At `700 K`, equilibrium constant for the reaction.
`H_(2)(g) + I_(2)(g) hArr 2HI(g)`
is `54.8`. If `0.5 "mol" L^(-1)` of `Hi(g)` is present at equilibrium at `700 K`. What are the concentration of `H_(2)(g)` and `I_(2)(g)` assuming that we initially started with `HI(g)` and allowed it to reach equilibrium at `700 K` ?

Answer

` 0.067 " mol " L^(-1) , 0.067 " mol " L^(-1)`

Explanation

Since , the equilibrium constant for the given reactions is 54.8 the value of the equilibrium constant for the reverse reactions would be equal to ` (1)/( 54.8 )`
Suppose we start with a moles of Hl . In a 1 litre container and allow it to attain equilibrium at 700 K . Suppose at total of x moles of Hl dissociate till equilibrium is attained . Therefore , we have
` {:(,, 2Hl(g) ,hArr, H_2(g) , +I_2 (g) ),( " Initial moles " ,,a " mol " L^(-1) ,, 0 ,2),( " Moles at equilibrium " ,, (a-x)" mol " L^(-1) ,, (x)/(2) " mol " L^(-) , (x)/(2) " mol" L^(-1)) :}`
According to the law of equilibrium
` " " K= ( [H_2][I_2])/( [Hl]^(2)) `
Given that `[Hl] = a- x = 0.5 " mol " L^(-1) ` Therefore
` " " (1)/( 54.8 ) = ( (x)/(2) .(x)/(2))/( (0.5) ^(2))`
or , `" " x^(2) = ((0.5) ^(2) xx 4) /( 54.8 ) = 0.018`
or, ` " " x= 0.134`
Therefore , at equilibrium ,
` [H_2] =(x)/(2) = (0.134)/( 2) = 0.067 " mol " L^(-1)`
and ` " " [I_2] = (x)/(2) = (0.134)/( 2)= 0.067 " mol " L^(-1) `