LiCl.3NH3<----->LiCl.NH3 + 2 NH3, Kp=4 atm^2 at 40 degrees Celcius. 
A 5L vessel contains 0.1 mol LiClNH3. How many moles of NH3 should be added to flask at this temperature to derive backward reaction to completion?    [0.728 moles]

Dear Student,

The computation are expressed below,

LiCl.3NH3(s)LiCl.NH3(s)+2NH3(g)we have to compute for,                        LiCl.NH3(s)+2NH3(g)LiCl.3NH3(s)Initial moles     0.1                      a                   0Final moles        0                       a-0.2           0.1at equn, KP'=1(PNH3')14=1(PNH3')P'NH3=2 atmPV=nRT2×5=n×0.0820×313n=0.38960.40a-0.2=0.10a=0.30Initial moles of NH3=0.30                                                                                                                
0.728 mole answer is for Kp = 9 sq.atm and T=37 C so kindly check.


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