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Chapter 13: Problem 94

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

Short Answer

Expert verified
The synthesis of ammonia gas from nitrogen gas and hydrogen gas is optimized by: a. Running the reaction at an elevated temperature, which increases the rate of reaction due to increased molecular collisions. However, a balance must be maintained to avoid shifting the equilibrium unfavorably. b. Removing ammonia from the reaction mixture as it forms, which shifts the equilibrium position towards more ammonia production according to Le Chatelier's principle. c. Using a catalyst to increase the reaction rate without altering the equilibrium position, as it lowers the activation energy for successful collisions. d. Running the reaction at high pressure to shift the equilibrium towards ammonia formation, given that the product side has fewer moles of gas than the reactant side.

Step by step solution

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a. Running the reaction at an elevated temperature

Running the reaction at an elevated temperature increases the rate of reaction due to the collision theory. The average kinetic energy of the molecules increases with temperature, leading to more successful collisions and higher reaction rates. This will effectively increase the rate of conversion of the reactants (nitrogen and hydrogen) into the desired product (ammonia), thus maximizing ammonia yield. However, it's important to note that according to the Le Chatelier's principle, an increase in temperature for an exothermic reaction will shift the equilibrium to the reactants, decreasing ammonia yield. Therefore, controlling the temperature is essential to find an optimal balance between reaction rate and equilibrium position.
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b. Removing the ammonia from the reaction mixture as it forms

Removing the ammonia from the reaction mixture as it forms helps maximize the yield of ammonia by shifting the equilibrium position toward the products, according to Le Chatelier's principle. As the concentration of the product (ammonia) decreases by its constant removal, the equilibrium will shift to counteract the change and produce more ammonia, thus increasing the yield.
03

c. Using a catalyst

Using a catalyst in the synthesis of ammonia helps maximize the ammonia yield by increasing the rate of the reaction without altering the equilibrium position. Catalysts work by lowering the activation energy of the reaction, enabling more molecules to participate in successful collisions and react. This, in turn, significantly increases the reaction rate, and the faster formation of ammonia leads to a higher yield in a shorter time frame.
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d. Running the reaction at high pressure

Running the synthesis of ammonia at high pressure maximizes the yield by shifting the equilibrium according to Le Chatelier's principle. The reaction is as follows: \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) Increasing pressure favors the side with fewer moles of gas. In this case, there are 4 moles of gas on the reactant side and 2 moles of gas on the product side. When the reaction is run at high pressure, the equilibrium will shift toward the side with fewer moles, favoring the formation of ammonia, thus increasing its yield.

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Most popular questions from this chapter

Le Chatelier's principle is stated (Section 13.7\()\) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g) :\) $$\mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g)$$ A 2.00 -g sample of \(\mathrm{XY}\) (molar mass \(=165 \mathrm{g} / \mathrm{mol} )\) is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\) . The pressure is held constant at 0.967 \(\mathrm{atm} .\) As \(\mathrm{XY}\) begins to dissociate, the piston moves until 35.0 mole percent of the original \(\mathrm{XY}\) has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of \(25^{\circ} \mathrm{C}\) .

Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation: $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$ a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more D to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

The reaction to prepare methanol from carbon monoxide and hydrogen $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ is exothermic. If you wanted to use this reaction to produce methanol commercially, would high or low temperatures favor a maximum yield? Explain.

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ equilibrium constants \(\left(K_{\mathrm{p}} \text { values ) as a function of temperature }\right.\) are\(\begin{array}{ll}{300^{\circ} \mathrm{C},} & {4.34 \times 10^{-3}} \\ {500^{\circ} \mathrm{C},} & {1.45 \times 10^{-5}} \\\ {600^{\circ} \mathrm{C},} & {2.25 \times 10^{-6}}\end{array}\) Is the reaction exothermic or endothermic?
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