Calculate the degree of dissociation of HI at `450^(@)C` if the equilibrium constant for the dissociation reaction is `0.263`.

Answer

`0*51`

Explanation

Suppose we start with 1 mole of HI and x is the degree of dissociation. Then at equilibrium,
`[HI] = (1-x) //V, [H_(2)]= x//2 V and [I_(2)] = x//2" moles per litre".`
Put the values in the equation, `K_(c) = ([H_(2)][I_(2)])/([HI]^(2)` and calculate x.