Problem 70 - Chemistry

Question

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?

Step-by-Step Solution

Answer

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 1: 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: \( K_p = P_{NH_3} \times P_{H_2S} \).

Step 2: 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: \(K_p = \frac{n_{NH_3} \times R \times T}{V} \cdot \frac{n_{H_2S} \times R \times T}{V} \)

Step 3: 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: \(x^2 = K_p\frac{V^2}{R^2T^2}\)

Step 4: 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.

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant (denoted as \(K_p\) for gaseous reactions) reflects the ratio of the products to reactants at equilibrium. It specifically measures the balance of partial pressures of gases in a reaction that has reached a stable state, where neither products nor reactants are favored. For the reaction \( \mathrm{NH}_{4} \mathrm{SH}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \), the equilibrium constant expression depends only on the gases formed, \( \mathrm{NH}_3 \) and \( \mathrm{H}_2\mathrm{S}\), since solids do not appear in equilibrium expressions. Thus, \( K_p = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2S} \), showing the product of their partial pressures. This expression helps predict how much reactant and product you'll have at equilibrium. The equilibrium constant value of \( K_p = 0.10 \) provides information about the reaction's balance under set conditions.

Ideal Gas Law

The ideal gas law is an essential formula in chemistry represented by \( PV = nRT \), where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the moles of the gas.
• \(R\) is the ideal gas constant (0.0821 L · atm/mol · K).
• \(T\) is the temperature in Kelvin.

This law provides a link between the physical properties of gases; specifically, it helps convert between the number of moles and the gas's partial pressure in a given volume and temperature. In our example, for both \( \mathrm{NH}_3 \) and \( \mathrm{H}_2\mathrm{S}\), the partial pressure can be derived using \( P = \frac{nRT}{V} \). By substituting this into the \( K_p \) expression, we can relate the change in moles of reactants or products in a reaction to measurable pressures, thus aiding in calculations needed for equilibrium problems.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It uses balanced chemical equations to understand the quantitative relationships within a reaction. For the decomposition reaction \( \mathrm{NH}_{4} \mathrm{SH}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \), stoichiometry tells us that the decomposition of one mole of \( \mathrm{NH}_{4} \mathrm{SH} \) yields one mole of \( \mathrm{NH}_3 \) and one mole of \( \mathrm{H}_2\mathrm{S}\). If \( x \) moles of \( \mathrm{NH}_{4} \mathrm{SH} \) decompose, then \( x \) moles of each gas will form. During calculations, stoichiometry is necessary to simplify \( K_p = \left(\frac{xRT}{V}\right)^2 \) to an expression where \( x \) (the amount of \( \mathrm{NH}_{4} \mathrm{SH} \) decomposed) becomes calculable by relating pressures and moles at equilibrium.

Partial Pressure

Partial pressure is the pressure that a single gas in a mixture would exert if it were alone in a container. It is crucial in determining how gases interact in a mixture, as seen in equilibrium reactions. Using Dalton's Law, for a particular gas \( \mathrm{NH}_3 \) or \( \mathrm{H}_2\mathrm{S}\) in our reaction, the partial pressures can be calculated as products of equilibrium processes. Given the gas's moles found by stoichiometry and converted via the ideal gas law, the partial pressures become: \( P_{NH_3} = \frac{n_{NH_3}RT}{V} \) and \( P_{H_2S} = \frac{n_{H_2S}RT}{V} \). These values are inserted into the equilibrium expression \( K_p = P_{NH_3} \times P_{H_2S} \). Understanding this allows one to manipulate the pressures in setups like our 10-L container, ensuring they align with the conditions of equilibrium—reflected in the given \( K_p \).