Question

Ammonium carbamate, NH4CO2NH2, is a salt of carbamic acid that is found in the blood and urine of mammals. When heated in a closed container, it decomposes and establishes the following equilibrium: NH4CO2NH2(s) ⇌ 2NH3(g) + CO2(g) When a 7.80-g sample of solid ammonium carbamate was introduced into a 5.00-L evacuated, sealed container, and the sample allowed to decompose until it reached equilibrium, the total gas pressure of the equilibrium mixture was 0.27 atm at 25oC. (a) Determine the partial pressure of ammonia (PNH3) and carbon dioxide (PCO2), respectively, at equilibrium. (b) What is the Kp for this reaction at 25oC? (c) How many grams of solid ammonium carbamate remains at equilibrium?

Answer 1

Initial moles of solid ammonium carbamate

= mass/molecular weight

= 7.80g / 78 g/mol

= 0.10 mol

Initial concentration = 0.10 mol / 5.00 L = 0.02 mol/L

Initial pressure = 0.02 mol/L x 0.0821 L-atm/mol-K x (25 + 273)K

= 0.489 atm

The balanced reaction with ICE TABLE

NH4CO2NH2(s) = 2NH3(g) + CO2(g)

I 0.489

C - x +2x +x

E (0.489 - x) 2x x

At equilibrium

Part a

Total gas pressure = 0.27 atm

2x + x = 0.27

x = 0.09 atm

PNH3 = 2*0.09 = 0.18 atm

P CO2 = 0.09 atm

Part b

Kp = PCO2 x PNH3²

Kp = 0.09 x 0 18 x 0.18 = 0.002916

= 2.92 x 10^-3

Part c

P (NH4CO2NH2) = 0.489 - 0.09 = 0.399 atm

Concentration = 0.399 atm / (0.0821 x 298)

= 0.0163 mol/L

Moles of NH4CO2NH2 remains = 0.0163 mol/L x 5.00 L

= 0.08154 mol

Mass of NH4CO2NH2 remains = 0.08154 mol x 78 g/mol

= 6.36 g