Chapter 15, Problem 15.132QP

About 75% 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:

$\begin{array}{l}{\text{CH}}_{\text{4}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)\rightleftarrows \text{CO}\left(g\right)+{\text{3H}}_{\text{2}}\left(g\right)\\ \Delta H^{\circ }=\text{206kJ/mol}\end{array}$

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

$\begin{array}{l}{\text{CH}}_{\text{4}}\left(g\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\rightleftarrows \text{CO}\left(g\right)+2{\text{H}}_{\text{2}}\left(g\right)\\ \Delta H^{\circ }=35.7\text{kJ/mol}\end{array}$

(a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (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:

To calculate and analyze the temperature, equilibrium constants $\left(\text{Kp, Kc}\right)$ values of given primary and secondary stages of methane $\text{(CO}\text{and}{\text{H}}_{\text{2}}\text{)}$ formation reactions with two different temperatures at ${800}^{\text{ο}}\text{C,}{\text{1000}}^{\text{ο}}\text{C}$.

Concept Introduction:

Temperature affect in equilibrium: This process chemical shifts changes (or) towards the product or reactant, which can be determined by studying the reaction and deciding whether it is exothermic or endothermic.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Kp and Kc: This equilibrium constants of gaseous mixtures, these difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system.

Equilibrium concentration: If Kc and the initial concentration for a reaction and calculate for both equilibrium concentration, and using the (ICE) chart and equilibrium constant and derived changes in respective reactants and products.

Free energy: The change in the standard energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy $(\text{ΔH)}$ of the system minus the change in the respective product of the temperature at the times of entropy of the system, the following equation $\text{G=H-(TS) or (ΔG=ΔH-TΔS)}$.

Answer to Problem 15.132QP

The equilibrium temperatures, partial pressure (Kp) and molar constant (Kc) values are given the statement of primary, secondary stages of equilibrium formation reaction is shown below.

$\begin{array}{l}\text{Primary}\text{Reaction}\text{(a)}\\ {\text{CH}}_{\text{4}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O}\rightleftharpoons \text{CO(g)}{\text{+3H}}_{\text{2}}\text{(g)}\text{--------}\Delta \text{H}°\text{=}\text{206}\text{KJ/mol}\\ \text{Secondary}\text{Reaction}\text{(b)}\\ {\text{CH}}_{\text{4}}\text{(g)}\text{+}\frac{1}{2}{\text{O}}_{\text{2}}\rightleftharpoons \text{CO(g)}{\text{+2H}}_{\text{2}}\text{(g)}\text{--------}\Delta \text{H}°\text{=}35.7\text{KJ/mol}\\ [Temprature\text{decresing]}-------[Equilibrium\text{shifts}\text{to left}]\\ [Appliedforlow\text{pressure}]------[Moreproductsareformed]\\ \text{Kp=}\text{1}\text{.4}\times {\text{10}}^{\text{5}}\\ {\text{K}}_p=\frac{{\text{(P}}_{\text{CO}}){(P_{{\text{H}}_{\text{2}}})}^3}{(P_{{\text{CH}}_{\text{4}}})(P_{H_{2}O})}\\ \text{The}\text{equilibrium}\text{pressure}\\ {\text{P}}_{{\text{CH}}_{\text{4}}}\text{=}\text{2}\text{atm,}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\text{2}\text{atm,}{\text{P}}_{\text{CO}}\text{=13}\text{atm,}{\text{P}}_{{\text{H}}_{\text{2}}}\text{=39}\text{atm}\end{array}$

Explanation of Solution

To Identify: The temperature and pressures are identified given the statement of primary and secondary stages of equilibrium reactions with respective temperatures.

Write and analyze the following equilibrium reactions.

$\begin{array}{l}\text{Primary}\text{Reaction}\text{(a)}\\ {\text{CH}}_{\text{4}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O}\rightleftharpoons \text{CO(g)}{\text{+3H}}_{\text{2}}\text{(g)}\text{--------}\Delta \text{H}°\text{=}\text{206}\text{KJ/mol}\\ \text{Secondary}\text{Reaction}\text{(b)}\\ {\text{CH}}_{\text{4}}\text{(g)}\text{+}\frac{1}{2}{\text{O}}_{\text{2}}\rightleftharpoons \text{CO(g)}{\text{+2H}}_{\text{2}}\text{(g)}\text{--------}\Delta \text{H}°\text{=}35.7\text{KJ/mol}\\ [Temprature\text{decresing]}-------[Equilibrium\text{shifts}\text{to left}]\\ [Appliedforlow\text{pressure}]------[Moreproductsareformed]\end{array}$

Given the primary and secondary stages of equilibrium reactions (a and b) involved some equilibrium steps are fallows.

Effect of temperatures: Given the both reactions are endothermic (heat absorption) and it equilibrium moved into left side, when temperature decreased, so this reaction get

$(+\text{ΔH)}$ Positive values and its equilibrium constant (Kc) become a lesser values. The le

Chatelier’s principle the products would be favored at high temperature.

If the reactions indeed the steam-reforming process is carried out at very high temperature in the range of ${800}^{\text{ο}}\text{C,}{\text{1000}}^{\text{ο}}\text{C}$. These formation processes are very interesting to note that in a plant that user natural gas as methane $\left({\text{CH}}_{\text{4}}\right)$ for both hydrogen generation and heating, about one-third of the gas is burned to maintain the high temperature.

Analyzing for pressure: Given the both reactions are more moles of products than reactants, therefore we expects products to be favored at low pressure. The both reactions are carried out at high pressure. The particular reasons are that the hydrogen gas produces is used captivity, for example if applied for high pressure leads to higher yields of ammonia formation process.

To find: Calculate the molar constant $(Kc)$ and partial pressure $(Kp)$ values for given the statement of primary and secondary stages of methane (CH₄) dissociation reaction.

Calculate and analyze the molar mass and $(Kp\&Kc)$ values with respective statement (b) in 1 and 2.

Let us consider the equilibrium constant values are statement

$\begin{array}{l}\text{Statement}\text{(b)}\text{i}\\ \text{Now we calculate the (Kp) values of equilibrium reaction}\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)\text{Kp}\text{=}\text{Kc}{\text{(RT)}}^{\text{Δn}}-----[1]\\ \text{Kp}\text{=}\text{Kc}{\text{[0}}^{\text{.0821}}\\ \text{Kp}\text{=}18\\ \text{Primary}\text{Reaction}\text{(a)}\\ {\text{CH}}_{\text{4}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O}\rightleftharpoons \text{CO(g)}{\text{+3H}}_{\text{2}}\text{(g)}\\ \text{Here}\text{Δn}\text{=}4\text{-2=}2\text{(Product - Reactant)}\\ \text{and}\text{given}\text{Temprature}{587}^{\text{0}}\text{C}\\ \text{T}\text{=}\text{(800+ 273)}\text{K}\\ =1073\text{K}\\ \text{The respactive values are substituted for equation }\\ \text{Kp}\text{=}(18){(}^{\text{0}}\\ \text{Kp}=1.4\times {10}^5\end{array}$

Let us considers the (ICE) equilibrium method

$\begin{array}{l}\text{Hare, }\\ {\text{CH}}_{\text{4}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O}\rightleftharpoons \text{CO(g)}{\text{+3H}}_{\text{2}}\text{(g)}\\ \text{Initial (atm): }151500\\ \text{Change (atm): }\text{-x}-x+x+3x\\ ----------------------------\\ \text{Eqilibrium (atm):}15-x15-xx+3x\\ \text{Solving for the equilibrium constant: }\\ {\text{K}}_p=\frac{{\text{(P}}_{\text{CO}}){(P_{H_2})}^3}{(P_{C{H}_4})(P_{H_{2}O})}\text{-----------[1]}\\ \text{The equilibrium pressure values (ICE) are substituted above the equation (1)}\\ 1.4\times {10}^5=\frac{(x){(3x)}^2}{(15-x)(15-x)}=\frac{27x^2}{{(15-x)}^2}------[2]\\ \text{Taking}\text{the square}\text{root}\text{of both sides}\text{above equation}\\ 3.7\times {10}^2=\frac{5.2x^2}{15-x}-------[3]\\ \text{The mathematical}\text{expressed}\text{as}\text{equation}\text{(3)}\\ \text{5}{\text{.2x}}^{\text{2}}+(3.7\times {10}^{2}x)-(5.6\times {10}^3)=0\\ Solvingthe\text{ quadratic}\text{euation,}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ a=52,b=3700,c=-56000\\ x=\frac{-3700\pm \sqrt{{(3700)}^2-4(52)(-5600)}}{2(52)}\\ x=\frac{-3700\pm \sqrt{{(3700)}^2-(-5600)4(52)}}{104}\\ x=\frac{-3700\pm \sqrt{25338000}}{104}\\ =\sqrt{25338000}=20\sqrt{63345}\\ x=\frac{-3700+20\sqrt{63345}}{104}\\ x=\left(\frac{5(\sqrt{63345-185})}{26}\right);x=-\left(\frac{5(\sqrt{63345-185})}{26}\right)\\ Solvedabovevalueswegetthe=13\\ x=13atm\\ \text{Here the other solution for (x) is negative and physically impossible}\\ \text{The }\text{equilibrium preassure is }\\ {\text{P}}_{{\text{CH}}_{\text{4}}}=(15-x)x=13\\ {\text{P}}_{{\text{CH}}_{\text{4}}}=2atm\\ {\text{P}}_{H_{2}O}=(15-13)=2atm\\ {\text{P}}_{H_2}=3(13)=39atm\end{array}$

The given the $\left({\text{CO and H}}_{\text{2}}\right)$ formation reaction the respective reactant to give two different products all exists in the same phase and this equilibrium reaction expression contains same conditions like methane gas converted into $\left({\text{CO and H}}_{\text{2}}\right)$ gas phase, so this equilibrium reactions has heterogeneous. The equilibrium constant can also be represented by Kp, were the “P” partial pressure. The each partial pressure values are derived given the methane equilibrium reaction equation at and different temperatures the simple calculation method as showed above.

Conclusion

The each reactant and product partial pressure values are calculated given the methane decomposition reaction and corresponding temperature at $800\text{and}{1000}^{\text{0}}\text{C}$.