Chemistry: Equilibrium Math Question 9

Question

The equilibrium constant for dissociation of ammonium chloride:

NH₄Cl(s) ↔ NH₃(g) + HCl(g)

is equal to K_P = 0.013312 at 550.0 K. If 1.5 g of ammonium chloride is placed in a 1.0 L vessel and charged with 1.9 bar of HCl(g) what is the final equilibrium pressure of NH₃(g)?

Answer 1

1.5 g NH4Cl x 1 mol/53.5 g = 0.028 moles

PV = nRT

P = nRT/V = (0.028)(0.0821)(550)/1 = 1.26 atm (used 1 bar = 1 atm instead of 0.9869)

We have the thermal decomposition of NH₄Cl at 550K:

NH₄Cl(s) ===> NH₃(g) + HCl(g)

1.26..................0..............1.9.......I

-x.....................+x.............+x........C

1.26-x..............x................x.........E

Kp = (NH3)(HCl)/(NH4Cl) = 0.013312

0.013312 = (x)(x)/(1.26 - x)

x² + 0.013312x - 0.01683 = 0

x = 0.1232 atm = equilibrium pressure of NH₃(g)