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A mixture of 0.3 mole of H(2) and 0.3 mo...

A mixture of 0.3 mole of `H_(2)` and 0.3 mole of `I_(2)` is allowed to react in a 10 lit vessel at `500^(@)C`. If `K_(C)` of `H_(2) +I_(2)

A

0.15 mole

B

0.24 mole

C

0.03 mole

D

0.06 mole

Text Solution

Verified by Experts

The correct Answer is:
D

`K_(C)=((2x // 10)^(2))/(((0.3Lx)/(10))^(2))=((2x)/(0.3-x))^(2)`
`64=((2x)/(0.3-x))^(2) implies x=0.24`
`:.` unreacted `I_(2)=0.3-0.24=0.06`
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