Question

For the equilibrium $LiCl.3NH_{3}\left(\right.s\left.\right)\rightleftharpoons LiCl.NH_{3}\left(\right.s\left.\right)+2NH_{3}\left(\right.g\left.\right);$ $K_{P}=9atm^{2}$ at $37^\circ C$ . A 5 L vessel contains 0.1 mole of $LiCl.NH_{3}$ . How many mole of $NH_{3}$ should be added to the flask at this temperature derive the backward reaction for completion? (use R = 0.082 atm-L/mol K)

• A. 0.49
• B. 0.59
• C. 0.69
• D. 0.79

Answer 1

Answer: (D)

$\underset{\underset{0.1}{-}}{L i C l . 3 N H_{3} \left(\right. s \left.\right)}\rightleftharpoons\underset{\underset{-}{0.1}}{L i C l . N H_{3} \left(\right. s \left.\right)}+\underset{\underset{a - 0.2}{a}}{2 N H_{3} \left(\right. g \left.\right)}$

$K_{P}=9=P_{N H_{3}}^{2}$

$P_{N H_{3}}=3$

PV = nRT

$3\times 5=\left(\right.a-0.2\left.\right)0.082\times 310$

$a=0.79$