/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 "Old-fashioned "smelling salts" ... [FREE SOLUTION] | 91Ó°ÊÓ

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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

Le Châtelier's principle is stated (Section \(12-7\) ) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease, and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}} \approx 1 \times 10^{-31}\) for the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$. a. Calculate the concentration of NO, in molecules/cm \(^{3}\) that can exist in equilibrium in air at \(25^{\circ} \mathrm{C} .\) In air, \(P_{\mathrm{N}_{2}}=0.8 \mathrm{atm}\) and \(P_{\mathrm{O}_{2}}=0.2 \mathrm{atm}\). b. Typical concentrations of NO in relatively pristine environments range from \(10^{8}\) to \(10^{10}\) molecules/cm \(^{3}\). Why is there a discrepancy between these values and your answer to part a?

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of 0.30 atm, were allowed to come to equilibrium at this temperature?

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.

At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}:\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 mol/L. Calculate the value of \(K\) for this reaction.

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