In a 500 mL flask, the degree of dissociation of `PCI_(5)` at equilibrium is 40% and the initial amount is 5 moles. The value of equilibrium constant in `mol L^(-1)` for the decomposition of `PCI_(5)` is

Options

A. 2.33
B. 2.66
C. 5.32
D. 4.66

Answer

B

Explanation

`PCl_(5) hArr PCl_(3) + Cl_(2)`
Given
C = 5 moles
`alpha = 40% = 0.4`
Volume = 500 mL = 0.5 L
`K_(c) = ([PCl_(3)][Cl_(2)]/[PCl_(5)]) = (c.alpha)(c.alpha)/c(1- alpha) = calpha^(2)/(1-alpha)`
` = (5/0.5 xx (0.4)^(2))/1 - 0.4 = 8/3 = 2.66 mol L^(-1)`