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)]xx[I_(2)])`
As 1 mole of `H_(2) ` reacts with 1 mole of `I_(2)`, after equilibrium , `[H_(2)]=[I_(2)]`
Hence `K=([HI]^(2))/([I_(2)]^(2))" "i.e., (HI)/([I_(2)])=sqrt(k)=sqrt(47.6)`
`i.e., [H] gt [I_(2)]` or `[I_(2)]+1//2O_(2)(g)`