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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

16

B

43473

C

43481

D

23377

Text Solution

Verified by Experts

The correct Answer is:
D
`HI(g)hArr1//2H_(2)(g)+1//2I_(2)(g)`
`K=([H_(2)]^(1//2)[I_(2)]^(1//2))/([HI]_(4))=8`
`H_(2)(g)+I_(2)(q)hArr 2HI(g)`
`K'=([HI]^(2))/([H_(2)][I_(2)])=(1//8)^(2)`
`K'=1//64`
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