Question

Solid ammonium iodide decomposes to ammonia and hydrogen gasses at sufficiently high tempeatures.

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

The equilibrium constant for the decomposition at 400^oC is 0.215. Twenty grams of ammonium iodide are sealed in a 0.75-L flack and heated to 400^oC.

(a) What is the total pressure in the flask at equilibrium?

(b) How much solid NH4I is left after the decomposition?

Answer 1

T = 400 + 273 = 673 K

NH4I mass = 20 g

NH4I molar mass = 144.94 g/mol

moles of NH4I = 20 / 144.94 = 0.138

Kp = 0.215

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

                                     x                x

Kp = x^2

0.215 = x^2

x = 0.464 atm

total pressure = x + x = 0.464 + 0.464

total pressure = 0.928 atm

now calculate moles of each species

NH3 moles : form P V = n R T

0.464 x 0.75 = n x 0.0821 x 673

n = 0.0063

moles of HI is also 0.0063

1 mole NH4I --------------------> 1 mol NH3 and 1 mol HI

0.0063 moles NH4I gives ---------------> 0.0063 moles NH3 and 0.0063 moles HI

moles remians

0.138 - 0.0063 = 0.1317

NH4I mass remains = 0.1317 x 144.94 = 19.01 g

19.01 g NH4I left