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Chapter 13: Problem 72

Solid \(\mathrm{NH}_{4} \mathrm{HS}\) decomposes by the following endothermic process: $$\mathrm{NH}_{4} \mathrm{HS}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ a. What effect will adding more \(\mathrm{NH}_{3}(g)\) have on the equilibrium? b. What effect will adding more \(\mathrm{NH}_{4} \mathrm{HS}(s)\) have on the equilibrium? c. What effect will increasing the volume of the container have on the equilibrium? d. What effect will decreasing the temperature have on the equilibrium?

Short Answer

Expert verified
a) Adding more NH3(g) will shift the equilibrium towards the reactants (left), favoring the formation of NH4HS(s). b) Adding more NH4HS(s) will have no effect on the equilibrium position. c) Increasing the volume of the container will shift the equilibrium to the right, favoring the formation of NH3(g) and H2S(g). d) Decreasing the temperature will shift the equilibrium to the right, favoring the formation of NH3(g) and H2S(g).

Step by step solution

01

Understand Le Châtelier's Principle

Le Châtelier's principle states that when a system at equilibrium is subjected to a change in concentration, temperature, or volume, the equilibrium will shift to counteract the change and re-establish equilibrium. Based on this principle, we can predict how the given changes will affect the equilibrium.
02

Adding more NH3 (g)

According to Le Châtelier's principle, adding more NH3 will cause the equilibrium to shift in the direction that decreases its concentration. In this case, it will shift to the left, favoring the formation of NH4HS(s) and consuming the excess NH3. So, adding more ammonia will shift the equilibrium towards the reactants.
03

Adding more NH4HS (s)

Since NH4HS is a solid, its concentration does not affect the equilibrium. Therefore, adding more NH4HS will have no effect on the equilibrium position.
04

Increasing the volume of the container

Increasing the volume of the container will cause the pressure to decrease. According to Le Châtelier's principle, the equilibrium will shift to the direction with more moles of gas to counteract the pressure change. In this reaction, there are more moles of gas on the right side, so the equilibrium will shift to the right. This favors the formation of NH3(g) and H2S(g).
05

Decreasing the Temperature

The reaction is endothermic, which means it absorbs heat as it proceeds in the forward direction. Decreasing the temperature means removing heat from the system. In this case, the equilibrium will shift in the direction that supplies the heat, which is the forward direction. Therefore, the equilibrium will shift to the right, favoring the formation of NH3(g) and H2S(g).

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Most popular questions from this chapter

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?

The reaction $$\mathrm{NH}_{4} \mathrm{SH}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ has \(K_{\mathrm{p}}=0.10\) at \(27^{\circ} \mathrm{C} .\) What is the minimum amount of \(\mathrm{NH}_{4} \mathrm{SH}\) that must be present for this reaction to be at equilibrium in a 10.0 \(\mathrm{-L}\) container?

Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$2 \mathrm{A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g)$$ what is the value of \(K\) at the same temperature for the reaction $$\mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{B}(g)$$ What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with 1.50 atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D},\) what is the mole fraction of \(\mathrm{B}\) once equilibrium is reached?

At a particular temperature, \(K=3.75\) for the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$ If all four gases had initial concentrations of \(0.800 M,\) calculate the equilibrium concentrations of the gases.

The reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\) . If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of \(\mathrm{NO}\) at equilibrium.
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