1.

The ratio of the forward reaction is two times that of the revers reaction at a given temperature and identical concentration.K_("equilibrium") is

Answer»

2.5
`2.0`
0.5
1.5

Solution :The rate of a forward reaction is two times that of reverse reaction at a given temperature.
`(d[A])/(DT)=2(d[B])/(dt)`
`K_("EQUILIBRIUM")=(K_(1))/(K_(2))=(d[A]//dt)/(d[B]//dt)=2(d[B]//dt)/(d[B]//dt)=2`.
`thereforeK_("equilibrium")=2`.


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