Question

. . Ammonium carbamate (molar mass = 78.08 g/mol) decomposes according to the reaction shown below, producing ammonia and carbon dioxide. (NH4)(NH2CO2)(s) ⇌ 2NH3(g) + CO2(g) Kp = 0.400 at 298 K 10.2 g of ammonium carbamate is sealed in an empty 2.00-L vessel and allowed to come to equilibrium at 298 K.

a. What are the partial pressures of NH3 and CO2 at equilibrium? NH3:_______________________________ CO2:_______________________________

b. What mass of solid ammonium carbamate is in the flask at equilibrium? Answer:_______________________________

c. The temperature of the reaction vessel increased to 398 K; when equilibrium is re-established, the total pressure in the vessel is 4.22 atm. Calculate the values of Kp and Kc at this temperature. Kp:_______________________________ Kc:_______________________________

Answer 1

Moles of Ammonium Carbamate= mass/ molar mass = 10.2/78.08=0.131

The reaction is   . (NH4)(NH2CO2)(s) ⇌ 2NH3(g) + CO2(g)

Kp = [NH3]² [CO2] = 0.4

Let P= partial pressure of CO2 formed. PNH3= 2P

Kp = 2P*2P*P= 0.4

P=0.63 atm

Partial pressure of CO2= 0.63 atm and partial pressure of NH3= 2*0.63 = 1.26 atm

Total pressure = 0.63+1.26= 1.89 atm

Moles of mixture of CO2 and NH3= PV/RT = 1.89*2/(0.0821*298)=0.1545 moles

Moles of CO2= (0.63/1.89)* 0.1545 ( moles of CO2/ total moles = partial pressure of CO2/ total pressure)

Moles of CO2= 0.0515 and moles of NH3= 2*0.0515= 0.103

Moles of CO2 formed = moles of NH4NH2CO2 reacted = 0.0515

Moles of NH4NH2CO2 remaining = 0.131-0.0515= 0.0795 moles

Mass of NH4NH2CO2 remaining = 0.0795*78.08=6.201 gm

2. Moles of gas mixture at new equilibrium = 4.22*2/(0.0821*398)=0.258 moles

At equilibrium total pressure = P+2P= 3P= 4.22

P= 4.22/3 = 1.406 atm

Kp = 1.406*(2*1.406)²= 11.11

Kp = KC*(RT)deltan

Deltan= change in number of moles due to reaction = 2-0= 2

11.11= KC*(0.0821*398)²

KC = 0.010405