Question

4. The total pressure observed at equilibrium in the dissociation of solid ammonium carbamate at a certain temperature is 2.0
atm. The equilibrium constant kp is
(a) 4.185
(b) 1.185
(c) 2.276
(d) 1.072​

Answer 1

Answer:

option (b)

Explanation:

Solid Ammonium carbonate on dissociation gives NH₃ and CO₂ and water.

(NH₃)₂CO₃ ⇄ 2 NH₃ + CO₂ + H₂O

      -                  2x          x         -

We do not consider the concentrations of  (NH₃)₂CO₃ and H₂O as they are in solid and liquid state respectively.

The other two are used in finding Kp as they are gases.

Now, given that, at equilibrium  

pressure = 2.0 atm

2x + x = 2

3x = 2

x = 2/3

So, partial pressure of NH₃ = 2x = 4/3 atm

partial pressure of CO₂ = x = 2/3 atm

Kp = $\frac{4}{3}^{2}$ x $\frac{2}{3}$

= $\frac{16}{9}$ x $\frac{2}{3}$

= $\frac{32}{27}$

= 1.185

Hope it helps!