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The value of equilibrium constant of the...

The value of equilibrium constant of the reaction. `HI(g)hArr(1)/(2)H_2(g)+(1)/(2)I_2(g)is 8.0` The equilibrium constant of the reaction. `H_2(g)+I_2(g)hArr2HI(g)` will be

A

`1/8`

B

`(1)/(16)`

C

`(1)/(64)`

D

16

Text Solution

Verified by Experts

The correct Answer is:
C
`HI(g)1/2H_2(g)+1/2I_2(g)`
K=8.0
For reverse reaction & double the conversation =`(1)/(64)`
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