Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 250.°C, Kc = 1.58 × 10−8 for the following equilibrium:
NH2COONH4(s) ⇌ 2 NH3(g) + CO2(g)
If 11.51 g of NH2COONH4 is put into a 0.500−L evacuated container, what is the total pressure at equilibrium? atm

The total pressure at equilibrium is 0.07503 atm

Explanation:

The partial pressure of the product at equilibrium will be calculated as follows;

Kp = Kc[RT]³

given;

equilibrium constant Kc = 1.58 X 10⁻⁸

gas constant R = 0.0821 L.atm/mol.K

temperature T = (250 +273) = 523 k

Kp = 1.58 X 10⁻⁸ *(0.0821)³ *(523)³ = 1.251 X 10⁻³

NH₂COONH₄(s) ⇌ 2NH₃(g) + CO₂

NH₂COONH₄(s): Kp = 0, since it is in solid state

2NH₃(g) + CO₂: Kp = 1.251 X 10⁻³

I.C.E Analysis on the product

          2NH₃(g)       CO₂

I  :      0                  0

C :      2x                x

E :     (2x-0)          (x-0)

At equilibrium, E: (2x-0)(x-0) = 1.251 X 10⁻³

(2x)(x) = 1.251 X 10⁻³

2x² = 1.251 X 10⁻³

x²   = (1.251 X 10⁻³)/2

x²  = 6.255 X 10⁻⁴

x  = √(6.255 X 10⁻⁴)

x  = 0.02501 atm

Partial pressure of 2NH₃(g)  = 2x = 2(0.02501 atm) = 0.05002 atm

Partial pressure of  CO₂ = x = 0.02501 atm

Total pressure = P(NH₃(g)) +P(CO₂)

Total pressure = 0.05002 atm + 0.02501 atm = 0.07503 atm

Therefore, the total pressure at equilibrium is 0.07503 atm

2b2t

explanation:

2b2t