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.

P(total) = 46.08 atm

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

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

c. is the answer

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