1.

On a give condition, the equilibrium concentration of HI, H_(2) and I_(2) are 0.80,1.10 mole/litre. The equlibrium constant for the reaction H_(2)+I_(2)hArr 2HI will be

Answer»

64
12
8
`0.8`

SOLUTION :`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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