Question

For the equilibrium :

$LiCl.3{NH}_3\left(s\right)\rightleftharpoons LiCl.3{NH}_3\left(s\right)+2{NH}_3$

$K_p$ = 9${atm}^2$ at $40C$. A 5-L vessel contains 0.1 mole of $LiCl.3N{H}_3$

How many moles of $N{H}_3$ should be added to the flask at this temperature to derive the backward reaction for completion?

• A. Initial moles of ${NH}_3$= 1.467 moles.
• B. Initial moles of ${NH}_3$= 0.291 moles.
• C. Initial moles of ${NH}_3$= 0.05 moles.
• D. Initial moles of ${NH}_3$= 0.7837 moles.

Answer 1

The correct option is D Initial moles of ${NH}_3$= 0.7837 moles.

$\because$ $LiCl.3{NH}_3\left(s\right)\rightleftharpoons LiCl.3{NH}_3\left(s\right)+2{NH}_3\left(g\right)$
[$K_p$=$9{atm}^2$]
$LiCl.3{NH}_3\left(s\right)+2{NH}_3\rightleftharpoons LiCl.3{NH}_3\left(s\right)$
[$K_{p_1}=\frac{1}{9}{atm}^{-2}$]
0.1 (a + 0.2) Initial moles
0 a 0.1 Final Moles at equilibrium
$\therefore$ Initial moles of ${NH}_3$ should be (a+0.02) to bring in completion of reaction.
At
equilibrium $K_{p_1}$= $\frac{1}{{(P}_{{NH}_3}^{\prime }{)}^2}$ or $\frac{1}{9}$
=$\frac{1}{{(P}_{{NH}_3}^{\prime }{)}^2}$
$\therefore P_{{NH}_3}^{\prime }$ = 3 atm
$\because$ PV=nRT
$\therefore$ $3\times 5=n\times 0.0821\times 313$
$\therefore$ n = 0.5837 i.e. a = 0.5837
$\therefore$ Initial moles of ${NH}_3$ = a + 0.2 = 0.5837 + 0.2=0.7837 moles.