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Question

For the equilibrium
LiCl.3NH3(s)LiCl.NH3(s)+2NH3(g);
Kp=9 atm2 at 37C. A 5 L vessel contains 0.1 mole of LiCl.NH3. How many moles of NH3 should be added to the flask at this temperature to derive the backward reaction for completion?
Use : R=0.082 atm L mol1 K1

A
0.2
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B
0.59
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C
0.69
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D
0.79
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Solution

The correct option is D 0.79
LiCl.3NH3(s)LiCl.NH3(s)+2NH3(g);
[Kp=9 atm2]
Therefore,
LiCl.NH3(s)+2NH3(g)LiCl.3NH3(s);Initial moles:0.1α0Final moles at eqm.:0(α0.2)0.1
[Kp=19(atm)2]
Let, initial moles of NH3 is α for completion of reaction.
At equilibrium, Kp=1(PNH3)2 or 19=1(PNH3)2PNH3=3 atmPV=nRT3×5=n×0.0820×310n=0.59 i.e., (α0.2)=0.59

Initial moles of NH3=0.79

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