Answer:
(a) Kp = 0.126
(b) 48.1 %
(c) More of the solid will decompose, therefore it will decrease.
Step-by-step 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.