Question

Ammonium carbamate, NH4CO2NH2 decomposes as NH4CO2NH2(s) ⇋2NH3(g) + CO2(g)
Starting with only the solid, it is found that at 40 degrees celsius the total gas pressure (NH3 and CO2) is 0.363 atm. Calculate the equilibrium constant.

Answer 1

NH4CO2NH2(s) ⇋2NH3(g) + CO2(g)

only NH3 and CO2 are gases. NH4CO2NH2 is solid and will therefore be ignored.

Number of moles of NH3= 2

Number of moles of CO2= 1

Total moles= 2+1=3

Mole fraction of NH3= 2/3

Mole fraction of CO2=1/3

Total pressure given= 0.363 atm

So, partial pressure of NH3= mole fraction of NH3 * total pressure

= 2/3 * 0.363

=0.242 atm

Similarly partial pressure of CO2= 1/3 * 0.363

=0.121 atm

Kp= (p NH3) ( p Co2)

Kp= 0.242 * 0.121

Kp= 0.0292 atm

Answer 2

Answer:

Kp=0.007086

Explanation:

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

nNH3=2

nCO2=1

totmol=1+2= 3

mole fraction NH3=⅔ and NO2=⅓

Total pressure=.363

partial pressure NH3=mole fraction x total pressure

pp=⅔*.0363=.242atm

PpCO2=⅓*.0363=.0121atm

kp=(PNH3)^2(PCO2)

=(.0242)^2(0.121)

=0.058564*0.121

Kp=0.007086