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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
To maximize the yield of ammonia during its synthesis from nitrogen gas and hydrogen gas, the following conditions should be met: a. Choose an optimal temperature to balance reaction rate and equilibrium shift due to the exothermic nature of the reaction. b. Remove ammonia from the reaction mixture as it forms, which causes the equilibrium to shift to the right and produce more ammonia according to Le Chatelier's principle. c. Use a catalyst, such as iron, to increase the reaction rate without affecting the equilibrium. d. Run the reaction at high pressure, which favors the side of the reaction with fewer moles of gas and shifts the equilibrium to the right, enhancing ammonia production.

Step by step solution

01

Understand the reaction and its characteristics

The synthesis of ammonia gas, known as the Haber process, involves the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia (NH3). The balanced equation for this reaction is: \(N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \) The reaction is exothermic (releases heat) and is reversible. Therefore, factors such as temperature, pressure, and catalysts will have an effect on the yield of ammonia.
02

Explain the effect of elevated temperature

Running the reaction at an elevated temperature causes it to be more kinetically favorable. This means that the reaction will take place at a faster rate, which is beneficial for the production of ammonia. However, since the reaction is exothermic, increasing the temperature will also shift the equilibrium to the left, favoring the reverse reaction (formation of N2 and H2). Therefore, to maximize the yield of ammonia, a compromise must be reached between these competing factors and the optimal temperature chosen.
03

Explain the effect of removing ammonia as it forms

Removing ammonia from the reaction mixture as it forms allows the reaction to continue moving forward and producing more ammonia. This is because, according to Le Chatelier's principle, when the product of a reaction is removed, the equilibrium will shift to the right to compensate, favoring the forward reaction and producing more ammonia.
04

Explain the effect of using a catalyst

A catalyst does not affect the equilibrium of a reaction, as it speeds up both the forward and reverse reactions equally. However, catalysts reduce the activation energy required for the reaction, resulting in an increase in reaction rate. In the case of ammonia synthesis, the use of a catalyst (e.g., iron) allows the reaction to proceed at a faster rate, thus producing more ammonia in a shorter amount of time.
05

Explain the effect of running the reaction at high pressure

Higher pressure favors the side of the reaction with fewer moles of gas. In the case of ammonia synthesis, since there are fewer moles of gas on the right side of the equation (2 moles of NH3) compared to the left side (1 mole of N2 and 3 moles of H2), increasing the pressure will shift the equilibrium to the right, favoring the production of ammonia. Running the reaction at high pressure will maximize the yield of ammonia. In conclusion, to maximize the yield of ammonia, an optimal temperature should be chosen, ammonia should be removed from the reaction mixture as it forms, a catalyst should be used, and the reaction should be run at high pressure.

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Most popular questions from this chapter

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{K}\) at an initial pressure of 1.00 atm, where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)$$. At equilibrium, the partial pressure of \(S_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{K}\)

Nitrogen gas \(\left(\mathrm{N}_{2}\right)\) reacts with hydrogen gas \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) At \(200^{\circ} \mathrm{C}\) in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 atm of hydrogen gas. At equilibrium,the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the \(K_{\mathrm{p}}\) value for this reaction.

For the reaction:$$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 \mathrm{M},\) calculate the equilibrium \(\mathrm{O}_{3}\) concentration.

At a particular temperature, \(K=4.0 \times 10^{-7}\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$,In an experiment, 1.0 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is placed in a 10.0 -L vessel. Calculate the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) when this reaction reaches equilibrium.

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$, Calculate the concentrations of all species at equilibrium for each of the following cases. a. \(1.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(2.0 \mathrm{g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a \(1.0-\mathrm{L}\) flask. b. 1.0 mole of pure HOCl is placed in a 2.0 -L flask.

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