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The pressure of iodine gas at 1273 K is ...

The pressure of iodine gas at `1273 K` is found to be `0.112` atm whereas the expected pressure is `0.074` atm. The increased pressure is due to dissociation `I_(2) hArr 2I`. Calculate `K_(p)`.

A

`0.074`

B

`0.148`

C

`0.05`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
2
`{:(l_(2)(g),hArr,,2l(g),),(0.074,,,,),(0.074-P,,,2P,):}`
`0.074-P+2P=0.111`
`P=0.037`
`K_(p)=(4P^(2))/(0.074-P)=(4(0.037)^(2))/(0.037)=0.148`
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