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16 mol of PCl(5)(g) is placed in 4 dm^(-...

16 mol of `PCl_(5)(g)` is placed in 4 `dm^(-3)` closed vessel. When the temperature is raised to 500 K, it decompses and at equilibrium, 1.2 mol of `PCl_(5)(g)` remains. What is `K_(c)` value for the decomposition of `PCl_(5)(g)` to `PCl_(3)(g)` and `Cl_(2)(g)` at 500K.

A

`0.013 `

B

`0.050`

C

`0.033`

D

`0.067`

Text Solution

Verified by Experts

The correct Answer is:
C
1.6 mol of `PCl_5` is placed in `4 dm^3` closed vessel.
`PCl_(5(g)) hArr PCl_(3(g)) +Cl_(2(g))`
`{:(1.6 "mol", 0 ,0, "(Initially)"),("(1.6 –x)mol" ,"x mol" ,"x mol" ,("At equilibrium")):}`
Given that 1.6 – x=1.2
`:.x = 0.4` mol
Therefore, `[PCl_5]=(1.2)/4=0.3 ,[PCl_3]=(0.4)/4 =0.1`
& `[Cl_2]=(0.4)/4=0.1`
`:. K_c =([PCl_3 ][Cl_2])/([PCl_5])=(0.1 xx0.1)/(0.3)=0.033`
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