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` ?

Explanation

The given reactions can be written as
` " " 2Hl(g) hArr H_2(g) +I_2 (g) `
For this reactions, ` " " K = (1)/(54. 8) = 1.82 xx 10^(-2)`
" " (because for the reverse reactions , K = 54.8)
As the number of the moles of `H_2 and N_2 ` are equal , their concentration at equilibrium will also be equal.
Suppose `" " [H_2(g) ]= [I_2 (g) ] x " mol " L^(-1)`
Give that ` " " [Hl(g) ]=0.5" mol "L^(-1) `
` therefore " " K= ( [H_2(g)][I_2(g)])/([Hl(g)]^(2))`
or ` " " 1.82 xx 10 ^(-2) = (x xx x )/( (0.5)^(2)) `
or ` " "x= [1.82 xx 10 ^(-2) xx (0.5)^(2) ] ^(1//2) `
` " " = 0.068 " mol " L^(-1)`
Hence , at equilibrium ,
` " " [H_2(g) ] =[I_2(g) ] =0.068 " mol " L^(-1)`