At 540K, 0.10 mole of `PCI_(5)` are heated in a 8 litre flask. The pressure of the equilibrium mixture is found to be 1.0 atm. Calculate `K_(c) and K_(p)` for the reaction.

Options

A. `2 xx 10 ^(-2) "mol litre"^(-1) , 1. 69 atm`
B. `4 xx 10 ^(-2) "mol litre"^(-1), 1. 77 atm`
C. `2.5 xx 10 ^(-2)"mol litre"^(-1) , 1. 69 atm`
D. `4 xx 10 ^(-2)"mol litre"^(-1) , 2. 63 atm`

Answer

B

Explanation

`PCl _(5) hArr PCl _(3) + Cl _(2)`
`{:(0.1, 0, 0 , "mole before dissociation"),(0.1-x, x, x,"mole after dissociation"):}`
Given, volume of container = 8 litre
Now `K _(x) = ([PCl _(3)] [Cl _(2)] )/( [ PCl _(5) ] ) = ((x)/(8) xx (x)/(8))/( ( (0. - x))/( 8)) = (x^(2))/( 8(0.1 -x )) ...(i)`
Also PV = nRT for the equilibrium mixture at `50 K , 1 xx 8= (0.1 + x) xx 0. 0 82 xx 540`
`therefor x = 0.08 " "...(2)`
Thus form Eq (1) and (2) `K _(c) = (0.08 xx 0.08)/( 8 ( 0.1 - 0. 08)) = 4 xx 10 ^(-2) "mol liter"^(-1) `
Also `K _(P) = K _(c) [RT ] ^( Delta n ) = 4 xx 10 ^(-2) (0.08 xx 540) = 1. 72 atm because Deta n = + 1`