For the equilibrium LiCl.3NH3 (s) → LiCl. NH3 (s) + 2NH3, KP = 9 atm2 at 40°C. A five litre vessel contains 0.1 mole of LiCl.NH3. How many minimum moles of NH3  should be added to the flask at this temperature to derive the backward reaction for completion?

Q: For the equilibrium
LiCl.3NH3 (s) $ \rightleftharpoons $ LiCl. NH3 (s) + 2NH3, KP = 9 atm2 at 40°C. A five litre vessel contains 0.1 mole of LiCl.NH3. How many minimum moles of NH3  should be added to the flask at this temperature to derive the backward reaction for completion?

Solution: LiCl.3NH3 (s) $ \rightleftharpoons $ LiCl.NH3(s) + 2NH3(g)

KP = 9 atm , Vol = 5 l , Temperature = 40 °C

Number of moles of LiCl.NH3 = 0.1

2NH3(g)  + LiCl.NH3(s)   $ \rightleftharpoons $   LiCl.3NH3(s)

a–2x                      0.1–x                           x

0.1–x = 0        ⇒ x = 0.1

a–2x = a–0.2

$\large K_P’ = \frac{1}{K_P} = \frac{1}{9} $

$\large P_{NH_3} = \frac{(a-0.2)0.0821 \times 313}{5}= 5.1395 (a-0.2) $

$\large K_P’ = \frac{1}{(5.1395)^2 (a-o.2)^2}  $

$\large \frac{1}{9} = \frac{1}{(5.1395)^2 (a-o.2)^2}  $

⇒ a = 0.7837

∴ Minimum number of moles of NH3 required = 0.7837