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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 \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=206 \mathrm{~kJ} / \mathrm{mol}\) The secondary stage is carried out at about \(1000^{\circ} \mathrm{C},\) in the presence of air, to convert the remaining methane to hydrogen: \(\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftarrows \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\) (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (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?

Short Answer

Expert verified
(a) High temperature favors products, lower pressure aids reaction but high pressure is economically practical. (b) For (i), calculate \( K_P \) using the formula, for (ii) use equilibrium expression to find gas pressures at equilibrium.

Step by step solution

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01

Understanding the Primary Stage Reaction

The primary stage reaction is \( \mathrm{CH}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g) + 3 \mathrm{H}_{2}(g) \) with \( \Delta H^{\circ} = 206 \mathrm{~kJ/mol} \). Since the reaction is endothermic, increasing the temperature will favor the formation of products. According to Le Châtelier's principle, increasing pressure shifts equilibrium to the side with fewer moles of gas, which is the reactant side. Thus, a lower pressure would favor product formation, but practically, this reaction is carried out at high pressure for economic reasons.
02

Understanding the Secondary Stage Reaction

The secondary reaction is \( \mathrm{CH}_{4}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \) with \( \Delta H^{\circ} = 35.7 \mathrm{~kJ/mol} \). It is also endothermic, thus higher temperatures favor product formation. At lower pressures, the equilibrium shifts to produce more gas because there are more moles on the products' side, similar to the primary stage.
03

Calculate \( K_P \) for the Primary Stage

The relationship between \( K_P \) and \( K_c \) is given by \( K_P = K_c(RT)^{\Delta n} \), where \( R = 0.0821 \) L atm K\(^-1\) mol\(^{-1}\), \( T = 1073 \) K, and \( \Delta n = 3 \). Therefore, \( K_P = 18 \times (0.0821 \times 1073)^3 \). Calculate the value to find \( K_P \).
04

Initial Conditions to Set Up ICE Table for Primary Stage

Start with initial partial pressures: \( P_{\mathrm{CH}_4} = P_{\mathrm{H}_2O} = 15 \) atm, and \( P_{\mathrm{CO}} = P_{\mathrm{H}_2} = 0 \) atm. Set up an ICE (Initial, Change, Equilibrium) table to calculate changes in pressures: \( x \) atm change for methane and steam, \( x \) atm for CO, and \( 3x \) atm for \( H_2 \).
05

Solve for \( x \) Using Equilibrium Expression

From the ICE table, express the equilibrium pressures as: \( P_{\mathrm{CH}_4} = 15 - x \), \( P_{\mathrm{H}_2O} = 15 - x \), \( P_{\mathrm{CO}} = x \), and \( P_{\mathrm{H}_2} = 3x \). Set up the equilibrium expression using \( K_P \): \( K_P = \frac{x \cdot (3x)^3}{(15-x)^2} \) and solve for \( x \) using the determined \( K_P \).
06

Calculate Equilibrium Pressures

Solve the equation from the previous step for \( x \). Substitute \( x \) back into the expressions for the equilibrium pressures: \( P_{\mathrm{CH}_4} = 15 - x \), \( P_{\mathrm{H}_2O} = 15 - x \), \( P_{\mathrm{CO}} = x \), and \( P_{\mathrm{H}_2} = 3x \) to find the pressures of all gases at equilibrium.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Production
Hydrogen production via steam reforming is a critical process in the industry, supplying about 75% of hydrogen used industrially. This method involves the conversion of methane, which is a primary component of natural gas, into hydrogen. The steam reforming process is implemented in two main stages: primary reforming and secondary reforming.

During the primary reforming stage, methane reacts with steam over a nickel catalyst at high temperatures, approximately 800°C. This reaction produces hydrogen and carbon monoxide in the following balanced chemical equation:
\[ \mathrm{CH}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g) + 3 \mathrm{H}_{2}(g) \]
Industrial conditions optimize hydrogen yield by controlling temperature and pressure based on equilibrium principles.

The secondary reforming step involves reacting any leftover methane with oxygen at even higher temperatures (around 1000°C) to further enhance hydrogen production. Together, these two stages maximize the efficiency of hydrogen extraction from methane.
Equilibrium Constant
The equilibrium constant, represented as \(K_{c}\) in terms of concentrations or \(K_{P}\) in terms of partial pressures, is a fundamental concept in understanding chemical equilibria. For the steam reforming process, each stage of the reaction has its own equilibrium constant.

In the primary stage of steam reforming, the equilibrium constant \(K_{c}\) is given as 18 at 800°C. This value indicates the ratio of the concentrations of products to reactants when the reaction has reached a state of equilibrium. To find \(K_{P}\), representing the partial pressures, we utilize the equation:
\[ K_{P} = K_{c}(RT)^{\Delta n} \]
where \(R\) is the gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\)), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas, calculated by subtracting the moles of reactants from the moles of products.

The equilibrium constant provides a quantitative measure for the tendency of the system to favor products or reactants under a given set of conditions.
Le Châtelier's Principle
Le Châtelier's principle is an essential theory for predicting the direction of shift in equilibrium in response to changes in concentration, temperature, pressure, or volume. For the steam reforming process, this principle helps in optimizing conditions to favor product formation.

Since both the primary and secondary steam reforming reactions are endothermic, increasing the temperature will cause the equilibrium to shift towards the formation of products, as adding heat can be considered analogous to adding a reactant in the context of an endothermic reaction.
  • Increased Temperature: Shifts equilibrium to the right, favoring product formation.
  • Decreased Pressure: As it involves more moles of gas on the product side, reducing pressure favors product formation.
  • Concentration Changes: Altering concentrations of reactants or products will shift the equilibrium to offset these changes, according to the principle.

Understanding this principle helps in adjusting conditions in industrial processes to maximize hydrogen yield efficiently.
Endothermic Reaction
Endothermic reactions absorb heat from their surroundings, a critical characteristic relevant to the steam reforming process for hydrogen production. During the primary reforming stage, the conversion of methane and water into carbon monoxide and hydrogen is highly endothermic, with a change in enthalpy (\(\Delta H^{\circ}\)) of 206 kJ/mol.

This means that heat input is necessary to drive the reaction forward, favoring product formation. The secondary stage reaction is also endothermic, though to a lesser extent, with a \(\Delta H^{\circ}\) of 35.7 kJ/mol.

Here's why this is crucial in industrial applications:
  • Heat Supply: Adequate heat must be supplied to maintain the reaction's progress.
  • Temperature Control: Proper temperature management ensures the reaction remains in the product-favorable direction.
  • Energy Consideration: While energy input is higher, the process is finely controlled to optimize hydrogen output efficiently.

Understanding endothermic processes helps in strategic operational decisions, ensuring maximum yield and cost-effectiveness in hydrogen production.

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Most popular questions from this chapter

Consider the following reaction at equilibrium: $$\mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g)$$ From the data shown here, calculate the equilibrium constant (both \(K_{P}\) and \(K_{\mathrm{c}}\) ) at each temperature. Is the reaction endothermic or exothermic? $$ \begin{array}{clc} \text { Temperature }\left({ }^{\circ} \mathbf{C}\right) & {[\mathbf{A}](\boldsymbol{M})} & {[\mathbf{B}](\boldsymbol{M})} \\ \hline 200 & 0.0125 & 0.843 \\ 300 & 0.171 & 0.764 \\ 400 & 0.250 & 0.724 \end{array} $$

Consider the reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At a certain temperature, the equilibrium concentrations $$ \text { are }[\mathrm{NO}]=0.31 M,\left[\mathrm{H}_{2}\right]=0.16 M,\left[\mathrm{~N}_{2}\right]=0.082 M, \text { and } $$ \(\left[\mathrm{H}_{2} \mathrm{O}\right]=4.64 \mathrm{M} .\) (a) Write the equilibrium expression for the reaction. (b) Determine the value of the equilibrium constant.

When heated at high temperatures, iodine vapor dissociates as follows: $$ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) $$ In one experiment, a chemist finds that when 0.054 mole of \(\mathrm{I}_{2}\) was placed in a flask of volume \(0.48 \mathrm{~L}\) at \(587 \mathrm{~K},\) the degree of dissociation (i.e., the fraction of \(\mathrm{I}_{2}\) dissociated) was \(0.0252 .\) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the reaction at this temperature.

At room temperature, solid iodine is in equilibrium with its vapor through sublimation and deposition. Describe how you would use radioactive iodine, in either solid or vapor form, to show that there is a dynamic equilibrium between these two phases.

Consider the following equilibrium process: \(\mathrm{PCl}_{5}(g) \rightleftarrows \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \quad \Delta H^{\circ}=92.5 \mathrm{~kJ} / \mathrm{mol}\) Predict the direction of the shift in equilibrium when (a) the temperature is raised, (b) more chlorine gas is added to the reaction mixture, (c) some \(\mathrm{PCl}_{3}\) is removed from the mixture, (d) the pressure on the gases is increased, (e) a catalyst is added to the reaction mixture.

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