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Question

Consider the following equilibrium in a closed container:
PCl5(g)PCl3(g)+Cl2(g)
PCl5 gas at a certain pressure is introduced in the container at 27 oC. However, the total pressure at equilibrium at 207 oC was found to be double the initial value. The % dissociation of PCl5 at 207 oC is: %

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Solution

As per Gay-Lussac's law,
P1T1=P2T2
If,
T1=27 oC=300 KP1=P (say)
For T2=207 oC=480 KP2= ?
P2=[T2T1]×P1=480300×P

PCl5(g)PCl3(g)+Cl2(g)P00 t=0 at 27 oC480P30000 t=0 at 207 oC480P300PPP t=t at 207 oC

Total pressure at equilibrium at 207 oC is
480P300P +2P = 2P (given)
8P5+P = 2P
P = 2P5
% dissociation of PCl5 at 207 oC = P[480P300]×100=2P58P5×100=25%


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