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The degree of dissociation of I(2) molec...

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:

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)hArr2I(g)`

`rArr n_(T)=a+aalpha`
`K_(p)=P_(1)^(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_("new")` be the pressure when `alpha=0.2`
`rArr K_(p)=(4xx0.16)/0.84=(4xx0.04)/0.96P_("new")`
`rArr P_("new")=(4xx0.96)/0.84=4.57 "atm"`
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