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Dissociation equilibrium constant of HI ...

Dissociation equilibrium constant of HI is `2.06xx10^(-2)` at `458^(@)C`. At equilibrium, concentrations of HI and `I_(2)` are `0.36M` and `0.15M` respectively. What is the equilibrium concentration of `H_(2)` at `458^(@)C`.

Text Solution

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The correct Answer is:
`[H_(2)]=1.78xx10^(-2)M`
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