For the reaction ,H_(2) + I_(2) hArr 2 HI , K= 47*6. If the intial number of moles of each reactant and product is 1 mole , then at equilibrium

For the reaction ,H_(2) + I_(2) hArr 2 HI , K= 47*6. If the intial num

1.	For the reaction ,H_(2) + I_(2) hArr 2 HI , K= 47*6. If the intial number of moles of each reactant and product is 1 mole , then at equilibrium
Answer» `[I_(2)] = [H_(2)],[I_(2)] GT [HI]` `[I_(2)]lt [H_(2)], [I_(2)] = [HI]` `[I_(2)] = [H_(2)] , [I_(2)] lt [HI]` ` [I_(2)] gt [H_(2)] , [I_(2)] = [HI]` Solution :` K= ([HI]^(2))/([H_(2)][I_(2)])` As 1 mole of`H_(2) " reactswith1 moleofI_(2)`,EVEN after equilibrium`[H_(2)] = [I_(2)]`. HENCE, `K= ([HI]^(2))/([I_(2)]^(2)) ` or ` ([HI])/([I_(2)])= sqrt(K) = sqrt(47.6)` i.e., `[HI] gt [I_(2)] or [I_(2)]lt [Hi] `