The decomposition of ammonium hydrogen sulfide $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ is an endothermic process. A \(6.1589-\mathrm{g}\) sample of the solid is placed in an evacuated 4.000 - \(L\) vessel at exactly \(24^{\circ} \mathrm{C}\). After equilibrium has been established, the total pressure inside is \(0.709 \mathrm{~atm}\). Some solid \(\mathrm{NH}_{4} \mathrm{HS}\) remains in the vessel. (a) What is the \(K_{P}\) for the reaction? (b) What percentage of the solid has decomposed? (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel?

Step by step solution

01

Determining the molar mass of the compound

The first step is to determine the molar mass of ammonium hydrogen sulfide (NH4HS). The molar mass of N (Nitrogen) is \(14.01 g/mol\), of H (Hydrogen) is \(1.01 g/mol\), and of S (Sulfur) is \(32.07 g/mol\). Therefore, the molar mass of NH4HS = \(1*14.01 + 5*1.01 + 1*32.07 = 53.12 g/mol\).

02

Calculating the initial moles

The initial number of moles of NH4HS can be calculated by using the molar mass. Moles of NH4HS = mass / molar mass = \(6.1589 g / 53.12 g/mol = 0.1159 mol\).

03

Calculating number of moles at equilibrium

At equilibrium, the total pressure inside the vessel is 0.709 atm. This pressure is due to the formation of NH3 and H2S. Since the reaction is a 1:1:1 ratio, the pressure from each gas must also be 0.709 atm. Therefore, the moles for each gas at equilibrium can be calculated using the ideal gas law pv=nRT. n = pv/RT = \((0.709 atm * 4.000 L) / (0.0821 L.atm/mol.K * 297.15 K) = 0.1153 mol\). Since this is the sum of the moles of NH3 and H2S, the moles of NH4HS left is the initial moles - moles decomposed i.e., \(0.1159 mol - 0.1153 mol = 0.0006 mol\).

04

Calculate Kp

The equilibrium constant Kp for the reaction can be calculated using the expression for Kp = [NH3]*[H2S] = (Pressure of NH3 * Pressure of H2S). Since both gases have the same pressure, we get \(Kp = 0.709 * 0.709 = 0.5025\).

05

Determining the decomposition percentage

The percentage of the solid compound that has decomposed can be calculated by the ratio of the moles of gas form to the initial moles of solid, multiplied by 100%. Decomposition percentage = \(0.1153 mol / 0.1159 mol) * 100 = 99.48% \). So, we can say 99.48% of NH4HS has decomposed.

06

Effect of doubling the volume

According to Le Châtelier's Principle, a change in the system (here, an increase in volume) will shift the equilibrium to oppose the change. Thus, increasing the volume at constant temperature will shift the equilibrium to the side with more gas molecules. Therefore, we can say that the amount of NH4HS solid will decrease (i.e. more solid will decompose to increase the number of moles of gas and counteract the increase in volume).

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