1.

The rate of forward reaction is two times that of reverse reaction at a given temperaturee and identical concentration. K_("equilibriuim") is

Answer»

`2.5`
`2.0`
`0.5`
`1.5`

Solution :The RATE of forward reaction is two TIMES that of reverse reaction at a given temperature and identical concentration `K_("equlibrium")` is 2 because the reaction is REVERSIBLE. So `K=(K_(1))/(K_(2))=2/1=2.`


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