The decomposition of ammonium hydrogen sulfide
NH4HS (s) <=> NH3 (g) + H2S (g)
is an endothermic process. A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at exactly 24°C. After equilibrium has been established, the total pressure inside is 0.709 atm. Some solid NH4HS remains in the vessel.
(a) What is the KP 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?

(a) Kp = 0.126

(b) 48.1 %

(c) More of the solid will decompose, therefore it will decrease.

Explanation:

(a) We have here the equilibrium decomposition of NH₄HS, according to the equation:

NH₄HS (s) ⇄ NH₃ (g) + H₂S

with an equlibrium cosnstant, Kp, given by

Kp = pNH₃ x pH₂S

where pNH3 and pH₂S are  the partial pressures of NH₃ and H₂S .

Since 1 mol NH₃ is produced for every 1 mol H₂S, it follows that the partial pressure of NH₃ is equal to the partial pressure of H₂S ( from ideal gas law  the pressure is proportional to number of mol):

pNH₃ = pH₂S = 1/2 (0.709) atm = 0.355 atm

and

Kp = 0.355 x 0.355 = (0.3545) ² = 0.126

(b) To solve this part we need to do a calculation based on the stoichiometry of the reaction by calculating the number of moles, n,  a partial pressure of 0.355 atm represent.

From the ideal gas law we can calculate it:

PV = nRT ⇒ n = PV/RT

where  P = 0.355 atm, V = 4.000 L, T = ( 24 + 273 ) K, and R is the gas constant 0.08205 Latm/Kmol.

n = 0.355 atm x 4.000 L / 0.08205 Latm/Kmol x 297

n = 0.058 mol

Now we can relate this 0.058 mol of NH₃ ( or H₂S ) to the number of moles of moles of NH₄HS that must have decompesed. Since it is a 1: 1 ratio:

(1 mol NH₄HS / 1 mol NH₃ ) x 0.058 mol NH₃  = 0.058 mol NH₄HS

Having the molar mass of NH₄HS, 51.11 g/mol we can calculate its mass:

0.058 mol NH₄HS x 51.11 g/mol =  2.9644 g

Percentage decomposition is then equal to:

2.9644 / 6.1589 g x 100 g = 48.1 %

(c) If the volume of the vessel, effectively we are reducing the pressure by a half , and the system will react according to LeChatelier's principle by producing more gaseous products to reach equilibrium again. Therefore, more of the solid NH₄HS will decompose, and the amount of it will we reduced compared to the previous part.