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The decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) is endothermic. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ (a) Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. (b) If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a sealed flask in which this equilibrium exists, how is the equilibrium affected? (c) What if some additional \(\mathrm{NH}_{3}\) is placed in a sealed flask containing an equilibrium mixture? (d) What will happen to the partial pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{~S}\) is removed from the flask?

Short Answer

Expert verified
(a) Increase produces more gases. (b) Equilibrium shifts right. (c) Equilibrium shifts left. (d) Partial pressure of \( \mathrm{NH}_{3} \) increases.

Step by step solution

01

Understand the Reaction

The given reaction is an endothermic decomposition of ammonium hydrogen sulfide, \( \mathrm{NH}_{4} \mathrm{HS} \), into ammonia (\( \mathrm{NH}_{3} \)) and hydrogen sulfide (\( \mathrm{H}_{2} \mathrm{S} \)) gases. This is represented as:\[\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g}) + \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\]The reaction is reversible and the system can reach an equilibrium.
02

Apply Le Chatelier’s Principle to Temperature Change

(a) If the temperature is increased, Le Chatelier's principle states that the equilibrium will shift to counteract the change. Because the reaction is endothermic, increasing temperature will cause the equilibrium to shift to the right, favoring the production of \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{~S} \) to absorb the added heat.
03

Effect of Adding More Reactant

(b) Adding more \( \mathrm{NH}_{4} \mathrm{HS} \) to the flask increases the concentration of the reactant, and according to Le Chatelier's principle, the equilibrium will shift to the right to consume the excess \( \mathrm{NH}_{4} \mathrm{HS} \), producing more \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{~S} \).
04

Effect of Adding a Product

(c) Adding more \( \mathrm{NH}_{3} \) shifts the equilibrium to the left in order to consume the added \( \mathrm{NH}_{3} \). This results in the formation of more \( \mathrm{NH}_{4} \mathrm{HS} \) and a decrease in \( \mathrm{H}_{2} \mathrm{~S} \).
05

Effect of Removing a Product

(d) Removing \( \mathrm{H}_{2} \mathrm{~S} \) causes the equilibrium to shift to the right to replace the removed\( \mathrm{H}_{2} \mathrm{~S} \). This offset will increase the partial pressure of \( \mathrm{NH}_{3} \) as well, as more of it will be produced.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a great tool to understand how chemical equilibria respond to changes. When a system at equilibrium is disturbed, this principle states that the system will adjust to minimize the disturbance and re-establish equilibrium.
  • If you add more reactants or remove products, the equilibrium will shift to favor the formation of products.
  • If you add more products or remove reactants, the equilibrium will shift to favor the formation of reactants.

For instance, when temperature changes in endothermic reactions, which require heat, increasing the temperature can be seen as adding a reactant (heat), which will shift the equilibrium toward the right, producing more products.

Another aspect is the response to pressure changes in gaseous reactions, where increasing pressure favors the side with fewer moles of gas. Overall, Le Chatelier's Principle helps predict the direction a reaction will shift under different conditions.
Endothermic Reactions
Endothermic reactions are processes that absorb energy from their surroundings, typically in the form of heat. They are characterized by having heat as a reactant. This means that endothermic reactions require an input of energy to occur.

Common everyday examples include photosynthesis and melting ice. In the decomposition of ammonium hydrogen sulfide (\( \mathrm{NH}_{4} \mathrm{HS} \)), the reaction is endothermic, so when the temperature is raised, it provides more heat (energy) to the system.
  • Higher temperatures supply more energy, driving the reaction towards producing more products, as dictated by Le Chatelier's Principle.
  • This makes the decomposition reaction more favorable, resulting in more ammonia (\( \mathrm{NH}_{3} \)) and hydrogen sulfide (\( \mathrm{H}_{2} \mathrm{S} \)).

Understanding endothermic reactions is crucial for predicting how altering temperature can influence chemical equilibrium states.
Partial Pressure
Partial pressure is a concept used to describe the pressure exerted by a single gas component in a mixture of gases. Each gas in a mixture exerts pressure independently of the others, and the total pressure is the sum of the partial pressures.
The partial pressure of each gas is proportional to its mole fraction in the gas mixture and is vital in expressing equilibrium constants for reactions involving gases.

In a sealed environment, where the equilibrium of the decomposition of ammonium hydrogen sulfide is established, changes in the amounts of gases like ammonia (\( \mathrm{NH}_{3} \)) can affect the total or partial pressures. For example, removing hydrogen sulfide (\( \mathrm{H}_{2} \mathrm{S} \)) shifts the equilibrium to produce more products (including ammonia), thus increasing its partial pressure.
  • This follows Le Chatelier's Principle, as the system adjusts to counteract the loss of \( \mathrm{H}_{2} \mathrm{S} \).

By understanding partial pressures, you can predict how changes in conditions affect the gaseous components of a chemical reaction.

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Most popular questions from this chapter

Write the equilibrium constant expression for each of these heterogeneous systems. (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+\mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{C}(\mathrm{s})+2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{~N}_{2}(\mathrm{~g})\) (c) \(\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

The formation of hydrogen sulfide from the elements is exothermic. $$ \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{8} \mathrm{~S}_{8}(\mathrm{~s}) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=-20.6 \mathrm{~kJ} / \mathrm{mol} $$ Predict the effect of each of these changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each change is made in a constant-volume system. (a) Adding more sulfur (b) Adding more \(\mathrm{H}_{2}\) (c) Raising the temperature

These amounts of \(\mathrm{CO}(\mathrm{g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{~g})\), and \(\mathrm{H}_{2}(\mathrm{~g})\) are introduced into a \(10.00-\mathrm{L}\) flask. The flask is sealed and heated to a very high temperature. \begin{tabular}{lcccc} \hline & \(n_{\mathrm{CO}}(\mathrm{mol})\) & \(n_{\mathrm{H}_{2} \mathrm{O}}(\mathrm{mol})\) & \(n_{\mathrm{CO}_{2}}(\mathrm{~mol})\) & \(n_{\mathrm{H}_{2}}(\mathrm{~mol})\) \\ \hline Case a & 1.0 & 0.10 & 0.10 & 0.10 \\ Case b & \(10 .\) & 1.0 & 1.0 & 1.0 \\ Case c & \(10 .\) & \(10 .\) & 1.0 & 1.0 \\ Case d & 5.62 & 0.381 & 1.75 & 1.75 \\ \hline \end{tabular} The equilibrium constant for the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) has the value \(K_{\mathrm{c}}=4.00\) at this temperature. For which cases will the concentration of CO increase as equilibrium is attained, and in which cases will the concentration of CO decrease?

Assume you place \(0.010 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) in a sealed \(2.0-\mathrm{L}\) flask at \(50 .{ }^{\circ} \mathrm{C}\). After the system reaches equilibrium, \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.00090 \mathrm{M} .\) Calculate the value of \(K_{\mathrm{c}}\) for this reaction. $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$

Many common nonmetallic elements exist as diatomic molecules at room temperature. When these elements are heated to \(1500 . \mathrm{K},\) the molecules break apart into atoms. A general equation for this type of reaction is \(\mathrm{E}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{E}(\mathrm{g})\) where E stands for an atom of each element. Equilibrium constants for dissociation of these molecules at \(1500 . \mathrm{K}\) are \begin{tabular}{lcll} \hline Species & \(\kappa_{c}\) & Species & \multicolumn{1}{c} {\(K_{c}\)} \\ \hline \(\mathrm{Br}_{2}\) & \(8.9 \times 10^{-2}\) & \(\mathrm{H}_{2}\) & \(3.1 \times 10^{-10}\) \\ \(\mathrm{Cl}_{2}\) & \(3.4 \times 10^{-3}\) & \(\mathrm{~N}_{2}\) & \(1 \times 10^{-27}\) \\ \(\mathrm{~F}_{2}\) & 7.4 & \(\mathrm{O}_{2}\) & \(1.6 \times 10^{-11}\) \\ \hline \end{tabular} (a) Assume that \(1.00 \mathrm{~mol}\) of each diatomic molecule is placed in a separate \(1.0-\mathrm{L}\) container, sealed, and heated to \(1500 . \mathrm{K}\). Calculate the equilibrium concentration of the atomic form of each element at \(1500 . \mathrm{K}\). (b) From these results, predict which of the diatomic elements has the lowest bond dissociation energy, and compare your results with thermochemical calculations and with Lewis structures.

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