For the reaction `PCl_(5)hArr PCl_(3)+Cl_(2)` in gaseous phase, `K_(c)=4`. In a 2L flask, there are 2 moles each of `PCl_(3)` and `Cl_(2)` and 0.5 mole of `PCl_(5)` . Equilibrium concentration of `PCl_(5)` is

Options

A. 0.25 mol `L^(-1)`
B. 0.125 mol `L^(-1)`
C. 0.75 mol `L^(-)`
D. 1.00 mol `L^(-1)`

Answer

A

Explanation

`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",(0.5)/(2),,(2)/(2),,(2)/(2)),("At eqm. ",(0.5-x)/(2),,(2+x)/(2),,(2+x)/(2)):}`
`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(((2+x)/(2))^(2))/(((0.5-x)/(2))^(2))`
`4=((2+x)^(2))/(2(0.5-x))`
`8(0.5-x)=(2+x)^(2)`
`4-8x=4+x^(2)+4x`
`x^(2)+12x=0`
`x(x+12)=0`
`x=0,x=-12`
`:. [PCl_(5)]_("eq")=(0.5-0)/(2)=0.25"mol"L^(-1)`