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At 1000K, the pressure of iodine gas is ...

At `1000K`, the pressure of iodine gas is found to be `0.1 atm` due to partial dissociation of `I_(2)(g)` into `I(g)`. Had there been no dissociation, the pressure would have been `0.07 atm`. Calculate the value of `K_(p)` for the reaction:
`I_(2)(g)hArr2I(g)` .

Text Solution

Verified by Experts

Analysing in terms of pressure directly:

rArr Total pressure at equilibrium
`=(0.07-p)+2p=0.1(given)`
`rArr p=0.03 "atm"`
`K_(p)=(p_(1))^(2)/p_(I_(2))=((2p)^(2))/((0.07-p))=((2xx0.03)^(2))/(0.07-0.03)`
Substituting value of `p`, we get
`rArr K_(p)=0.09 "atm"` units
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