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Old-fashioned "smelling salts" consist of ammonium carbonate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammonium carbonate $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

Short Answer

Expert verified
As the temperature increases, the smell of ammonia will increase. This is because the decomposition of ammonium carbonate is endothermic, and according to Le Chatelier's principle, the equilibrium will shift towards the production of more ammonia gas when the temperature is increased.

Step by step solution

01

Identify the endothermic reaction

The reaction for the decomposition of ammonium carbonate is endothermic, which means it absorbs heat: \[ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \]
02

Apply Le Chatelier's principle to the reaction

According to Le Chatelier's principle, if we increase the temperature, the equilibrium will shift in the direction that absorbs the heat. In this case, that is the direction of the endothermic reaction, which leads to the formation of ammonia (NH₃), carbon dioxide (CO₂), and water (H₂O) gases.
03

Determine the effect on the smell of ammonia

As the temperature increases, the equilibrium shifts in the direction of the endothermic reaction, producing more ammonia (NH₃), carbon dioxide (CO₂), and water (H₂O) gases. Since ammonia is responsible for the smell, an increase in its concentration directly corresponds to an increase in the smell.
04

Conclusion

As the temperature increases, the smell of ammonia will also increase due to a shift in the equilibrium position towards the production of more ammonia gas, according to Le Chatelier's principle.

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Key Concepts

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

Endothermic Reactions
An endothermic reaction is a type of chemical reaction that requires an input of energy in the form of heat to proceed. Simply put, it absorbs heat from its surroundings. When a reaction is endothermic, the products are at a higher energy level than the reactants, meaning energy has been taken in during the process.

In the case of the decomposition of ammonium carbonate, the reaction is endothermic. This means that the breaking down of ammonium carbonate into ammonia, carbon dioxide, and water vapor requires heat.

You can think of endothermic reactions like a baking process in the kitchen. Just as you need heat for your cake to rise, an endothermic reaction needs heat to proceed. In this context, adding heat to the system drives the reaction forward, enabling more of the reactants to transition into products.
Equilibrium Shift
In chemical reactions, equilibrium refers to the state where the rates of the forward and reverse reactions are equal, meaning the concentration of reactants and products remains constant over time. Le Chatelier's Principle helps us predict how a system at equilibrium will respond to changes in temperature, pressure, or concentration of components.

For endothermic reactions like the decomposition of ammonium carbonate, an increase in temperature causes the equilibrium to shift towards the direction where heat is absorbed, i.e., the endothermic direction. This is because the system looks to counteract the added heat by favoring the process that absorbs it.

In summary, when you increase the temperature in our reaction involving ammonium carbonate, the system compensates by shifting the equilibrium to produce more ammonia, carbon dioxide, and water vapor, enhancing the reaction's progression forward.
Ammonium Carbonate Decomposition
The decomposition of ammonium carbonate is an illustrative example of an endothermic reaction. The process involves a solid transforming into gas products, specifically ammonia, carbon dioxide, and water vapor.

The chemical equation representing this decomposition is:
\[\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \] This reaction is crucial for understanding how temperature impacts chemical systems. As the reaction absorbs heat, increasing the temperature leads to more ammonia gas formation, which is detectable due to its strong smell.

Old-fashioned smelling salts often contain ammonium carbonate. When subjected to higher temperatures, they decompose more readily, releasing ammonia vapor. This increase in ammonia results in a noticeable, stronger smell. Thus, adjusting the thermal conditions can directly influence the effectiveness and intensity of reactions involving ammonium carbonate.

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Most popular questions from this chapter

In a given experiment, \(5.2\) moles of pure NOCl was placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5} $$ a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that is present at equilibrium. b. Calculate the equilibrium concentrations for all species.

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is \(93.71 \%\) carbon by mass, and a \(0.256\) -mole sample of naphthalene has a mass of \(32.8 \mathrm{~g}\). What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene \((s) \rightleftharpoons\) naphthalene \((g)\) $$ K=4.29 \times 10^{-6}(\text { at } 298 \mathrm{~K}) $$ If \(3.00 \mathrm{~g}\) solid naphthalene is placed into an enclosed space with a volume of \(5.00 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\), what percentage of the naphthalene will have sublimed once equilibrium has been established?

For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .{ }^{\circ} \mathrm{C}\). $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ If a \(20.0-\mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\) is put into a \(10.0\) - \(\mathrm{L}\) container and heated to \(800 .{ }^{\circ} \mathrm{C}\), what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of \(K\) for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of \(K\) for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.

Consider the following exothermic reaction at equilibrium: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

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