The degree of dissociation of `I_(2)` molecule of `1000^(@)C` and underatmosperic is 40% by volume .if the disscoiation is reduced to 20% at he same temp ., total equilibrium pressure on the gas is:

Options

A. 1.57 atm
B. 2.57 atm
C. 3.57 atm
D. 4.57 atm

Answer

D

Explanation

`I_(2)(g) rarr 2I(g)`
`implies n_(T) = a + a alpha`
`K_(p) = p_(I)^(2)/p_(I_(2)) = (2alpha/1+alpha P)^(2)/((1-alpha/1+alpha)p) = 4alpha^(2)p/(1-alpha^(2))`
Given: `alpha = 0.4` at P = 1.0 atm
`Let P_(n ew)` be the pressure when `alpha = 0.2`
`implies K_(p) = 4 xx 0.16/0.84 = 4 xx 0.4/0.96 P_(n ew) implies P_(new) = 4 xx 0.96/0.84 = 4.57 atm`