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On a given condition, the equilibrium co...

On a given condition, the equilibrium concentration of `HI , H_(2)` and `I_(2)` are `0.80` , `0.10` and `0.10` mole/litre. The equilibrium constant for the reaction `H_(2) + I_(2) hArr 2HI` will be

A

64

B

12

C

8

D

`0.8`

Text Solution

Verified by Experts

The correct Answer is:
A
`H_(2)+I_(2)hArr2HI,[HI]=0.08,[H_(2)]=0.10,[I_(2)]=0.10`
`K_(c)=([HI]^(2))/([H_(2)][I_(2)])=(0.08xx0.80)/(0.10xx0.10)=64`
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