Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24{ }^{\circ} \mathrm{C}\). The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) taken together) is 0.614 atm. What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The equilibrium constant (Kp) for this reaction at \(24{ }^{\circ} C\) is approximately 0.094.

Step by step solution

01

Write the equilibrium expression

For the given reaction: \(NH_{4} SH(s) \rightleftharpoons NH_{3}(g) + H_{2} S(g)\) We need the equilibrium expression using Kp: \(K_p = \frac{P_{NH_3} \times P_{H_2S}}{1}\) Note that we don't include the solid NH4SH in the expression.
02

Identifying the initial conditions

Initially, there are no gaseous reaction products: Initial pressure of NH3 = 0 Initial pressure of H2S = 0
03

Deduce the changes in pressure

Let x be the pressure change of NH3 and H2S during the reaction: NH3 pressure change: +x H2S pressure change: +x This means, the pressure of NH3 and H2S should be equal at the equilibrium.
04

Find the partial pressures at equilibrium

Since the total pressure at equilibrium is 0.614 atm, and the change in pressure for both gases is equal (both are x), we can write: \(x + x = 0.614 \space atm\) Solving this gives us x: \(2x = 0.614 \space atm\) \(x = 0.307 \space atm\) Thus, at equilibrium: \(P_{NH_3} = 0.307 \space atm\) \(P_{H_2S} = 0.307 \space atm\)
05

Calculate the Kp value

Now we can calculate the Kp value using the partial pressures at equilibrium: \(K_p = \frac{P_{NH_3} \times P_{H_2S}}{1}\) \(K_p = \frac{0.307 \times 0.307}{1}\) \(K_p = 0.094169 \) So, the equilibrium constant (Kp) for this reaction at \(24{ }^{\circ} C\) is approximately 0.094.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial for interpreting the behavior of reactions, like the dissociation of solid ammonium hydrosulfide into ammonia and hydrogen sulfide gases. Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, the concentrations of reactants and products remain constant over time, but not necessarily equal.

When solid NH_{4}SH is introduced into a closed container, it will dissociate until it reaches the point where the rate at which it forms gas is equal to the rate at which the gas recombines to form the solid. At this point, the system is said to be in equilibrium. It's essential to note that these dynamic processes are happening simultaneously and are not static.
Partial Pressure
When dealing with gases, the concept of partial pressure becomes important in the context of chemical equilibrium. Partial pressure is defined as the pressure contributed by a single gas in a mixture of gases. It's directly proportional to its mole fraction in the mixture and to the total pressure.

In the given exercise, ammonia (NH_{3}) and hydrogen sulfide (H_{2}S) are produced in equal amounts and together they create a total pressure of 0.614 atm at equilibrium. This implies that the partial pressure of each gas is half of the total pressure because in the reaction for every mole of NH_{4}SH that dissociates, one mole of NH_{3} and one mole of H_{2}S are produced. Thus, their partial pressures must be equal, as exemplified by the calculation in step 4 of the provided solution.
Le Chatelier's Principle
Le Chatelier's principle is an important concept to predict the effect of a change in conditions on a chemical equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. This principle applies to changes in concentration, pressure, volume, and temperature.

For example, if the pressure over the equilibrium mixture of the gases were increased, Le Chatelier's principle predicts that the system would respond by favoring the reaction that produces fewer gas molecules. In the case of the dissociation of solid ammonium hydrosulfide, since both the forward and the reverse reactions produce and consume gas molecules in a 1:1 ratio, a pressure change does not shift the position of equilibrium. However, if one gas were removed or added to the system, the equilibrium would indeed shift to restore the balance.

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Most popular questions from this chapter

At \(2000^{\circ} \mathrm{C}\) the equilibrium constant for the reaction $$2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ is \(K_{c}=2.4 \times 10^{3}\). If the initial concentration of \(\mathrm{NO}\) is \(0.175 \mathrm{M}\) what are the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2},\) and \(\mathrm{O}_{2} ?\)

Consider the hypothetical reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g) .\) A flask is charged with 0.75 atm of pure \(A\), after which it is allowed to reach equilibrium at \(0^{\circ} \mathrm{C}\). At equilibrium the partial pressure of \(\mathrm{A}\) is 0.36 atm. (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?\) (c) What could we do to maximize the yield of \(\mathrm{B}\) ?

Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$\begin{aligned} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) & \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c} &=2.0 \\ 2 \mathrm{NO}(g) & \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c} &=2.1 \times 10^{30} \end{aligned}$$

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) .\) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium- constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

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