Question

Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 202°C, Kc = 0.008 for the following equilibrium: If 6.82 g of NH2COONH4 is put into a 0.5-L evacuated container, what is the total pressure (in atm) at equilibrium? Enter a number to 2 decimal places.

Answer 1

P(total) = 46.08 atm

Step-by-step explanation:

Given data:

Mass of NH₂COONH₄ = 6.8 g

Kc = 0.008

Temperature = 202°C = 202+273 = 475 K

Total pressure at equilibrium = ?

Solution:

Equilibrium equation:

NH₂COONH₄ ⇄ 2NH₃ + CO₂

Formula:

Kp = Kc(RT)³

Kp = 0.008 (0.0821 atm.L/mol.K )³(475K)³

Kp = 471.56

NH₃ CO₂

Initial concentration 0 0

Change in concentration 2x x

equilibrium concentration 2x x

AS

471.56 = 2x (x)

471.56 = 2x²

x² = 471.56 /2

x² = 235.78

x = 15.36

Pressure of ammonia = 2x = 2(15.36) = 30.72 atm

Pressure of carbon dioxide = x = 15.36 atm

Total pressure:

P(total) = P(NH₃) + P(CO₂)

P(total) = 30.72 atm + 15.36 atm

P(total) = 46.08 atm