For the equilibrium LiCl.3NH3(.s.)leftharpoons LiCl.NH3(.s.)+2NH3(.g.); KP=9atm2 at 37° C . A 5 L vessel contains 0.1 mole of LiCl.NH3 . How many mole of NH3 should be added to the flask at this temperature derive the backward reaction for completion? (use R = 0.082 atm-L/mol K)

Question

For the equilibrium LiCl.3NH3(.s.)leftharpoons LiCl.NH3(.s.)+2NH3(.g.); KP=9atm2 at 37° C . A 5 L vessel contains 0.1 mole of LiCl.NH3 . How many mole of NH3 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

Tardigrade Admin · Accepted Answer

underset underset0.1-L i C l . 3 N H3 (. s .)leftharpoons underset underset-0.1L i C l . N H3 (. s .)+ underset underseta - 0.2a2 N H3 (. g .) KP=9=PN H32 PN H3=3 PV = nRT 3× 5=(.a-0.2.)0.082× 310 a=0.79

Q. For the equilibrium LiCl.3NH3​(s)⇌LiCl.NH3​(s)+2NH3​(g); KP​=9atm2 at 37∘C . A 5 L vessel contains 0.1 mole of LiCl.NH3​ . How many mole of NH3​ should be added to the flask at this temperature derive the backward reaction for completion? (use R = 0.082 atm-L/mol K)

0.49

0.59

0.69

0.79

0.1−​LiCl.3NH3​(s)​⇌−0.1​LiCl.NH3​(s)​+a−0.2a​2NH3​(g)​ KP​=9=PNH3​2​ PNH3​​=3 PV = nRT 3×5=(a−0.2)0.082×310 a=0.79

$0.1 - L i Cl .3 N H_{3} (s) ⇌ - 0.1 L i Cl . N H_{3} (s) + a - 0.2 a 2 N H_{3} (g)$

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

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

PV = nRT

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

$a = 0.79$