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Question

Find the total pressure to terms of Kp for 50% dissociation of PCl5 at 250C.

A
KP
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B
32KP
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C
3KP
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D
12KP
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Solution

The correct option is C 3KP
PCl5(C)PCl3+Cl2
T=0100(Iniitial moles)T=eq.(112)1212(moles at equlibrium)=0.50.50.5
Total moles at equilibrium
= 0.5 + 0.5 + 0.5 = 1.5
Kp=PPCl2.PCl2PPCl5
=0.51.5×p0.51.5×P0.51.5×P(P=Total pressure)
Kp=13P
P=3Kp

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