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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.
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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

At a particular temperature, a \(3.0-\mathrm{L}\) flask contains 2.4 moles of \(\mathrm{Cl}_{2}, 1.0\) mole of \(\mathrm{NOCl}\) , and \(4.5 \times 10^{-3}\) mole of NO. Calculate \(K\) at this temperature for the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\) . After equilibrium is reached the total pressure is 1.5 atm and 16\(\%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g) .\) a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C} .\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g) .\) c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{atm} ) ?\)

For the reaction $$\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s)$$ \(K=400\) . at \(35.0^{\circ} \mathrm{C} .\) If 2.00 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and \(\mathrm{NH}_{4} \mathrm{HS}\) are placed in a \(5.00-\mathrm{L}\) vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium?

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The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.
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