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.

Options

A. `0.013`
B. `0.050`
C. `0.033`
D. `0.067`

Answer

C

Explanation

`1.6` mol of `PCl_(5)` is placed in 4 `dm^(3)` closed vessel. `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`
`{:(1.6mol,0,0,("initially")),((1.6-x)mol,xmol, xmol, ("At equlibrium")):}`
Given that `1.6-x=1.2`
`thereforex=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`
`thereforeK_(C)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.1xx0.1)/(0.3)=0.033.`