For the reaction, `H_(2) + I_(2)hArr 2HI, K = 47.6` . If the initial number of moles of each reactant and product is 1 mole then at equilibrium

Options

A. `[I_(2)] = [H_(2)],[I_(2)] gt [HI]`
B. `[I_(2)]lt [H_(2)], [I_(2)] = [HI]`
C. `[I_(2)] = [H_(2)] , [I_(2)] lt [HI]`
D. ` [I_(2)] gt [H_(2)] , [I_(2)] = [HI]`

Answer

C

Explanation

` K= ([HI]^(2))/([H_(2)][I_(2)])`
As 1 mole of `H_(2) " reacts with 1 mole of I_(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] `