1.

If the equilibrium constant of the reaction 2HIhArrH_(2)+I_(2)is 0.25, then the equlibrium constnat of the reaction H_(2)+I_(2)hArr2HI would be

Answer»

`1.0`
`2.0`
`3.0`
`4.0`

Solution :`K_(1)` for REACTION `2HIhArr H_(2)+I_(2)is 0.25 K_(2)` for reaction `H_(2)+I_(2)hArr2HI "will be"K_(2)=(1)/(K_(1))=(1)/(0.25)=4` Because `II^(nd)` reaction is reverse of `I^(ST).`


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