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The rate of forward reaction is two time...

The rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration. `K_(equilibrium)` is

A

`2.5`

B

`2.0`

C

`0.5`

D

`1.5`

Text Solution

Verified by Experts

The correct Answer is:
B
The rate of forward reaction is two times that of reverse reaction at a gives temperature and identical concetration `K_(equilibrium)` is 2 because the reaction is revessible. So `K = (K_(1))/(K_(2)) = (2)/(1) = 2`
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