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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: $$ \begin{aligned} \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g) &+3 \mathrm{H}_{2}(g) \\ \Delta H^{\circ} &=206 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\) in the presence of air, to convert the remaining methane to hydrogen: \(\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)\) $$ \Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol} $$ (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) For the primary stage, high temperature and high pressure favor product formation; for the secondary stage, high pressure favors product formation. (b) \( K_p = 3.03 \times 10^6 \). The equilibrium pressures can be found by substituting these values into the equilibrium constant expression and solving for x.

Step by step solution

01

Predict the conditions favoring product formation

To answer this, it must be noted that according to Le Chatelier's Principle, an increase in temperature will favor the endothermic direction and an increase in pressure will favor the side with fewer moles of gas. (a) For the primary stage, the reaction is endothermic (\( \Delta H^{\circ} = 206 \)kJ/mol) and has fewer moles of gas on the product side (4 vs 2). So higher temperatures and higher pressures would favor product formation. (b) For the secondary stage, there's no given information about the nature of the reaction (endothermic or exothermic). However, the number of moles of gas decreases from reactants to products (1.5 to 3), thus higher pressures would favor the product side.
02

Calculate Kp for the primary stage

To calculate the pressure equilibrium constant (Kp) from the concentration equilibrium constant (Kc), the ideal gas law can be used in combination with the relationship between Kp and Kc, \(K_p = K_c(RT)^{\Delta n}\), where R is the ideal gas constant (0.0821 L atm / mol K), T is the temperature in Kelvin (800°C + 273.15 = 1073.15 K), and \( \Delta n \) is the change in moles of gas in the reaction (3+1-1-1 = 2). Therefore, \( K_p = 18 \times (0.0821 \times 1073.15)^2 = 3.03 \times 10^6 \).
03

Find the pressures of all gases at equilibrium

The initial pressures of methane (P\_CH4) and water (P\_H2O) are given as 15 atm each, and that of CO (P\_CO) and H2 (P\_H2) can be assumed to be zero. At equilibrium, let x be the pressure change of methane and steam. Because methane and steam each produces 1 mole of CO and 3 moles of H2, the pressures at equilibrium will be P\_CH4 = 15 - x, P\_CO = x, P\_H2O = 15 - x, P\_H2 = 3x. Now plug these into the equation of Kp, \( K_p = \frac{P\_CO P\_H2^3}{P\_CH4 P\_H2O} \), and solve for x to get the equilibrium pressures.

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Key Concepts

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

Le Chatelier's Principle
When studying chemical reactions and how they reach equilibrium, one fundamental concept is Le Chatelier's Principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, such as temperature, pressure, or concentration, the system adjusts to counteract the imposed change and a new equilibrium is established.

This principle can be applied to industrial processes such as the steam-reforming process for producing hydrogen. An understanding of Le Chatelier's Principle helps us predict how changes in conditions will affect the production yield of hydrogen. For instance, since the primary stage in the steam-reforming process is endothermic, increasing the temperature will shift the equilibrium to the right, producing more hydrogen and carbon monoxide according to the principle. Similarly, increasing the pressure will shift the equilibrium toward the side with fewer moles of gas, according to our principle.
Equilibrium Constant (Kc and Kp)
The equilibrium constant is a number that expresses the ratio of the concentrations (Kc) or the partial pressures (Kp) of the products to reactants at equilibrium for a particular reaction at a constant temperature. Kc is used when concentrations are measured in molarity, and Kp is used when measuring gaseous reactants and products in terms of their partial pressures.

For the primary stage of the steam-reforming process, the equilibrium constant Kc is given as 18 at 800°C. To convert this to the pressure equilibrium constant (Kp), we use the formula \(K_p = K_c(RT)^{\Delta n}\), where R is the ideal gas constant, T is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas during the reaction. The ability to calculate and understand these constants is crucial in predicting how the reaction will respond to changes in conditions and in optimizing industrial hydrogen production.
Industrial Hydrogen Production
Hydrogen is a key component in various industrial processes, and about 75% of it is produced through the steam-reforming process. This process involves the reaction of methane with steam under high temperatures and pressures to yield hydrogen and carbon monoxide.

In the primary stage of steam reforming, optimizing temperature and pressure conditions is crucial to maximize hydrogen production. As the reaction is endothermic, industrial reactors are operated at high temperatures to shift the equilibrium toward hydrogen production, which aligns with Le Chatelier's Principle. The secondary stage, though not described as endothermic or exothermic in our exercise, also considers the molar volume of gases and uses a higher pressure to push the equilibrium towards the desired hydrogen-rich products.
Endothermic Reactions
Endothermic reactions are those that absorb heat from their surroundings. This property is a core aspect in the steam-reforming process of hydrogen production. During the primary stage, the reaction absorbs 206 kJ of energy per mole of methane, demonstrating its endothermic nature.

According to Le Chatelier's Principle, the system will react to an increase in temperature by shifting the equilibrium to absorb the excess heat, thus favoring the endothermic reaction direction. Understanding this concept is crucial for operators in the hydrogen production industry as it directly impacts the efficiency of the reaction and, ultimately, the yield of hydrogen gas. This is why the primary reforming stage operates at high temperatures such as 800°C, enhancing the reaction conditions for hydrogen production.

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Most popular questions from this chapter

Iodine is sparingly soluble in water but much more so in carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\). The equilibrium constant, also called the partition coefficient, for the distribution of \(\mathrm{I}_{2}\) between these two phases $$ \mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) $$ is 83 at \(20^{\circ} \mathrm{C}\). (a) A student adds \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\) to \(0.200 \mathrm{~L}\) of an aqueous solution containing \(0.032 \mathrm{~g}\) \(\mathrm{I}_{2} .\) The mixture is shaken and the two phases are then allowed to separate. Calculate the fraction of \(\mathrm{I}_{2}\) remaining in the aqueous phase. (b) The student now repeats the extraction of \(I_{2}\) with another 0.030 \(\mathrm{L}\) of \(\mathrm{CCl}_{4}\). Calculate the fraction of the \(\mathrm{I}_{2}\) from the original solution that remains in the aqueous phase. (c) Compare the result in (b) with a single extraction using \(0.060 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\). Comment on the difference.

Consider the dissociation of iodine: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ A \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is heated at \(1200^{\circ} \mathrm{C}\) in a 500 -mL flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in problem \(15.65(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.

A sealed glass bulb contains a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases. When the bulb is heated from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\), what happens to these properties of the gases: (a) color, (b) pressure, (c) average molar mass, (d) degree of dissociation (from \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) ), (e) density? Assume that volume remains constant. (Hint: \(\mathrm{NO}_{2}\) is a brown gas; \(\mathrm{N}_{2} \mathrm{O}_{4}\) is colorless.)

A sample of pure \(\mathrm{NO}_{2}\) gas heated to \(1000 \mathrm{~K}\) decomposes: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{P}\) is \(158 .\) Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is 0.25 atm at equilibrium. Calculate the pressure of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in the mixture.

What effect does an increase in pressure have on each of these systems at equilibrium? (a) \(\mathrm{A}(s) \rightleftharpoons 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftharpoons \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftharpoons \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\) The temperature is kept constant. In each case, the reacting mixture is in a cylinder fitted with a movable piston.

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