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Question

PCl5(g)PCl3(g)+Cl2(g). If 1 mole of PCl5 be put in a container of volume V litre and at equilibrium, x moles of it was decomposed, find its Kp and Kc at equilibrium pressure of P atm.

A
Kc=x2(1x)V and KP=x2P1x2
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B
Kc=x4(1x)V and KP=x4P1x2
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C
Kc=x(1x)V and KP=xP1x2
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D
None of the above
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Solution

The correct option is A Kc=x2(1x)V and KP=x2P1x2
The given reaction is PCl5(g)PCl3(g)+Cl2(g)
KP=PPCl3PCl2PPCl5and Kc=[PCl3][Cl2][PCl5]
PCl5PCl3+Cl2Initial moles100Moles at equilbrium1xxxConcentration at equilbrium1xVxVxV
Total moles at equilbrium=1x+x+x=1+x
Kc=(xV)2(1xV)=x2(1x)V
Partial pressure of PPCl3=XPCl3Peqb,PPCl5=XPCl5Peqb,PCl2=XCl2Peqb
Peqb=P,XPCl3=x1+x,XCl2=x1+x,XPCl5=1x1+x

so, KP=PPCl3PCl2PPCl5
KP=(x1+xP)2(1x1+x)P=x2P1x2

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