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Chapter 13: Problem 70

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?

Short Answer

Expert verified
The minimum amount of NH₄SH that must be present for the reaction to be at equilibrium in a 10.0-L container is approximately 0.000415 moles.

Step by step solution

01

Write the Equilibrium Constant Expression

For the given chemical reaction, \( NH_4SH \leftrightharpoons NH_3(g) + H_2S(g) \). The Kₚ expression will be: \( K_p = \frac{P_{NH_3} \times P_{H_2S}}{P^0}\), where Pₙₕ₃, Pₕ₂ₛ, and P^0 are the partial pressures of NH₃, H₂S, and NH₄SH, respectively. However, since NH₄SH is a solid, its partial pressure is not included in the expression, so we have: \( K_p = P_{NH_3} \times P_{H_2S} \).
02

Use the Ideal Gas Law to Convert Moles to Partial Pressure

Ideal gas law formula: \( PV = nRT \) Here, we are interested in partial pressures. So, we can write equations for NH₃ and H₂S in terms of moles: \(P_{NH_3} = \frac{n_{NH_3} \times R \times T}{V} \) and \(P_{H_2S} = \frac{n_{H_2S} \times R \times T}{V} \) Since the equilibrium Kₚ is given in terms of partial pressure, we can substitute in this relationship to get our equation in terms of moles: \(K_p = \frac{n_{NH_3} \times R \times T}{V} \cdot \frac{n_{H_2S} \times R \times T}{V} \)
03

Use Stoichiometry to Relate Moles to Initial Amount of NH₄SH, and Simplify the Equation

Let x moles of NH₄SH decompose, then x moles of NH₃ and H₂S will be formed at equilibrium. So, we can write: \(K_p = \frac{x \times R \times T}{V} \cdot \frac{x \times R \times T}{V}\) Simplifying the equation we get: \(x^2 = K_p\frac{V^2}{R^2T^2}\)
04

Solve for x (Number of Moles of NH₄SH)

Now we plug in the given values: \( K_p = 0.10 \), \( V = 10.0 \, L \), \( R = 0.0821\, L \cdot atm \cdot K^{-1} \cdot mol^{-1} \), and \( T = (27 + 273.15)\, K = 300.15\, K \). So, \( x^2= 0.10\frac{(10.0)^2}{(0.0821)^2(300.15)^2}\) Now, solve for x, which will give us the moles of NH₄SH needed: \( x = \sqrt{0.10 \cdot \frac{10^2}{0.0821^2 \cdot 300.15^2}} \) \( x = 0.000415 \, mol \) The minimum amount of NH₄SH that must be present for the reaction to be at equilibrium in a 10.0-L container is approximately 0.000415 moles.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C},\) gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that 12.5\(\%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is 0.900 atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 -L vessel, calculate the equilibrium concentrations of all species.

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.

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction $$3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ At equilibrium, the concentrations are $\left[\mathrm{H}_{2}\right]=5.0 M,\left[\mathrm{N}_{2}\right]=$ \(8.0 M,\) and \(\left[\mathrm{NH}_{3}\right]=4.0 \mathrm{M} .\) What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

Consider the following reaction at \(725^{\circ} \mathrm{C} :\) $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ At equilibrium, a \(4.50-\mathrm{L}\) container has 2.6 \(\mathrm{g}\) of carbon, \(\mathrm{CO}_{2}\) at a partial pressure of \(0.0020 \mathrm{atm},\) and a total pressure of 0.572 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(725^{\circ} \mathrm{C}\)
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