`PCl _(5) (g) hArr PCl _(3) (g) + Cl _(2) (g).` The equilibrium, `K_(c)` for the dissociation of `PCl_(5)` , is `4.0 xx 10^(-2)` at `250^(@)C` in a 3.0L flask when equilibrium concentration of `Cl_(2)` is 0.15 mol/L. What was the pressure of `PCl_(5)` before any dissociation? `(R=0.082 L-atm K^(-1)mol^(-1))`

Options

A. `37.0 atm`
B. `30.59 atm`
C. `24.05 atm`
D. `6.745 atm`

Answer

B

Explanation

`PCl _(5) (g) hArr PCL _(3) (g) + Cl _(2) (g)`
`{:(a, -, -, "initial conc"), (a-x, x, x,"equi conc"):}`
`K _(c) = ( x ^(2))/( a -x) 0.04 = ((0 . 15 )^(2))/( a - 0. 15) a = 0. 71 25`
`PV =nRT P = (0.7125) (0.0821) 523 = 30.59` atm