`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,"equal conc."):}`
`K_(c) = (x^(2))/(a-x), 0.4 = (0.15)^(2), a = 0.7125`
`PV = nRT, P = (0.7125)(0.0821) 523 = 30.59` atm