Question

For the equilibrium:
$LiCl.3N{H}_3\left(s\right)\rightleftharpoons LiCl.N{H}_3\left(s\right)+2N{H}_3\left(g\right)K_p=9at{m}^2$

At $27{}^{o}C$, a 8.2 L vessel contains 0.1 mol of $LiCl.N{H}_3\left(s\right)$. How many moles of $N{H}_3$ should be added to the flask at this temperature to drive the backward reaction to completion? (Take R= 0.082 atm-L/molK)

Answer 1

Say 'a' mole of $N{H}_3$ is added:
$LiCl.N{H}_3\left(s\right)+2N{H}_3\left(g\right)\rightleftharpoons LiCl.3N{H}_3\left(s\right)$;
Initially
0.1 a
At completion,
0 a-0.2
Since, $K_p=\frac{1}{9at{m}^2}=\frac{1}{P_{N{H}_3}^2}$
So, $P_{N{H}_3}=3atm$
Now applying,
PV= nRT
$3\times 8.2=(a-0.2)\times 0.082\times 300$
a = 1.2 mol