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The equilibrium constant for a reaction ...

The equilibrium constant for a reaction
`A+2B hArr 2C` is `40`. The equilibrium constant for reaction `C hArr B+1//2 A` is

A

`1//40`

B

`(1//40)^(1//2)`

C

`(1//40)^(2)`

D

`40`

Text Solution

Verified by Experts

The correct Answer is:
B
For the reaction,
`2C rarr A+2B, K'=(1)/(K)`
and for the reaction
`C rarr (1)/(2)A +B, K''=[(1)/(K)]^((1)/(2))`
Thus equilibrium constant `=[(1)/(40)]^(1//2)`
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