The concentration of` H_2, I_2` and HI at 731K respectively `0.92 xx 10^(-2), 0.20 xx 10^(-2)` and `2.96 xx 10^(-2) "mol L"^(-1)`, calculate equilibrium constant.
`H_(2(g)) + I_(2(g)) hArr 2HI_((g))`

Explanation

The given reaction of synthesis of HI is as under.
`H_(2(g)) + I_(2(g)) hArr 2HI_((g))`
and its `K_c` can represent as under
`K_c=([HI]^2)/([H_2][I_2])`
where, [HI] =`0.92xx10^(-2) "mol L"^(-1)`
`[I_2]=0.20xx10^(-2) "mol L"^(-1)`
`[HI]=2.96xx10^(-2) "mol L"^(-1)`
`=((2.96xx10^(-2))^2("mol L"^(-1))^2)/((0.92xx10^(-2))(0.20xx10^(-2))("mol L"^(-1))( mol L^(-1)))`
=47.62