\text{P}}_{{\text{CH}}_{\text{4}}}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}}\\ \\ \text{1}{\text{.4×10}}^{\text{5}}\text{=}\frac{\text{x}{\left(\text{3x}\right)}^{\text{3}}}{\left(\text{15-x}\right)\left(\text{15-x}\right)}\text{=}\frac{\text{27}{\text{x}}^{\text{4}}}{{\left(\text{15-x}\right)}^{\text{2}}}\end{array}$

Taking the square root of both sides,

$\begin{array}{l}\text{3}{\text{.7×10}}^{\text{2}}\text{=}\frac{\text{5}\text{.2}{\text{x}}^{\text{2}}}{\text{15-x}}\\ \text{5}\text{.2}{\text{x}}^{\text{2}}\text{+}\left(\text{3}{\text{.7×10}}^{\text{2}}\text{x}\right)\text{-}\left(\text{5}{\text{.6×10}}^{\text{3}}\right)\text{=}\text{0}\end{array}$

The value of x can be obtained by solving the quadratic equation

  $\text{x}\text{=}\text{13}\text{atm}$

The pressure at equilibrium can be calculated as follows:

$\begin{array}{l}{\text{P}}_{{\text{CH}}_{\text{4}}}\text{=}\text{15-x}\text{=}\text{15-13}\text{=}\text{2}\text{atm}\\ {\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\text{15-x}\text{=}\text{15-13}\text{=}\text{2}\text{atm}\\ {\text{P}}_{\text{CO}}\text{=}\text{13}\text{atm}\\ {\text{P}}_{{\text{H}}_{\text{2}}}\text{=}\text{3}\left(\text{13}\right)\text{atm}\text{=}\text{39}\text{atm}\end{array}$