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"Old-fashioned "smelling salts" consist of ammonium carbon=ate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammomium 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 due to the shift in equilibrium towards the right side, leading to an increased production of ammonia gas \(\mathrm{NH}_{3}(g)\).

Step by step solution

01

Understand the reaction

As given, the reaction for the decomposition of ammonium carbonate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s)\) is: \[ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \] The words "smelling salts" imply that the smell of ammonia (\(\mathrm{NH}_{3}\)) is being released, and we know this is an endothermic reaction, meaning it absorbs heat from the surroundings.
02

Le Chatelier's Principle

Le Chatelier's Principle states that when a system at equilibrium experiences a change in conditions, the equilibrium will shift to counteract the change and minimize its effect. In other words, if we increase the temperature, the reaction will try to counteract that change by shifting towards the side that absorbs heat - the endothermic side.
03

Impact of increasing temperature

Since the decomposition reaction is endothermic (absorbing heat), when the temperature is increased, the equilibrium will shift in the direction that absorbs this heat to counteract the change. In this case, the equilibrium would shift towards the right side, producing more ammonia gas \(\mathrm{NH}_{3}(g)\), carbon dioxide gas \(\mathrm{CO}_{2}(g)\), and water gas \(\mathrm{H}_{2} \mathrm{O}(g)\).
04

Conclusion

As the temperature increases, the smell of ammonia will increase due to the shift in equilibrium towards the right side, leading to an increased production of ammonia gas \(\mathrm{NH}_{3}(g)\).

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Key Concepts

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

Le Chatelier's Principle
Understanding how a chemical system responds to changes is essential for predicting the outcome of reactions. Le Chatelier's Principle is a fundamental concept that helps us comprehend the behavior of a system in chemical equilibrium under stress. It posits that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

This principle can be applied in various scenarios, such as changes in concentration, pressure, and particularly temperature. For instance, in the decomposition of ammonium carbonate, increasing the temperature causes the equilibrium to shift in favor of the endothermic reaction—where heat is absorbed—leading to the formation of more ammonia, carbon dioxide, and water vapor.

Application to Everyday Chemical Reactions

In everyday chemical reactions, like the use of smelling salts that release ammonia, Le Chatelier's Principle allows us to predict that raising the temperature will intensify the smell due to increased production of ammonia gas, a key component in smelling salts.
Endothermic Reactions
When a chemical reaction requires heat to proceed and absorbs energy from its surroundings, it is termed endothermic. Unlike exothermic reactions that release heat, endothermic reactions result in a temperature drop in the surroundings.

The decomposition of ammonium carbonate is a classic example of an endothermic reaction. The system absorbs heat to break the bonds in solid ammonium carbonate, producing gases like ammonia and carbon dioxide in the process.

Identifying Endothermic Reactions

Endothermic reactions are identified through certain cues such as a cooling sensation or the requirement of continuous heat to maintain the reaction. In the laboratory, endothermic processes are often observed with a thermometer showing a temperature decrease in the immediate environment of the reaction vessel.
Chemical Equilibrium
Chemical equilibrium is the state in which the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products. It's crucial to recognize that equilibrium does not mean the reactants and products are present in equal amounts, but that their ratios are constant over time in a closed system.

In the context of ammonium carbonate decomposition, the system reaches equilibrium when the production of ammonia, carbon dioxide, and water vapor occurs at the same rate as their recombination to form ammonium carbonate.

Equilibrium in Dynamic Systems

Chemical equilibrium is dynamic—although it might appear static on a macroscopic level, reactants and products continuously interconvert at a molecular level. This concept is essential in industries such as pharmaceuticals and environmental engineering, where precise control over reaction conditions can affect the yield and efficiency of chemical processes.

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Most popular questions from this chapter

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned}2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\\P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\\P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm}\end{aligned}$$. Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$.For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCI, 0.10 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.10 mole of \(\mathrm{H}_{2} \mathrm{O}\) b. A \(2.0-\) L flask contains 0.084 mole of HOCI, 0.080 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.98 mole of \(\mathrm{H}_{2} \mathrm{O}\) c. A 3.0 -L flask contains 0.25 mole of HOCI, 0.0010 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.56 mole of \(\mathrm{H}_{2} \mathrm{O}\).

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.

Write the equilibrium expression \((K)\) for each of the following gas-phase reactions. a. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\). b. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\). c. \(\operatorname{SiH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)+2 \mathrm{H}_{2}(g)\). d. \(2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g)\).

Suppose the reaction system $$\mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional UO \(_{2}(s)\) is added to the system. b. The reaction is performed in a glass reaction vessel; \(\mathrm{HF}(g)\) attacks and reacts with glass. c. Water vapor is removed.

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