What is the total pressure at equilibrium?

Question

At 250 °C, Ke = 1.58 x 10‒8 for the following equilibrium: NH2COONH4 (s)  ⇌  2 NH3 (g) + CO2 (g)

If 7.80 g of NH2COONH4 is put into a 0.500-L evacuated container, what is the total pressure at equilibrium?

Answer 1

7.80 g NH₂COONH₄ x 1 mol/78 g = 0.10 moles

NH₂COONH₄ (s)  ⇌  2 NH₃ (g) + CO₂ (g)

0.1 mol.......................0.................0..........Initial

-x..............................+2x.............+x.........Change

0.1-x........................2x.................x.........Equilibrium

Keq = 1.58x10_-8 = [NH₃]²[CO₂]

1.58x10^-8 = (2x)²(x) = 4x³

x³ = 3.95x10^-9

x = 1.58x10^-3

There are 3x moles of products so total moles of products = 3 x 1.58x10-3 moles = 4.74x10-3 moles

PV = nRT

P = nRT/V = (4.74x10^-3)(0.0821)(523)/0.5

P = 0.407 atm