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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.

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 路 atm/mol 路 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 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 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鈥攔eflected in the given \( K_p \).