Ammonium Iodide dissociates reversibly to ammonia and hydrogen iodide: NH4I(s) N

Question

Ammonium Iodide dissociates reversibly to ammonia and hydrogen iodide: NH4I(s) NH3(g) + HI(g) Kp = 0.215 at 400 degree C If 150 g of ammonium iodide is placed into a 3.00 L vessel and heated to 400 degree C, calculate the partial pressure of ammonia when equilibrium is reached. Hint: Ammonium Iodide is a solid, is it used in the equilibrium calculation? Initial Ci Delta Final Cf Equilibrium Values

Explanation / Answer

NH4 NH3 + HI 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.

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