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The degree of dissociation of I(2) "mole...

The degree of dissociation of `I_(2)` "mole"cule at `1000^(@)C` and under `1.0 atm` is `40%` by volume. If the dissociation is reduced to `20%` at the same temperature, the total equilibrium pressure on the gas will be:

A

1.57 atm

B

2.57 atm

C

3.57 atm

D

4.57 atm

Text Solution

Verified by Experts

The correct Answer is:
D
`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`
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