Question

Ammonium iodide dissociates reversibly to ammonia and hydrogen iodide:
NH4I(s) ↔ NH3(g) + HI(g)
At 400ºC, Kp = 0.215. If 150 g of ammonium iodide is placed into a 3.00-L vessel and heated to 400ºC, calculate the partial pressure of ammonia when equilibrium is reached

help plz.. atleast set me up so i know where to go or how to solve

Answer 1

NH4I(s) ==> NH3(s) + HI(g)

Kp = PNH3*PHI = 0.215

mole fraction NH3 = 0.5
mole fraction HI = 0.5
(Since moles NH3 = moles HI at equilibrium, then each will be just 1/2 of the total which makes mole fraction of each 0.5.)

PNH3 = XNH3*Ptotal = 0.5*Ptotal
PHI = XHI*Ptotal= 0.5*Ptotal

Now substitute into Kp expression the partial pressures of PNH3 and PHI and you are left with only one unknown. Solve for Ptotal. After finding Ptotal, use that back in the PNH3 = XNH3*Ptotal to find PNH3.
Repost as a new problem if you run into trouble. This question is far from the top and I may miss it.