Question

14. Ammonium carbamate dissociates as
NH2,COONH4(s) = 2NH3(g) + CO2(g)
In a closed vessel containing ammonium carbamate
in equilibrium, ammonia is added such that the
partial pressure of NH3, now equals to the original
total pressure. The ratio of total pressure now to
the original pressure is

ans is 31/27

plz solve the question......
​

Answer 1

Answer:

31/27

Explanation:

NH₂COONH₄(s) = 2NH₃(g) + CO₂(g)

Step 1:

Let the initial partial pressure be “P”.  

So, according to the balanced equation given, the partial pressure of NH₃ is “2P” and CO₂ is “P” .

The initial/original total pressure becomes = 2P + P = 3P……. (i)

∴ Kp = [ρNH₃]² [ρCO₂]  

Or, Kp = [2P]² * [P] …….. (ii)

Step 2:

Since after adding NH₃, the partial pressure of NH₃ now becomes equal to the initial total partial pressure i.e., 3P and partial pressure of CO₂ will be “ P’ ”.

Total final pressure = 3P + P’ …… (iii)

∴ Kp = [3P]² * [P’] …….. (iv)

Step 3:

Equating (ii) & (iv), we get

[2P]² * [P] = [3P]² * [P’]

Or, 4 P = 9 P’

Or, P’ = 4P / 9 ….. (v)

Step 4:

Thus,

The ratio of the total final pressure to the initial/original total pressure  is

= [3P + P’] / 3P ...... [from (i) & (iii)]

= [3P + (4P/9)] / 3P ........ [from (v)]

= [27P + 4P] / 27P

= 31/27

Answer 2

Answer:

31/27

Explanation:

NH2COONH4(s)⇋2NH3(g)+CO2(g)

Initial 2P P

K

p=(pNH3) 2(pCO2)

Kp=(2P)2(P).........(i)

Now, in second case P

t(initial)=3PNH2COONH4(s)⇋2NH3(g)+C02(g)

Initial 3P P'

Kp=(3P)²(P′).........(ii)

From equation (i) and (ii)

(2P)2(P)=(3P)2(P′)P′= 94P

PT(old)/PT(new)= 3P/3P+P′ = 3P3P+94P=27/31