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