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List four ways in which the yield of ammonia in the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \Delta H^{\circ}<0 $$ can be improved for a given amount of \(\mathrm{H}_{2}\). Explain the principle behind each way.

Short Answer

Expert verified
Lower temperature, higher pressure, use a catalyst, and continuously remove ammonia.

Step by step solution

01

Understanding the Reaction

The given reaction is \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \) and is exothermic (\( \Delta H^{\circ} < 0 \)). This means that heat is released during the formation of ammonia.
02

Le Chatelier's Principle - Temperature

According to Le Chatelier's Principle, since the reaction is exothermic, lowering the temperature shifts the equilibrium towards the formation of more ammonia. This improves the yield of ammonia.
03

Le Chatelier's Principle - Pressure

The reaction shows a decrease in the number of gas molecules ons shifting from 4 moles (1 mole \( \mathrm{N}_{2} \) and 3 moles \( \mathrm{H}_{2} \)) to 2 moles (\( 2 \mathrm{NH}_{3} \)). Increasing the pressure shifts the equilibrium toward the formation of ammonia, thus increasing the yield.
04

Adding a Catalyst

While a catalyst does not shift the position of equilibrium, it speeds up both the forward and reverse reactions equally. Therefore, a catalyst like iron can be used to reach equilibrium faster, thus improving the production rate of ammonia.
05

Removing Ammonia

Removing ammonia from the reaction mixture will shift the equilibrium to the right, according to Le Chatelier’s principle. This will favour the formation of more ammonia, and improve the yield.

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Key Concepts

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

Equilibrium in Chemical Reactions
Chemical reactions can reach a state of balance known as equilibrium, where the rates of the forward and reverse reactions are equal. This means the concentrations of reactants and products remain constant over time, although they are not necessarily equal.
At equilibrium, even though the reactions continue to occur, there is no net change in the amounts of reactants and products.
  • Equilibrium is dynamic, not static.
  • It can be reached from any direction — starting from reactants, products, or both.
Le Chatelier's Principle helps predict how changes in temperature, pressure, and concentration affect the position of equilibrium.
When a system at equilibrium experiences a change, it will adjust to counteract the change and restore equilibrium.
Exothermic Reactions
Exothermic reactions are those that release energy, typically in the form of heat, as they proceed. In the reaction of nitrogen and hydrogen to form ammonia, energy is released, hence the negative sign in the enthalpy change (\( \Delta H < 0 \)).
This means the equilibrium can be affected by temperature changes. Lowering the temperature of an exothermic reaction shifts the equilibrium toward the products, increasing the yield.
Because the reaction releases heat, lowering the temperature counterbalances this release, promoting the formation of more product.
  • Exothermic processes release heat from the system to the surroundings.
  • This type of reaction often feels warm or hot to the touch.
  • Lowers the system's internal energy.
In industrial settings, controlling temperature is crucial to maximize the production of desired products.
Catalysts in Chemical Reactions
Catalysts are substances that speed up chemical reactions without being consumed in the process. They work by providing an alternative pathway with a lower activation energy for the reaction. In the ammonia synthesis reaction, using a catalyst like iron allows the reaction to reach equilibrium faster.
While catalysts accelerate reaching equilibrium, they do not change the position of the equilibrium itself.
This means they help achieve the same end product more quickly by speeding up both forward and reverse reactions equally.
  • Catalysts lower the activation energy required for a reaction.
  • They remain unchanged after the reaction.
  • Widely used in industry to increase efficiency and reduce costs.
Using a catalyst is a key factor in industrial chemical processes, improving the rate at which products are generated without altering the balance of substances at equilibrium.

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Most popular questions from this chapter

Iodine, \(\mathrm{I}_{2}\), is a blue-black solid, but it easily vaporizes to give a violet vapor. At high temperatures, this molecular substance dissociates to atoms: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ An absent-minded professor measured equilibrium constants for this dissociation at two temperatures, \(700^{\circ} \mathrm{C}\) and \(800^{\circ} \mathrm{C}\). He obtained the \(K_{p}\) values 0.01106 and \(0.001745,\) but he couldn't remember which value went with what temperature. Can you please help him assign temperatures to the equilibrium constants? State your argument carefully.

An equilibrium mixture of \(\mathrm{SO}_{3}, \mathrm{SO}_{2},\) and \(\mathrm{O}_{2}\) at $$ 727^{\circ} \mathrm{C} \text { is } 0.0160 \mathrm{M} \mathrm{SO}_{3}, 0.0056 \mathrm{M} \mathrm{SO}_{2} \text { , and } 0.0021 \mathrm{M} \mathrm{O}_{2} $$ What is the value of \(K_{c}\) for the following reaction? $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$

In the contact process, sulfuric acid is manufactured by first oxidizing \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\), which is then reacted with water. The reaction of \(\mathrm{SO}_{2}\) with \(\mathrm{O}_{2}\) is $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ A 2.000-L flask was filled with \(0.0400 \mathrm{~mol} \mathrm{SO}_{2}\) and \(0.0200 \mathrm{~mol} \mathrm{O}_{2}\). At equilibrium at \(900 \mathrm{~K},\) the flask contained \(0.0296 \mathrm{~mol} \mathrm{SO}_{3}\). How many moles of each substance were in the flask at equilibrium?

The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ equals 49 at \(230^{\circ} \mathrm{C}\). If \(0.400 \mathrm{~mol}\) each of phosphorus trichloride and chlorine are added to a 4.0-L reaction vessel, what is the equilibrium composition of the mixture at \(230^{\circ} \mathrm{C} ?\)

Ammonium hydrogen sulfide, \(\mathrm{NH}_{4} \mathrm{HS}\), is unstable at room temperature and decomposes: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ You have placed some solid ammonium hydrogen sulfide in a closed flask. Which of the following would produce less hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S},\) which is a poisonous gas? a. Removing some \(\mathrm{NH}_{3}\) from the flask b. Adding some \(\mathrm{NH}_{3}\) to the flask c. Removing some of the \(\mathrm{NH}_{4} \mathrm{HS}\) d. Increasing the pressure in the flask by adding helium gas Explain each of your answers.

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