Ammonium carbonate (solid) decomposes when heated to produce three gaseous products - ammonia, water, and carbon dioxide.

Question

1. Write the balanced equation for the reaction.
2. Suppose that ammonium carbonate is heated to 500 K in a sealed vessel. At equilibrium there is still some solid in the vessel, and the partial pressure of the ammonia gas is 4.00 atm. Find both the Kp and the Kc of the reaction at 500 K.

3. It is found that the mass loss of the ammonium carbonate on being heated is 1.000 gram. What is the volume of the sealed vessel?

Answer 1

(NH4)2CO3(s) ==> 2NH3(g) + H2O(g) + CO2(g)

equilibrium:
NH3 = 4.00 atm
Therefore, CO2 = 2.00 atm
and H2O = 2.00 atm
P_total = 4.00 + 2.00 + 2.00 = 8.00 atm.

Kp =PNH3^2*PCO2*PH2O = (4)^2*2*2 = 64
Kc = Kp(RT)^{-delta n}
= 64(0.08206*773)^-4 = 3.95E-6

3.
moles (NH4)2CO3 = 1.000 g/molar mass = 0.01041. Therefore, moles NH3 = 0.02082, moles H2O = 0.01041, moles CO2 = 0.01041 and total moles = 0.04163.
Use PV = nRT to calculate volume.
8*V = 0.04163*0.08206*773, solve for V.