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2 "mole" of PCl(5) were heated in a clos...

`2 "mole"` of `PCl_(5)` were heated in a closed vessel of `2 litre` capacity. At equilibrium `40%` of `PCl_(5)` dissociated into `PCl_(3)` and `Cl_(2)`. The value of the equilibrium constant is:

A

0.266

B

0.53

C

2.66

D

5.3

Text Solution

Verified by Experts

The correct Answer is:
A
`PCl_(5) hArr PCl_(3) + Cl_2`
`{:(("at t = 0 moles") ,2, 0, 0),(("at t = eq. moles"),(2xx60)/100,(2xx40)/100,(2xx60)/100):}`
Concentration `= ("Moles")/("Vol.")`
`{:((1.2)/2,(0.8)/2,(0.8)/2),(0.6,0.4,0.4):}`
`K_C = ([PCl_3][Cl_2])/([PCl_5])=(0.4 xx0.4 )/(0.6) = (0.8)/3 = 0.266`
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