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For reaction : H(2)(h)+I(2)(g) hArr 2HI(...

For reaction : `H_(2)(h)+I_(2)(g) hArr 2HI(g)` at certain temperature, the value of equilibrium constant is `50`. If the volume of the vessel is reduced to half of its original volume, the value of new equilibrium constant will be

A

`25`

B

`50`

C

`100`

D

Unpredictable

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(,H_(2),+,I_(2),hArr,2HI),("Initial",1,,1,,0),("Final",(1-alpha)/v,,(1-alpha)/v,,(2alpha)/v):}`
Since `Deltan=0" " [2-(1+1)]`
`:. K` does not depend on volume.
So on reducing volume to half K does not change.
`:. K=50`
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